The mass flow rate of air in the gas-turbine cycle.

Question

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Answer 1

Question

Chapter 10.9, Problem 109RP

(a)

To determine

The mass flow rate of air in the gas-turbine cycle.

(a)

Expert Solution

Answer to Problem 109RP

The mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

Explanation of Solution

Show the T-s diagram as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 10.9, Problem 109RP

Express Prandtl number at state 8s.

Pr8s=P8sP7Pr7 (I)

Here, pressure at state 8s is P8s, pressure at state 7 is P7 and Prandtl number at state 7 is Pr7.

Express enthalpy at state 8.

h8=h7+h8s−h7ηC (II)

Here, enthalpy at state 7 is h7, enthalpy at state 8s is h8s and the compressor efficiency is ηC.

Express Prandtl number at state 10s.

Pr10s=P10sP9Pr9 (III)

Here, pressure at state 10s is P10s, pressure at state 9 is P9 and Prandtl number at state 9 is Pr9.

Express enthalpy at state 10.

h10=h9−ηT(h9−h10s) (IV)

Here, enthalpy at state 9 is h9, compressor of the gas and steam turbine is ηT and enthalpy at state 10s is h10s.

Express enthalpy at state 1.

h1=hf@10 kPa (V)

Here, enthalpy of saturation liquid at pressure of 10 kPa is hf@10 kPa.

Express specific volume at state 1.

v1=vf@10 kPa (VI)

Here, specific volume of saturation liquid at pressure of 10 kPa is vf@10 kPa.

Express initial work input.

wpI,in=v1(P2−P1) (VII)

Here, pressure at state 2 and 1 is P2 and P1 respectively.

Express enthalpy at state 2.

h2=h1+wpI,in (VIII)

Express quality at state 4s.

x4s=s4s−sf@3 MPasfg@3 MPa (IX)

Here, entropy at state 4s is s4s, entropy at saturation liquid and evaporation at pressure of 3 MPa is sf@3 MPa and sfg@3 MPa respectively.

Express enthalpy at state 4s.

h4s=hf@3 MPa+x4shfg@3 MPa (X)

Here, enthalpy at saturation liquid and evaporation at pressure of 3 MPa is hf@3 MPa and hfg@3 MPa respectively.

Express enthalpy at state 4.

h4=h3−ηT(h3−h4s) (XI)

Here, enthalpy at state 3 is h3 and enthalpy at state 4s is h4s.

Express quality at state 6s.

x6s=s6s−sf@10 kPasfg@10 kPa (XII)

Here, entropy at state 6s is s6s, entropy at saturation liquid and evaporation at pressure of 10 kPa is sf@10 kPa and sfg@10 kPa respectively.

Express enthalpy at state 6s.

h6s=hf@10 MPa+x6shfg@10 MPa (XIII)

Here, enthalpy at saturation liquid and evaporation at pressure of 10 MPa is hf@10 MPa and hfg@10 MPa respectively.

Express enthalpy at state 6.

h6=h5−ηT(h5−h6s) (XIV)

Here, enthalpy at state 5 is h5 and enthalpy at state 6s is h6s.

Express the mass flow rate of air in the gas-turbine cycle from energy balance equation.

m˙air=h3−h2h10−h11m˙s (XV)

Here, enthalpy at state 10 is h10, enthalpy at state 11 is h11, enthalpy at state 3 and 2 is h3 and h2 respectively, and mass flow rate of steam is m˙s.

Conclusion:

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 7 and Prandtl number at state 7 corresponding to temperature at state 7 of 290 K.

h7=290.16 kJ/kgPr7=1.2311

Substitute 8 for P8sP7 and 1.2311 for Pr7 in Equation (I).

Pr8s=8(1.2311)=9.849

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XVI)

Here, the variables denote by x and y is Prandtl number at state8s and enthalpy at state 8s respectively.

Show the enthalpy at state 8s corresponding to Prandtl number as in Table (1).

Prandtl number at state 8s Pr8s	Enthalpy at state 8s h8s(kJ/kg)
9.684 (x1)	523.63 (y1)
9.849 (x2)	(y2=?)
10.37 (x3)	533.98 (y3)

Substitute 9.684, 9.849 and 10.37 for x1,x2 and x3 respectively, 523.63 kJ/kg for y1 and 533.98 kJ/kg for y3 in Equation (XVI).

y2=(9.849−9.684)(533.98 kJ/kg−523.63 kJ/kg)(10.37−9.684)+523.63 kJ/kg=526.12 kJ/kg=h8s

Thus, enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 is,

h8s=526.12 kJ/kg

Substitute 290.16 kJ/kg for h7, 526.12 kJ/kg for h8s and 0.85 for ηC in Equation (II).

h8=290.16 kJ/kg+[526.12 kJ/kg−290.16 kJ/kg0.85]=567.76 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 9 and Prandtl number at state 9 corresponding to temperature at state 9 of 1400 K.

h9=1515.42 kJ/kgPr9=450.5

Here, enthalpy at state 9 is h9 and Prandtl number at state 9 is Pr9.

Substitute 18 for P10sP9 and 450.5 for Pr9 in Equation (III).

Pr10s=18(450.5)=56.3

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 10s corresponding to Prandtl number at state 10s of 56.3 using an interpolation method.

Show the enthalpy at state 10s corresponding to Prandtl number as in Table (2).

