thermodynamcis

Power Cycles

The ideal Rankine Cycle is composed of

the following processes:

1 - 2: Isentropic expansion in the

engine; S=C

2 3: Constant pressure rejection of heat in the condenser; P=C

3 B: Adiabatic pumping; S=C

B 1: Constant pressure addition of heat in the steam generator, P=C

Note:

a). In the ideal cycle, the state of steam

leaving the steam generator and entering the

engine are the same as well as the state of

feedwater leaving the pump and entering the

steam generator. This means that there is no

pressure drop and no heat leakage in the

steam line and feedwater line.

b). The quantity of the working substance

within the system is constant. This implies

that there are no leakages in the system.

Heat added, QA

Energy Balance:

Ein = Eout

QA + hB = h1

QA = h1 - hB

Heat Rejected, QR

Energy Balance:

Ein = Eout

h2 = h3 + QR

QR = h2 h3

Engine Work, W

Energy Balance:

Ein = Eout

h1 = h2 + W

W = h1 h2

Considering the change in Kinetic Energy,

h1 + K1 = h2 + K2 + W

W = h1 h2 + K1 K2

Pump Work, WP

Exact Pump Work:

Ein = Eout

h3 + WP = hB

WP = hB h3

Approximate Pump Work

The state of feedwater leaving the pump is that of a compressed liquid. Very often, compressed liquid tables are not available, hence, the properties of a compressed liquid are not easily obtainable. Therefore, the exact pump work is difficult to determine.

The following assumptions are made in the determination of the approximate pump work.

Water is practically an incompressible liquid. Therefore, v3 = vB

The change in internal energy is negligible.

uB = u3


Energy Balance:

Ein = Eout u3 + Wf3 + WP + = uB + WfB

WP = WfB Wf3 WP = PB vB P3v3 WP = v3 (PB - P3)

Net Cycle Work, WNET

WNET = Gross Work Pump Work WNET = W WP

WNET = h1 h2 - WP

Another method of determining the net cycle

work is obtaining it from the T-s Diagram.

WNET = area 123B1

= area (123B1) - area (23bc2)

= QA - QR

= (h1 - hB ) (h2 h3 )

= h2 h1 (hB h3 )

= h1 h2 WP Thermal Efficiency, eC

Steam Rate, m

Steam rate is the mass of steam used to

perform a unit work or the mass flow rate of

steam consumed to produce a unit of power.

For good design, a lower value of steam rate

is desired. A lower value of steam rate means

that a smaller quantity of steam is needed to

develop the desired power output.

Let:

P = power output, kW

W = work done by a kg of steam, kJ/kg

M = steam rate, kg/kWh

By definition:

But:

Then;

For the Ideal Rankine Cycle:

The ideal Rankine Engine is either a steam

turbine or a steam engine. The difference

between an ideal Rankine Engine and an

ideal Rankine Cycle is that an ideal engine

does not include pump work since it is

concerned only with all the processes

occurring inside the engine. On the other

hand, the ideal cycle must include pump

work because the pump is needed for the

completion of the cycle. And for the engine

to be ideal, the expansion process should

be isentropic.

Work, W

W = h1 h2

Thermal Efficiency, ee The definition of thermal efficiency does not

directly applies to an engine because no heat is added to it. Instead, it is charged with the enthalpy of steam entering the engine and credited with the enthalpy of saturated liquid at the condensing temperature.

When applied to an engine, the thermal efficiency, e=W/QA becomes e=W/EC where EC is energy chargeable against the engine.

EC = enthalpy of steam entering the engine enthalpy of saturated liquid at the

condensing temperature

Steam Rate, me

Heat Rate, HR

Heat rate is the energy chargeable per unit

of work or the rate of energy chargeable per

unit of power.

Heat Rate, HR

Relation between ee and HR

As can be seen from the resulting

equation (previous slide), the

thermal efficiency is inversely

proportional to the heat rate. This

means that the lower the heat rate,

the higher the thermal efficiency

and the higher the heat rate, the

lower the thermal efficiency.

Any presence of the following conditions will

transform the ideal cycle into an actual one.

1. Pressure drop in the steam generator.

2. Pressure drop in the steam line (1-1).

3. Pressure drop in the condenser.

4. Pressure drop in the feedwater line (B-B)

5. Heat losses in the steam line.

6. Heat losses in the turbine/engine.

7. Irreversible adiabatic expansion in the turbine.

8. Inefficient pump.

9. Subcooled condensate.

Heat Added, QA

QA = h1 - hB

Heat Rejected, QR

QR = h2 - h3

Engine Work, W

Irreversible adiabatic expansion from 1 2

W = h1 - h2

Polytropic expansion from 1 2

Ein = Eout

h1 = h2 + W + QLoss

W = h1 h2 - QLoss

Pump Work, WP

where: nm = pump mechanical efficiency

Actual Cycle Thermal Efficiency, ec

If during the expansion process, the steam

undergoes a process other than isentropic

process, the engine is said to be an actual one.

Engine Analysis

Work, W

W = h1 - h2 Energy Chargeable against the engine, EC

EC = h1 - hf3

Thermal Efficiency, ee

1. Turbo-generator is a generator driven by a turbine.

2. Ideal work, W, is the work done by the steam during a reversible adiabatic expansion process in the turbine.

3. Indicated or actual fluid work, W1 or W, is the work done by the steam during an irreversible adiabatic expansion or polytropic expansion process in the turbine.

4. Brake work, WB, is the useful work, i.e., the available work at the engine shaft.

5. Combined Work, WK, is the electrical energy available at the generator outlet.

Ideal Thermal Efficiency, e

Indicated Thermal Efficiency, ei

Brake Thermal Efficiency, eb

Combined Thermal Efficiency, ek

Note: e> ei > eb > ek

Ideal Steam Rate, m

Indicated Steam Rate, m1

Brake Steam Rate, mb

Combined Steam Rate, mk

Indicated Engine Efficiency, ni

Brake Engine Efficiency, nb

Combined Engine Efficiency, nk

Turbine/Engine Mechanical Efficiency, nme

Generator Efficiency, ng

Ideal Heat Rate, HR

HR = (m) (EC)

Indicated Heat Rate, HRi HRi = (mi)(EC)

Brake Heat Rate, HRb HRb = (mb)(EC)

Combined Heat Rate, HRk HRk = (mk)(EC)

3 1, p. 71

Steam is generated at 4.10 MPa and 440C

and condensation occurs at 0.105 MPa.

