Thermodynamic Basic

7/30/2019 Thermodynamic Basic

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UNESCO-NIGERIA TECHNICAL & VOCATIONAL

EDUCATION REVITALISATION PROJECT-PHASE II

THERMODYNAMICS II

COURSE CODE: MEC213

YEAR II- SEMESTER III

THEORY

Version 1: December 2008

NATIONAL DIPLOMA IN


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MECHANICAL ENGINEERING TECHNOLOGY

THERMODYNAMICS II (MEC 213)

Table of contents

WEEK 1: 1.1.0:THERMAL EFFICIENCY

1.1 Definition And Units

1.2:Mechanical Efficiency ( )m of an engine

1.3: Thermal Efficiency ( )m of an engine

1.4: Work Examples

WEEK 2: 2.0: MEASUREMENT OF INDICATED POWER

2.1 I.P of Multi cylinder Engines

2.2 Worked Examples

2.3 Energy balance account for an I.C. engine regarded as

operating in a closed system:

2.4 Worked Examples

WEEK 3: 3.0 REVERSIBILITY AND IRREVESIBILIT OF

THERMODYAMIC CYCLES/PROCESSES:

3.1. A Reversible cycle

3.2. An Irreversible cycle

3.3 Second Law of Thermodynamics

3.3.1 Clauses version of 2nd law of thermodynamics:


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3.3.2 Lord Kelvins version 2nd law of thermodynamics:

3.4 Definition of heat engines

3.21 Element of a heat engine cycle

3.3 Deduction from the 2nd Law of thermodynamics

3.5 Element of a heat engine cycle

3.6. Deduction from the 2nd

Law of thermodynamics

WEEK 4: 4.0 THE CARNOT CYCLE EFFICIENCY

4.1 Carnot Principle

4.2 Thermal Efficiency of a Carnot cycle Engine

4.3 Worked Examples

4.3 Exercises and Solutions

WEEK 5: 5.0 THE SUN SOLAR ENERGY

5.1 Solar Energy -- Energy from the Sun

5.2 Photovoltaic Energy

5.3 Some advantages of photovoltaic systems

5.4 Solar Thermal Heat

5.5 Energy from the Sun

WEEK 6: 6.0 APPLICATION OF SOLAR ENERGY

6.1 Architecture and urban planning

6.1.1 Agriculture and horticulture

6.2 Solar lighting


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6.3 Solar thermal

6.3.1 Heating, cooling and ventilation

6.4. Water treatment

6.5 Cooking

6.6 Electrical generation

6.6 Concentrating solar power

6.7 Experimental solar power

6.8 Solar vehicles6.9 Energy storage methods

6.10 Development, deployment and economics

WEEK 7: 7.0 ENTROPY

7.1 Property Diagrams Involving Entropy

7.1.1 The T-S Diagram

7.2 Determination of Dryness Fraction from Area in T-S

Diagram

7.3 For a Perfect Gas

7.4 The h-S Diagram

WEEK 8: 8.0 WORKED EXAMPLES

8.1 Isentropic Efficiency

8.2 IsentropicEfficiency of Turbine

8.4 Worked Examples


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8.5 QUIZ ONE (ENTROPY)

WEEK 9: 9.0 PURE SUBSTANCES

9.1 Properties and State

9.2 Property Diagram for Phase-Change Processes

9.2.1 P-V Diagram for Pure Substance

9.3 Two property rules for pure substances

WEEK 10: 10.0 Pressure Law

10.1 Ideal Gas Law

10.2 Worked Examples

WEEK 11: 11.0 IDEALAND REAL GASES


11.2 Exothermic and Endothermic Reactions

11.3 QUIZ (THREE)

WEEK 12: 12.0 FUELS AND COMBUSTION

12.1 Classification of Fuels into Solid, Liquid and Gaseous

12.2 Combustion Equations

12.3 QUIZ( ONE)

WEEK 13: 13.0 STOICHIOMETRIC AIR-FUEL RATIO


13.2 QUIZ (TWO)


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WEEK 14: 14.0 CALORIFIC VALUE OF FUELS

14.0.1 Gross Calorific Value at Constant Volume (Qgross,v)

14.0.2 Net Calorific Value at Constant Volume (Qnet, v)

14.0.3 Gross Calorific Value at Constant Pressure (Qgross,p)

14.0.4 Net Calorific Value at Constant Pressure (Qnet,p)

14.1 Practical Determination of Calorific Value

14.2 Solid Fuel using the bomb calorimeter

14.3 Gas Calorimeter


WEEK 15: 15.0 DENSITY OF GAS MIXTURE

15.1 Atmospheric and Ecological Pollution

15.2 Carbon monoxide (CO)

15.3 Carbon dioxide (CO2)

15.4 Sulphuric dioxide (SO2)

15.5 Oxides of Nitrogen (NO)

15.6 Volatile Organic Compounds (VOCs)

15.7 Photochemical Smog

15.8 Particulates

15.9 Ecological Considerations

15.10 Ecological Considerations


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WEEK ONE

1.0 THERMAL EFFICIENCY:

1.1 Definition and Units:

Power: power is defined as the rate of doing work, and a work rate of one joule per second is named the

WATT.

The Brake power (b.p.): of an engine is the power available for work at the output shaft of the engine,

and is measured using some form of brake. It also means useful work output of an engine. The indicated

power (i.p.): of an engine is the power developed in its cylinder, and is measured by a form of pressure

indicator connected to the cylinder head. The indicated power is always greater than the brake power of

an engine, because there will always be a reduction of power between the cylinder and the output shaft

due to friction between the moving parts and the pumping power required to change the cylinder.

i. p.= b.p + friction power [i.p.= b.p+ f.p.]

1.2 Mechanical Efficiency ( )m of an engine is the rate.

It measures the efficiency with which an engine coverts the power developed in its cylinders into useful

power available at the output shaft. An average value of mechanical efficiency for a petrol engine

running at normal speed would be 0.80 or 80%.

1.3 Thermal Efficiency

An engine cannot covert all the heat energy of the fuel into work, for some inevitable reasons. Hence

the thermal efficiency of an engine is the measure of how efficiently the fuel is being used in the engine.

Definition: It is defined as the ratio of the work done per second to the Heat supplied from the fuel per

second.

t = Work done per second


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Heat supplied from the fuel per second

There are two cases to be considered: they are the indicated thermal efficiency and the brake thermal

efficiency. Take the case of the (I.C.E) internal combustion Engine first. In this case the thermal

efficiencies are follows:

Indicated thermal Efficiency ( t ) = Energy equivalent of the i.p per sec. (J/s, watts)

Energy supplied by fuel/sec

Indicated thermal efficiency = work done per second (J/S) _

Heat supplied per second (J/kg)

But Heat supplied per second = kg of fuel supplied per sec. x Calorific value of fuel (J/kg)

it = i. p. (watts) _

kg of fuel/ sec x C. V. (J/kg)

Alternatively:

Indicated t = Work done/ hour = i. p x 3600 _

Heat supplied/ hour kg of fuel/hr x c.v

But kg of fuel/ hour = specific fuel consumption (kg/kwh)

Indicated power in kw


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it = 3600 x103 _

Specific fuel consumption x C. V.

Similarly: Brake thermal Efficiency:

Brake t = b. p. (watts) _

kg of fuel/sec x c. v.

1.4 WORKED EXAMPLES

Example (1.0):

An oil engine developing 37.5kw uses 9.0kg of oil per hour of calorific value 54MJ/kg. 8.5kg are used to

overcome friction in the engine itself.

Determine:

(a) brake power (b) specific fuel consumption (brake basis) (c) Mechanical Efficiency (d) Indicated

thermal Efficiency of the Engine.

Sol:

Given that: i.p = 37.5kw; fuel consumption = 9.0kg/ hr

Friction power (fp) = 8.5kw;

Calorific value = 45MJ/kg (45 x 10

6

J/kg )

(a) b.p = i.p f.p = 37.5 8.5 = 29.0kw


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(b) Fuel consumption = 9.0 kg/hr;

but brake power = 29.0kw

specific fuel consumption (sfc) in kg/kwh (Brake basis)

i.e. s. f. c. = Fuel consumption = 9.0kg/hr

Brake power 29.0 kw = 0.310kg/kwh

(c) Mech. Efficiency ( m ) = b.p = 29.0 = 0.773

i.p 37.5 = 77.3%

(d) Indicated thermal Efficiency ( )it =c.v.x/sec.

)(.

fuelofkg

wattspi

=6

3

10x45x3600

0.910x5.37

=6

3

10x45x9.0

3600x10x5.37

= 33.32%

1.5 MEASUREMENT OF BRAKE POWER

Any method for measuring the brake horse power of an engine involves the application of a torque

which resists the motion of the crank shaft. One method is of use the simple rope brake as throw below:


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A rope makes one complete turn round the rim of a fly wheel keyed to the Engine crankshaft. The rope

carries a dead load at one end and is hooked to a spring balance at the other, the direction of the

rotation being against pull of the dead load. The engine is started with the load off and increasing load

may be applied by adding weights to the dead load hanger.

Let W = dead load on brake (kgf)

S = spring balance reading (kgf)

D = Diameter of brake wheel (m)

d = diameter of rope (m)

By using a brake or dynamometer, it is possible to determine the useful work Output of an engine.

Electrical dynamometer setup showing engine, torque measurement arrangement and tachometer

A dynamometer consists of an absorption (or absorber/driver) unit, and usually includes a means for

measuring torque and rotational speed. An absorption unit consists of some type of rotor in a housing.

