Thermodynamics 2 - Rankine Cycle.pptx

THERMODYNAMICS 2

Power Cycles

RANKINE CYCLE 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

SCHEMATIC DIAGRAM

P – V DIAGRAM

T – S DIAGRAM

RANKINE CYCLENote: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.

CYCLE ANALYSIS Heat added, QA

Energy Balance: Ein = Eout

QA + hB = h1

QA = h1 - hB

CYCLE ANALYSIS Heat Rejected, QR


h2 = h3 + QR

QR = h2 – h3

CYCLE ANALYSIS 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

ENERGY BALANCE

CYCLE ANALYSIS Pump Work, WP

Exact Pump Work: Ein = Eout

h3 + WP = hB

WP = hB – h3

CYCLE ANALYSIS 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

CYCLE ANALYSIS Approximate Pump Work 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

CYCLE ANALYSIS Another method of determining the net

cycle work is obtaining it from the T-s Diagram.

WNET = area 1–2–3–B–1

= area (1–2–3–B–1) - area (2–3–b–c–2)

= QA - QR

= (h1 - hB ) – (h2 – h3 )

= h2 – h1 – (hB – h3 )

= h1 – h2 – WP

Thermal Efficiency, eC

RANKINE CYCLE

CYCLE ANALYSIS 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.

CYCLE ANALYSIS Let:

P = power output, kW

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

M = steam rate, kg/kWh

By definition:

But:

CYCLE ANALYSISThen;

For the Ideal Rankine Cycle:

THE IDEAL RANKINE ENGINE 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.

ENGINE ANALYSIS 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.

ENGINE ANALYSIS EC = enthalpy of steam entering the

engine – enthalpy of saturated liquid at the condensing temperature

Steam Rate, me

ENGINE ANALYSIS Heat Rate, HR Heat rate is the energy chargeable per

unit of work or the rate of energy chargeable per unit of power.

ENGINE ANALYSIS Heat Rate, HR

Relation between ee and HR

ENGINE ANALYSIS

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.

THE ACTUAL RANKINE CYCLE 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.

SCHEMATIC DIAGRAM

CYCLE ANALYSIS 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

ENERGY BALANCE

CYCLE ANALYSIS Pump Work, WP’

where: nm = pump mechanical efficiency

Actual Cycle Thermal Efficiency, ec’

THE ACTUAL RANKINE ENGINE 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’

DEFINITION OF TERMS1. 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.

TURBO-GENERATOR SET

THERMAL EFFICIENCIES Ideal Thermal Efficiency, e

Indicated Thermal Efficiency, ei

Brake Thermal Efficiency, eb

Combined Thermal Efficiency, ek

Note: e> ei > eb > ek

STEAM RATES Ideal Steam Rate, m

Indicated Steam Rate, m1

Brake Steam Rate, mb

Combined Steam Rate, mk

ENGINE EFFICIENCIES Indicated Engine Efficiency, ni

Brake Engine Efficiency, nb

Combined Engine Efficiency, nk

Turbine/Engine Mechanical Efficiency, nme

Generator Efficiency, ng

HEAT RATE 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)

EXAMPLE PROBLEM3 – 1, p. 71 Steam is generated at 4.10 MPa and

440°C 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?

T – S DIAGRAM

SOLUTION@ P1 = 4.10 MPa

@ P2 = 0.105 MPa

h2 = hf2 - x2hfg2 = 423.24+(0.925)(2254.4)h2 = 2508.6 kJ/kgh3 = hf @ 0.105 MPa = 423.24 kJ/kgvf3 = vf @ 0.105 MPa = 0.0010443 m3/kg

P1 = 4.10 MPa h1 = 3305.7 kJ/kg

T1 = 440°C 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

SOLUTIONWP = 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

THE REHEAT CYCLE 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.

T – S DIAGRAM OF RANKINE CYCLE

T – S DIAGRAM OF REHEAT CYCLE

THE REHEAT CYCLE 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

SCHEMATIC DIAGRAM

T – S DIAGRAM OF REHEAT CYCLE (1 STAGE OF REHEATING)



QB + hB = h1

QB = h1 - hB

ENERGY BALANCE

CYCLE ANALYSIS 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



h4 = h5 + QR

QR = h4 – h5


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

ENERGY BALANCE

CYCLE ANALYSIS Pump Work, WP

Approximate Pump Work WP ≈ vf5 (PB – P5)

Exact Pump Work: Ein = Eout

h5 + WP = hB

WP = hB – h5

CYCLE ANALYSIS Net Cycle Work, WNET

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

NET CYCLE WORK

CYCLE ANALYSIS 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 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

ENGINE ANALYSIS 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 + ΣQRH

For the given engine: EC = h1 - h5 + QRH

= h1 - h5 + (h3 – h2)