Prandtl number at state 10s Pr10s	Enthalpy at state 10s h10s(kJ/kg)
52.59 (x1)	843.98 (y1)
56.3 (x2)	(y2=?)
57.60 (x3)	866.08 (y3)

Use excels and substitutes the values from Table (II) in Equation (XVI) to get,

h10s=860.35 kJ/kg

Here, enthalpy at state 10s is h10s.

Substitute 1515.42 kJ/kg for h9, 0.90 for ηT and 860.35 kJ/kg for h10s in Equation (IV).

h10=1515.42 kJ/kg−(0.90)(1515.42 kJ/kg−860.35 kJ/kg)=925.86 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 11 corresponding to temperature at state 11 of 520 K.

h11=523.63 kJ/kg

Here, enthalpy at state 11 is h11.

Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 10 kPa.

hf@10 kPa=191.81 kJ/kgvf@10 kPa=0.00101 m3/kg

Substitute 191.81 kJ/kg for hf@10 kPa in Equation (V).

h1=191.81 kJ/kg

Substitute 0.00101 m3/kg for vf@10 kPa in Equation (VI).

v1=0.00101 m3/kg

Substitute 0.00101 m3/kg for v1, 15 MPa for P2 and 10 kPa for P1 in Equation (VII).

wpI,in=(0.00101 m3/kg)(15 MPa−10 kPa)=(0.00101 m3/kg)(15,000 kPa−10 kPa)kJkPa⋅m3=15.14 kJ/kg

Substitute 191.81 kJ/kg for h1 and 15.14 kJ/kg for wpI,in in Equation (VIII).

h2=191.81 kJ/kg+15.14 kJ/kg=206.95 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 3 of 15 MPa and temperature at state 3 of 450°C.

h3=3157.9 kJ/kgs3=6.1428 kJ/kg⋅K

Here, enthalpy and entropy at state 3 is h3 and s3 respectively.

Due to throttling process, entropy at state 3 is equal to entropy at state 4s.

s3=s4s=6.1428 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 3 MPa(3000 kPa).

sf@3 MPa=2.6454 kJ/kg⋅Ksfg@3 MPa=3.5402 kJ/kg⋅Khf@3 MPa=1008.3 kJ/kghfg@3 MPa=1794.9 kJ/kg

Substitute 6.1428 kJ/kg⋅K for s4s, 2.6454 kJ/kg⋅K for sf@3 MPa and 3.5402 kJ/kg⋅K for sfg@3 MPa in Equation (IX).

x4s=6.1428 kJ/kg⋅K−2.6454 kJ/kg⋅K3.5402 kJ/kg⋅K=0.9880

Substitute 0.9880 for x4s, 1008.3 kJ/kg for hf@3 MPa and 1794.9 kJ/kg for hfg@3 MPa in Equation (X).

h4s=1008.3 kJ/kg+(0.9880)(1794.9 kJ/kg)=2781.7 kJ/kg

Substitute 3157.9 kJ/kg for h3, 0.90 for ηT and 2781.7 kJ/kg for h4s in Equation (XI).

h4=3157.9 kJ/kg−(0.90)(3157.9 kJ/kg−2781.7 kJ/kg)=2819.4 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 5 of 3 MPa and temperature at state 5 of 500°C.

h5=3457.2 kJ/kgs5=7.2359 kJ/kg⋅K

Here, enthalpy and entropy at state 5 is h5 and s5 respectively.

Due to throttling process, entropy at state 5 is equal to entropy at state 6s.

s5=s6s=7.2359 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 10 kPa.

sf@10 kPa=0.6492 kJ/kg⋅Ksfg@10 kPa=7.4996 kJ/kg⋅Khf@10 kPa=191.81 kJ/kghfg@10 kPa=2392.1 kJ/kg

Substitute 7.2359 kJ/kg⋅K for s6s, 0.6492 kJ/kg⋅K for sf@10 kPa and 7.4996 kJ/kg⋅K for sfg@10 kPa in Equation (XII).

x6s=7.2359 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.8783

Substitute 0.8783 for x6s, 191.81 kJ/kg for hf@10 kPa and 2392.1 kJ/kg for hfg@10 kPa in Equation (XIII).

h6s=191.81 kJ/kg+(0.8783)(2392.1 kJ/kg)=2292.7 kJ/kg

Substitute 3457.2 kJ/kg for h5, 0.90 for ηT and 2292.7 kJ/kg for h6s in Equation (XIV).

h6=3457.2 kJ/kg−(0.90)(3457.2 kJ/kg−2292.7 kJ/kg)=2409.1 kJ/kg

Substitute 3157.9 kJ/kg for h3, 206.95 kJ/kg for h2, 925.862 kJ/kg for h10, 523.63 kJ/kg for h11, and 30 kg/s for m˙s in Equation (V).

m˙air=3157.9 kJ/kg−206.95 kJ/kg925.862 kJ/kg−523.63 kJ/kg(30 kg/s)=220.1 kg/s

Hence, the mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

(b)

To determine

The rate of total heat input.

(b)

Expert Solution

Answer to Problem 109RP

The rate of total heat input is 227,700 kW.

Explanation of Solution

Express the rate of total heat input.

Q˙in=m˙air(h9−h8)+m˙s(h5−h4) (XVII)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 1515.42 kJ/kg for h9, 567.76 kJ/kg for h8, 3457.2 kJ/kg for h5 and 2819.4 kJ/kg for h4 in Equation (XVII).