The rotor is coupled to the engine or other equipment under test and is free to rotate at whatever

speed is required for the test. Some means is provided to develop a braking torque between

dynamometer's rotor and housing. The means for developing torque can be frictional, hydraulic,

electromagnetic etc. according to the type of absorption/driver unit.

W

D

d

wheel

SSpring

Dead weight

Rope Brake


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Work done = force x distance

If a braking force F is applied at the rim of a wheel of radius r, the

product, Fr, which is the work done is called the resisting torque, T.

If the wheel is kept tuning at N rev/s against this braking force; the work done during one revolution of

the brake per second = Resisting force x Distance through which the force is over come per second.

Work done per second = F x 2r x N

= 2N x Fr (but Fr = torque T)

Work done per second = 2NT = T

Work done/sec = T (Watts)

Hence brake power = useful work output of an engine

= T (Watts).

Where T = resisting torque (Nm) and

= angular velocity (radians/Second)

Now, consider the load to be acting along the centre line of the rope:-

Resisting torque due to deed load = 9.81 x W

+

2

dDNm

r

n rev/s

Braking force, f(N)


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where 9.81(N) = g (i.e. acceleration due to gravity)

Torque due to spring pull = 9.81 x S

+

2

dDNm

Effective braking torque, T = 9.81 x W

+

2

dD- 9.81 x S

+

2

dD

Effective braking, T = 9.81 x W

+

2

dD 9.81 x S

+

2

dD

= 9.81(W- S)

+

2

dDNm

and b. p = T (watts)

Note: simple rope brake can only be safely used on

Low speed engine which have their speed kept reasonably constant by a governor. For high speed

engine, a widely used brake is the Heenan and fromde Hydralic Dynamometer.

Example:

In a test on a single-cylinder gas engine using a simple rope brake, the following reading were taken:-

Dead load 29kgf, spring balance 4kgf: speed 284 rev/min; diameter of brake wheel is 1.05m, diameter

of rope = 20mm. calculate the b. p. being developed by the engine.

Solution:

Braking torque = 9. 81(W - S)

+

2

dDNm

= 9. 81 (29 - 4)

+

2

02.005.1Nm

= 9. 81 (29 - 4)

2

07.1 = 9. 81 x 25 x 0.535

T = 131 .2 Nm

b. p = T = T x 2N =60x1000

284x2x2.131 =


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=3

3

10x60

10x117..234

= 3. 902kw

Note: Tachometer- is used for measuring angular velocity of rotating objects usually measured inrev/min.

Planimeter is used for measuring area under a graph

WEEK TWO

2.0 MEASUREMENT OF INDICATED POWER:

In order to determine the indicated power of an engine, it is necessary to know the work conditions

which exist in the cylinders. The indicated mean Effective Pres sure (imep) may be obtained direct from

the indicator diagrams and from a knowledge of the (imep) of an engine the i.p can be calculated. An

indicate diagrams in a pressure- volume graph of the condition in the cylinder throughout a complete

cycle.

A typical diagram from a gas engine is shown below


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Thus, The area under an expansion curve, i.e. when volume is increasing , represents work done by the

gases (i.e. Positive work).

The area under a compression curve i.e. when volume is decreasing represents work done on the gases

(i.e. Negative Work).

By adding algebraically the area for each operation throughout the cylinder, the diagram may be seen to

consist of two enclosed area, the larger of which represent the work done by the engine each cycle. The

smaller enclosed area is called the pumping loop and represents a loss of work from the engine resulting

from the necessity to clean and recharge the cylinder. We shall ignore for the moment, the pumpingloop, because it is usually too small to measure on a normal diagram.

Mean Effective Pressure: the average net pressure which acting on the piston area for one Stroke and

does the same work as that represented by the indicator diagram is known as the Mean Effective

pressure, Pm. The enclosed area of the diagram is irregular in shape but a rectangle of equal area and

having the same base length would have a height equal to the Mean Effective pressure. Hence to

determine the mean Effective pressure from given indicator diagram measure the enclosed area using a

planimeter and then.

Mean Effective pressure,( )

( ) 1

No.Springx

mmdiagramoflength

mmdiagramofArea

2

BasePm =

Now, the indicated power can be calculated using the mean effective pressure follow:

Useful work

Ex ansion

Compression

Exhaust

Pumping loss (-ve)

Suctiond

bc

aP

VO


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Indicated power (i.p) = Work done per second inside the cylinder the cylinder

= (Nm/s =J/s =Watt)

= Work done/cycle x Number of cycles/sec.

= (Nm =J) (Watt)

Now work done per cycle = Area of indicate diagram

= Pm x Vs

= Pm x (A x L)

I. P = Pm x L x A x N (J/s)

= Pm LAN (watts)

Where:

Pm = Mean Effective Pressure N/m2

A = Area of piston M2

( )

L = Length of Stroke, m

N = Number of working strokes/ sec (Cycles)

Note:

For a four-stroke cycle, no of cycles/sec = engine speed. For a two stroke cycle engine, No. of cycle/sec

= the engine speed.

For a gas engine on four-stroke cycle, n =2

speedEngine.

Where there are misses, n =2

speedEngine- No. of misses/sec.

2.1 I.P of Multi cylinder Engines

For engine having more than one cylinder, the most accurate method is to measure the i.p of each

cylinder separately. The i.p of the engine is the sum of the i.ps of the separate cylinders.

Thus, i.pEngine = i.p one cylinder x Number of cylinders.


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2.2 Worked Examples:

Example 1:

A four-cylinder petrol engine at 1200rev/min gave 25.3KW b.p. when one cylinder was cut out the b.p.

decreased to 17.6 KW. Estimate the ; p of the Engine

Solution:

Given that: The b.p. of 4-cylinder Engine 25.3kw

The b.p. of 3-cylinder the Engine =17.6kw

i.p. = 4(25.3-17.6)

= 30.8kw

Example 2

In a text in a singlecylinder oil engine operating on the four stocker cycle and fitted with a simple ropebrake, the following reading were taken.

Brake which diameter = 0.7m, speed 450rev/min, rope ria. 20mm, dead weight on rape 21kgf, spring

balance reading 3.4kgf, area of indicator diagram = 404mm2 , length of indicator diagram 65mm, spring

No, 140kN/m2

per mm, bore 100mm, strake 150mm, engine used 0.75kg/h of oil of calorific value

45mJ/kg. calculate the b.p.; i.p.; mechanical efficiency and indicate thermal efficiency of the engine.

Solution :-

Given Data on above

(a) b.p. = T But T = 9. 81(W - S)

+

2

dDNm


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= 9.81 (21-3.4)

+310x2

20700

T= 62.16Nm & = 2N

( )60

450x2x62.16N2xT! ==T

b.p. = 2.93KW

(b) diagramof

No.SpringxdiagramofArea

PbutN,AP.. mm LengthLpi ==

=1

10x140x

65

404 3

=2KN/m2.870

LANmPi.p =

= ( )2x60

450x0.1

4x10x150x10x2.870

23-3

KW.8443i.p =

(c) Mechanical Efficiency (m) =

pi

pb

.

.

=84.3

93.2= 0.763

m = 76.3%


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(d) Indicated thermal Efficiency (it)

it = c.v.xfuel/secof

)(.

kg

wattspi

=6

10x45x3600

75.0

)(3844 watts

=6

10x45x0.75

3600x3844

Indicated 1 = 0.410 = 41.0%

2.3 Energy balance account for an I.C. engine regarded as operating in a closed

system:

To determine the energy balance account of an engine operating in a closed system, the first law of

thermodynamics is applied. The law states that the heat flow across the boundary is equal to the work

flow across the boundary.

Heat outflow to Cooling water, Qc

Heat inflow from

fuel supplied Qf

Heat outflow by radiation, and

out and correction, QrHeat outflow to Exploit

Qe Energy balance Account

Work outflow to brake, WEngine


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Applying the 1st law, WQQQQ ecrf =

or eerf QQQWQ +++

This statement may be shown in the form of a balance sheet (often called a Heat or Energy Balance

Account), the left hand side showing the heat energy supplied and right hand side showing its

distribution. The balance may be based on 1kg of fuel, or on a basis of time, and the items may be

expressed as heat quantities and as percentage of energy supplied.

Fir I.C. E, the table is usually as follows:

(a) Energy supplied per second, Qf= mass of fuel used/sec. x c.v.

= kg of fuel used/sec. x c.v

(b) Energy distributed per second

(i) Work outflow, W = b.p (watts).

(ii) Heat flow to cooling water, Qc = m c (tout tin) watts

Where m = flow of cooling water kg/s

C = specific heat capacity

tout = leaving temperature of cooling water

tin = Entry temperature of cooling water

(iii) Heat flow to exhaust and surrounding, Qe + Qr

An addition to the balance sheet which is often asked for is the heat to friction. This accounts for the

reduction of power between the cylinder and the output shaft of an engine due to friction between the

moving parts. This heat to friction = (i.p b.p.) watts.


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This is already included as heat flowing from the engine partly in the cooling water, partly in the exhaust

gases and partly in the radiation to the surroundings.

2.4 WORKED EXAMPLES:

Example 3:

The following results were obtained during the trial of a four-stroke national oil engine of cylinder bore

200mm and stroke 400mm.

Effective brake wheel diameter = 1.6m; speed = 258rev/min

Effective brake load = 47kgf, area of indicator diagram = 320mm2

Spring No. = 110 KN/m2 per mm; length of diagram = 65mm; fuel used/hour = 3.2kg/hr; c.v. of fuel =

45MJ/kg; cooling water = 380C. (Given Cw = 4187J/kgK.)