(a) For a Rankine engine operating between

these limits, compute the thermal efficiency

and the heat rate.

(b) Considering that a Rankine cycle occurs

between the same limits, determine QA, QR,

WNET, and eC.

(c) What mass flow rate is required for a net

output of 30, 000 kW?

@ P1 = 4.10 MPa

@ P2 = 0.105 MPa

h2 = hf2 - x2hfg2 = 423.24+(0.925)(2254.4)

h2 = 2508.6 kJ/kg

h3 = hf @ 0.105 MPa = 423.24 kJ/kg

vf3 = vf @ 0.105 MPa = 0.0010443 m3/kg

P1 = 4.10 MPa h1 = 3305.7 kJ/kg

T1 = 440C s1 = 6.8911 kJ/kg-K

hf2 = 423.24 kJ/kg sf2 = 1.3181 kJ/kg-K

hfg2 = 2254.4 kJ/kg sfg2 = 6.0249 kJ/kg-K

WP = vf3(PB - P3) = (0.0010443)(4100-105)

WP = 4.17 kJ/kg

hB = h3 + WP = 423.24 + 4.17 = 427.4 kJ/kg

(a) Rankine Engine

W = h1 h2 = 3305.7 2508.6 = 797.1 kJ/kg

EC = h1 hf2 = 3305.7 423.24 = 2882.5 kJ/kg

ee = W/EC

ee = (797.1 / 2882.5) x100%= 27.65%

m = 3600/W

m = 3600/ 797.1 = 4.156 kg/kWh

(b) Rankine Cycle

QA = h1 hB = 3305.7 427.4 = 2878.3 kJ/kg

QR = h2 h3 = 2508.6 423.24 = 2085.4 kJ/kg

WNET = QA QR = 2878.3 2085.4 = 729.9 kJ/kg

or WNET = W WP = 797.1 4.17 = 792.9 kJ/kg

eC = WNET /QA = (792.9/2878.3)x100% = 27.55%

(c) Steam flow rate

= (30,000 kW)/ WNET kJ/kg = 30,000 kJ/s / 792.9 kJ/kg = 37.84 kg/s

Moisture is harmful to the blades of the

turbine. It causes erosion and cavitations of

the turbine blades. As have been observed in

the previous cycle, the moisture content

increases during the later stages of the

expansion process. One solution to this

problem is by reheating the steam after

partial expansion in the turbine. Reheating

minimizes the efficiency of the cycle.

Steam is usually withdrawn and reheated by

few degrees before the saturation point.

The ideal reheat cycle with one stage of reheating is composed of the following processes:

1-2: Partial isentropic expansion in the turbine, S = C

2-3: Constant pressure resuperheating in the reheater, P = C

3-4: Complete isentropic expansion in the turbine, S = C

4-5: Constant pressure rejection of heat in the condenser, P = C

5-B: Adiabatic pumping process, S = C

B-1: Constant pressure addition of heat in the boiler, P = C

Heat Added, QA

Energy Balance:

Ein = Eout QB + hB = h1

QB = h1 - hB

Reheater

Energy Balance:

Ein = Eout QRH + h2 = h3 QRH = h3 h2

Therefore:

QA = QB + QRH QA = h1 - hB + h3 h2

For a given number of stages of reheating,

Where: n = number of reheaters

Heat Rejected, QR

Energy Balance:

Ein = Eout

h4 = h5 + QR

QR = h4 h5

Engine Work, W

Energy Balance:

Ein = Eout h1 + h3 = h2 + h4 + W W = h1 h2 + h3 - h4

Another means of determining engine work is by getting the sum of the work done by the steam during the different stages of expansion.

W = W1-2 + W3-4

W = h1 h2 + h3 - h4

Pump Work, WP


WP vf5 (PB P5)

Exact Pump Work:

Ein = Eout

h5 + WP = hB

WP = hB h5


WNET = Engine Work Pump Work

WNET = h1 h2 + h3 - h4 WP

Another method:

WNET = QA - QR

= (h1 - hB + h3 h2) (h4 h5)

= h1 h2 + h3 h4 (hB h5)

= h1 h2 + h3 h4 WP

Thermal Efficiency, ec

WNET = h1 h2 + h3 - h4 WP QA = h1 - hB + h3 h2

But

hB = h5 + WP QA = h1 h2 + h3 h5 WP

Steam Rate, mc mc = 3600/ WNET ,in kg/kWh

The ideal reheat engine ignores the pressure

drop in the reheater. The engine is an ideal

one whether there is or there is no pressure

drop in the reheater for as long as the

expansion process is an isentropic one.

Engine Analysis

Work, W

W = h1 h2 + h3 - h4 Steam Rate, me

me = 3600/ W

Energy Chargeable against the engine, EC General equation which is applicable to a

reheat engine only.

EC = enthalpy of steam entering the

engine - enthalpy of saturated liquid at

the condensing temperature + QRH For the given engine:

EC = h1 - h5 + QRH = h1 - h5 + (h3 h2)

= h1 - h2 + h3 h5

Any presence of the following conditions will make the ideal reheat cycle an actual one.

Pressure drop in the boiler/steam generator.

PB P1; P1< PB Pressure drop in the steam line (1 1).

P1 < P1; P1 P1 t1 < t1; t1 t1

Pressure drop in the reheater. P2 < P3; P2 P3

Pressure drop in the condenser.

P4 P5; P5< P4

Irreversible adiabatic expansion process.

s2 s1 and s4 s3 QLoss = 0, but

s2 > s1 and s4 > s3 Polytropic expansion process.

QLoss = 0 and s1 s2 and s4 s3 Heat losses in the steam line (1 1)

Inefficient pump.

Pump efficiency < 100%

Pressure drop in the feedwater line.