Q˙in={(220.1 kg/s)[(1515.42−567.76) kJ/kg]+(30 kg/s)[(3457.2−2819.4) kJ/kg]}=227,700 kJ/s[kWkJ/s]=227,700 kW

Hence, the rate of total heat input is 227,700 kW.

(c)

To determine

The thermal efficiency of the combined cycle.

(c)

Expert Solution

Answer to Problem 109RP

The thermal efficiency of the combined cycle is 48.2%.

Explanation of Solution

Express the rate of total heat output.

Q˙out=m˙air(h11−h7)+m˙s(h6−h1) (XVIII)

Express the thermal efficiency of the combined cycle.

ηth=1−Q˙outQ˙in (XIX)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 523.63 kJ/kg for h11, 290.16 kJ/kg for h7, 2409.1 kJ/kg for h6 and 191.81 kJ/kg for h1 in Equation (XVIII).

Q˙out={(220.1 kg/s)[(523.63−290.16) kJ/kg]+(30 kg/s)[(2409.1−191.81) kJ/kg]}=117,900 kJ/s[kWkJ/s]=117,900 kW

Substitute 117,900 kW for Q˙out and 227,700 kW for Q˙in in Equation (XIX).

ηth=1−117,900 kW227,700 kW=0.482=48.2%

Hence, the thermal efficiency of the combined cycle is 48.2%.

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Answer 2

Question

Chapter 10.9, Problem 109RP

(a)

To determine

The mass flow rate of air in the gas-turbine cycle.

(a)

Expert Solution

Answer to Problem 109RP

The mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

Explanation of Solution

Show the T-s diagram as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 10.9, Problem 109RP

Express Prandtl number at state 8s.

Pr8s=P8sP7Pr7 (I)

Here, pressure at state 8s is P8s, pressure at state 7 is P7 and Prandtl number at state 7 is Pr7.

Express enthalpy at state 8.

h8=h7+h8s−h7ηC (II)

Here, enthalpy at state 7 is h7, enthalpy at state 8s is h8s and the compressor efficiency is ηC.

Express Prandtl number at state 10s.

Pr10s=P10sP9Pr9 (III)

Here, pressure at state 10s is P10s, pressure at state 9 is P9 and Prandtl number at state 9 is Pr9.

Express enthalpy at state 10.

h10=h9−ηT(h9−h10s) (IV)

Here, enthalpy at state 9 is h9, compressor of the gas and steam turbine is ηT and enthalpy at state 10s is h10s.

Express enthalpy at state 1.

h1=hf@10 kPa (V)

Here, enthalpy of saturation liquid at pressure of 10 kPa is hf@10 kPa.

Express specific volume at state 1.

v1=vf@10 kPa (VI)

Here, specific volume of saturation liquid at pressure of 10 kPa is vf@10 kPa.

Express initial work input.

wpI,in=v1(P2−P1) (VII)

Here, pressure at state 2 and 1 is P2 and P1 respectively.

Express enthalpy at state 2.

h2=h1+wpI,in (VIII)

Express quality at state 4s.

x4s=s4s−sf@3 MPasfg@3 MPa (IX)

Here, entropy at state 4s is s4s, entropy at saturation liquid and evaporation at pressure of 3 MPa is sf@3 MPa and sfg@3 MPa respectively.

Express enthalpy at state 4s.

h4s=hf@3 MPa+x4shfg@3 MPa (X)

Here, enthalpy at saturation liquid and evaporation at pressure of 3 MPa is hf@3 MPa and hfg@3 MPa respectively.

Express enthalpy at state 4.

h4=h3−ηT(h3−h4s) (XI)

Here, enthalpy at state 3 is h3 and enthalpy at state 4s is h4s.

Express quality at state 6s.

x6s=s6s−sf@10 kPasfg@10 kPa (XII)

Here, entropy at state 6s is s6s, entropy at saturation liquid and evaporation at pressure of 10 kPa is sf@10 kPa and sfg@10 kPa respectively.

Express enthalpy at state 6s.

h6s=hf@10 MPa+x6shfg@10 MPa (XIII)

Here, enthalpy at saturation liquid and evaporation at pressure of 10 MPa is hf@10 MPa and hfg@10 MPa respectively.

Express enthalpy at state 6.

h6=h5−ηT(h5−h6s) (XIV)

Here, enthalpy at state 5 is h5 and enthalpy at state 6s is h6s.

Express the mass flow rate of air in the gas-turbine cycle from energy balance equation.

m˙air=h3−h2h10−h11m˙s (XV)

Here, enthalpy at state 10 is h10, enthalpy at state 11 is h11, enthalpy at state 3 and 2 is h3 and h2 respectively, and mass flow rate of steam is m˙s.

Conclusion:

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 7 and Prandtl number at state 7 corresponding to temperature at state 7 of 290 K.

h7=290.16 kJ/kgPr7=1.2311

Substitute 8 for P8sP7 and 1.2311 for Pr7 in Equation (I).

Pr8s=8(1.2311)=9.849

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XVI)

Here, the variables denote by x and y is Prandtl number at state8s and enthalpy at state 8s respectively.

Show the enthalpy at state 8s corresponding to Prandtl number as in Table (1).

