Chapter 5.pdf

Thermodynamics

Chapter 5: The First Law ofThermodynamics

Joule’s Experiment

During 1843-1849, Joule conducted several

experiments that subsequently led to the

formulation of the first law of thermodynamics.

Thermodynamics – Chapter 5

James Prescott Joule

Joule's Paddle-Wheel Apparatus,

1849, London Science Museum


� Experiment consists of two process cycle.

� In process 1-2, work W was done on the system

by means of a paddle wheel.

� Heat was transferred from the fluid to the

atmosphere in process 2-1 till the system returns

to the original temperature.


� The amount of work was measured by the

change in potential energy of a weight mg

falling through a height Z.

� The amount of heat transfer Q from the fluid to

the atmosphere was measured.

� By repeating the experiments for different

systems, Joule found that net work input W was

always proportional to the net amount of heat

transfer Q from the system measured in their

conventional units.

δ δJ Q W=∫ ∫� �Thermodynamics – Chapter 5

� Joule found that in the English system of units,

in which historically heat is measured in British

thermal unit (Btu) and work in foot pound

force (ft lbf), the mechanical equivalent of heat

is

J = 778.17 ft lbf/Btu

� In SI units, both heat and work are expressed in

same unit.


Mechanical Equivalent of Heat (J)

Joule experimented on the amount of

mechanical work needed to raise the

temperature of a pound of water by

one degree Fahrenheit and found a

consistent value of 778.17 foot

pound force (4.186 J/cal).


Observations From Joule’s Experiment

� Net work input W is always proportional to the

net amount of heat transfer Q from the system.

� Since the system at the end is restored to its

original state, so, algebraic sum of heat and work

interactions during a thermodynamic cycle is

zero (when same units are used).

� This experiment led to the formulation of the first

law of thermodynamics.


Statement of First Law of thermodynamics

For a system undergoing a thermodynamic cycle

the net heat transfer during the cycle (cyclic

integral of heat added) is equal to the net work

done during the cycle (cyclic integral of work

done).

δ δQ W=∫ ∫� �


� The symbol δδδδ is used to indicate that W and Q

are inexact differentials.

� The constraints of this law: This law applies to a

closed system and only to a thermodynamic

cycle.

� Both heat and work have same units.

� The first law of thermodynamics is also known as

the conservation of energy principle.


Some aspects of the first law

First Law of thermodynamics for a

Change in State of a Control mass

2 1 2 1

1 2 1 2

A B A BQ Q W Wδ δ δ δ+ = +∫ ∫ ∫ ∫

Path 1-A-2-B-1

Path 1-C-2-B-1

( ) ( )2 2

1 1

δ δ δ δA C

Q W Q W− = −∫ ∫

Subtracting and rearranging

2 1 2 1

1 2 1 2

C B C BQ Q W Wδ δ δ δ+ = +∫ ∫ ∫ ∫

P

V2 2 2 2

1 1 1 1

A C A CQ Q W Wδ δ δ δ− = − ⇒∫ ∫ ∫ ∫


� It is seen that for any process between states 1and 2, the quantity (δQ – δW) is always thesame.

� It is independent of the path followed for thischange of state.

� So, it is a point or state function and hence aproperty. This property is called energy (E) ofthe control mass.

� Thus we can write,

------------ (1)dE Q Wδ δ= −


2 1 1 2 1 2E E Q W− = −

Since E is a property, its derivative can be written as

dE. When Eq. (1) is integrated from an initial state 1

to a final state 2,

where E1 and E2 are the initial and final values of

energy of the control mass.

� For a cycle, initial and final state is identical, so

energy increase is zero.

� For isolated system W and Q = 0. Hence E2 = E1


Significance of the Property E

• It represents all the energy of the system

(macroscopic and microscopic forms) in the

given state.

• Macroscopic: Kinetic or Potential energy of the

system as a whole with respect to the chosen

coordinate frame.

• Microscopic: Energy associated with motion and

position of the molecules, and the structure of the

atom is called internal energy (U).