PB < PB

Heat Added, QA

QA = QB + QRH

QA = (h1 hB) + (h3 h2)

Heat Rejected, QR QR = h4 - h5

Engine Work, W

W = h1 h2 + h3 h4

If irreversible adiabatic expansion process or,

W = h1 h2 + h3 h4 QLoss

If polytropic expansion process


WNET = W WP

Actual Pump Work, WP

Thermal efficiency, eC

If the expansion process is no longer

isentropic, the engine is said to be an actual

one.

Engine Analysis

Work, W

W = h1 h2 + h3 h4 Energy Chargeable against the engine, EC

EC = h1 hf5 + QRH = h1 - hf5 + (h3 h2)

Thermal efficiency, ee

Steam Rate, me

Heat Rate, HR

HR = (me)(EC)

3 5, p. 88

In a reheat cycle steam at 8.0 MPa and 485C enters the turbine and expands to 1.4 MPa. At this point, the steam is withdrawn and passed through a reheater. It re-enters the turbine at 1.3 MPa and 720C. Expansion now occurs to the condenser pressure of 0.006 MPa. For the cycle and 1 kg of steam, determine (a) QA , (b) WNET and (c) eC . For the engine, determine (d) W, (e) eC and (f) the steam flow for an engine output of 40,000 kW.

h1 = h @ 8.0 MPa and 485C = 3361 kJ/kg

h2 = h @ 1.4 MPa and S2 = S1 = 2891 kJ/kg

h3 = h @ 1.3 MPa and 720C = 3968 kJ/kg

h4 = h @ 0.006 MPa and S4 = S3 = 2526 kJ/kg

h5 = hf @ 0.006 MPa = 151.53 kJ/kg

vf5 = vf @ 0.006 MPa = 1.0064x10-3 m3/kg

WP = vf5 (PB P5) = (1.0064x10-3)(8000-6) = 8.05 kJ/kg

hB5 = h5 + WP = 151.53 + 8.05 = 159.58 kJ/kg

(a) QA = h1 hB5 + h3 h2 = 3361 159.58 + 3968 2891

= 4278.4 kJ/kg

(b) W = h1 h2 + h3 h4 = 3361 2891 + 3968 2526 = 1912 kJ/kg

WNET = W WP = 1912 8.05 = 1904 kJ/kg (c) eC = WNET /QA = (1904/4278.4)x 100% = 44.5%

(d) W = 1912 kJ/kg

(e) EC = h1 h2 + h3 hf5 = 3361 -2891 + 3968 -151.83

= 4286.5 kJ/kg

ee = W/EC

= (1912/4286.5)x100% = 44.6%

(f) m = 3600/W

= 3600/1912 = 1.88 kg/kWh

Steam flow rate = (40,000 kWh)(1.88 kg/kWh)

= 75,200 kg/h or 20.89 kg/s

Introduction

The thermal efficiency of a simple power plant is less than fifty percent (50%). This means that more than half of the heat added to the water in the boiler is just wasted and rejected in the condenser. In order to utilize some of these heats that would have been wasted and rejected in the condenser, part of the throttle steam is extracted or bled for feedwater heating after it has partially expanded in the turbine. The extraction/ bled points occur near the saturation state. The process of heating feedwater in this manner is called regeneration and the cycle governing it is the REGENERATIVE CYCLE.

Increase in thermal efficiency

By definition, e=WNET/Q . Examining the equation, the two ways of increasing the thermal efficiency are (a) by increasing the net cycle work and (b) by reducing the heat supplied, QA . The temperature of feedwater entering the boiler in the regenerative cycle (tB5) is higher than that of the original Rankine cycle (tB). Since the feedwater enters the boiler at a relative high temperature, a smaller quantity of heat is needed to transform it to steam than without the regenerative feedwater heating. This in effect tend to increase the thermal efficiency.

Increase in thermal efficiency

It is true that the net work done per kilogram of the throttle steam in the regenerative cycle is less than that of the Rankine cycle as the consequence of the extraction of steam for feedwater heating. This tends to decrease the thermal efficiency. But the rate of decreased in the heat supplied, QA is faster than the reduction rate in the net cycle work, WNET. Therefore, the net result of this is an increase in thermal efficiency.

Decrease in the moisture content during the

later stages of expansion.

It is a fact that the quality of exhaust steam

for both cycles are the same, i.e., x2

(Rankine cycle) = x3 (Regenerative cycle).

But the quantity of exhaust steam decreases

in the regenerative cycle as the result of the

bleeding process. Therefore, the moisture

content decreases.

Plant Layout of Regenerative Cycle With One

Stage of Extraction for Feedwater Heating

Basis: 1kg of throttle steam

Mass of Bled Steam, m

Mass Balance:

min = mout

mB4 + m = 1

Energy Balance:

Ein = Eout

mh2 + mB4hB4 = m5h5

mh2 + (1 - m)hB4 = (1)h5

Alternate Method:

Heat Balance:

Heat from bled steam = Heat to feedwater

m(h2 h5) = mB4(h5 hB4)

m(h2 h5) = (1-m)(h5 hB4)

The condensate pump work is often small so that it

can be neglected. Neglecting condensate pump work,

hB4 = h4

It can now be said that for any feedwater heater

using direct contact type (open heaters).

mass of bled steam = mass of feedwater leaving

the heater

(General Equation)

Neglecting pump works

h6 hB6 h5 hB5

Applying the general equation for

determining the quantity of bled steam,

Heat Supplied, QA

Pump:

Energy Balance:

Ein = Eout WP2 + h5 = hB5

Where: WP2 = vf5 (PB5 P5)

Boiler:

Energy Balance:

Ein = Eout QA + hB5 = Eout

QA = h1 - hB5 But:

hB5 = h5 + WP2 Therefore:

QA = h1 - h5 - WP2

Heat Rejected, QR

Energy Balance:

Ein = Eout

(1 - m)h3 = QR + (1 - m)h4

QR = (1 - m)(h3 h4)

Engine Work, W

Energy Balance:

Ein = Eout h1 = mh2 + (1 - m)h3 + W

W = h1 - mh2 - (1 - m)h3

= h1 - mh2 - (1 - m)h3 + h2 - h2 = (h1 - h2) + (1 - m)h2 - (1 - m)h3 W= (h1 - h2) + (1 - m)(h2 h3)

Another method:

W = W (stage work)

W = W1-2 + W2-3 = (h1 - h2) + (1 - m)(h2 h3)

Total Pump Work, WP

WP = WP1 + WP2

WP1 = vf4 (PB4 P4)

WP2 = vf5 (PB5 P5)

Approximate Total Pump Work

WP = vf4 (PB5 P4)


WNET = W WP

WNET = (h1 - h2) + (1 - m)(h2 h3) - WP

Thermal efficiency, eC

Engine Analysis

Work, W

W = (h1 - h2) + (1-m)(h2 h3)

Energy Chargeable, EC

The engine is charged with the enthalpy of steam entering

the engine and credited with the enthalpy of feedwater

leaving the last heater assuming that all the bled steam

are used for feedwater heating.