= h1 - h2 + h3 – h5

THE ACTUAL REHEAT CYCLE 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’

THE ACTUAL REHEAT CYCLE 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

SCHEMATIC DIAGRAM – 1 STAGE OF REHEATING

T – S DIAGRAM

HEAT LOSS IN THE STEAM LINE

CYCLE ANALYSIS 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 Net Cycle Work, WNET’

WNET’ = W’ – WP’

CYCLE ANALYSIS Actual Pump Work, WP’

Thermal efficiency, eC’

THE ACTUAL REHEAT ENGINE 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’)

ENGINE ANALYSIS Thermal efficiency, ee’

Steam Rate, me’

Heat Rate, HR’HR’ = (me’)(EC’)

EXAMPLE PROBLEM3 – 5, p. 88 In a reheat cycle steam at 8.0 MPa and

485°C 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 720°C. 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.

T – S DIAGRAM

SOLUTIONh1 = h @ 8.0 MPa and 485°C = 3361 kJ/kg

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

h3 = h @ 1.3 MPa and 720°C = 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

SOLUTION(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/kgWNET = 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/kgee = W/EC

= (1912/4286.5)x100% = 44.6%(f) m = 3600/W = 3600/1912 = 1.88 kg/kWhSteam flow rate = (40,000 kWh)(1.88 kg/kWh)= 75,200 kg/h or 20.89 kg/s

THE REGENERATIVE CYCLE 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.

T – S DIAGRAM (RANKINE CYCLE)

T – S DIAGRAM OF REGENERATIVE CYCLE WITH ONE

STAGE OF REHEATING

EFFECTS OF REGENERATIVE FEEDWATER HEATING 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.

EFFECTS OF REGENERATIVE FEEDWATER HEATING 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.

EFFECTS OF REGENERATIVE FEEDWATER HEATING 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.

SCHEMATIC DIAGRAM

Plant Layout of Regenerative Cycle With One Stage of Extraction for Feedwater

Heating

T – S DIAGRAM

CYCLE ANALYSIS 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

MASS BALANCE

CYCLE ANALYSIS 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).

CYCLE ANALYSISmass 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,

REGENERATIVE CYCLE WITH 2-STAGES OF EXTRACTION FOR

FEEDWATER HEATING

CYCLE ANALYSIS 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

ENERGY BALANCE

ENERGY BALANCE


Energy Balance:Ein = Eout

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

QR = (1 - m)(h3 – h4)


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 = ΣW (stage work)W = W1-2 + W2-3

= (h1 - h2) + (1 - m)(h2 – h3)

ENERGY BALANCE

CYCLE ANALYSIS Total Pump Work, ΣWP

ΣWP = WP1 + WP2

WP1 = vf4 (PB4 – P4)

WP2 = vf5 (PB5 – P5) Approximate Total Pump Work

ΣWP = vf4 (PB5 – P4)

Net Cycle Work, WNET

WNET = W – ΣWP

WNET = (h1 - h2) + (1 - m)(h2 – h3) - ΣWP

Thermal efficiency, eC

THE IDEAL REGENERATIVE ENGINE 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

THE ACTUAL REGENERATIVE CYCLE 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.

SCHEMATIC DIAGRAM

Plant Layout of Actual Regenerative Cycle with One Stage of Extraction for Feedwater Heating

CYCLE ANALYSIS 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’)

CYCLE ANALYSIS Pump Work, WP’

WP’ = ΣWP

= WP1’ + WP2’

Net Cycle Work, WNET’

WNET’ = W’ – WP’ Thermal Efficiency, EC’

EXAMPLE PROBLEM3 – 10, p.107 Steam is delivered to an engine at 5.4

MPa and 600°C. Before condensation at 31°C, 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.

T – S DIAGRAM

SOLUTIONh1 = h @ 5.40 MPa and 600°C = 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 @ 31°C = 129.97 kJ/kg

h5 = hf @ 0.60 MPa = 670.56 kJ/kg

vf4 = vf @ 31°C = 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

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

ΣWP = 5.42 kJ/kg

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

SOLUTIONHeat Balance:

(1-m)(h5 – h4) = m(h2 – h5)

m = 0.1898 kg of throttle steamWNET = W – WP = (h1 - h2) + (1 - m)(h2 – h3) – ΣWP

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/kgee = W/(h1 – h2) = [1326.2/(3663.3-670.56)]x100% =

44.3%m = 3600/1362.2 = 2.71 kg/kWh

THE REHEAT-REGENERATIVE CYCLE 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.

CYCLE ANALYSIS

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.

CYCLE ANALYSIS 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.