( ) ( ) ----- (2)dE dU d KE d PE= + +

The first law of thermodynamics for a change of state

( ) ( ) --- (3)dE dU d KE d PE Q Wδ δ= + + = −

where2

and( )

( )2

( ) ( )

d mVd KE mVdV

d PE mg dZ d mg Z

= =

= =


• In absence of electrical, magnetic, surface tension

effects

• i.e., for a simple compressible system

E = U + KE + PE

Integrating for a change of state from state 1 to state

2 with constant g,

2 1 1 2 1 2

2 2

2 1

2 1 2 1 1 2 1 22 2

E E Q W

mV mVU U mgZ mgZ Q W

− = −

− + − + − = −


--- (4)dE dU mVdV mgdZ Q Wδ δ= + + = −

Three Observations

� The property E, the energy of the control mass,

was found to exist.

� The net change of the energy of the control mass

is always equal to the net transfer of energy across

the boundary as heat and work.


� This equation can give only changes in internal

energy, kinetic energy, and potential energy.

in out System

Net Energy Transfer by heat and work Change in U, K.E and P.E energies

− = ∆14243 14243E E E

Internal Energy – A Thermodynamic Property

• Total energy contained by a thermodynamic system

• It is an extensive property and can be one of the

independent properties of a pure substance

• The symbol U designates the internal energy of a

given mass of a substance and u as the specific

internal energy.

• In the liquid – vapor saturation region,

(1 ) ( )

liq vap

lq f vap g

f g f g f

f fg

U U U

mu m u m u

u x u xu u x u u

u u xu

= +

= +

= − + = + −

= +


Internal energy

Calculate the specific internal energy of

saturated steam having a pressure of 0.6

MPa and quality of 95%

u=uf + xufg

= 669.9 + 0.95 (1897.5) = 2472/5 kJ/kg

(E 5.4) Determine the missing property (P, T or

x) and v for water at

(a) T=300oC, u = 2780 kJ/kg

From table B.1.1, Psat = 8581kPa, ug = 2563 kJ/kg. So

given u > ug so the state is in the superheated vapor

region at some P < Psat

Searching through table B 1.3 at 300 C (superheated

vapor), u = 2780 kJ/kg at 1600 < P < 1800 kPa.

By linear interpolation, P = 1648 kPa and v = 0.1542 m3/kg

x is undefined (superheated).

Problem analysis and solution technique

1. What is the control mass/volume?

2. What is the initial state (known properties)?

3. What do we know about the final state?

4. What is the process that takes place? Is anything

constant or zero? Any functional relation between

properties?

5. Would a P-v, T-v diagram be useful?

6. What is the thermodynamic model for the behavior of

the substance (steam tables, ideal gas law, etc.)?

7. Use of all of the above information to analyze and solve

the problem

Enthalpy -The Thermodynamic Property

• Consider a control mass

undergoing quasi-equilibrium

constant-pressure process.

• Assume there is no change in

kinetic or potential energy .

• Work done during the process

is that associated with the

boundary movement.


Taking the gas as our control mass and applying the

first law,

1 2 2 1 1 2Q U U W= − +

The work can be calculated from the relation,

2 2

1 2 2 1

1 1

( )W PdV P dV P V V= = = −∫ ∫Therefore,

1 2 2 1 2 2 1 1

2 2 2 1 1 1( ) ( )

Q U U PV PV

U PV U PV

= − + −

= + − +


� Heat transfer during the process is given in terms

of the change in the quantity U + PV between

the initial and final states.

� Because all these quantities are thermodynamic

properties, that is, functions only of the state of

the system, their combination must have these

same characteristics.

� Therefore, it is convenient to define a new

extensive property, the enthalpy,

H = U + PV


� The heat transfer in a constant-pressure, quasi-

equilibrium process is equal to the change in

enthalpy, which includes both the change in U and

the work for this particular process

� Many tables and charts of thermodynamics properties

give values for enthalpy but not for the internal

energy. In that case, u can be determined by

u = h – pv

� Is this valid when analyzing system processes that do

not occur at constant pressure?

� Remember enthalpy is a state function so it can be used to

calculate u irrespective of the process

The enthalpy of a substance in a saturation state and

with a given quality is found in the same way as the

specific volume and internal energy.

(1 )

liq vap

lq f vap g

f g

f fg

H H H

mh m h m h

h x h xh

h h xh

= +

= +

= − +

= +

For substances for which compressed-liquid tables

are not available, the enthalpy is taken as that

saturated liquid at the same temperature.