Calculate: (a) Mechanical Efficiency

(b) Indicated thermal efficiency, and (c) Draw up an energy balance account on a basis of 1

minute.

Solution:

Mechanical Efficiency m =pi

pb

.

.

But b.p = T where T = 9.81 x 47 x2

6.1= 368.9N

and = 2N = 260

258= 27.02

b.p = T = 368.9 x 27.02

= 9.967 KW


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i.p = Pm LAN;

Pm = 320 x 110 x 103

= 541 x 103N/m

2

i.p = Pm LAN = 541 x 103 x 0.4 x 4

x (0.2)

2

x 60x2

258

= 14.62 KW

Hence m =pi

pb

.

.=

62.14

967.9= 0.6817

m = 68.17%

(b) Indicated t =c.v.xfuel/sec.of

)(.

kg

wattspi

=6

3

10x45x2.3

3600x10x62.14

i.t = 0.3655 or 36.55%

(c)

Energy supplied or

input/min.

Joules % Energy distribution/min. Joules %

Kg of fuel/s x c.v.

60

2.3x 45 x 106

2.4 x 106

100 (i) To useful work = b.p x

60

(ii) To cooling water = mc

(tout tin) = 2.3 x 4187

x 38

(iii)To exhaust and

radiation (by diff.) [2.4

598.0 x 103

365.9 x 103

24.92

15.25


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x 106 (598.0 +

365.9)103].

1436 x 103

59.83

Total 2400 x 103

100 Total 2400 x 103

100

2.5 EXERCISES:

Q1. The following observations were recorded during a four stroke singlecylinder oil engine:

duration of trial =30 min; oil consumption =4.4kg; calorific value of oil = 42MJ/Kg; Area of indicator

diagram = 850mm2,

length of diagram = 80mm; spring rating = 56KN/m2

per mm; effective brake

wheel diameter = 1.5m; speed = 200 rpm, brake load= 135kgf; spring balance reading = 18kgf; length

of stroke = 450mm; cooling water flow = 11kgf/min; temperature rise of cooling water =360C. If the

engine is 57.2% efficient.

Calculate:

a. The cylinder bore of the engine

b. The brake specific fuel consumption

c. Indicated thermal efficiency

d. Draw up a heat balance sheet in MJ/min for the engine (Cw=4.18KJ/kgk)

Q2. The following results were obtained during a test on a singlecylinder, double acting steam

engine fitted with a simple rope brake: swept volume =7.85 liters, speed =300rpm, brake load

=136kg; spring balance reading = 90N, indicated mean effective pressure = 2.32 bar; steam

consumption =0.056kg/s, condenser cooling water =113kg/min; temperature rise of condenser


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cooling water =11K; Hot well temperature =380C; calorific value of fuel =2.514MJ/Kg

(Cw=4.18KJ/kgk).

Determine

a The effective diameter of the brake wheel, if the efficiency of the engine is 80.5%

b The brake thermal efficiency

c Specific steam consumption on indicated basis

d Draw up a complete energy balance account for the test on a basis of MJ/min.

Q3 .An oil engine on a four stroke cycle has a swept volume of 14 liters and a mean effective

pressure of 5.67 bars. Its rated speed is 6.6rps and is tested at this speeds speed against a brake

which has a torque arm of 0.7m. The net brake load is 755.5N and the fuel consumption is

0.0025Kg/s. calorific value of fuel =44MJkg-1

, cooling water circulation =9.0kg/min at an inlet

temperature of 380C and an outlet temperature of 710C, and the energy rejected through the

exhaust pipe is 33.6KW. (Cw=4.18KJg-1

K-1

)

Calculate: (a) the engine torque

(b) The Mechanical efficiency

(c) Brake specific fuel consumption

(d) Draw up an overall energy balance in KJ/s and as

2.6 SOLUTIONS:

Q1. Given that; Duration = 30min; f c = 4.4kg; c v = 42 MJ/kg;

Aid = 850mm2; Lid = 80mm; spring No. = 56KN/M

2/mm


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Wheel dia = 1.5m; N = 200rpm; W = 135kgf;

S = 18kgf; stroke = 450mm; cooling water = 11kg/mm=in.

Temp. rise = 360C; %2.57=m

(a) ,.

ip

pbm = but b.p = ( )

( )Nx

dDSWT 2

2

+=

= ( )( )

60

2002

2

5.11818581.9 xx

1802920200275.011781.9 =xxxx

= KW030.18

1

..,

4.

2NoSP

xLid

AidPbutNx

DxLxPLANPpi mmm ===

=2

KN/m5951

56x

80

850=

` Ip =2x60

200x

4

D2x0.45x10x595

3

6

2

3 10x23x168

480xi.pD

200xx0.45x10x594

2x60x4x==

ip

Sincei.p

b.p=m , KW31.52

0.572

18.03

b.p.

m

===

pi

(b) kg/kwh0.48818.03

2x4.4

b.p

f.c... ===csfb


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(c)( )

VCxsfuelofKg

Wattspiit

./

..=

=6

3

10424.4

60301052.31

xx

xxx

= 30. 72%

(d)

Energy

supplied/min

MJ % Energy Distributed / min MJ %

Qf = kg of oil/s x C.V

4.4kg x 42 x 106

30min

6.16 100

(i) to useful work, w =b.p x 60

= 18.03 x 60

(ii) To cooling water, Qc MCw

t = 11 x 4.187 x 36.

(iii) To surrounding Qr

Qf (W + Qc)

6.16 - (1.082 + 1.658)

1.082

1.658

3.420

17.56

26.92

55.52

Total 6.16 100 Total 6.160 100.00


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Q2. Given That: VS = 7.85 Litres; N = 300rpm; W = 136kg; S = 90N

Pm = 2.32bar; f.c = 0.056kg/s, cooling water =

113kg/min; Hot well temp. = 380C; Dt = 11K;

C.V. = 2.514 MJ/Kg; Cw = 4.18KJ/kgk; m = 80.5%

(a) NxVxPLANPi.pbut,i.p

b.p smmm ===

=60

300x

10

7.85x10x32.2

3

5

Since it is double acting; i.p for the engine = 9.114 x 2

i.p = 18.228KW

Hence b.p = m x i.p = 0.805 x 18.228

= 14.674KW

But ( ) ;22

. NxD

SWgpb e = let De rep. effective dia

= (9.8 x 136 90) x De x 2 x 5

DexxDe

x 22.543,19522

1244674.14 ==

.75m022.543,19

674,14e ==D


28/136

Effective dia of brake wheel, De = 0.75m

(b) bt =( )

610x2.514x0.056

46741

C.Vxfuel/SofKg

. =

wattspbs

= 0.1042 = 10.42

(c) s. f .c (i) =pi

cf=

pi

hrSteamofkg

powerIndicated

nconsumptioSteam

.

/.

=

kwhkgx

/06.11228.18

3600056.0=

(d) Energy balance account

Energy supplied/min MJ % Energy Distributed / min MJ %

Qf= kg of fuell/s x C.V

0.056 x 60 x 2.514 x 106

30min 8.45 100

(i) to useful work, w = b.p x 60

= 14.67 x 60

(ii) To cooling water, Qc Qc =

MCw t 113 x 4.18 x 11

0.88

5.20

10.41

61.54


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(iii) To Hot well Qw MCw t =

0.056 x 60 x 4.18 x 38 = 0.056 x

60 x 4.18

(iv) To surrounding Qr

Qf (W + Qc + Qw)

8.45 (0.88 + 5.20 + 0.534)

0.534

1.836

6.32

21.73

Total 8.45 100 Total 8.450 100.00

Q3. Given that; Vs = 14 litres, Pm = 5.67bar, N = 6.6 rps; r= 0.7m;

W(F) = 755.4N; f.c. = 0.0025kg//s; c.v = 44MJ/kg

Cooling water = 9kg/min; ti = 380C; to = 710C

Qe = 33.6KW; Cw = 4.18KJ /kgk;

(a) Engine torgue, T = Fr = 755.4 x 0.7

= 528.76 Nm = 528.8Nm

(b) m = b.p but b.p = T = T x 2 N = 528.8 x 2 x 6.6

= 21.927kw = 22.0kw

And i.p = PmLAN = Pm x Vs x N


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=2

6.6

10

141067.5

5xxx

= 26.2 kw.

m =pipb.. = 0.84 = 840%

(c) bsfc =22.0

90

1

3600x

22.0

0.0025

.

.=

pb

cf

= Kg/Kwh409.0

Energy

supplied/min

MJ % Energy Distributed / min KJ %

Qf = kg of fuel/s x

C.V

0.0025 x 44 x 106

110 100

(i) to useful work, w = b.p 22.0

(ii) To cooling water, Qc MCw

t

9 x 4.18 x (71 38)

60

(iii) To exhaust Qe

Qe = 33.6kw given

(iv) To surrounding, Qr

Qf (w + Qc + Qe)

110 (22 + 20.75 + 33.6)

22.00

20.7

33.6

33.7

20.0

18.82

30.55

30.63

Total 110 100 Total 110 100.00


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WEEK THREE

3.0 REVERSIBILITYÀNDÌRREVESIBILITYÒF THERMODYAMIC

CYCLES/PROCESSES:

The thermodynamic cycles in general are classified into two, namely:-

1. Reversible or ideal cycle and

2. Irreversible or natural or real cycle

3,1. A Reversible cycle: A thermodynamically reversible cycle consists of reversible processes

only and a reversible process is one which is performed in such a way that at the end of the

process, both the system and the surroundings may be restored to their initial states.