SCHEMATIC DIAGRAM – CASE 1

T – S DIAGRAM (CASE 1)

CYCLE ANALYSIS

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

CYCLE ANALYSIS 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 = ΣW of stages of expansion

= W1-2 + W2-3 + W3-4

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

ENERGY BALANCE

CYCLE ANALYSIS Total Pump Work, ΣWP

ΣWP = WP1 + WP2

Exact formula:ΣWP = vf5 (PB6 – P6) + vf7 (PB7 – P7)

Approximate formula:ΣWP = vf5 (PB7 – P6)

WNET = Engine work – Pump work

= W – ΣWP

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

- vf5 (PB7 – P6)

CYCLE ANALYSIS 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 + ΣQReheaters

= (h1 – h7 - WP2)+(1 - m)(h4 – h3)

ENERGY BALANCE

CYCLE ANALYSIS 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.


CYCLE ANALYSIS

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


CYCLE ANALYSIS 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

LAST HEATER

SECOND HEATER

FIRST HEATER

CYCLE ANALYSIS 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)

ENERGY BALANCE

ENERGY BALANCE

CYCLE ANALYSIS 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

ENERGY AND MASS BALANCE

CYCLE ANALYSIS 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

CYCLE ANALYSIS 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:



LAST HEATER

SECOND HEATER

THIRD HEATER

CYCLE ANALYSIS 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)


CYCLE ANALYSISWNET = 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]


CYCLE ANALYSIS 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) Thermal Efficiency, eC

CYCLE ANALYSIS 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.



CYCLE ANALYSIS Mass of Bled Steam Open Heater no. 1

Open Heater no. 2

CYCLE ANALYSIS 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

OrW = W1-2 + W3-4 + W4-5

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


EXAMPLE PROBLEM3 – 14, p.133

Steam at 5 MPa and 365°C enters a turbine and expands until it becomes saturated. The steam is withdrawn and reheated to 330°C. After expansion in the turbine to 150°C, 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.

T – S DIAGRAM

SOLUTIONh1 = h @ 5 MPa and 365°C = 3108 kJ/kg

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

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

h4 = h @ 0.28 MPa and 150°C 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

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

ΣWP = 5.06 kJ/kg

SOLUTION Heater:

= 0.1264 kgW = (h1 – h2) + (h3 – h4) + (1 – m1)(h4 – h5)

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

= 1060.5 kJ/kgWNET = W – ΣWP = 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

INCOMPLETE EXPANSION CYCLE 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

T – S DIAGRAM

CYCLE ANALYSIS Engine Work, W Recalling isentropic and isometric

processes.

Isentropic Isometric

CYCLE ANALYSIS Neglecting ΔP and ΔK, 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)

CYCLE ANALYSIS For incomplete-expansion with zero

clearance: W = W1-2 + W2-3

W = h1 – h2 + v2(P2 – P3)


Ein = Eout

QA + hB = h1

QA = h1 - hB

CYCLE ANALYSIS 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

PUMP ENERGY BALANCE

CYCLE ANALYSIS Thermal Efficiency, eC

THE INCOMPLETE-EXPANSION ENGINE Energy Chargeable, EC

EC = h1 + hf3

Thermal Efficiency, e

Mean Effective Pressure, MEP or Pm

Ideal Mean Effective Pressure

ENGINE ANALYSIS Indicated Mean Effective Pressure,

MEPI

Brake Mean Effective Pressure, MEPB

Combined Mean Effective Pressure, MEPK

ENGINE ANALYSIS Steam Rate, m Ideal Steam Rate, m

Indicated Steam Rate, mI

Brake Steam Rate, mB

Combined Steam Rate, mK

ENGINE ANALYSIS Thermal Efficiency, e Indicated Thermal Efficiency, eI

Brake Thermal Efficiency, eB

Combined Thermal Efficiency, eK

ENGINE ANALYSIS Engine Efficiency, n Indicated Engine Efficiency, nI

Brake Engine Efficiency, nB

ENGINE ANALYSIS Combined Engine Efficiency, nK

Mechanical Efficiency, nm

Generator Efficiency, ng

ENGINE ANALYSIS Approximate Enthalpy of Exhaust

Steam, he’

Where:

W’ = actual work W’ = WI, indicated work

Energy Balance:

h1 = Qloss + WI + he’

he’ = h1 - Qloss - WI

ENERGY BALANCE

RELATIONSHIP BETWEEN EFFICIENCIES AND STEAM RATES eI = nIe

Proof:

eB = nBe

Proof:

RELATIONSHIP BETWEEN EFFICIENCIES AND STEAM RATES

eK = nKe

Proof:

eK = ngnmeI

Proof: eK = ngnmeI

RELATIONSHIP BETWEEN EFFICIENCIES AND STEAM RATES nm = mI/mB

Proof:

nm = eB/eI

Proof:

RELATIONSHIP BETWEEN EFFICIENCIES AND STEAM RATES

ng = eK/eB

Proof:

nK = nInmng

Proof:

RELATIONSHIP BETWEEN EFFICIENCIES AND STEAM RATES mK = mI/nmng

Proof:

EXAMPLE PROBLEM3 -19, p. 154 Steam at 1.10 MPa and 250°C 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.