Prandtl number at state 8s Pr8s	Enthalpy at state 8s h8s(kJ/kg)
9.684 (x1)	523.63 (y1)
9.849 (x2)	(y2=?)
10.37 (x3)	533.98 (y3)

Substitute 9.684, 9.849 and 10.37 for x1,x2 and x3 respectively, 523.63 kJ/kg for y1 and 533.98 kJ/kg for y3 in Equation (XVI).

y2=(9.849−9.684)(533.98 kJ/kg−523.63 kJ/kg)(10.37−9.684)+523.63 kJ/kg=526.12 kJ/kg=h8s

Thus, enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 is,

h8s=526.12 kJ/kg

Substitute 290.16 kJ/kg for h7, 526.12 kJ/kg for h8s and 0.85 for ηC in Equation (II).

h8=290.16 kJ/kg+[526.12 kJ/kg−290.16 kJ/kg0.85]=567.76 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 9 and Prandtl number at state 9 corresponding to temperature at state 9 of 1400 K.

h9=1515.42 kJ/kgPr9=450.5

Here, enthalpy at state 9 is h9 and Prandtl number at state 9 is Pr9.

Substitute 18 for P10sP9 and 450.5 for Pr9 in Equation (III).

Pr10s=18(450.5)=56.3

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 10s corresponding to Prandtl number at state 10s of 56.3 using an interpolation method.

Show the enthalpy at state 10s corresponding to Prandtl number as in Table (2).

Prandtl number at state 10s Pr10s	Enthalpy at state 10s h10s(kJ/kg)
52.59 (x1)	843.98 (y1)
56.3 (x2)	(y2=?)
57.60 (x3)	866.08 (y3)

Use excels and substitutes the values from Table (II) in Equation (XVI) to get,

h10s=860.35 kJ/kg

Here, enthalpy at state 10s is h10s.

Substitute 1515.42 kJ/kg for h9, 0.90 for ηT and 860.35 kJ/kg for h10s in Equation (IV).

h10=1515.42 kJ/kg−(0.90)(1515.42 kJ/kg−860.35 kJ/kg)=925.86 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 11 corresponding to temperature at state 11 of 520 K.

h11=523.63 kJ/kg

Here, enthalpy at state 11 is h11.

Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 10 kPa.

hf@10 kPa=191.81 kJ/kgvf@10 kPa=0.00101 m3/kg

Substitute 191.81 kJ/kg for hf@10 kPa in Equation (V).

h1=191.81 kJ/kg

Substitute 0.00101 m3/kg for vf@10 kPa in Equation (VI).

v1=0.00101 m3/kg

Substitute 0.00101 m3/kg for v1, 15 MPa for P2 and 10 kPa for P1 in Equation (VII).

wpI,in=(0.00101 m3/kg)(15 MPa−10 kPa)=(0.00101 m3/kg)(15,000 kPa−10 kPa)kJkPa⋅m3=15.14 kJ/kg

Substitute 191.81 kJ/kg for h1 and 15.14 kJ/kg for wpI,in in Equation (VIII).

h2=191.81 kJ/kg+15.14 kJ/kg=206.95 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 3 of 15 MPa and temperature at state 3 of 450°C.

h3=3157.9 kJ/kgs3=6.1428 kJ/kg⋅K

Here, enthalpy and entropy at state 3 is h3 and s3 respectively.

Due to throttling process, entropy at state 3 is equal to entropy at state 4s.

s3=s4s=6.1428 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 3 MPa(3000 kPa).

sf@3 MPa=2.6454 kJ/kg⋅Ksfg@3 MPa=3.5402 kJ/kg⋅Khf@3 MPa=1008.3 kJ/kghfg@3 MPa=1794.9 kJ/kg

Substitute 6.1428 kJ/kg⋅K for s4s, 2.6454 kJ/kg⋅K for sf@3 MPa and 3.5402 kJ/kg⋅K for sfg@3 MPa in Equation (IX).

x4s=6.1428 kJ/kg⋅K−2.6454 kJ/kg⋅K3.5402 kJ/kg⋅K=0.9880

Substitute 0.9880 for x4s, 1008.3 kJ/kg for hf@3 MPa and 1794.9 kJ/kg for hfg@3 MPa in Equation (X).

h4s=1008.3 kJ/kg+(0.9880)(1794.9 kJ/kg)=2781.7 kJ/kg

Substitute 3157.9 kJ/kg for h3, 0.90 for ηT and 2781.7 kJ/kg for h4s in Equation (XI).

h4=3157.9 kJ/kg−(0.90)(3157.9 kJ/kg−2781.7 kJ/kg)=2819.4 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 5 of 3 MPa and temperature at state 5 of 500°C.

h5=3457.2 kJ/kgs5=7.2359 kJ/kg⋅K

Here, enthalpy and entropy at state 5 is h5 and s5 respectively.

Due to throttling process, entropy at state 5 is equal to entropy at state 6s.

s5=s6s=7.2359 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 10 kPa.

sf@10 kPa=0.6492 kJ/kg⋅Ksfg@10 kPa=7.4996 kJ/kg⋅Khf@10 kPa=191.81 kJ/kghfg@10 kPa=2392.1 kJ/kg

Substitute 7.2359 kJ/kg⋅K for s6s, 0.6492 kJ/kg⋅K for sf@10 kPa and 7.4996 kJ/kg⋅K for sfg@10 kPa in Equation (XII).

x6s=7.2359 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.8783

Substitute 0.8783 for x6s, 191.81 kJ/kg for hf@10 kPa and 2392.1 kJ/kg for hfg@10 kPa in Equation (XIII).

h6s=191.81 kJ/kg+(0.8783)(2392.1 kJ/kg)=2292.7 kJ/kg

Substitute 3457.2 kJ/kg for h5, 0.90 for ηT and 2292.7 kJ/kg for h6s in Equation (XIV).

h6=3457.2 kJ/kg−(0.90)(3457.2 kJ/kg−2292.7 kJ/kg)=2409.1 kJ/kg

Substitute 3157.9 kJ/kg for h3, 206.95 kJ/kg for h2, 925.862 kJ/kg for h10, 523.63 kJ/kg for h11, and 30 kg/s for m˙s in Equation (V).

m˙air=3157.9 kJ/kg−206.95 kJ/kg925.862 kJ/kg−523.63 kJ/kg(30 kg/s)=220.1 kg/s

Hence, the mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

(b)

To determine

The rate of total heat input.