3.2. An Irreversible cycle: A cycle will be considered thermodynamically irreversible if any of

the processes constituting the cycle is irreversible. Thus, in an irreversible cycle, the initial

conditions are not restored at the end of the cycle.

In actual practice most of the processes are irreversible to some degree. The main causes for the

irreversibility are: (a)

mechanical and fluid friction (b) unrestricted expansion (c) Heat

transfer with a finite temperature difference.

3.3 SECOND LAW OF THERMODYNAMICS:

Like the first law, the second law of thermodynamics is a very important law and it is based

upon observable phenomena. It states that: Heat will not transfer up the gradient of

temperature of its accord. This does not mean that heat cannot be made to transfer up the

gradient of temperature. It can, but it has to be facilitated by the aid of external energy.


32/136

External energy, however it is not required for the transfer heat from a higher temperature

gradient to a lower one it being a purely natural phenomena.

This law imposes limit upon the actual quantity of heat which can be extracted from any given

heat energy supply and hence, by the principle of conservation of energy, imposes a limit uponthe amount of work which can expected theoretically. This limit appears when the temperature

of the heat energy supply becomes the same as its surroundings at which condition heat

exchange ceases.

3.3.1 Clauses version of 2nd

law of thermodynamics:

He stated that: It is impossible for a self acting machine, unaided by an external agency to

covey heat from a body at a low temperature to one at a higher temperature. This implies that

although the machine may have its own heat energy content in the form of high and low

temperature quantities, heat energy transfer from the low temperature to the high temperature

is impossible unless some external energy supply is used to run the machine. Actually this is the

principle of the Refrigeration and heat pump. Note, that if this is the case, the machine ceases to

be self-acting.

3.3.2 Lord Kelvins version 2nd

law of thermodynamics:

He stated that: We cannot transfer heat into work merely by cooling a body already below the

temperature of the coldest surrounding objects.

This implies that when a body researches the temperature of the coldest surroundings objects,

no further natural extraction of heat is possible, and hence no further work can be performed.

Another def.:- The thermodynamic engine is a device in which energy is supplied in the form of

heat and some of this energy is transformed into work.

3.4 Definition of heat engines:


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Since the heat is defined as that of transfer of energy which results from a difference in

temperature, then a heat engine must be an engine in which a transfer of heat occurs.

If heat is introduced into a system and as a result of a cycle process. Some work appears from

that system, together with some heat rejection from the system, then this is a heat engine. Inpractice, such as an engine, or plant really, is the closed circuit stream terbium point of a power

station. This can be illustrated as follow:

On the other hand, the open circuit, internal combustion engine since as a petrol engine in

strictly not a heat engine for fuel and air are admitted, which must cross the system boundary,

combustion is internal, as the name implies, and combustion products and heat are rejected,

with some work crossing the system boundary. However, thermodynamic engine are mostly

referred to as heat engines.

3.5 Element of a heat engine cycle:

The essential element of a thermodynamic cycle involving a heat engine are:

1. A working substance: A medium for receiving and rejecting heat, and doing work, it is the

substance that undergo the change of state e.g. steam, air, etc.

2. A source of heat (A hot body or heat Reservior

This is where heat may be added to the working substance.

3. A heat sink (cold body): This is the body to which heat may be rejected by the working

substance. In practice a heat sink in a natural receiver such as the atmosphere, a lake, a

lagoon, a river or ocean.

4. An Engine: An engine in that device in which the working substance may do work or have

work done on it.

When the temperature of any heat energy supply has fallen then the heat energy is said to have

been degraded. Complete degradation occurs when thermal equilibrium has been established.

3.6. Deduction from the 2nd

Law of thermodynamics:

1. From this second law of thermodynamics, it follows that in order to run all the various engine

and devices in use today, and hence maintain the present state of civilization a source of supply


34/136

of fuel is absolutely necessary. It is by burning fuel that the various working substances have

their temperature put up above their surroundings thus enabling them to release their heat

energy in a natural manner in accordance with 2nd

law. Developments in the field of nuclear

physics have shown the way of obtaining a further supply of heat energy. This will augment the

dwindling coal, oil and natural gas stocks of world. For instance, it has been seriously forecasted

that world crude oil stock could be seriously low, if not completely exhausted by the end of this

20th century.

2. By virtue of this 2nd law, it is essential that all fuels should be used as efficiently as possible in

order that fuel stock may be preserved for as long as possible. It must always be remembered

that when once the heat energy from any particular fuel has been degraded, then further heat

energy is only obtainable at the experience of further fuel.

WEEK 4

4.0 THE CARNOT CYCLE EFFICIENCY

4.1 Carnot Principle:

Reversibility, as it applies to the thermodynamic engine, was discussed by a Frenchman, Sadi

Carnot, in a paper entitled. Reflections on the motive power of heat which was published in

1824. In the paper, Carnot conceived of an engine working on thermodynamically reversible

processes, and from this concept deduced what has since been called Carnots principle. This

states that: No engine can be more efficient than a reversible engine working between the

same limits of temperature. The principle of thermodynamic engine is that it receives heat at

some high temperature from a heat source. The engine then converts some of this heat into

work and then rejects the reminder into a sink


35/136

Consider a thermodynamic reversible engine R working between the temperature limits of

source T1 and sink T2. If the engine receives Q units of heat from the source and temperature T1.

It will convert W units of this heat into work and then reject (Q-W) units of heat into sink at

lower temperature T2. as shown in fig (a).

Assume that source other engine E can be found which is more efficient than the reversible

engine R. Since it is more efficient, this engine E will require less heat supplied to perform the

same amount of work, W. If this engine is used to drive a Carnot heat pump then it will need less

heat than the Carnot engine to produce the work needed to drive the heat pump. This would

mean that the combination would be returning more heat to the source at the higher

temperature T1 than is being taken out. But then would directly contravene the second law of

thermodynamics which says that heat cannot flow from a lower to a higher temperature

without the aid of work from an outside source, and so we conclude that no such more efficient

engine exists. Now we have achieved a very important step in the investigation into engine

efficiencies becomes in have identified an engine with the best possible efficiency i.e. one

consisting only of reversible operation. The next step is to find out the efficiency of the Carnot

engine.

4.2 Thermal Efficiency of a carnot cycle Engine:

Source T

Sink T

Q W

W

Q

Source T

Sink T

(Q W)

Q

QE

Q dQ - W

Q dQ

Carnot Carnot

1

2

Isothermal expansion

T1 = T2

Adiabatic com resssion

P1

P2

P


36/136

By calculating the thermal efficiency of this cycle it is possible to establish the maximum possible

efficiency between the temperature limits taken. The fig (c) above shows the Carnot cycle

illustrated on a P-V diagram. The cycle is made up of four reversible processes i.e. put together

that they form a closed cycle. Thus by proceeding round the cycle, it is possible to return to the

original state and hence the admit of repetition. The processes are as follows:

1-2: Isothermal expansion

Pressure falls from P1 to P2

Volumes increase from V1 to V2

Temperature remain constant at T1 =T2

1

21

1

211 lnln

V

VmRT

V

VVPdonework ==

1

2

1 lnV

VmRTreveivedHeat =


37/136

For Isothermal, Q = W.

2-3: Adiabatic expansion

Pressure falls from P2 to P3

Volume increase from V2 to V3

Temperature falls from T2 to T3

Work done =1

3322

VPVP=

( )

1

32

TTmR

For the adiabatic, Q = O.

No heat transfer during this process.

3-4: Isothermal compression

Pressure increase from P3 to p4

Volume reduced from V3 to V4

Temperature remain constant at T3=T4

Work done =

4

3

33

3

4

33 lnlnV

VVP

V

VVP =

=

4

3

3 lnV

VmRT

For the Isothermal, Q = W

Heat rejected =

4

3

3 lnV

VmRT


38/136

4-1: Adiabatic compression

Pressure increases from P4 to P1

Volume reduced from V4 to V1

Temperature increases from T4 to T1

Work done =1

1144

VPVP=

( )

1

4411

VPVP

=( )

1

41

TTmRT

For the adiabatic, Q = O

No heat transfer during this process. Note that this process returns the gas its original state 1

The work done during this cycle may be determined by summing the areas beneath the various

processing taking the expansions as positive areas and compression as negative areas.

Thus: work done /cycle = area (1-2) + area (2-3) - area (3-4)

-area (4-1)

= area 1234 =area enclose by cycle

OR

Work done/cycle

( ) ( )1-

T-TmR-

V

VInmRT-

1-

T-TmR

V

VInmRT 41

4

33

32

1

21

+= . (1)

But from the Isothermal process; 4321 TTandTT ==


39/136

( ) ( )

1-

T-TmR

1-

T-TmR 4132

=

Hence, from equation (1);

4

33

1

21

V

VlnmRT-

V

VlnmRTdone/cycleork =W .. (2)

For the Adiabatic, 1-4;

1

1

4

4

1

=

V

V

T

T.(3)

For the Adiabatic

1

2

3

3

2;32

=

VV

TT (4)

But 4321 TTandTT ==

3

2

4

1

T

T

T

T= (5)

Hence from Equation (3) and (4);

4

3

1

2

2

3

1

4

V

V

V

Vor

V

V ==

V

V..(6)

Substituting Equation (6) in equation (2);

( )1

331

V

VInT-TmRdone/cycleork =W (7)

This is positive work done and this is always the case if the processes of a cycle proceed in a clockwise

direction. External work can thus be obtained from such cycles. If the processes proceed in an anti

clockwise direction then the work done is negative, in which case equation (7) now becomes;


40/136

( )1

231

V

VlnT-TmR-done/cycleork =W .(8)

Negative work means that external work must be put in to carry out such cycle

Now.

receivedHeat

rejectedHeat-receivedHeathermal =T

From the above analysis

1

21

1

33

1

21

VVlnmRT

lnV

VInmRT

V

VmRT

th

=

=

( )

1

21

1

231

V

VInmRT

V

VInTTmRT

4

3

1

2 V

V

V

VSince = from equation (6)

1

21

T

T-Thermal = T (9)

=tempAbs.

tempabs.Min.-tempabs..