EC = Enthalpy of steam entering the turbine - Enthalpy of

feedwater leaving the last heater

For the given cycle

EC = h1 h5

Thermal efficiency, ee

Any presence of the following conditions will make an ideal cycle an actual one.

Pressure drop in the boiler.

P1< PB5 Pressure drop in the steam line (1-1)

P1< P1 Pressure drop in the condenser.

P4< P3 Pressure drop in the bled steam line.

P2< P2 Pressure drop in the feedwater line.

PB5< PB5 Heat losses in the steam lines (1-1) and (2-2). Heat losses in the turbine

Inefficient Pump

Heat losses in the heaters.

Plant Layout of Actual Regenerative Cycle with One Stage

of Extraction for Feedwater Heating

Heat Added, QA

QA = h1 hB5

Heat Rejected, QR

QR = (1-m)(h3 - h4)

Mass of Bled Steam

Engine Work, W

W = (h1 h2)+(1-m)(h2 h3)

Pump Work, WP WP = WP

= WP1 + WP2


WNET = W WP

Thermal Efficiency, EC

3 10, p.107

Steam is delivered to an engine at 5.4 MPa

and 600C. Before condensation at 31C,

steam is extracted for feedwater heating at

0.60 MPa.

For an ideal cycle, find (a) the amount of

steam extracted (b) W and (c) e. For an

ideal engine and the same states, compute

(d) W and e and (e) steam rate.

h1 = h @ 5.40 MPa and 600C = 3663.3 kJ/kg

h2 = h @ 0.6 MPa and S2 = S1 = 2987 kJ/kg

h3 = h @ 0.004469 MPa and S3 = S1 = 2187 kJ/kg

h4 = hf @ 31C = 129.97 kJ/kg

h5 = hf @ 0.60 MPa = 670.56 kJ/kg

vf4 = vf @ 31C = 1.0064x10-3 m3/kg

vf5 = vf @ 0.60 MPa = 1.1006x10-3 m3/kg

WP2 = vf5 (PB5 P5) = (1.1006x10-3)(5400-600)

WP2 = 5.3 kJ/kg

WP = vf4 (PB5 P4) = (1.0064x10-3)(5400-4.496)

WP = 5.42 kJ/kg

hB5 = h5 + WP = 670.56 + 5.3 = 675.86 kJ/kg

Heat Balance:

(1-m)(h5 h4) = m(h2 h5)

m = 0.1898 kg of throttle steam

WNET = W WP = (h1 - h2) + (1 - m)(h2 h3) WP WNET = 3663.3 2978 + (1 - 0.1898)(2978 2187)-5.42

WNET = 1320.75 kJ/kg

QA = h1 hB5 = 3663.3 675.86 = 2987.44 kJ/kg

eC = WNET /QA = (1320.75 / 2987.44)x100% = 44.2%

W = h1 h2 + (1 - m)(h2 h3)

W = 3663.3 2978 + (1 0.1898)(2978 2187)

W = 1326.3 kJ/kg

ee = W/(h1 h2) = [1326.2/(3663.3-670.56)]x100% = 44.3%

m = 3600/1362.2 = 2.71 kg/kWh

In this cycle, the reheat cycle and the

regenerative cycle are combined to attain

the following objectives.

1. Further improvement in the overall thermal

efficiency.

2. Further reduction in the moisture content

of steam during the latter part of the

expansion process.

The solution to a reheat-regenerative cycle

problems differs from that of the previous

cycles namely: reheat cycle, Rankine cycle,

and regenerative cycle. It does not follow

fixed set of formulas nor there is a fixed

pattern. Each problem requires a particular

solution although the same laws, definitions,

and principles are still being used. Problems

can best be solved by energy balance and

mass balance.

CASE 1

Assume an ideal reheat-regenerative cycle:

after some expansion, steam is extracted for

feedwater heating; after further expansion,

there is a reheat; then expansion to exhaust.

Write the equations for (a) the quantity of

extracted steam, (b) the net work, and (c)

the thermal efficiency. The equations should

refer to a T-s Diagram with named points.

Mass of Bled Steam, m

General Equation:

mass of bled steam = mass of feedwater leaving the

heater

Neglecting condensate pump work (Pump 1) hB6 h6

Net Cycle Work, WNET Engine Work, W

First Method:

Energy Balance:

Ein = Eout h1 + (1 - m)h4 = mh2 + (1 - m)h3 + (1 - m)h5 + W

W = h1 + (1 - m)h4 - mh2 + (1 - m)(h3 + h5)

Second Method:

W = W of stages of expansion

= W1-2 + W2-3 + W3-4 = h1 h2 + (1 - m)(h2 h3) + (1 - m)(h4 - h5)

Total Pump Work, WP WP = WP1 + WP2

Exact formula:

WP = vf5 (PB6 P6) + vf7 (PB7 P7)

Approximate formula:

WP = vf5 (PB7 P6)

WNET = Engine work Pump work

= W WP = h1 h2 + (1 - m)(h2 h3) + (1 - m)(h4 - h5)

- vf5 (PB7 P6)

Thermal Efficiency, e

Heat Added, QA

QB = h1 hB7

QB = h1 h7 - WP2

But:

WP2 = vf7 (PB7 P7)

QRH = (1 - m)(h4 h2)

General Equation:

QA = QBoiler + QReheaters

= (h1 h7 - WP2)+(1 - m)(h4 h3)

CASE 2

Assume an ideal reheat-regenerative cycle, with, first, an extraction for feedwater heating, then later a single reheating, and finally, two extraction points for feedwater heating. Sketch the energy diagram and write equations for (a) the quantity of steam extracted at each point, (b) the work from QA and QR and the turbine work, and (c) the thermal efficiency of the cycle. The equation should refer to a T-s diagram with named points.