(b)

Expert Solution

Answer to Problem 109RP

The rate of total heat input is 227,700 kW.

Explanation of Solution

Express the rate of total heat input.

Q˙in=m˙air(h9−h8)+m˙s(h5−h4) (XVII)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 1515.42 kJ/kg for h9, 567.76 kJ/kg for h8, 3457.2 kJ/kg for h5 and 2819.4 kJ/kg for h4 in Equation (XVII).

Q˙in={(220.1 kg/s)[(1515.42−567.76) kJ/kg]+(30 kg/s)[(3457.2−2819.4) kJ/kg]}=227,700 kJ/s[kWkJ/s]=227,700 kW

Hence, the rate of total heat input is 227,700 kW.

(c)

To determine

The thermal efficiency of the combined cycle.

(c)

Expert Solution

Answer to Problem 109RP

The thermal efficiency of the combined cycle is 48.2%.

Explanation of Solution

Express the rate of total heat output.

Q˙out=m˙air(h11−h7)+m˙s(h6−h1) (XVIII)

Express the thermal efficiency of the combined cycle.

ηth=1−Q˙outQ˙in (XIX)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 523.63 kJ/kg for h11, 290.16 kJ/kg for h7, 2409.1 kJ/kg for h6 and 191.81 kJ/kg for h1 in Equation (XVIII).

Q˙out={(220.1 kg/s)[(523.63−290.16) kJ/kg]+(30 kg/s)[(2409.1−191.81) kJ/kg]}=117,900 kJ/s[kWkJ/s]=117,900 kW

Substitute 117,900 kW for Q˙out and 227,700 kW for Q˙in in Equation (XIX).

ηth=1−117,900 kW227,700 kW=0.482=48.2%

Hence, the thermal efficiency of the combined cycle is 48.2%.

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Answer 3

Question

Chapter 10.9, Problem 109RP

(a)

To determine

The mass flow rate of air in the gas-turbine cycle.

(a)

Expert Solution

Answer to Problem 109RP

The mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

Explanation of Solution

Show the T-s diagram as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 10.9, Problem 109RP

Express Prandtl number at state 8s.

Pr8s=P8sP7Pr7 (I)

Here, pressure at state 8s is P8s, pressure at state 7 is P7 and Prandtl number at state 7 is Pr7.

Express enthalpy at state 8.

h8=h7+h8s−h7ηC (II)

Here, enthalpy at state 7 is h7, enthalpy at state 8s is h8s and the compressor efficiency is ηC.

Express Prandtl number at state 10s.

Pr10s=P10sP9Pr9 (III)

Here, pressure at state 10s is P10s, pressure at state 9 is P9 and Prandtl number at state 9 is Pr9.

Express enthalpy at state 10.

h10=h9−ηT(h9−h10s) (IV)

Here, enthalpy at state 9 is h9, compressor of the gas and steam turbine is ηT and enthalpy at state 10s is h10s.

Express enthalpy at state 1.

h1=hf@10 kPa (V)

Here, enthalpy of saturation liquid at pressure of 10 kPa is hf@10 kPa.

Express specific volume at state 1.

v1=vf@10 kPa (VI)

Here, specific volume of saturation liquid at pressure of 10 kPa is vf@10 kPa.

Express initial work input.

wpI,in=v1(P2−P1) (VII)

Here, pressure at state 2 and 1 is P2 and P1 respectively.

Express enthalpy at state 2.

h2=h1+wpI,in (VIII)

Express quality at state 4s.

x4s=s4s−sf@3 MPasfg@3 MPa (IX)

Here, entropy at state 4s is s4s, entropy at saturation liquid and evaporation at pressure of 3 MPa is sf@3 MPa and sfg@3 MPa respectively.

Express enthalpy at state 4s.

h4s=hf@3 MPa+x4shfg@3 MPa (X)

Here, enthalpy at saturation liquid and evaporation at pressure of 3 MPa is hf@3 MPa and hfg@3 MPa respectively.

Express enthalpy at state 4.

h4=h3−ηT(h3−h4s) (XI)

Here, enthalpy at state 3 is h3 and enthalpy at state 4s is h4s.

Express quality at state 6s.

x6s=s6s−sf@10 kPasfg@10 kPa (XII)

Here, entropy at state 6s is s6s, entropy at saturation liquid and evaporation at pressure of 10 kPa is sf@10 kPa and sfg@10 kPa respectively.

Express enthalpy at state 6s.

h6s=hf@10 MPa+x6shfg@10 MPa (XIII)

Here, enthalpy at saturation liquid and evaporation at pressure of 10 MPa is hf@10 MPa and hfg@10 MPa respectively.

Express enthalpy at state 6.

h6=h5−ηT(h5−h6s) (XIV)

Here, enthalpy at state 5 is h5 and enthalpy at state 6s is h6s.