Max

Max.(10)

From Equation (9); Thermal

1

31T

T= . (11)

And from Equation (3), (4) and (5);

1-

1

2

3

1-

1

4

4

1 V

V

V

rVT

T=

=

=

When r = adiabatic compression and expansion volume ratio.


41/136

From Equation (11);

1-yr

1-1hermal =T 12)

This thermal efficiency gives the maximum possible thermal efficiency obtained between any this given

temperature limit. It is new possible to suggest that if any engine working between the same

temperature limits heat a thermal efficiency lower than this, then thermal efficiency improvement is

theoretically possible. Most engine have a thermal efficiency much lower than the carnot efficiency.

The militants aim should be an attempt to reach an efficiency as near 100% as possible.

4.3 W0RKED EXAMPLES:

The overall volume expansion ratio of a carnot cycle is 15. The temperature limit of the cycle are 260 0C

and 210. Determine;

(a) The volume rates of the isothermal and adiabatic processes.

(b) The thermal efficiency of the cycle.(take = 1.4)

SOLUTIONS:

Given that: T1=2600C; T4 = 21

0C; = 1.4; overall

Vol. expansion ratio = 15

(a) for the Adiabatic ;

1

2

3

1

1

4

3

2

4

1

=

==

V

V

V

V

T

T

T

T


42/136

1-

1

3

21-

1

4

1

2

3

1

4 TT

V

V

=

==

T

T

V

V

= 0.41

1-1.4

1

1.812294

533=

= 4.42812.125 =

Volume ratio of adiabatic = 4.42 = ra

Volume ratio of isothermals =

=

41

13

4

3

VV

VVV

V

Isothermal vol. ratio = 3.394.42

15

ratiovol.

ratiovol.==

Adiabatic

Overall

(b) Thermal Efficiency = 100%xT

T-

1

41T

100%x533

294-533

=

100%x533

239=

%8.44=

Alternatively 14.1142.4

1111 == r

14.142.4

11

=


43/136

= 05521812.1

11 =

= %8.44448.0 =

4.3 EXERCISES AND SOLUTIONS

Q1 (a). In a carnot cycle process, whose temperature limits are 3000C and 50

0C, a total of 230g of gas

was taken in. If the hyperbolic expansion volume ratio is 2.5

Determine:

i. The isentropic expansion volume ratio

ii. The thermal efficiency of the cycle

iii. The work done per cycle. (Take R=0.28KJ/kgk; = 1.4).

b. 0.5kg of air is first expanded isothermally at a temperature of 2350C from 3.5MN/m

2to 2.1

MN/m2 and further expanded adiabatically to 140KN/m2 . The air is then cooled at constant pressure and

is finally returned to is initial state by adiabatic compression.

Calculation:-

i. The external work done by the air per cycle

ii. The thermal efficiency of cycle. (For the air, take = 1.4; Cp =1KJ/Kgk; R=287J/Kgk)

iii.

Q2. A carnot cycle works with isentropic compression ratio of 5, and hyperbolic expansion ratio of 2.

The volume of air at the beginning of the hyperbolic expansion is 0.3m3 . If the maximum temperature

and pressure is limited to 550K and 21 bar respectively.

Determine;

a. Minimum temperature in the cycle.

b. Thermal efficiency of the cycle


44/136

c. Pressure at all the salient points

d. Work done per cycle (take ratio of specific heats as 1.4)

Q3 (a) A carnot engine working between 3770C and 37

0C produce

150KJ of work. Find:

(i) the thermal efficiency of the engine

(ii) the heat supplied during the process

(b) A carnot engine operates between two reservoirs at temperature T1 and T3. The work output of

the engine is 0.6 times the heat rejected. The difference in temperatures between source and the sink is

2000C. Calculate:

(i) the thermal efficiency of the engine

(ii) the source and the sink temperatures

SOLUTIONS:

Q1.(a) Given that; t1 = 3000C; t3 = 50

0C; M = 230g (0.23kg)

Ri = 2.5 (i.e T1 = 573k; T3 = 323k); R = 0.28kg/K

Let

(i) Isewtropic expansion vol ratio = ra

For adiabatic process [ ] 11

3

3

1

1

4

3

2

4

1

=

=

==

r

rr

raV

V

V

V

T

T

T

T

=4

1

T

Tr

vr

a

11


45/136

ra =14.1

1

323

573

= (1.774) 4.0

1

ra = (1.774) 2.5 = 4.19

ii.1

%100

573

333573

1

31 xT

TTt

=

=

= %6.431

%100573250 =x

iii. WD/cycle = MR (T1 T3) In

1

2

V

V= 0.23 x 0.28 (573 323) ln 2.5

=0.23 x 0.28 x 250 x 0.916

= 14.75KJ

OR WD/cycle = MRT1 In

1

2

V

Vx

th = 0.23 x 0.28 x 573 x In 2.5 x 0.436

= 0.23 x 0.28 x 573 x 0.916 x 0.436

= 14.74 KJ

(b) Given that; M = 0.5kg; T1 = T2 = 508K; P1 = 3.5 MN/M2; P2 = 2.1MN/M2; P3 = 140KN/M

2 =

1.4; CP = IKJ/kgk; R = 28TJ/kgk.

(i) External work done/cycle = Q2 - Q4 = (QS Q


46/136

MRT1In

1

2

V

V- MCp (T3 T4)

P1V1 = MRT1

V1 = MRT1 = 0.5 x 287 x 508 = 72898

P1 3.5 x 106

3.5 x 106

V1 = 2.08

For isothermal process, P1V1 =P2V2

V2 =2

11

P

VP=

1

08.2

1.2

5.3x

= 3.466

Q12 = MRT1 In1

2

V

V= 0.5 X 0.287 X 508 x In (1.666)

= 0.5 X 0.287 X 508 X 0.510

= 37.18KJ

1

3

2

3

2

=

PP

TT

1

2

323

=

P

PTT But T2 = 508K (Since T1 = T2)


47/136

= 5084.1

4.0

1.2

14.0

= 508 x (0.04) 0.286

= 234.2k

1

4

1

4

1

=

P

P

T

T

( ) 286.04.1

4.01

1

414 04.0508

5.3

14.0508 =

=

=

P

PTT

= 202.3k

WD/Cycle = mRT1 ln mcpV

V

1

2 - mCp (T3 - T4)

= 37.18 0.5 x 1 x (234.2 202.3)

= 37.18 0.5 x 31.9 = 37.19 15.95

= 21.23KJ

Q2 Given that, ra = 5, ri = 2, V1, = 0.3m3; T1 = 550k;

P1 = 21bar = 21 x 105 N/M2; 4.1==CpCv

But 51

4 ==V

Vra and 2

1

2 ==V

Vri


48/136

(a) Minimum temp. in the cycle let the min. temp. = T3 or T4

First consider Isentropic (adiabetic) compn process (4 1)

T4 =9036.1

1T =9036.1

550= k289

= 160C

T3 or T4 = 289K or 160C

(b) Thermal efficiency of the cycle

1

31

T

TTt

= =

550

289550

= 0.4745

= 47.45%

(c) Pressure at all the salient points, let P2, P3 and P4 represent pressures at points 2,3, and 4respectively.

For isothermal expansion process (1 2);

,VP 2211 =VP 2

121x

2

11

2

112 ===

V

VxP

V

VPP

1.90365V

1-1.41

1

4

4

1 ==

=

VT

T


49/136

= 10.5 bar

For isotropic expansion process (2 3)

( ) 4.14.1

3

233322 2.05.10

5

110.5

V

VP2xP,VP =

=

==

VP

= 1.103 bar

Since ,52

2

1

4 ==V

V

V

Vhence

5

1

3

2 =V

V

For isentropic compression process (4 1)

==

4

11

1411144PP;VP

V

VVP

( ) 4.11.4

4 0.2x215

121 =

=P

= 2.206 bar

(d) Work done/cycle, W = Heat Supplied Heat rejected.

W = Qs Qr = Q12 Q34

For Heat supplied,

1

2112

VVLnVP=Q

= 21 x 106

x 0.3 x In 2 = 436.68KJ


50/136

For Heat Rejected Q34 =

4

333

V

VInVP

= 1.103 x 105 x 3.0 x In 2

= 299.26 KI

Work done/cycle, W = 436.68 299.36

= 207.32KJ

Q3. (a) Given that; t1 = 3770C; t3 = 37

0C, W = 150 KJ

(i) Thermal efficiency,t

=1

31

TTT but T1 = 650K; T3 = 300K

=650

310650

= 0.523 = 52.3%

(ii) Heat supplied during the process Q12 =

t

W

Since t =12

Q

VV

SuppliedHeat

doneWork

Q12 =

t

W

=

523.0

150= 286.8KJ

(b) Given that; work output; W = 0.6Qr = 0.6Q34

Temp. difference, T1 T3 = 2000C (200k)


51/136

(i) Thermal efficiency,t =

Supplied

done

Heat

Work=

s

r

Q

Q-sQ

t =rQ+WD

WD =rQ6.0

6.0+r

r

QQ

=

r

r

Q

Q

6.1

6.0

t =

6.1

6.0= 375.0

= 37.5%

(ii) Source and sink temperatures, let T1, = Source temp.