Engineering Thermodynamics

(5.14) 2 kg of water at 120oC with a quality of 25%

has its temperature raised to 20oC in a

constant volume process. What are the new

quality and specific internal energy ?

State 1, at 120oC

1 1 11 1

3

( )

= 0.001061+ 0.25 (0.89152 - 0.001061)

= 0.22367 m /

f g fx

kg

ν ν ν ν= + −

State 2 has same ν at 140oC


( )2

2

2 2

2

2

2

0.22367 - 0.00108 0.4386

0.50849 - 0.00108

588.72 + 0.4386 1961.3 = 1448.9 kJ/kg

f

fg

f fg

x

u u xu

ν ν

ν

−= = =

= +

= ×


(5.15) 2 kg of water at 200 kPa with a quality of

25% has its temperature raised to 20oC in a

constant pressure process. What is the change

in enthalpy ?

State 1, at 200 kPa

State 2 has same pressure

1 11 1= 504.7 + 0.25 2201.6

= 1055.1 kJ /

f fgh h x h

kg

= + ×

o

220 120.2 + 20 = 140.2 C

satT T= + =

So state 2 is superheated vapor


( )2

140.2 - 120.2 2706.3 + 2768.5 - 2706.3

150 120.2

2748 kJ/kg

h =−

=

2 1 2748 - 1055.1

= 1693 kJ/kg

h h− =

Specific Heat

� It takes different amounts of energy to raise thetemperature of identical masses of differentsubstances by 1 degree.

� This is because different substances havedifferent energy storing capabilities.


� It is thus desirable to have a property that enables

us to compare the energy storage capabilities of

various substances. This property is the specific

heat.

� It is defined as the energy required to raise the

temperature of a unit mass of a substance by one

degree.


� The SI unit of specific heat is kJ/kg.K.

� Denoting specific heat by C, we have by

definition

Mathematical Model

1q Qc

T m T

δ δ

δ δ

= =


Specific Heat at Constant Volume and

Pressure

� Cv is the energy required to raise the temperature

of a unit mass of a substance by one degree as the

volume is maintained constant.

� Cp is the energy required to do the same at

constant pressure is the specific heat at constant

pressure


� Which is greater in value? Cp or Cv?

� The specific heat at constant pressure will alwaysbe greater than that at constant volume becauseat constant pressure the system can changevolume and energy for this expansion must besupplied.


Specific Heat at Constant Volume

From definition of specific heat

1V

V V

q Qc

T m T

δ δ

δ δ

= =

Since for constant vo 0lum

e P dV

dU Q Wδ δ

=

= − (from First Law)

1V

V V V

q u Uc

T T m T

δ

δ

∂ ∂ = = =

∂ ∂


Specific Heat at Constant Pressure

From definition of specific heat

1P

P P

q Qc

T m T

δ δ

δ δ

= =

1 2 2 1 2 2 1 1

2 2 2 1 1 1 2 1

Since for constant pressure process

(

first

( )

)

w)

(

la

Q U U PV PV

U PV U PV H H

= − + −

= + − + = −

Therefore, 1

P

P P P

q h Hc

T T m T

δ

δ

∂ ∂ = = =

∂ ∂


Observations

� Both expressions for specific heats contain

thermodynamic properties and can be written as

cv = f (u, T, ν) and cP = f (h, T, P )

� cv and cp are state functions and hence are

properties.

� cv is a measure of the variation of internal energy

of a substance with temperature.

� cP is a measure of the variation of enthalpy of a

substance with temperature


� Note the equations for cP and cv are property

relations and as such independent of the type of

process. They are valid for any substance

undergoing any process.

Sketch showing two ways in which a given ∆Umay be achieved


dh du c dT≈ ≈

2 1 2 1 2 1( )h h u u c T T− − −� �


Specific Heat for Solids and Liquids

Also, for both of these phases, the specific volumeis very small, such that in many cases

Where c is either the constant-volume or constantpressure specific heat, as the two would be nearlythe same.

Since both of phases are nearly incompressible,

( )dh du d Pv du vdP Pdv= + = + + du vdP≈ +

� In general, u depends on the two independent propertiesspecifying the state

�For a low-density gas, however u depends primarily on T andmuch less on the second term, P or v.