Express the mass flow rate of air in the gas-turbine cycle from energy balance equation.

m˙air=h3−h2h10−h11m˙s (XV)

Here, enthalpy at state 10 is h10, enthalpy at state 11 is h11, enthalpy at state 3 and 2 is h3 and h2 respectively, and mass flow rate of steam is m˙s.

Conclusion:

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 7 and Prandtl number at state 7 corresponding to temperature at state 7 of 290 K.

h7=290.16 kJ/kgPr7=1.2311

Substitute 8 for P8sP7 and 1.2311 for Pr7 in Equation (I).

Pr8s=8(1.2311)=9.849

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XVI)

Here, the variables denote by x and y is Prandtl number at state8s and enthalpy at state 8s respectively.

Show the enthalpy at state 8s corresponding to Prandtl number as in Table (1).

Prandtl number at state 8s Pr8s	Enthalpy at state 8s h8s(kJ/kg)
9.684 (x1)	523.63 (y1)
9.849 (x2)	(y2=?)
10.37 (x3)	533.98 (y3)

Substitute 9.684, 9.849 and 10.37 for x1,x2 and x3 respectively, 523.63 kJ/kg for y1 and 533.98 kJ/kg for y3 in Equation (XVI).

y2=(9.849−9.684)(533.98 kJ/kg−523.63 kJ/kg)(10.37−9.684)+523.63 kJ/kg=526.12 kJ/kg=h8s

Thus, enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 is,

h8s=526.12 kJ/kg

Substitute 290.16 kJ/kg for h7, 526.12 kJ/kg for h8s and 0.85 for ηC in Equation (II).

h8=290.16 kJ/kg+[526.12 kJ/kg−290.16 kJ/kg0.85]=567.76 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 9 and Prandtl number at state 9 corresponding to temperature at state 9 of 1400 K.

h9=1515.42 kJ/kgPr9=450.5

Here, enthalpy at state 9 is h9 and Prandtl number at state 9 is Pr9.

Substitute 18 for P10sP9 and 450.5 for Pr9 in Equation (III).

Pr10s=18(450.5)=56.3

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 10s corresponding to Prandtl number at state 10s of 56.3 using an interpolation method.

Show the enthalpy at state 10s corresponding to Prandtl number as in Table (2).

Prandtl number at state 10s Pr10s	Enthalpy at state 10s h10s(kJ/kg)
52.59 (x1)	843.98 (y1)
56.3 (x2)	(y2=?)
57.60 (x3)	866.08 (y3)

Use excels and substitutes the values from Table (II) in Equation (XVI) to get,

h10s=860.35 kJ/kg

Here, enthalpy at state 10s is h10s.

Substitute 1515.42 kJ/kg for h9, 0.90 for ηT and 860.35 kJ/kg for h10s in Equation (IV).

h10=1515.42 kJ/kg−(0.90)(1515.42 kJ/kg−860.35 kJ/kg)=925.86 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 11 corresponding to temperature at state 11 of 520 K.

h11=523.63 kJ/kg

Here, enthalpy at state 11 is h11.

Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 10 kPa.

hf@10 kPa=191.81 kJ/kgvf@10 kPa=0.00101 m3/kg

Substitute 191.81 kJ/kg for hf@10 kPa in Equation (V).

h1=191.81 kJ/kg

Substitute 0.00101 m3/kg for vf@10 kPa in Equation (VI).

v1=0.00101 m3/kg

Substitute 0.00101 m3/kg for v1, 15 MPa for P2 and 10 kPa for P1 in Equation (VII).

wpI,in=(0.00101 m3/kg)(15 MPa−10 kPa)=(0.00101 m3/kg)(15,000 kPa−10 kPa)kJkPa⋅m3=15.14 kJ/kg

Substitute 191.81 kJ/kg for h1 and 15.14 kJ/kg for wpI,in in Equation (VIII).

h2=191.81 kJ/kg+15.14 kJ/kg=206.95 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 3 of 15 MPa and temperature at state 3 of 450°C.

h3=3157.9 kJ/kgs3=6.1428 kJ/kg⋅K

Here, enthalpy and entropy at state 3 is h3 and s3 respectively.

Due to throttling process, entropy at state 3 is equal to entropy at state 4s.

s3=s4s=6.1428 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 3 MPa(3000 kPa).

sf@3 MPa=2.6454 kJ/kg⋅Ksfg@3 MPa=3.5402 kJ/kg⋅Khf@3 MPa=1008.3 kJ/kghfg@3 MPa=1794.9 kJ/kg

Substitute 6.1428 kJ/kg⋅K for s4s, 2.6454 kJ/kg⋅K for sf@3 MPa and 3.5402 kJ/kg⋅K for sfg@3 MPa in Equation (IX).

x4s=6.1428 kJ/kg⋅K−2.6454 kJ/kg⋅K3.5402 kJ/kg⋅K=0.9880

Substitute 0.9880 for x4s, 1008.3 kJ/kg for hf@3 MPa and 1794.9 kJ/kg for hfg@3 MPa in Equation (X).

h4s=1008.3 kJ/kg+(0.9880)(1794.9 kJ/kg)=2781.7 kJ/kg

Substitute 3157.9 kJ/kg for h3, 0.90 for ηT and 2781.7 kJ/kg for h4s in Equation (XI).

h4=3157.9 kJ/kg−(0.90)(3157.9 kJ/kg−2781.7 kJ/kg)=2819.4 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 5 of 3 MPa and temperature at state 5 of 500°C.

h5=3457.2 kJ/kgs5=7.2359 kJ/kg⋅K

Here, enthalpy and entropy at state 5 is h5 and s5 respectively.