T3 = Sink temp.

But t =1

31

T

TT =

1

200

T

t

T

2001 =

0.375 =

1

200T

T1 = ( )Ck = 3.2603.533375.0

200

Since 20031 =TT

20013

= TT

= C= 3.602003.260


52/136

WEEK FIVE

5.O: THE SOLAR ENERGY:

5.1 Solar Energy -- Energy from the Sun

has produced energy for billions of years. Solar energy is the suns rays (solar radiation) that reach the

earth.

Solar energy can be converted into other forms of energy, such as heat and electricity. In the1830s, the British astronomer John Herschel used a solar thermal collector box (a device that

absorbs sunlight to collect heat) to cook food during an expedition to Africa. Today, people use

the sun's energy for lots of things.

Solar energy can be converted to thermal (or heat) energy and used to:

Heat water for use in homes, buildings, or swimming pools.

Heat spaces inside greenhouses, homes, and other buildings.

Solar energy can be converted to electricity in two ways:

Photovoltaic (PV devices) or solar cells change sunlight directly into electricity. PV

systems are often used in remote locations that are not connected to the electric grid.

They are also used to power watches, calculators, and lighted road signs.

Solar Power Plants - indirectly generate electricity when the heat from solar thermal

collectors is used to heat a fluid which produces steam that is used to power generator.

Out of the 15 known solar electric generating units operating in the United States at theend of 2006, 10 of these are in California, and 5 in Arizona. No statistics are being

collected on solar plants that produce less than 1 megawatt of electricity, so there may be

smaller solar plants in a number of other states.

The major disadvantages of solar energy are:

The amount of sunlight that arrives at the earth's surface is not constant. It depends on

location, time of day, time of year, and weather conditions.

Because the sun doesn't deliver that much energy to any one place at any one time, alarge surface area is required to collect the energy at a useful rate.

5.2 PHOTOVOLTAIC ENERGY

Photovoltaic energy is the conversion of sunlight into electricity. A photovoltaic cell, commonlycalled a solar cell or PV, is the technology used to convert solar energy directly into electricalpower. A photovoltaic cell is a nonmechanical device usually made from silicon alloys.


53/136

Sunlight is composed of photons, or particles of solar

energy. These photons contain various amounts ofenergy corresponding to the different wavelengths of

the solar spectrum. When photons strike a

photovoltaic cell, they may be reflected, pass right

through, or be absorbed. Only the absorbed photonsprovide energy to generate electricity. When enough

sunlight (energy) is absorbed by the material (asemiconductor), electrons are dislodged from the

material's atoms. Special treatment of the material

surface during manufacturing makes the front surface

of the cell more receptive to free electrons, so theelectrons naturally migrate to the surface.

When the electrons leave their position, holes are

formed. When many electrons, each carrying a

negative charge, travel toward the front surface of thecell, the resulting imbalance of charge between thecell's front and back surfaces creates a voltagepotential like the negative and positive terminals of a

battery. When the two surfaces are connected throughan external load, electricity flows.

The photovoltaic cell is the basic building block of a

photovoltaic system. Individual cells can vary in sizefrom about 1 centimeter (1/2 inch) to about 10

centimeter (4 inches) across. However, one cell only produces 1 or 2 watts, which isn't enough

power for most applications. To increase power output, cells are electrically connected into a

packaged weather-tight module. Modules can be further connected to form an array. The termarray refers to the entire generating plant, whether it is made up of one or several thousand

modules. The number of modules connected together in an array depends on the amount of

power output needed.

The performance of a photovoltaic array is dependent upon sunlight. Climate conditions (e.g.,

clouds, fog) have a significant effect on the amount of solar energy received by a photovoltaicarray and, in turn, its performance. Most current technology photovoltaic modules are about 10

percent efficient in converting sunlight. Further research is being conducted to raise this

efficiency to 20 percent.

The photovoltaic cell was discovered in 1954 by Bell Telephone researchers examining thesensitivity of a properly prepared silicon wafer to sunlight. Beginning in the late 1950s,

photovoltaic cells were used to power U.S. space satellites (learn more about the history of

photovaltaic cells). The success of PV in space generated commercial applications for this

technology. The simplest photovoltaic systems power many of the small calculators and wristwatches used everyday. More complicated systems provide electricity to pump water, power

communications equipment, and even provide electricity to our homes.


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5.3 Some advantages of photovoltaic systems are:

1. Conversion from sunlight to electricity is direct, so that bulky mechanical generatorsystems are unnecessary.

2. PV arrays can be installed quickly and in any size required or allowed.

3. The environmental impact is minimal, requiring no water for system cooling andgenerating no by-products.

Photovoltaic cells, like batteries, generate direct current (DC) which is generally used for small

loads (electronic equipment). When DC from photovoltaic cells is used for commercial

applications or sold to electric utilities using the electric grid, it must be converted to alternatingcurrent (AC) using inverters, solid state devices that convert DC power to AC.

Historically, PV has been used at remote sites to provide electricity. In the future PV arrays may

be located at sites that are also connected to the electric grid enhancing the reliability of the

distribution system.

5.4 SOLAR THERMAL HEAT

Solar thermal(heat) energy is often used for heating swimming pools, heating water used inhomes, and space heating of buildings. Solar space heating systems can be classified as passive

or active.

Passive space heating is what happens to your car on a hot summer day. In buildings, the air is

circulated past a solar heat surface(s) and through the building by convection (i.e. less dense

warm air tends to rise while more dense cooler air moves downward) . No mechanical equipmentis needed for passive solar heating.

.


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5.5: Solar energy technology:

The parabolic dish engine system, which concentrates solar power

Solar energy is the light and radiant heat from the Sun that influences Earth's climate and

weather and sustains life. Solar power is sometimes used as a synonym for solar energy or morespecifically to refer to electricity generated from solar radiation. Since ancient times solar energy

has been harnessed for human use through a range of technologies. Solar radiation along with

secondary solar resources such as wind and wave power, hydroelectricity and biomass accountfor most of the available flow ofrenewable energy on Earth.

Solar energy technologies can provide electrical generation by heat engine or photovoltaicmeans, space heating and cooling in active and passive solar buildings; potable water via

distillation and disinfection, daylighting, hot water, thermal energy for cooking, and high

temperature process heat for industrial purposes.

Insolation and Solar radiation

About half the incoming solar energy reaches the earth's surface.


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The Earth receives 174 petawatts (PW) of incoming solar radiation (insolation) at the upperatmosphere.

[1]Approximately 30% is reflected back to space while the rest is absorbed by

clouds, oceans and land masses. The spectrum of solar light at the Earth's surface is mostly

spread across the visible and near-infrared ranges with a small part in the near-ultraviolet.[2]

The absorbed solar light heats the land surface, oceans and atmosphere. The warm air containingevaporated water from the oceans rises, driving atmospheric circulation or convection. When this

air reaches a high altitude, where the temperature is low, water vapor condenses into clouds,which rain onto the earth's surface, completing the water cycle. The latent heat of water

condensation amplifies convection, producing atmospheric phenomena such as cyclones and

anti-cyclones. Wind is a manifestation of the atmospheric circulation driven by solar energy.[3]

Sunlight absorbed by the oceans and land masses keeps the surface at an average temperature of

14 C.[4]

The conversion of solar energy into chemical energy via photosynthesis produces food,

wood and the biomass from which fossil fuels are derived.[5]

Solar radiation along

with secondary solarresources such as wind

and wave power,

hydroelectricity andbiomass account for

99.97% of the available

renewable energy on

Earth.[10][11]

The totalsolar energy absorbed by

Earth's atmosphere,

oceans and land massesis approximately 3,850 zettajoules (ZJ) per year.[12] In 2002, this was more energy in one hour

than the world used in one year.[13][14]

Photosynthesis captures approximately 3 ZJ per year in

biomass.[15]

The amount of solar energy reaching the surface of the planet is so vast that in oneyear it is about twice as much as will ever be obtained from all of the Earth's non-renewable

resources of coal, oil, natural gas, and mined uranium combined.[16]

From the table of resources it would appear that solar, wind or biomass would be sufficient to supply all

of our energy needs, however, the increased use of biomass has had a negative effect on global

warming and dramatically increased food prices by diverting forests and crops into biofuel

production.[17] As intermittent resources

WEEK SIX

Yearly energy resources & annual energy consumption (TWh)

Solar energy absorbed by atmosphere, oceans and Earth[6]

751,296,000.0

Wind energy (technical potential)[7]

221,000.0

Electricity (2005)[8]

-45.2

Primary energy use (2005) [9] -369.7


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6.0 APPLICATION OF SOLAR ENERGY

Average insolation showing land area (small black dots) required to replace the total world energy

supply with solar electricity. Insolation for most people is from 150 to 300 W/m

2

or 3.5 to 7.0kWh/m

2/day.

Solar energy refers primarily to the use ofsolar radiation for practical ends. All other renewableenergies other than geothermal derive their energy from energy received from the sun.

