Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781260048766
Author: CENGEL
Publisher: MCG
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Chapter 10.9, Problem 98RP

Consider a steam power plant that operates on a regenerative Rankine cycle and has a net power output of 150 MW. Steam enters the turbine at 10 MPa and 500°C and the condenser at 10 kPa. The isentropic efficiency of the turbine is 80 percent, and that of the pumps is 95 percent. Steam is extracted from the turbine at 0.5 MPa to heat the feedwater in an open feedwater heater. Water leaves the feedwater heater as a saturated liquid. Show the cycle on a T-s diagram, and determine (a) the mass flow rate of steam through the boiler and (b) the thermal efficiency of the cycle. Also, determine the exergy destruction associated with the regeneration process. Assume a source temperature of 1300 K and a sink temperature of 303 K.

(a)

Expert Solution
Check Mark
To determine

The mass flow rate of steam through the boiler.

Answer to Problem 98RP

The mass flow rate of steam through the boiler is 159.7kg/s.

Explanation of Solution

Draw the Ts diagram of the given regenerative Rankine cycle as shown in Figure 1.

Thermodynamics: An Engineering Approach, Chapter 10.9, Problem 98RP

Here, water (steam) is the working fluid of the regenerative Rankine cycle. The cycle involves two pumps.

Write the formula for work done by the pump during process 1-2.

wpI,in=v1(P2P1)ηP (I)

Here, the specific volume is v, the pressure is P, the isentropic efficiency of the pump is ηP, and the subscripts 1 and 2 indicates the process states.

Write the formula for enthalpy (h) at state 2.

h2=h1+wpI,in (II)

Write the formula for work done by the pump during process 3-4.

wpII,in=v3(P4P3)ηP (III)

Here, the specific volume is v, the pressure is P, and the subscripts 3 and 4 indicates the process states.

Write the formula for enthalpy (h) at state 4.

h4=h3+wpII,in (IV)

At state 6s:

The steam expanded to the pressure of 0.5MPa(500kPa) and the steam is at the state of saturated mixture.

The quality of water at state 6s is expressed as follows.

x6s=s6ssf,6ssfg,6s (V)

The enthalpy at state 6s is expressed as follows.

h6s=hf,6s+x6shfg,6s (VI)

Here, the enthalpy is h, the entropy is s, the quality of the water is x, the suffix f indicates the fluid condition, the suffix fg indicates the change of vaporization phase; the subscript 6s indicates the ideal process state 6s.

The isentropic efficiency for the process 5-6 is expressed as follows.

ηT=h5h6h5h6sh6=h5ηT(h5h6s) (VII)

At state 7s:

The steam enters the condenser at the pressure of 10kPa and at the state of saturated mixture.

The quality of water at state 7s is expressed as follows.

x7s=s7ssf,7ssfg,7s (VIII)

The enthalpy at state 7s is expressed as follows.

h7s=hf,7s+x7shfg,7s (IX)

Here, the subscript 7s indicates the ideal process state 7s.

The isentropic efficiency for the process 5-7 is expressed as follows.

ηT=h5h7h5h7sh7=h5ηT(h5h7s) (X)

Here, the subscript 6s, and 7s indicates the ideal process states; all other subscripts such as 1, 2, 3, 4, 5, 6 and 7 indicates the actual process states.

Write the formula for heat in (qin) and heat out (qout) of the cycle.

qin=h5h4 (XI)

qout=(1y)(h7h1) (XII)

Here, the mass fraction steam extracted from the turbine to the inlet mass of the boiler (m˙6/m˙3) is y.

Write the general equation of energy balance equation.

E˙inE˙out=ΔE˙system (XIII)

Here, the rate of net energy inlet is E˙in, the rate of net energy outlet is E˙out and the rate of change of net energy of the system is ΔE˙system.

At steady state the rate of change of net energy of the system (ΔE˙system) is zero.

ΔE˙system=0

Refer Equation (XIII).

Write the energy balance equation for open feed water heater.

E˙inE˙out=0E˙in=E˙outm˙inhin=m˙outhoutm˙6h6+m˙2h2=m˙3h3 (XIV)

Rewrite the Equation (XIV) in terms of mass fraction y.

yh6+(1y)h2=1h3yh6+h2yh2=h3y(h6h2)=h3h2y=h3h2h6h2 (XV)

Write the formula for net work output of the cycle.

wnet=qinqout (XVI)

Write the formula for mass flow rate of steam.

m˙=W˙netwnet (XVII)

Here, the net power output of the cycle is W˙net.