Due to throttling process, entropy at state 5 is equal to entropy at state 6s.

s5=s6s=7.2359 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 10 kPa.

sf@10 kPa=0.6492 kJ/kg⋅Ksfg@10 kPa=7.4996 kJ/kg⋅Khf@10 kPa=191.81 kJ/kghfg@10 kPa=2392.1 kJ/kg

Substitute 7.2359 kJ/kg⋅K for s6s, 0.6492 kJ/kg⋅K for sf@10 kPa and 7.4996 kJ/kg⋅K for sfg@10 kPa in Equation (XII).

x6s=7.2359 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.8783

Substitute 0.8783 for x6s, 191.81 kJ/kg for hf@10 kPa and 2392.1 kJ/kg for hfg@10 kPa in Equation (XIII).

h6s=191.81 kJ/kg+(0.8783)(2392.1 kJ/kg)=2292.7 kJ/kg

Substitute 3457.2 kJ/kg for h5, 0.90 for ηT and 2292.7 kJ/kg for h6s in Equation (XIV).

h6=3457.2 kJ/kg−(0.90)(3457.2 kJ/kg−2292.7 kJ/kg)=2409.1 kJ/kg

Substitute 3157.9 kJ/kg for h3, 206.95 kJ/kg for h2, 925.862 kJ/kg for h10, 523.63 kJ/kg for h11, and 30 kg/s for m˙s in Equation (V).

m˙air=3157.9 kJ/kg−206.95 kJ/kg925.862 kJ/kg−523.63 kJ/kg(30 kg/s)=220.1 kg/s

Hence, the mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

(b)

To determine

The rate of total heat input.

(b)

Expert Solution

Answer to Problem 109RP

The rate of total heat input is 227,700 kW.

Explanation of Solution

Express the rate of total heat input.

Q˙in=m˙air(h9−h8)+m˙s(h5−h4) (XVII)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 1515.42 kJ/kg for h9, 567.76 kJ/kg for h8, 3457.2 kJ/kg for h5 and 2819.4 kJ/kg for h4 in Equation (XVII).

Q˙in={(220.1 kg/s)[(1515.42−567.76) kJ/kg]+(30 kg/s)[(3457.2−2819.4) kJ/kg]}=227,700 kJ/s[kWkJ/s]=227,700 kW

Hence, the rate of total heat input is 227,700 kW.

(c)

To determine

The thermal efficiency of the combined cycle.

(c)

Expert Solution

Answer to Problem 109RP

The thermal efficiency of the combined cycle is 48.2%.

Explanation of Solution

Express the rate of total heat output.

Q˙out=m˙air(h11−h7)+m˙s(h6−h1) (XVIII)

Express the thermal efficiency of the combined cycle.

ηth=1−Q˙outQ˙in (XIX)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 523.63 kJ/kg for h11, 290.16 kJ/kg for h7, 2409.1 kJ/kg for h6 and 191.81 kJ/kg for h1 in Equation (XVIII).

Q˙out={(220.1 kg/s)[(523.63−290.16) kJ/kg]+(30 kg/s)[(2409.1−191.81) kJ/kg]}=117,900 kJ/s[kWkJ/s]=117,900 kW

Substitute 117,900 kW for Q˙out and 227,700 kW for Q˙in in Equation (XIX).

ηth=1−117,900 kW227,700 kW=0.482=48.2%

Hence, the thermal efficiency of the combined cycle is 48.2%.

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Answer 4

Question

Chapter 10.9, Problem 109RP

(a)

To determine

The mass flow rate of air in the gas-turbine cycle.

(a)

Expert Solution

Answer to Problem 109RP

The mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

Explanation of Solution

Show the T-s diagram as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 10.9, Problem 109RP

Express Prandtl number at state 8s.

Pr8s=P8sP7Pr7 (I)

Here, pressure at state 8s is P8s, pressure at state 7 is P7 and Prandtl number at state 7 is Pr7.

Express enthalpy at state 8.

h8=h7+h8s−h7ηC (II)

Here, enthalpy at state 7 is h7, enthalpy at state 8s is h8s and the compressor efficiency is ηC.

Express Prandtl number at state 10s.

Pr10s=P10sP9Pr9 (III)

Here, pressure at state 10s is P10s, pressure at state 9 is P9 and Prandtl number at state 9 is Pr9.

Express enthalpy at state 10.

h10=h9−ηT(h9−h10s) (IV)

Here, enthalpy at state 9 is h9, compressor of the gas and steam turbine is ηT and enthalpy at state 10s is h10s.

Express enthalpy at state 1.

h1=hf@10 kPa (V)

Here, enthalpy of saturation liquid at pressure of 10 kPa is hf@10 kPa.

Express specific volume at state 1.

v1=vf@10 kPa (VI)

Here, specific volume of saturation liquid at pressure of 10 kPa is vf@10 kPa.

Express initial work input.

wpI,in=v1(P2−P1) (VII)

Here, pressure at state 2 and 1 is P2 and P1 respectively.

Express enthalpy at state 2.

h2=h1+wpI,in (VIII)

Express quality at state 4s.

x4s=s4s−sf@3 MPasfg@3 MPa (IX)

Here, entropy at state 4s is s4s, entropy at saturation liquid and evaporation at pressure of 3 MPa is sf@3 MPa and sfg@3 MPa respectively.