T – S DIAGRAM

SOLUTIONGiven:

@ 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 = 205°C S1 = 6.6659 kJ/kg-K

vf2 = 0.0010613

hf2 = 508.03 sf2 = 1.5386

vfg2 = 0.86444 hfg2 = 2199.8 sfg2 = 5.5803

SOLUTIONW = h1 – h2 + v2(P2 – P3)

W = 2834.2 – 2529.2 + (0.7953)(205 – 105)W = 384.5 kJ/kgee = W/(h1 – hf3) = [384.5/(2834.2-423.24)] x100%

ee = 15.95%

m = 3600/W = 3600/384.5 = 0.363 kg/kWhWI = 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 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.

SCHEMATIC DIAGRAM

EXAMPLE PROBLEM3 – 24, p.166

A 30,000 kW existing plant has the following throttle conditions:

PS = 1.50 MPa; TS = 260°C

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.

SOLUTIONhs = h @1.5 MPa, 260°C = 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/kWhWK = 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

ON MOLLIER’S CHART BY CUT AND TRY METHOD

SOLUTIONh0 :

@ 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 = 416°C

PROBLEM SOLVING15 – 174 (Sta. Maria) A reheat-regenerative engine receives steam at 207 bar

and 593°C, expanding it to 38.6 bar, 343°C. At this point, the steam passes through a reheater and reenters the turbine at 34.5 bar, 593°C, hence expands to 9 bar, 492°C, 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 171°C 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

SCHEMATIC DIAGRAM

T – S DIAGRAM

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

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

h1 = 3511.3 kJ/kg

@Pt. 2: P2 = 3.86 MPa; T2 = 343°C

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

h2 = 3010.0 kJ/kg

@Pt. 3: P3 = 3.45 MPa; T3 = 593°C


h3 = 3662.5 kJ/kg T(°C) S (kJ/kg-K)

580 7.3880

593 S3

600 7.4409

SOLUTIONInterpolate:

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

@Pt. 4: P4 = 0.90 MPa; T4 = 492°C


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

SOLUTION@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= 742.833205.4m + 163.40 – 163.40m = 742.83

3042m = 579.43m = 0.19 kg


SOLUTION 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%

PROBLEM SOLVING16 – 174-175 (Sta. Maria) In a 35,000 kW turbo-generator that receives steam at

6.9 MPa and 370°C, 11% of the throttle steam is actually extracted at 2MPa, 215°C; with the remainder being reheated to 1.8 MPa and 315°C; 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 205°C. 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% ?

SCHEMATIC DIAGRAM

T – S DIAGRAM



h1 = 3077.6 kJ/kg

S1 = 6.3314 kJ/kg-K

@Pt. 2: P2 = 2 MPa; T2 = 215°C


h2 = 2807.2 kJ/kg

S2 = 6.3566 kJ/kg-K

@Pt. 3: P3 = 1.8 MPa; T3 = 315°C


h3 = 3063.2 kJ/kg

S3 = 6.88105 kJ/kg-K

SOLUTION@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


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


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

SOLUTION@Pt. 6: P6 = 0.005 MPa; Saturated Liquid

h6 = hf6 = 137.82 kJ/kg

S6 = Sf6 = 0.4764 kJ/kg-K


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

SOLUTION@Pt. 7: P7 = 0.724 MPa; Saturated Liquid

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

SOLUTION 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 = 995.6172/0.95W = 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%

PROBLEM SOLVING17 – 175, p.175 There are developed 25,000 kW by a reheat-

regenerative engine (turbo-generator) which receives steam at 4.2 MPa, 313°C and exhaust at 0.007 MPa. At 1.90 MPa and 215°C, 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 270°C 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 207°C. 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.

SCHEMATIC DIAGRAM

T – S DIAGRAM



h1 = 2990.35 kJ/kg

@Pt. 2: P2 = 1.9 MPa; T2 = 215°C

Tsat = 209.84°C; 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


Tsat = 207.15°C; 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


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


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

SOLUTION@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


h6 = hf6 = 163.40 kJ/kg

S6 = Sf6 = 0.5592 kJ/kg-K




h7 = hf7 = 827.29 kJ/kg

S7 = Sf7 = 2.2778 kJ/kg-K




h8 = hf8 = 897.02 kJ/kg

S8= Sf8 = 2.4233 kJ/kg-K



SOLUTION % of Extracted Steam:

m1 = 0.035

m2 = 0.234

SOLUTION 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.

Documents

Thermodynamics 2 - Rankine Cycle.pptx