Note: Pump 1 is condensate pump

Pump 4 is main boiler feedwater pump

P2 = P11 = PB10 PB11 = P1

P5 = P10 = PB9 P7 = P8

P6 = P9 = PB8 P3 = P4

Mass of Bled Steam

Neglecting condensate pump work and pump

works between heaters.

Last Heater:

Second Heater:

First Heater:

hB8 h8 hB9 h8 hB10 h10

Work

Cycle Work, WNET

QB = h1 hB11 QB = h1 h11 WP4

QB = h1 h11 vf11 (PB11 P11) QRH = (1 - m1)(h4 h3)

QA = QBoiler + QReheaters

= h1 h11 vf11 (PB11 P11) + (1-m1)(h4 h3)

QR = (1 - m1 - m2 - m3)(h7 h8)

Therefore, net cycle work based from QA and QR is,

WNET = QA QR = h1 h11 vf11 (PB11 P11) + (1-m1)(h4 h3)

- (1 - m1 - m2 - m3)(h7 h8)

Turbine Work, W

First Method:

Energy Balance:

Ein = Eout h1 + (1 - m1)h4 = m1h2 + (1 - m1)h3 + m2h5 + m3h6

+ (1 - m1 - m2 - m3)h7 + W

W = h1 + (1 - m1)h4 - m1h2 - (1 - m1)h3 - m2h5 - m3h6 - (1 - m1 - m2 - m3)h7

Second Method:

W = W1-2 + W2-3 + W3-4 + W4-5 + W5-6 + W6-7

W=(h1 h2) + (1 m1)(h2 h3) + (1 - m1)(h4 - h5)

+ (1 - m1 - m2) (h5 h6) + (1 - m1 - m2 - m3)(h6 h7)

Thermal Efficiency, eC

CASE 3

The same as Case 2 except that the three

extraction points occur after the reheating

Mass of Bled Steam

Last Heater:

Second Heater:

Third Heater:

Work

Cycle Work, WNET QB = h1 hB11

QB = h1 h11 WP4 QB = h1 h11 vf11 (PB11 P11)

QRH = h3 h2

QA = QBoiler + Qreheaters

QA = h1 h11 vf11 (PB11 P11) + (h3 h2)

QR = (1 - m1 - m2 - m3)(h7 h8)

(Same as in Case 2)

WNET = QA QR = h1 h11 vf11 (PB11 P11) + (h3 h2)

- (1 - m1 - m2 - m3)(h7 h8)

Turbine Work

Ein = Eout h1 + h3 = h2 + m1h4 + m2h5 + m3h6

+ (1 - m1 - m2 - m3)h7 + W

W = h1 + h3 [h2 + m1h4 + m2h5 + m3h6 + (1 - m1 - m2 - m3)h7]

Another Method:

W + W5-6 + W6-7 W = (h1 h2) + (h2 h3) + (1 - m1)(h4 - h5)

+ (1 - m1 - m2) (h5 h6)

+ (1 - m1 - m2 - m3)(h6 h7)


CASE 4

Assume an ideal reheat-regenerative cycle:

after some expansion, part of the steam is

extracted for feedwater heating; the

remainder are withdrawn and reheated to

the original temperature; after further

expansion, a second extraction occurs; then

expansion to exhaust. Write the equations

for (a) the quantity of steam extracted and

(b) the turbine work.

Mass of Bled Steam

Open Heater no. 1

Open Heater no. 2

Turbine Work, W

Ein = Eout

h1 + (1 - m1)h3 = m1h2 + (1 - m1)h2 + m2h4

+ (1 - m1 - m2)h5 + W

W = (h1 + h2) + (1 - m1)h3 - m2h4 - (1 - m1 - m2)h5 Or

W = W1-2 + W3-4 + W4-5 W = (h1 - h2) + (1 - m1)(h3 - h4) + (1 - m1 - m2) (h4- h5)

3 14, p.133

Steam at 5 MPa and 365C enters a turbine

and expands until it becomes saturated. The

steam is withdrawn and reheated to 330C.

After expansion in the turbine to 150C, m1

kg is extracted for feedwater heating. The

remaining steam expands to the condenser

pressure of 0.016 MPa. For 1 kg of steam,

find WNET, eC, ee, and the ideal steam rate.

h1 = h @ 5 MPa and 365C = 3108 kJ/kg

h2 = h @ S2 = S1 and saturated = 2786 kJ/kg

h3 = h @ 1.25 MPa and 330C = 3110 kJ/kg

h4 = h @ 0.28 MPa and 150C and S4 = S3 = 2762 kJ/kg

h5 = h @ 0.016 MPa and S5 = S4 = 2315 kJ/kg

h6 = hf @ 0.016 MPa = 231.56 kJ/kg

h7 = hf @ 0.28 MPa = 551.48 kJ/kg

vf6 = vf @ 0.016 MPa = 1.0147x10-3 m3/kg

vf7 = vf @ 0.28 MPa = 1.0709x10-3 m3/kg

WP2 = vf7 (P1 P7) = (1.0709x10-3)(5000-280)

WP2 = 5.05 kJ/kg

hB7 = h7 + WP2 = 551.48 + 5.05 = 556.5 kJ/kg

WP = vf6 (P1 P6) = (1.0147x10-3)(5000-16)

WP = 5.06 kJ/kg

Heater:

= 0.1264 kg

W = (h1 h2) + (h3 h4) + (1 m1)(h4 h5)

= (3106 2786) + (3110 2762) + (1- 0.1264)(2762 2315)

= 1060.5 kJ/kg

WNET = W WP = 1060.5 - 5.06 = 1055.4 kJ/kg

QA = h1 hB7 + h3 h2 = 3108 551.48 + 3100 2786 = 2875.5 kJ/kg

EC = h1 hf7 + h3 h2 = 3108 551.48 + 3100 2786 = 2880.5 kJ/kg

eC = WNET /QA = (1055.4 / 2875.5)x100% = 36.70%

ee = W/EC = (1060.5/2880.5)x100% = 36.82%

m = 3600/1060.5 = 3.39 kg/kWh

The ideal incomplete-expansion cycle is composed of the following processes:

1-2: Isentropic expansion process in the engine, s = C

2-3: Constant volume rejection of heat process in the engine, V = C

3-4: Constant pressure rejection of heat process in the condenser, P = C

4-B: Reversible adiabatic pumping process,

s = C

B-1: Constant pressure addition of heat process in the boiler, P = C

Engine Work, W

Recalling isentropic and isometric processes.