Solar technologies are broadly characterized as either passive or active depending on the waythey capture, convert and distribute sunlight. Active solar techniques use photovoltaic panels,

pumps, and fans to convert sunlight into useful outputs. Passive solar techniques include

selecting materials with favorable thermal properties, designing spaces that naturally circulateair, and referencing the position of a building to the Sun. Active solar technologies increase the

supply of energy and are considered supply side technologies, while passive solar technologiesreduce the need for alternate resources and are generally considered demand side technologies.

6.1 Architecture and urban planning

Passive solar building design andUrban heat island

Darmstadt University of Technology won the 2007 Solar Decathlon in Washington, D.C. with this passive

house designed specifically for the humid and hot subtropical climate

Sunlight has influenced building design since the beginning of architectural history. Advancedsolar architecture and urban planning methods were first employed by the Greeks and Chinese,

who oriented their buildings toward the south to provide light and warmth.

The common features ofpassive solar architecture are orientation relative to the Sun, compact

proportion (a low surface area to volume ratio), selective shading (overhangs) and thermal mass.When these features are tailored to the local climate and environment they can produce well-lit

spaces that stay in a comfortable temperature range. Socrates' Megaron House is a classic

example of passive solar design. The most recent approaches to solar design use computermodeling tying together solar lighting, heating and ventilation systems in an integrated solar


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design package. Active solar equipment such as pumps, fans and switchable windows can

complement passive design and improve system performance.

6.1.1 Agriculture and horticulture

Agriculture, Horticulture, andGreenhouse

Greenhouses like these in the Netherlands' Westland municipality grow vegetables, fruits and flowers.

Agriculture seeks to optimize the capture of solar energy in order to optimize the productivity of

plants. Techniques such as timed planting cycles, tailored row orientation, staggered heightsbetween rows and the mixing of plant varieties can improve crop yields. While sunlight is

generally considered a plentiful resource, the exceptions highlight the importance of solar energyto agriculture. During the short growing seasons of the Little Ice Age, French and English

farmers employed fruit walls to maximize the collection of solar energy. These walls acted as

thermal masses and accelerated ripening by keeping plants warm. Early fruit walls were builtperpendicular to the ground and facing south, but over time, sloping walls were developed to

make better use of sunlight. In 1699, Nicolas Fatio de Duillier even suggested using a tracking

mechanism which could pivot to follow the Sun Applications of solar energy in agriculture asidefrom growing crops include pumping water, drying crops, brooding chicks and drying chicken

manure More recently the technology has been embraced by vinters, who use the energy

generated by solar panels to power grape presses.

Greenhouses convert solar light to heat, enabling year-round production and the growth (in

enclosed environments) of specialty crops and other plants not naturally suited to the localclimate. Primitive greenhouses were first used during Roman times to produce cucumbers year-

round for the Roman emperor Tiberius. The first modern greenhouses were built in Europe in the

16th century to keep exotic plants brought back from explorations abroad. Greenhouses remainan important part of horticulture today, and plastic transparent materials have also been used to

similar effect in polytunnels and row covers.

6.2 Solar lighting


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Daylighting features such as this oculus at the top of the Pantheon in Rome have been in use since

antiquity.

The history of lighting is dominated by the use of natural light. The Romans recognized a right tolight as early as the 6th century and English law echoed these judgments with the PrescriptionAct of 1832. In the 20th century artificial lighting became the main source of interior

illumination but daylighting techniques and hybrid solar lighting solutions are ways to reduceenergy consumption.

Daylighting systems collect and distribute sunlight to provide interior illumination. This passive

technology directly offsets energy use by replacing artificial lighting, and indirectly offsets non-solar energy use by reducing the need for air-conditioning. Although difficult to quantify, the use

ofnatural lighting also offers physiological and psychological benefits compared to artificial

lighting. Daylighting design implies careful selection of window types, sizes and orientation;exterior shading devices may be considered as well. Individual features include sawtooth roofs,

clerestory windows, light shelves, skylights and light tubes. They may be incorporated into

existing structures, but are most effective when integrated into a solar design package that

accounts for factors such as glare, heat flux and time-of-use. When daylighting features areproperly implemented they can reduce lighting-related energy requirements by 25%.

6.3 Solar thermal

Solar thermal energy

Solar thermal technologies can be used for water heating, space heating, space cooling andprocess heat generation.

(a)Water heating

Solar hot waterandSolar combisystem


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Solar water heaters facing the Sun to maximize gain

Solar hot water systems use sunlight to heat water. In low geographical latitudes (below

40 degrees) from 60 to 70% of the domestic hot water use with temperatures up to 60 C can beprovided by solar heating systems. The most common types of solar water heaters are evacuated

tube collectors (44%) and glazed flat plate collectors (34%) generally used for domestic hot

water; and unglazed plastic collectors (21%) used mainly to heat swimming pools.

6.3.1 Heating, cooling and ventilation

Solar heating, Thermal mass, Solar chimney, andSolar air conditioning

MIT's Solar House #1, built in 1939, used seasonal thermal storage for year-round heating.

In the United States, heating, ventilation and air conditioning (HVAC) systems account for 30%(4.65 EJ) of the energy used in commercial buildings and nearly 50% (10.1 EJ) of the energyused in residential buildings. Solar heating, cooling and ventilation technologies can be used to

offset a portion of this energy.

Thermal mass is any material that can be used to store heatheat from the Sun in the case of

solar energy. Common thermal mass materials include stone, cement and water. Historically they

have been used in arid climates or warm temperate regions to keep buildings cool by absorbing

solar energy during the day and radiating stored heat to the cooler atmosphere at night. Howeverthey can be used in cold temperate areas to maintain warmth as well. The size and placement of

thermal mass depend on several factors such as climate, daylighting and shading conditions.

When properly incorporated, thermal mass maintains space temperatures in a comfortable rangeand reduces the need for auxiliary heating and cooling equipment.

A solar chimney (or thermal chimney, in this context) is a passive solar ventilation system

composed of a vertical shaft connecting the interior and exterior of a building. As the chimney


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warms, the air inside is heated causing an updraft that pulls air through the building.

Performance can be improved by using glazing and thermal mass materials in a way that mimicsgreenhouses

6.4 Water treatment

Solar still, Solar water disinfection, Solar desalination, andSolar Powered Desalination Unit

Application of SODIS technology in Indonesia to water disinfection

Solar distillation can be used to make saline or brackish water potable. The first recordedinstance of this was by 16th century Arab alchemists.

[49]A large-scale solar distillation project

was first constructed in 1872 in the Chilean mining town of Las Salinas. The plant, which had

solar collection area of 4,700 m, could produce up to 22,700 L per day and operated for

40 years. Individual still designs include single-slope, double-slope (or greenhouse type),vertical, conical, inverted absorber, multi-wick, and multiple effect. These stills can operate in

passive, active, or hybrid modes. Double-slope stills are the most economical for decentralized

domestic purposes, while active multiple effect units are more suitable for large-scale

applications.

Small scale solar powered sewerage treatment plant

Solar energy may be used in a water stabilisation pond to treat waste water without chemicals orelectricity. A further environmental advantage is that algae grow in such ponds and consume

carbon dioxide in photosynthesis.


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6.5 Cooking

Solar cooker

The Solar Bowl in Auroville, India, concentrates sunlight on a movable receiver to produce steam for

cooking.

Solar cookers use sunlight for cooking, drying and pasteurization. They can be grouped intothree broad categories: box cookers, panel cookers and reflector cookers. repositioned to track

the Sun. The solar bowl is a concentrating technology employed by the Solar Kitchen in

Auroville, India, where a stationary spherical reflector focuses light along a line perpendicular tothe sphere's interior surface, and a computer control system moves the receiver to intersect this

line. Steam is produced in the receiver at temperatures reaching 150 C and then used for process

heat in the kitchen.[60]

Unglazed transpired collectors (UTC) are perforated sun-facing walls used for preheating

ventilation air. UTCs can raise the incoming air temperature up to 22 C and deliver outlettemperatures of 4560 C. The short paybackperiod of transpired collectors (3 to 12 years)

makes them a more cost-effective alternative than glazed collection systems.[67]

As of 2003, over

80 systems with a combined collector area of 35,000 m had been installed worldwide, including

an 860 m collector in Costa Rica used for drying coffee beans and a 1,300 m collector inCoimbatore, India used for drying marigolds.

6.6 Electrical generation

Sunlight can be converted into electricity using photovoltaics (PV), concentrating solar power(CSP), and various experimental technologies. PV has mainly been used to power small and

medium-sized applications, from the calculator powered by a single solar cell to off-grid homespowered by a photovoltaic array. For large-scale generation, CSP plants like SEGS have been the

norm but recently multi-megawatt PV plants are becoming common. Completed in 2007, the

14 MW power station in Clark County, Nevada and the 20 MW site in Beneixama, Spain arecharacteristic of the trend toward larger photovoltaic power stations in the US and Europe.


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11 MW Serpa solar power plant in Portugal

A solar cell, or photovoltaic cell (PV), is a device that converts light into direct current using thephotoelectric effect. The first solar cell was constructed by Charles Fritts in the 1880s. The

earliest significant application of solar cells was as a back-up power source to the Vanguard Isatellite, which allowed it to continue transmitting for over a year after its chemical battery was

exhausted. The successful operation of solar cells on this mission was duplicated in many other

Soviet and American satellites, and by the late 1960s, PV had become the established source of

power for them. Photovoltaics went on to play an essential part in the success of earlycommercial satellites such as Telstar, and they remain vital to the telecommunications

infrastructure today.