Internal Energy, Enthalpy and Specific

Heat Relations for Ideal Gases


� For an ideal gas, Pv = RT and u = f (T )

0 0 0 or = ⇒ = =

v v v

duc du c dT dU mc dT

dT


� The relation between the internal energy and thetemperature can be given by

∂ =

∂ v

v

uc

T

� Because the internal energy of an ideal gas is nota function of specific volume, for an ideal gas

Where the subscript 0 denotes the specific heat ofan ideal gas

� From definition of enthalpy, we have

� Since R is a constant and u = f (T), it follows that

h = f (T )

0 0 0 or

p

p

p p p

hc

T

dhc dh c dT dH mc dT

dT

∂ =

∂

= ⇒ = =


� The relation between the enthalpy and thetemperature can be given by

h u pv u RT= + = +

Pressure – volume diagram for an ideal gas

Since u and h are f (T ) for an ideal gas, lines of

constant T are also lines of constant u and constant h

No matter what the path, (1-2, 1-2’, 1-2”), change in u

(and h) remains the same.

0 0( ), ( )

v pc f T c f T= =


� For an ideal gases,


� The ideal gas specific heat for a given substance

is often called the zero-pressure specific heat.

� The zero-pressure, constant-volume specific heat

is given by Cvo.

� The zero-pressure, constant-pressure specific

heat is given by Cpo.

( ) ( )0 0

and v p

du c T dT dh c T dT= =

•Heat capacity for some gases as function of temperature.

•These values are determined by the techniques of statistical thermodynamics.

• The principal factor causing c to vary is molecular vibration.

• Complex molecules have multiple vibration modes, so greater T dependency


The change in internal energy or enthalpy for an ideal

gas during a process from state 1 to 2,

To carry out the integrations relation for cP0 as

functions of T is required. There are three possibilities

to examine.

1) Assume constant specific heat, that is, no temperature

dependence (Table A.5).

True for some cases (fig) and a reasonable approximation if

Cp in the relevant T range is used


2 1 0 2 1( )p

h h h C T T∆ = − = −

( )2 1 0ph h h C T dT∆ = − = ∫

2) Use an analytical equation cP0 as a function oftemperature.

� Because the results of specific-heat calculations fromstatistical thermodynamics don’t lend themselves toconvenient mathematical forms, these results have beenapproximated empirically.

� The equations for cP0 as a function of temperature arelisted in Table A.6 (Page 7) for a number of gases.


3) To integrate the results of the calculations ofstatistical thermodynamics from an arbitraryreference temperature to any other temperatureT and to define a function


0

2 1

2 1

0 0

0

2 1 0 0

T

T p

T

T T

p p T T

T T

h C dT

h h C dT C dT h h

=

− = − = −

∫

∫ ∫

and it is seen that the reference temperaturecancels out. This function hT is listed for air inTable A.7 (Page 8).

Specific Heat Relations for Ideal Gases

� It has been noted that for an ideal gas


0 0

0 0

Differentation and substituting for and

= + = +

= +

= +

− =

p v

p v

dh du

h u pv u RT

dh du RdT

C dT C dT RdT

C C R

� The ratio of two specific heats of a gas is called

the specific heat ratio/heat capacity

ratio/adiabatic index and denoted by k or γ where

k = γγγγ = cP0 / cv0

� This ratio also is a function of temperature, but

the variation is mild.

� For monatomic gases, its value is essentially

constant at 1.667.

� Many diatomic gases, including air, have a

specific ratio of about 1.4 at room temperature.


(5.82 /5.83) Use the ideal gas air table A.7 toevaluate the heat capacity Cp at 300 K as a slopeof the curve h(T) by ∆h/∆T. How much larger isit at 1000 K and 1500 K.


320 290

0300 K

320 290

320.58 290.431.005 kJ/kg.K

30

p

h hdh hc

dT T

−∆⇒ = = =

∆ −−

= =

1050 950

01000 K

1050 950

1103.48 989.441.14 kJ/kg.K

100

p

h hdh hc

dT T

−∆⇒ = = =

∆ −−

= =


1550 1450

01500 K

1550 1450

1696.45 1575.41.21 kJ/kg.K

100

p

h hdh hc

dT T

−∆⇒ = = =

∆ −−

= =

Notice an increase of 14%, 21% respectively.