At state 1: (Pump I inlet)

The water exits the condenser as a saturated liquid at the pressure of 10kPa. Hence, the enthalpy, specific volume, and entropy at state 1 is as follows.

h1=hf@10kPav1=vf@10kPas1=sf@10kPa

Refer Table A-5, “Saturated water-Pressure table”.

The enthalpy (h1), specific volume (v1) and entropy (s1) at state 1 corresponding to the pressure of 10kPa is 191.81kJ/kg, 0.001010m3/kg, and 0.6492kJ/kgK respectively.

At state 3: (Pump II inlet)

The water exits the open feed water heater as a saturated liquid at the pressure of 0.5MPa(500kPa). Hence, the enthalpy, specific volume, and entropy at state 3 is as follows.

h3=hf@500kPav3=vf@500kPas3=sf@500kPa

Refer Table A-5, “Saturated water-Pressure table”.

The enthalpy (h3), specific volume (v3), and entropy (s3) at state 3 corresponding to the pressure of 0.5MPa(500kPa) is 640.09kJ/kg, 0.001093m3/kg, and 1.8604kJ/kgK respectively.

At state 5:

The steam enters the turbine as superheated vapor.

Refer Table A-6, “Superheated water”.

The enthalpy (h5) and entropy (s5) at state 5 corresponding to the pressure of 10MPa(10000kPa) and the temperature of 500°C is as follows.

h5=3375.1kJ/kgs5=6.5995kJ/kgK

Refer Figure 1.

s5=s6s=s7s=6.5995kJ/kgK

At state 6s:

The steam expanded to the pressure of 0.5MPa(500kPa) and in the state of saturated mixture.

Refer Table A-5, “Saturated water-Pressure table”.

Obtain the following properties corresponding to the pressure of 0.5MPa(500kPa).

hf,6s=640.09kJ/kghfg,6s=2108.0kJ/kgsf,6s=1.8604kJ/kgKsfg,6s=4.9603kJ/kgK

At state 7s:

The steam enters the condenser at the pressure of 10kPa and at the state of saturated mixture.

Refer Table A-5, “Saturated water-Pressure table”.

Obtain the following properties corresponding to the pressure of 10kPa.

hf,7s=191.81kJ/kghfg,7s=2392.1kJ/kgsf,7s=0.6492kJ/kgKsfg,7s=7.4996kJ/kgK

Conclusion:

Substitute 0.001010m3/kg for v1, 10kPa for P1, 500kPa for P2, and 0.95 for ηP in Equation (I).

wpI,in=(0.001010m3/kg)(500kPa10kPa)0.95=0.5209kPam3/kg×1kJ1kPam3=0.52kJ/kg

Substitute 191.81kJ/kg for h1, and 0.52kJ/kg for wpI,in in Equation (II).

h2=191.81kJ/kg+0.52kJ/kg=192.33kJ/kg

Substitute 0.001093m3/kg for v3, 500kPa for P3, 10000kPa for P4, and 0.95 for ηP in Equation (III).

wpII,in=(0.001093m3/kg)(10000kPa500kPa)0.95=10.93kPam3/kg×1kJ1kPam3=10.93kJ/kg

Substitute 640.09kJ/kg for h3, and 10.93kJ/kg for wpII,in in Equation (IV).

h4=640.09kJ/kg+10.93kJ/kg=651.02kJ/kg

From Figure 1.

s5=s6s=s7s=6.5995kJ/kgK

Substitute 6.5995kJ/kgK for s6, 1.8604kJ/kgK for sf,6s, and 4.9603kJ/kgK for sfg,6s in Equation (V).

x6s=6.5995kJ/kgK1.8604kJ/kgK4.9603kJ/kgK=0.9554

Substitute 640.09kJ/kg for hf,6s, 2108.0kJ/kg for hfg,6s, and 0.9554 for x6s in

Equation (VI).

h6s=640.09kJ/kg+0.9554(2108.0kJ/kg)=640.09kJ/kg+2013.9832kJ/kg=2654.0732kJ/kg

Substitute 3375.1kJ/kg for h5, 0.80 for ηT, 2654.0732kJ/kg for h6s in Equation (VII).