The high cost of solar cells limited terrestrial uses throughout the 1960s. This changed in the

early 1970s when prices reached levels that made PV generation competitive in remote areas

without grid access. Early terrestrial uses included powering telecommunication stations, off-

shore oil rigs, navigational buoys and railroad crossings. These off-grid applications have provenvery successful and accounted for over half of worldwide installed capacity until 2004.

Building-integrated photovoltaics cover the roofs of the increasing number of homes.

The 1973 oil crisis stimulated a rapid rise in the production of PV during the 1970s and early1980s.

[77]Economies of scale which resulted from increasing production along with

improvements in system performance brought the price of PV down from 100 USD/watt in 1971

to 7 USD/watt in 1985.[78]

Steadily falling oil prices during the early 1980s led to a reduction in


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funding for photovoltaic R&D and a discontinuation of the tax credits associated with the Energy

Tax Act of 1978. These factors moderated growth to approximately 15% per year from 1984through 1996.

6.7 Concentrating solar power

Concentrating solar power

Solar troughs are the most widely deployed and the most cost-effective CSP technology.

Concentrated sunlight has been used to perform useful tasks since the time ofancient China. Alegend claims that Archimedes used polished shields to concentrate sunlight on the invadingRoman fleet and repel them from Syracuse. Auguste Mouchout used a parabolic trough to

produce steam for the first solar steam engine in 1866, and subsequent developments led to the

use of concentrating solar-powered devices for irrigation, refrigeration and locomotion.

The PS10 concentrates sunlight from a field of heliostats on a central tower.

A solar trough consists of a linear parabolic reflector that concentrates light onto a receiver

positioned along the reflector's focal line. The reflector is made to follow the Sun during the

daylight hours by tracking along a single axis. Trough systems provide the best land-use factor of

any solar technology.

A solar power tower uses an array of tracking reflectors (heliostats) to concentrate light on a central

receiver atop a tower. Power towers are less advanced than trough systems but offer higher efficiency

and better energy storage capability The Solar Two in Barstow, California and the Planta Solar 10 in

Sanlucar la Mayor, Spain are representatives of this technology.


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6.8 Solar vehicles

Solar vehicle, Electric boat, andSolar balloon

Australia hosts the World Solar Challenge where solar cars like the Nuna3 race through a 3,021 km

(1,877 mi) course from Darwin to Adelaide.

Development of a solar powered car has been an engineering goal since the 1980s. The WorldSolar Challenge is a biannual solar-powered car race, where teams from universities and

enterprises compete over 3,021 kilometres (1,877 mi) across central Australia from Darwin to

Adelaide. In 1987, when it was founded, the winner's average speed was 67 kilometres per hour(42 mph) and by 2007 the winner's average speed had improved to 90.87 kilometres per hour

(56.46 mph). The North American Solar Challenge and the planned South African SolarChallenge are comparable competitions that reflect an international interest in the engineering

and development of solar powered vehicles.

Some vehicles use solar panels for auxiliary power, such as for air conditioning, to keep the

interior cool, thus reducing fuel consumption.

Helios UAV in solar powered flight


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In 1974, the unmanned Sunrise IIplane made the first solar flight. On 29 April 1979, the SolarRisermade the first flight in a solar powered, fully controlled, man carrying flying machine,reaching an altitude of 40 feet (12 m). In 1980, the Gossamer Penguin made the first piloted

flights powered solely by photovoltaics. This was quickly followed by the Solar Challenger

which crossed the English Channel in July 1981. In 1990 Eric Raymond in 21 hops flew from

California to North Carolina using solar power. Developments then turned back to unmannedaerial vehicles (UAV) with the Pathfinder(1997) and subsequent designs, culminating in the

Helios which set the altitude record for a non-rocket-propelled aircraft at 29,524 metres(96,860 ft) in 2001. TheZephyr, developed by BAE Systems, is the latest in a line of record-

breaking solar aircraft, making a 54-hour flight in 2007, and month-long flights are envisioned

by 2010.

A solar balloon is a black balloon that is filled with ordinary air. As sunlight shines on the

balloon, the air inside is heated and expands causing an upward buoyancy force, much like anartificially heated hot air balloon. Some solar balloons are large enough for human flight, but

usage is generally limited to the toy market as the surface-area to payload-weight ratio is

relatively high.

[120]

6.9 Energy storage methods

Thermal mass, Thermal energy storage, Phase change material, Grid energy storage, and V2G

Solar Two's thermal storage system generated electricity during cloudy weather and at night.

Storage is an important issue in the development of solar energy because modern energy systemsusually assume continuous availability of energy.

[123]Solar energy is not available at night, and

the performance of solar power systems is affected by unpredictable weather patterns; therefore,storage media or back-up power systems must be used.

Thermal mass systems can store solar energy in the form of heat at domestically usefultemperatures for daily or seasonal durations. Thermal storage systems generally use readily

available materials with high specific heat capacities such as water, earth and stone. Well-designed systems can lower peak demand, shift time-of-use to off-peakhours and reduce overallheating and cooling requirements. Phase change materials such as paraffin wax and Glauber's

salt are another thermal storage media. These materials are inexpensive, readily available, and

can deliver domestically useful temperatures (approximately 64 C). The "Dover House" (inDover, Massachusetts) was the first to use a Glauber's salt heating system, in 1948.


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Solar energy can be stored at high temperatures using molten salts. Salts are an effective storage

medium because they are low-cost, have a high specific heat capacity and can deliver heat attemperatures compatible with conventional power systems. The Solar Two used this method of

energy storage, allowing it to store 1.44 TJ in its 68 m storage tank with an annual storage

efficiency of about 99%.Off-grid PV systems have traditionally used rechargeable batteries to

store excess electricity. With grid-tied systems, excess electricity can be sent to the transmissiongrid. Net metering programs give these systems a credit for the electricity they deliver to the

grid. This credit offsets electricity provided from the grid when the system cannot meet demand,effectively using the grid as a storage mechanism.

Pumped-storage hydroelectricity stores energy in the form of water pumped when energy isavailable from a lower elevation reservoir to a higher elevation one. The energy is recovered

when demand is high by releasing the water to run through a hydroelectric power generator.

6.10 Development, deployment and economics

Deployment of solar power to energy grids

Nellis Solar Power Plant, the largest photovoltaic power plant in North America

Beginning with the surge in coal use which accompanied the Industrial Revolution, energyconsumption has steadily transitioned from wood and biomass to fossil fuels. The earlydevelopment of solar technologies starting in the 1860s was driven by an expectation that coal

would soon become scarce. However development of solar technologies stagnated in the early

20th century in the face of the increasing availability, economy, and utility of coal andpetroleum.[130]

The 1973 oil embargo and 1979 energy crisis caused a reorganization of energy policies around

the world and brought renewed attention to developing solar technologies.

[131][132]

Deploymentstrategies focused on incentive programs such as the Federal Photovoltaic Utilization Program in

the US and the Sunshine Program in Japan. Other efforts included the formation of researchfacilities in the US (SERI, now NREL), Japan (NEDO), and Germany (Fraunhofer Institute for

Solar Energy Systems ISE).[133]

Between 1970 and 1983 photovoltaic installations grew rapidly, but falling oil prices in the early

1980s moderated the growth of PV from 1984 to 1996. Since 1997, PV development has


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accelerated due to supply issues with oil and natural gas, global warming concerns (see Kyoto

Protocol), and the improving economic position of PV relative to other energytechnologies.[citation needed] Photovoltaic production growth has averaged 40% per year since 2000

and installed capacity reached 10.6 GW at the end of 2007.[42]

Since 2006 it has been economical

for investors to install photovoltaics for free in return for a long term power purchase agreement.

50% of commercial systems were installed in this manner in 2007 and it is expected that 90%will by 2009.[134]

Nellis Air Force Base is receiving photoelectric power for about 2.2 /kWh and

grid power for 9 /kWh.[135][136]

Commercial solar water heaters began appearing in the United States in the 1890s.[137]

These

systems saw increasing use until the 1920s but were gradually replaced by cheaper and morereliable heating fuels.[138] As with photovoltaics, solar water heating attracted renewed attention

as a result of the oil crises in the 1970s but interest subsided in the 1980s due to falling

petroleum prices. Development in the solar water heating sector progressed steadily throughoutthe 1990s and growth rates have averaged 20% per year since 1999.[41] Although generally

underestimated, solar water heating is by far the most widely deployed solar technology with an

estimated capacity of 154 GW as of 2007.

[41]

Solar installations in recent years have also largely begun to expand into residential areas, with

governments offering incentive programs to make "green" en stallation (sized between 1.3 kWand 5 kW) is estimated at 18 to 23 years, considering such cost factors as parts, installation and

maintenance, as well as the average energy production of a system on an annual basis.

WEEK 7

7.0 Entropy

The first law of thermodynamics deals with the property energy and the

conservation of it. The second law leads to the definition of a new property

called entropy. Entropy is a somewhat abstract property, and it is difficult

to give a physical description of it. Entropy is best understood and


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appreciated by studying its uses in commonly encountered engineering

processes and this is what we intend to do. Entropy is a non conserved

property, and there is no such thing as a conservation of entropy principle.

The second law of thermodynamics of the leads to expressions that

involve inequalities, an irreversible (i.e. actual) heat engine, for example, is

less efficient than a reversible one operating between the same two

thermal energy reservoirs. Likewise, an irreversible refrigerator or a heat

pump has a lower coefficient of performance (COP) than a reversible one

operating between the same temperature limits.

Entropy i

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