(5.83/5.82) We want to find the change in u forcarbon dioxide between 600 K and 1200 K.

a) Find it from a constant Cvo from table A.5

b) Find it from a Cvo evaluated from equation inA.6 at the average T.

c) Find it from the values of u listed in table A.8


2 1 0 2 1(a) ( )

0.653 kJ/kg.K (1200 600) K

391.8 kJ/kg

vu u u C T T∆ = − ≅ −

= × −

=


1 2

avg

avg

2 3

0 0 1 2 3

2 3

0

0 0

2 1 0 2 1

1200 600(b) 900

2 2

9000.9

1000 1000

(KJ/kg)

0.45 1.67 0.9 1.27 0.9 0.39 0.9

1.2086 kJ/kg.K

1.2086 0.1889 1.0197 kJ/kg.K

( )

1.0197 kJ.kg

p

p

v p

v

T TT

T

C C C C C

C

C C R

u u u C T T

θ

θ θ θ

+ += = =

= = =

= + + +

= + × − × + ×

=

= − = − =

∆ = − = −

= ( ).K 1200 600 K 611.8 kJ/kg− =

The First Law as a Rate Equation

The first law can be expressed as a rate equation i.e.,either the instantaneous or average rate at whichenergy crosses the control surface as heat and workand the rate at which the energy of the control masschanges.

This form of the first law finds extensiveapplications in thermodynamics, fluid mechanics,and heat transfer. ( ) ( )

d KE d PEdUQ W

dt dt dt

dEQ W

dt

+ + = −

= −

& &

& &


)(

)(

.

.

power

HTrate

W

Q


(5.10) A pot of water is boiling on a stove supplying

325 W to the water. What is the rate of mass

(kg/sec) vaporizing, assuming a constant

pressure process ?

=

325 W0.144 g/sec

2257 kJ/kg

fg

fg

fg

Q dE W du PdV dH h dm

Q dmh

dt dt

dm Q

dt h

δ δ

δ

= + + = =

=

= = =

�

Assume that the process takes place at atmospheric

pressure 101 kPa, so T = 100oC


Airplane takeoff from an aircraft carrier is assisted

by a steam-driven piston/cylinder with an average

pressure of 1250 kPa. A 17,500 kg airplane should

be accelerated from zero to a speed of 30 m/sec with

30% of the energy coming from the steam piston.

Find the needed piston displacement volume.

Hint : Take airplane as control mass

No change in internal or potential energy. Find the

energy increase of the control mass


( )2 2

2 1 2

10 0.5 17500 30

2

7875 kJ

E E m V− = − = × ×

=

The work supplied by the piston is 30% of the

energy increase.

2 1

3

0.30( )

0.30 7875 = 2362.5 kJ

2362.5 = 1.89 m

1250

piston avg

avg

W PdV P V E E

WV

P

= = ∆ = −

= ×

∆ = =

∫


(5.35/5.34) A 100-L rigid tank contains nitrogen(N2) at 900 K, 3 MPa. The tank is now cooled to100 K. What are the work and heat transfer forthis process?

2 1

2 1 1 2 1 2

2 1 1 2

( )

C.V. : Nitrogen in tank.

Energy equation :

Process : constant an 0d

m m

m u u Q W

Vv v

mV W

=

− =

= == ⇒

−

=

1 1

31 1

State 1: superheated

Tab

900 K, 3 Mpa

0.090 m /kg, 691.7 kJ/kle B.6.2 : g

T P

v u

= = ⇒

= =


3

31

0.1 m1.111kg

0.090 m /kg

Vm

v= = =

32 2 1 @100K

3

3

2

100 K, 0.090 m /kg

200kPa : 0.1425 m /kg, 71.73 kJ/kg

400kPa

State 2:

superheated

look in T

: 0.06806 m /kg,

able B.6.2 at 1

69.30 kJ/k

00 K

interpolation give

g

341kPas : ,

gT v v v

v u

v u

P u

= = = >

= =

= =

=

⇒

2 70.0 kJ/kg=


( )1 2 2 1 1 2( )

1.111 kg 70.0 691.7 kJ/kg 0

= 690.7 kJ

Q m u u W= − +

= − +

−


(5.135/5.153) A small flexible bag contains 0.1 kgammonia at –10oC and 300 kPa. The bag material issuch that the pressure inside varies linearly withvolume. The bag is left in the sun with an incidentradiation of 75 W, losing energy with an average 25 Wto the ambient ground and air. After a while the bag isheated to 30oC at which time the pressure is 1000 kPa.Find the work and heat transfer in the process and theelapsed time.