Isentropic Isometric

Neglecting P and K, the area behind the curve on the PV plane represents the work of a steady

flow process.

For isentropic process:

WS = area behind the curve, area 1-2-a-b-1

WS = h1 h2

For isometric process:

WS = area behind the curve, area 2-3-c-d-2

WS = v2(P2 P3)

For incomplete-expansion with zero clearance:

W = W1-2 + W2-3

W = h1 h2 + v2(P2 P3)

Heat Added, QA

Ein = Eout

QA + hB = h1

QA = h1 - hB

Incomplete expansion engine almost always

operate with a low initial pressure, hence,

the pump work is very small so that it can be

neglected.

From the pump energy balance:

hB = h4 + WP

WP 0

hB h4 QA = h1 h4

Energy Chargeable, EC

EC = h1 + hf3


Mean Effective Pressure, MEP or Pm

Ideal Mean Effective Pressure

Indicated Mean Effective Pressure, MEPI

Brake Mean Effective Pressure, MEPB

Combined Mean Effective Pressure, MEPK

Steam Rate, m

Ideal Steam Rate, m

Indicated Steam Rate, mI

Brake Steam Rate, mB

Combined Steam Rate, mK


Indicated Thermal Efficiency, eI

Brake Thermal Efficiency, eB

Combined Thermal Efficiency, eK

Engine Efficiency, n

Indicated Engine Efficiency, nI

Brake Engine Efficiency, nB

Combined Engine Efficiency, nK

Mechanical Efficiency, nm

Generator Efficiency, ng

Approximate Enthalpy of Exhaust Steam, he

Where:

W = actual work

W = WI, indicated work

Energy Balance:

h1 = Qloss + WI + he he = h1 - Qloss - WI

eI = nIe

Proof:

eB = nBe

Proof:

eK = nKe

Proof:

eK = ngnmeI

Proof: eK = ngnmeI

nm = mI/mB

Proof:

nm = eB/eI

Proof:

ng = eK/eB Proof:

nK = nInmng

Proof:

mK = mI/nmng

Proof:

3 -19, p. 154

Steam at 1.10 MPa and 250C is delivered to the throttle of an engine. The steam expands to 0.205 MPa, where release occurs. Exhaust is at 0.105 MPa. A test of the engine showed an indicated steam consumption of 13.28 kg/kWh and a mechanical efficiency of 85%. Find (a) the ideal work and ideal thermal efficiency, (b) the ideal steam rate, (c) the brake and indicated works, (d) the brake thermal efficiency, (f) the MEP of the ideal engine and the indicated MEP.

Given:

@ P2 = 0.205 MPa

S2 = Sf2 + x2 Sfg2

S2 = 1.5386 + (x2) 5.5803

x2 = 0.9188 = 91.88%

h2 = hf2 + x2 hfg2

h2 = 508.03 + (0.9188) 2199.8

h2 = 2529.2 kJ/kg

v2 = vf2 + x2 vfg2

v2 = 0.0010613 + (0.9188) 0.86444

v2 = 0.7953 m3/kg

h4 = hf3 = hf @ 0.105 MPa = 423.24 kJ/kg

P1 = 1.10 MPa h1 = 2834.2 kJ/kg

T1 = 205C S1 = 6.6659 kJ/kg-K

vf2 = 0.0010613 hf2 = 508.03 sf2 = 1.5386

vfg2 = 0.86444 hfg2 = 2199.8 sfg2 = 5.5803

W = h1 h2 + v2(P2 P3)

W = 2834.2 2529.2 + (0.7953)(205 105)

W = 384.5 kJ/kg

ee = W/(h1 hf3) = [384.5/(2834.2-423.24)] x100%

ee = 15.95%

m = 3600/W = 3600/384.5 = 0.363 kg/kWh

WI = 3600/mI = 3600/13.28 = 271.1 kJ/kg

WB = nm WI = (0.85)(271.1) = 230.4 kJ/kg

eB = WB/(h1 hf3) = [230.4/(2834.5-423.24)]x100% = 9.56%

nB = WB/W = (230.4/384.5)x 100% = 59.92%

Ideal MEP = W/v2 = 384.5/0.7953 = 483.47 kPa

Indicated MEP = WI/v2 = 271.1/0.7953 = 340.88 kPa

Superposition or topping unit is a new set of

high-pressure equipment to be added or

topped into the existing system with the idea

of increasing the capacity of the whole

system and at the same time replacing the

old boiler (oil-fired) with a new high-pressure

steam generator (coal-fired). As shown in the

diagram, the new Hp turbine or the topping

unit is a non-condensing turbine and its

exhaust will be utilized by the old low-

pressure turbine.

3 24, p.166

A 30,000 kW existing plant has the following

throttle conditions:

PS = 1.50 MPa; TS = 260C

The steam rate of this plant is 5.67 kg/kWh.