h6=3375.1kJ/kg0.80(3375.1kJ/kg2654.0732kJ/kg)=3375.1kJ/kg576.8214kJ/kg=2798.2786kJ/kg2798.3kJ/kg

Substitute 6.5995kJ/kgK for s7s, 0.6492kJ/kgK for sf,7s, and 7.4996kJ/kgK for sfg,7s in Equation (VIII).

x7s=6.5995kJ/kgK0.6492kJ/kgK7.4996kJ/kgK=0.7934

Substitute 191.81kJ/kg for hf,7s, 2392.1kJ/kg for hfg,7s, and 0.7934 for x7s in

Equation (IX).

h7s=191.81kJ/kg+0.7934(2392.1kJ/kg)=191.81kJ/kg+1897.8921kJ/kg=2089.7021kJ/kg2089.7kJ/kg

Substitute 3375.1kJ/kg for h5, 0.80 for ηT, 2089.7kJ/kg for h7s in Equation (X).

h7=3375.1kJ/kg0.80(3375.1kJ/kg2089.7kJ/kg)=3375.1kJ/kg1028.32kJ/kg=2346.78kJ/kg2346.8kJ/kg

Substitute 640.09kJ/kg for h3, 192.33kJ/kg for h2, and 2798.3kJ/kg for h6 in Equation (XV).

y=640.09kJ/kg192.33kJ/kg2798.3kJ/kg192.33kJ/kg=447.762605.97=0.1718

Substitute 3375.1kJ/kg for h5, and 651.02kJ/kg for h4 in Equation (XI).

qin=3375.1kJ/kg651.02kJ/kg=2724.08kJ/kg2724.1kJ/kg

Substitute 2346.8kJ/kg for h7, 0.1718 for y, and 191.81kJ/kg for h1 in Equation (XII).

qout=(10.1718)(2346.8kJ/kg191.81kJ/kg)=0.8282(2154.99kJ/kg)=1784.7627kJ/kg=1784.7kJ/kg

Substitute 2724.1kJ/kg for qin, and 1784.7kJ/kg for qout in Equation (XVI).

wnet=2724.1kJ/kg1784.7kJ/kg=939.4kJ/kg

Substitute 150MW for W˙net and 939.4kJ/kg for wnet in Equation (XVII).

m˙=150MW939.4kJ/kg=150MW×103kJ/s1MW939.4kJ/kg=159.6764kg/s159.7kg/s

Thus, the mass flow rate of steam through the boiler is 159.7kg/s.

(b)

Expert Solution
Check Mark
To determine

The thermal efficiency of the cycle and the exergy destruction associated with the regeneration process.

Answer to Problem 98RP

The thermal efficiency of the cycle is 34.5% and the exergy destruction associated with the regeneration process is 39.25kJ/kg.

Explanation of Solution

Write the formula for thermal efficiency of the cycle (ηth).

ηth=1qoutqin (XVIII)

Write the formula for exergy destruction associated with the regeneration cycle.

xdestroyed, reheat=T0sgen=T0(mesemisi+qsurrT0)=T0[s3ys6(1y)s2] (XIX)

Here, the entropy generation is sgen, the entropy is s, the heat transferred to the surrounding is qsurr, the surrounding (sink) temperature is T0, and the subscripts 2, 3 and 6 indicates the process states.

Conclusion:

Substitute 2724.1kJ/kg for qin, and 1784.7kJ/kg for qout in Equation (XVIII).

ηth=11784.7kJ/kg2724.1kJ/kg=10.6551=0.3448×100=34.5%

Thus, the thermal efficiency of the cycle is 34.5%.

Consider the regeneration process- open feed water heater, process states 6,2,3.

Here, s1=s2=0.6492kJ/kgK.

Substitute 303K for T0, 1.8604kJ/kgK for s3, 0.1718 for y, 6.5995kJ/kgK for s6, and 0.6492kJ/kgK for s2 in Equation (XIX).

xdestroyed, reheat=303K(1.8604kJ/kgK(0.1718)(6.5995kJ/kgK)(10.1718)(0.6492kJ/kgK))=303K(0.1295kJ/kgK)=39.2475kJ/kg39.25kJ/kg

Thus, the exergy destruction associated with the regeneration process is 39.25kJ/kg.

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Chapter 10 Solutions

Thermodynamics: An Engineering Approach

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