Take Ammonia as constant mass.

2 1

2 1 1 2 1 2

Continuity Eq.

Energy equation: ( - )

Process: (linear in V)

m m m

m u u Q W

P A BV

= =

= −

= +

State 1: from table B.2 (page 692/720) compressedliquid P > Psat therefore take saturated liquid atsame temperature.

o 3

1

1

( 10 C) = 0.001534 m /kg,

= 133.96 kJ/kg

f

f

v v

u u

= −

=


State 2: Table B.2.1 (page no. 692/720) at 30oC :

P < Psat so superheated vapor

3

2 2

3

2 2

From superheated table B.2.2 (page no. 696)

= 0.13206 m /kg, = 1347.1 kJ/kg,

= = 0.0132 m

v u

V mv

⇒

Work is done while flexible bag bulge outwards atincreasing pressure, so we get

( ) ( )

1 2 1 2 2 1

1( ) ( )

2

1300 1000 0.1 0.13206 0.001534

2

8.484 J

W P P V V= + −

= + × −

=


Heat transfer is found from the energy equation

1 2 2 1 1 2

.

.

1 2

( )

= 0.1 (1347.1-133.96) + 8.484

= 121.314 + 8.484 = 129.8 kJ

75 25 50 Watts

129800/ 2596 sec = 43.3 min

50

net

net

Q m u u W

Q

t Q Q

= − +

= − =

= = =


(5.28/Q) Find the missing properties.

o 3

2a. H 0, 250 C, = 0.02 m / kg, ? ?T P uν= = =

( ) ( )

oTable B.1.1 at 250 C: 3973 kPa

/ 0.02 - 0.001251 / 0.04887

= 0.38365

1080.37 + 0.38365 1522

= 1664.28 kJ/kg

f g sat

f g

f fg

P

x

u u xu

ν ν ν

ν ν ν

< < ⇒ =

= − =

= + = ×


2c. H O 2 C, 100 kPa ? ?

oT P u ν= − = = =

(5.37/5.39) A cylinder fitted with a frictionlesspiston contains 2 kg of superheated refrigerantR-134a vapor at 350 kPa, 100oC. The cylinder isnow cooled so the R-134a remains at constantpressure until it reaches a quality of 75%.Calculate the heat transfer in the process


2 1

1 2 1 2

C.V.: R 134a

Energy Eq.

Process : P const.

( 2 1)

1W2 PdV P V

P(V2 V1) Pm(v )

2 v1

m m m

m u u Q W

= =

− = −

= ⌠ = ∆

= − = −

−

= =>


State 1: 350 kPa, 100oC Table B.5.2 (pg 711)

h1 = (490.48 + 489.52)/2

= 490 kJ/kg

State 2: 350 kPa, x2=0.75 Table B.5.1 (pg 709)

h2 = 206.75 + 0.75 ×194.57

= 352.7 kJ/kg (P=350.9 kPa, saturation table)


1Q2 = m(u2 - u1) + 1W2

= m(u2 - u1) + Pm(v2 - v1)

= m(h2 - h1)

1Q2 = 2 × (352.7 – 490)= -274.6 kJ

(5.57/5.51) A cylinder having a piston restrained bya linear spring (of spring constant 15 kN/m)contains 0.5 kg of saturated vapor water at 120°C,as shown in Fig. Heat is transferred to the water,causing the piston to rise. If the piston cross-sectional area is 0.05 m2, and the pressure varieslinearly with volume until a final pressure of 500kPa is reached. Find the final temperature in thecylinder and the heat transfer for the process.


C.V. Water in cylinder.