An additional 12,500 kW is wanted from this

superposed unit using an average indicated

efficiency of 78% and a mechanical-electrical

efficiency of 98%, estimate the steam

conditions of the superposed plant.

hs = h @1.5 MPa, 260C = 2946.7 kJ/kg

ms = (mk)(Pk) =(5.67)(30,000) = 170,100 kg

For the topping unit:

mk = ms/output = 170100/12500 = 13.6 kg/kWh

WK = 3600/ mk = 3600/13.6 = 264.7 kJ/kg

WI = W/nme = 264.7/0.96 = 275.73 kJ/kg

WI = ht hs 275.73 = ht 2946.7

ht = 3222.43 kJ

h0 :

@ point 0,

P0 = 1.5 MPa

h0 = 2868.93 kJ/kg

then;

S0 = 6.52 kJ/kg-K

@ point t,

St = S0 = 6.52 kJ/kg-K

ht = 3222.43 kJ/kg

then;

Pt = 6.0 MPa and Tt = 416C

15 174 (Sta. Maria)

A reheat-regenerative engine receives steam at 207 bar and 593C, expanding it to 38.6 bar, 343C. At this point, the steam passes through a reheater and reenters the turbine at 34.5 bar, 593C, hence expands to 9 bar, 492C, at which point the steam is bled for feedwater heating. Exhaust occurs at 0.07 bar. Beginning at the throttle (point 1), these enthalpies are known (kJ/kg)

For ideal engine, sketch the events on the T-s plane and for 1 kg of throttle steam, find (a) the mass of bled steam, (b) the work, (c) the efficiency, and (d) the steam rate. In the actual case, water enters the boiler at 171C and the brake engine efficiency is 75% (e) determine the brake work and the brake thermal efficiency. (f) Let the pump efficiency be 65%, estimate the enthalpy of the exhaust steam.

h1=3511.3 h2=3082.1 h4=3205.4 h5=2308.1 h7=723.59

h2=3010.0 h3=3662.5 h4=322.9 h6=163.4 h7=742.83

@Pt. 1: P1 = 20.7 MPa; T1 = 593C

Tsat = 368.635C; Tsat < T1 ; Therefore, SH

h1 = 3511.3 kJ/kg

@Pt. 2: P2 = 3.86 MPa; T2 = 343C

Tsat = 244.23C ; Tsat < T2 ; Therefore, SH

h2 = 3010.0 kJ/kg

@Pt. 3: P3 = 3.45 MPa; T3 = 593C


h3 = 3662.5 kJ/kg

T(C) S (kJ/kg-K)

580 7.3880

593 S3

600 7.4409

Interpolate:

S3 = 7.42239 kJ/kg-K = S4 = S5

@Pt. 4: P4 = 0.90 MPa; T4 = 492C


h4 = 3205.4 kJ/kg

S4 = 7.42239 kJ/kg-K

@Pt. 5: P5 = 0.007 MPa; S5 = 7.42239 kJ/kg-K

Sg = 8.2758 kJ/kg-K; S5 < Sg ; Therefore,WET

h5 = 2308.1 kJ/kg

@Pt. 6: P6 = 0.007 MPa; Saturated Liquid

h6 = hf6 = 163.40 kJ/kg

S6 = Sf6 = 0.5592kJ/kg-K

v6 = vf6 = 1.0074x10-3 m3/kg

@Pt. B6: PB6 = 0.90 MPa; SB6 = 0.5592kJ/kg-K

Sf = 0.6224 kJ/kg-K; SB6 < Sf; Therefore, SUBCOOLED


h7 = hf7 = 742.83 kJ/kg S7 = Sf7 = 2.0946 kJ/kg-K

v7 = vf7 = 1.212x10-3 m3/kg

@Pt. B7: PB7 = 20.7 MPa; SB7 = 2.0946 kJ/kg-K

Sf = 4.0762 kJ/kg-K; SB7 < Sf; Therefore, SUBCOOLED

Mass of Bled Steam

Ein = Eout mh4 + (1 m)h6 = 1h7

m(3205.4) + (1 m)163.40= 742.83 3205.4m + 163.40 163.40m = 742.83

3042m = 579.43

m = 0.19 kg

Work

Ein = Eout 1h1 + 1h3 = 1h2 + mh4 + (1 - m)h5 + W

3511.3 + 3668.5 = 3010 + (0.19)(3205.4) + (1 - 0.19) (2308.1) + W

W = 1685.213 kJ/kg

Efficiency, e

ee = W/EC EC = (h1 h7) + (h3 h2)

EC = 3511.3 - 742.83 + 3662.5 3010.0

EC = 3420.7 kJ/kg

ee = 1685.213 / 3420.7 = 49.26% Steam Rate, m

m = 3600/W

m = 3600 / 1685.213 = 2.14 kg/kWh

WB , eB WB = nB W

WB = (0.75)(1685.213) = 1263.91 kJ/kg

eB = WB /EC = 1263.91/3420.7 = 37%

16 174-175 (Sta. Maria) In a 35,000 kW turbo-generator that receives steam

at 6.9 MPa and 370C, 11% of the throttle steam is actually extracted at 2MPa, 215C; with the remainder being reheated to 1.8 MPa and 315C; then 20% of the throttle steam is actually extracted at 0.724 MPa, each extraction serving an open feedwater heater. The engine exhaust to a condenser pressure of 0.005 MPa and the temperature of the feedwater from the last heater is 205C. The combined steam rate of the turbo-generator unit is 4.898 kg/kWh and the generator efficiency is 95%. For the total throttle flow to an ideal engine, find (a) extracted steam for the last heater, (b) W, (c) e. For the actual engine, find (d) eK (e) nK. (f) What is the enthalpy of the actual exhaust when the pump efficiency is 60% ?

@Pt. 1: P1 = 6.9 MPa; T1 = 370C


h1 = 3077.6 kJ/kg S1 = 6.3314 kJ/kg-K

@Pt. 2: P2 = 2 MPa; T2 = 215C


h2 = 2807.2 kJ/kg S2 = 6.3566 kJ/kg-K

@Pt. 3: P3 = 1.8 MPa; T3 = 315C


h3 = 3063.2 kJ/kg S3 = 6.88105 kJ/kg-K

@Pt. 4: P4 = 0.724 MPa; S4 = 6.88105 kJ/kg-K

Sg = 6.69662 kJ/kg-K; S4 > Sg ; Therefore, SH

Double Interpolation:

@P = 0.72 MPa

P (MPa) S (kJ/kg-K)

0.72 6.6985

0.724 Sg

0.73 6.6938

S (kJ/kg-K) h(kJ/kg)

6.8717 2843.7

6.88105 h0.72

6.9185 2866.0

Interpolate:

h0.72 = 2848.155 kJ/kg

@P = 0.73 MPa

h0.73 = 2853.94 kJ/kg

S (kJ/kg-K) h(kJ/kg)