Continuity: m2 = m1 = m ;

Energy : m(u2 - u1) = 1Q2 - 1W2


Process: P2 = P1 + ksm/Ap2 (v2 - v1)

State 2: P2 = 500 kPa

500 kPa = 198.5+15 × 0.5/(0.05)2 (v2 - 0.89186)

⇒ v2 = 0.9924 m3/kg

State 1: 120oC, saturated vapor

Table B.1.1(page 674 )

=> v1 = 0.89186 m3/kg, u1 = 2529.24 kJ/kg

Table B.1.3 => T2 = 803°C; u2 = 3668 kJ/kg

W12 = ⌠ PdV = ((P1 + P2)/2) m(v2 - v1)

= ((198.5 + 500)/2) × 0.5 × (0.9924 - 0.89186)

= 17.56 kJ


1Q2 = m(u2 - u1) + 1W2

= 0.5 × (3668 - 2529.2) + 17.56 = 587 kJ


(5.59/5.60) A 10 m high open cylinder,with Acyl = 0.1m2 contains 20oC waterabove and 2 kg of 20oC water below a198.5 kg thin insulated floating piston.Assume standard g, Po. Now heat isadded to the water below the piston sothat it expands , pushing the piston up,causing the water on top to spill overthe edge. This process continues untilthe piston reaches the top of thecylinder. Find the final state of thewater below the piston (T, P, v) and theheat added during the process.

B

A


( )

2

o

1

3

1 1

(

State 1 is compressed liquid 20 C

From Table B.1.1 (Page no. 674)

0.001002 m / kg 83.94 kJ/kg

0.1 10 998 kg

0.001002mass above the piston

A

A f A f

totaltotal

H O at

T

v v u u

Vm

vm

=

≅ = ≅ =

×= = =

1 ) 1 996 kgB total Am m= − =

( )

2 ( 1 )

1

3

1 1 1

State 1A:

198.5 9.81 + 996 9.81 101 +

0.1 1000 218.2 kPa

0.001002 2 0.002004 m

P H O at B

A o

P

A A A

m g m gP P

A

V v m

+= +

× ×=

×=

= × = × =


( )2

3 322 2

198.5 9.81State 2: 101 +

0.1 1000

120.5 kPa

10.1 10 1 m 0.5 m /kg

2

A o

P

AA cyl A

A

mgP P

A

VV A h v

m

×= + =

×

=

= × = × = = = =

( )

2 2

2 2

2

o 3 3

2

2

From Table B.1.2 120.5 kPa by interpolation

104.8 C, 0.001047 m / kg, 1.43128 m / kg

439.3 kJ / kg, 2070.5 kJ / kg

0.5 0.001047 0.348

1.4312

A A

A A

A

A f fg

f fg

A

P

T v v

u u

x

u

⇒ =

= = =

= =

−= =

2 439.3 0.348 2070.5 1161 kJ/kgA

= + × =


( )( )

2 1 1 2 1 2

1 2 1 2 2 1

Energy equation ( )

1 ( ) ( )

21

= 218.2 120.5 1 0.002 169 kJ2

Am u u Q W

W PdV P P V V

⇒ − = −

= = + −

+ − =

∫

1 2 2 1 1 2 2 1 1 2( ) ( )

= 2(1161 - 83.94) + 169

= 2323.12 kJ

A A A AQ m u u W m u u W= − + = − +

(5.78/5.75) A car with mass 1275 kg drives at60 km/h when the brakes are applied quickly todecrease its speed to 20 km/h. Assume the brakepads are 0.5 kg mass with heat capacity of 1.1kJ/kg.K and the brake discs/drums are 4.0 kgsteel. Further assume both masses are heateduniformly. Find the temperature increase in thebrake assembly.

Car, loses Kinetic energy, gains Internal

energy, No Heat transfer, No wo

C.V:

rk, constant=m

2 1 1 2 1 2

2 2

car 2 1 brake 2 1

Energy Eqn. (5.11): E E 0 0

1m ( ) m ( ) 0

2

Q W

V V u u

− = − = −

− + − =

The brake system mass is two different kinds sosplit it, also use Cv from Table A.3 since we donot have a u table for steel or brake pad material.

steel pad V

2

2 2 2 2

car

2 2

0

C from Table A.3:

m C m C

1 1000m 60 20 ) m s

2 3600

kJ4 0 46 0 5 1 1 T

K

=1275 kg 0 5 3200 0 07716 m s

65 9 C

V

VT T

T

∆ + ∆

= −

× + × ∆

× × ×

⇒ ∆ =

(

( . . . )

. ( . )

.

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