6.8573 2842.6

6.88105 h0.73

6.9042 2865.0

Interpolate:

h4 = 2849.31 kJ/kg

@Pt. 5: P5 = 0.005 MPa; S5 = 6.88105 kJ/kg-K

Sg = 8.3951 kJ/kg-K; S5 < Sg ; Therefore, WET

S5 = Sf5 + x5 Sfg5

6.88105 = 0.4674 + (x5) 7.9187

x5 = 0.8088 = 80.88%

h5 = hf5 + x5 hfg5

h5 = 137.82 + (0.8088)(2423.7)

h5 = 2098.109 kJ/kg

P (MPa) h (kJ/kg)

0.72 2848.155

0.724 h4

0.73 2853.94




Sf = 0.4764 kJ/kg-K; SB6 < Sf ; Therefore, SUBCOOLED

P (MPa) S (kJ/kg-K)

0.72 2.0035

0.724 Sf

0.73 2.0091


hf = 703.176 kJ/kg

@Pt. 8: P8 = 2 MPa; Saturated Liquid

h8 = hf8 = 908.79 kJ/kg

S8 = Sf8 = 2.4474 kJ/kg-K

P (MPa) h (kJ/kg)

0.72 702.20

0.724 hf

0.73 704.64

W = (h1 h2) + (1 m1)(h3 h4) + (1 - m1 m2)(h4 h5)

W = (3077.6 2807.2) + (0.89)(3063.2 2849.31) + [1 0.11 (0.20)(0.89)](2849.31 2098.109)

W = 995.6172/0.95

W = 1048.02 kJ/kg

ee = W/EC EC = h1 hf8 + h3 h2

EC = 3077.6 908.79 + 3063.2 2807.2

EC = 2424.81 kJ/kg

ee = (1048.02 / 2424.81) x100%= 43.22%

mK = 3600/ WK

WK = 3600 / 4898 = 734.99 kJ/kg

eK = WK / EC eK = (734.99 / 2424.81)x100% = 30.31%

nK = WK / W

nK = (734.99 / 1048.02)x100% = 70.13%

17 175, p.175

There are developed 25,000 kW by a reheat-regenerative engine (turbo-generator) which receives steam at 4.2 MPa, 313C and exhaust at 0.007 MPa. At 1.90 MPa and 215C, part of the steam is extracted for feedwater heating and the remainder is withdrawn for reheating. The reheated steam enters the turbine at 1.8 MPa and 270C and expands to 1.38 MPa, where more steam is extracted for feedwater heating and the remainder expands to the condenser pressure of 0.007 MPa and an actually quality of 90%. Feedwater leaves the last heater at a temperature of 207C. The generator has an efficiency of 95%. For the ideal engine, find (a) the percentages of the extracted steam, (b) W, and (c) e. Let the actual extracted steam be 85% of those for the ideal engine and for the actual engine, find (d) the total throttle flow, if the break work equal the fluid work, (e) eK and (f) nK.

@Pt. 1: P1 = 4.2 MPa; T1 = 313C


h1 = 2990.35 kJ/kg @Pt. 2: P2 = 1.9 MPa; T2 = 215C

Tsat = 209.84C; Tsat < T2; Therefore, SH

h2 = 2813.4 kJ/kg S2 = 6.3905 kJ/kg-K

P (MPa) h (kJ/kg)

310 2982.1

313 h1

320 3009.6

@Pt. 3: P3 = 1.8 MPa; T3 = 270C

Tsat = 207.15C; Tsat < T3; Therefore, SH

h3 = 2959.5 kJ/kg

S3 = 6.6976 kJ/kg-K

@Pt. 4: P4 = 1.38 MPa; S4 = S3 = 6.6976 kJ/kg-K

Sg = 6.4743 kJ/kg-K; S4 > Sg ; Therefore, SH

Double Interpolation

@ 1.35 MPa

S (kJ/kg-K) h (kJ/kg)

6.6743 2881.9

6.6976 h1.35

6.6980 2893.9

Interpolate:

h1.35 = 2893.697 kJ/kg

@ 1.40 MPa

h1.40 = 2901.854 kJ/kg

S (kJ/kg-K) h (kJ/kg)

6.6778 2891.7

6.6976 h1.40

6.7012 2903.7

Interpolate:

h4 = 2898.59 kJ/kg

@Pt. 5: P5 = 0.007 MPa; S5 = 6.6976 kJ/kg-K

Sg = 8.2758 kJ/kg-K; S5 < Sg ; Therefore, WET

S5 = Sf5 + x5 Sfg5

6.6976 = 0.5592 + (x5) 7.7167

x5 = 0.7955 = 79.55%

h5 = hf5 + x5 hfg5

h5 = 163.40 + (0. 7955)(2409.1)

h5 = 2079.839 kJ/kg

P (MPa) h (kJ/kg)

1.35 2893.697

1.38 h4

1.40 2901.854

@Pt. 5: P5 = 0.007 MPa; x5 = 0.90

h5' = hf5 + x5 hfg5

h5' = 163.40 + (0. 90)(2409.1)

h5' = 2331.59 kJ/kg










h8 = hf8 = 897.02 kJ/kg S8= Sf8 = 2.4233 kJ/kg-K



% of Extracted Steam:

m1 = 0.035

m2 = 0.234

W = (h1 h2) + (1 m1)(h3 h4) + (1 - m1 m2)(h4 h5)

W = (2990.35 2813.4) + (0.965)(2959.5 2898.59) + (0.731)(2898.59 2079.839)

W = 834.24 kJ/kg

ee = W/EC EC = h1 hf8 + h3 h2

EC = 2990.35 897.02 + 2959.5 2813.4

EC = 2239.43 kJ/kg

ee = (834.24 / 2239.43) x100%= 37.25%

Thermodynamics 2

Section: 57003 Schedule: 7:00 9:40/ M - F

Submitted to:

Engr. Bienvenido D. Manuntag Jr.

Group 5 Group 6

Abangco, Kevin B. Arante, Kenneth V.

Honorario, James O. Axalan, Jimwell M.

Pedroso, Elvin Louie R. Corpuz, Adrian Lorenzo G.

Polo, Napoleon S. Dumalaon, Mark Lorenze R.

Sarmiento, Emmanuel G. Rocela, Noriel E.

Tagaya, Jerome M. Uyam, Jeffrey A. Jr.

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