Express enthalpy at state 4s.

h4s=hf@3 MPa+x4shfg@3 MPa (X)

Here, enthalpy at saturation liquid and evaporation at pressure of 3 MPa is hf@3 MPa and hfg@3 MPa respectively.

Express enthalpy at state 4.

h4=h3−ηT(h3−h4s) (XI)

Here, enthalpy at state 3 is h3 and enthalpy at state 4s is h4s.

Express quality at state 6s.

x6s=s6s−sf@10 kPasfg@10 kPa (XII)

Here, entropy at state 6s is s6s, entropy at saturation liquid and evaporation at pressure of 10 kPa is sf@10 kPa and sfg@10 kPa respectively.

Express enthalpy at state 6s.

h6s=hf@10 MPa+x6shfg@10 MPa (XIII)

Here, enthalpy at saturation liquid and evaporation at pressure of 10 MPa is hf@10 MPa and hfg@10 MPa respectively.

Express enthalpy at state 6.

h6=h5−ηT(h5−h6s) (XIV)

Here, enthalpy at state 5 is h5 and enthalpy at state 6s is h6s.

Express the mass flow rate of air in the gas-turbine cycle from energy balance equation.

m˙air=h3−h2h10−h11m˙s (XV)

Here, enthalpy at state 10 is h10, enthalpy at state 11 is h11, enthalpy at state 3 and 2 is h3 and h2 respectively, and mass flow rate of steam is m˙s.

Conclusion:

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 7 and Prandtl number at state 7 corresponding to temperature at state 7 of 290 K.

h7=290.16 kJ/kgPr7=1.2311

Substitute 8 for P8sP7 and 1.2311 for Pr7 in Equation (I).

Pr8s=8(1.2311)=9.849

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XVI)

Here, the variables denote by x and y is Prandtl number at state8s and enthalpy at state 8s respectively.

Show the enthalpy at state 8s corresponding to Prandtl number as in Table (1).

Prandtl number at state 8s Pr8s	Enthalpy at state 8s h8s(kJ/kg)
9.684 (x1)	523.63 (y1)
9.849 (x2)	(y2=?)
10.37 (x3)	533.98 (y3)

Substitute 9.684, 9.849 and 10.37 for x1,x2 and x3 respectively, 523.63 kJ/kg for y1 and 533.98 kJ/kg for y3 in Equation (XVI).

y2=(9.849−9.684)(533.98 kJ/kg−523.63 kJ/kg)(10.37−9.684)+523.63 kJ/kg=526.12 kJ/kg=h8s

Thus, enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 is,

h8s=526.12 kJ/kg

Substitute 290.16 kJ/kg for h7, 526.12 kJ/kg for h8s and 0.85 for ηC in Equation (II).

h8=290.16 kJ/kg+[526.12 kJ/kg−290.16 kJ/kg0.85]=567.76 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 9 and Prandtl number at state 9 corresponding to temperature at state 9 of 1400 K.

h9=1515.42 kJ/kgPr9=450.5

Here, enthalpy at state 9 is h9 and Prandtl number at state 9 is Pr9.

Substitute 18 for P10sP9 and 450.5 for Pr9 in Equation (III).

Pr10s=18(450.5)=56.3

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 10s corresponding to Prandtl number at state 10s of 56.3 using an interpolation method.

Show the enthalpy at state 10s corresponding to Prandtl number as in Table (2).

Prandtl number at state 10s Pr10s	Enthalpy at state 10s h10s(kJ/kg)
52.59 (x1)	843.98 (y1)
56.3 (x2)	(y2=?)
57.60 (x3)	866.08 (y3)

Use excels and substitutes the values from Table (II) in Equation (XVI) to get,

h10s=860.35 kJ/kg

Here, enthalpy at state 10s is h10s.

Substitute 1515.42 kJ/kg for h9, 0.90 for ηT and 860.35 kJ/kg for h10s in Equation (IV).

h10=1515.42 kJ/kg−(0.90)(1515.42 kJ/kg−860.35 kJ/kg)=925.86 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 11 corresponding to temperature at state 11 of 520 K.

h11=523.63 kJ/kg

Here, enthalpy at state 11 is h11.

Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 10 kPa.

hf@10 kPa=191.81 kJ/kgvf@10 kPa=0.00101 m3/kg

Substitute 191.81 kJ/kg for hf@10 kPa in Equation (V).

h1=191.81 kJ/kg

Substitute 0.00101 m3/kg for vf@10 kPa in Equation (VI).

v1=0.00101 m3/kg

Substitute 0.00101 m3/kg for v1, 15 MPa for P2 and 10 kPa for P1 in Equation (VII).

wpI,in=(0.00101 m3/kg)(15 MPa−10 kPa)=(0.00101 m3/kg)(15,000 kPa−10 kPa)kJkPa⋅m3=15.14 kJ/kg

Substitute 191.81 kJ/kg for h1 and 15.14 kJ/kg for wpI,in in Equation (VIII).

h2=191.81 kJ/kg+15.14 kJ/kg=206.95 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 3 of 15 MPa and temperature at state 3 of 450°C.

h3=3157.9 kJ/kgs3=6.1428 kJ/kg⋅K

Here, enthalpy and entropy at state 3 is h3 and s3 respectively.

Due to throttling process, entropy at state 3 is equal to entropy at state 4s.

s3=s4s=6.1428 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 3 MPa(3000 kPa).

sf@3 MPa=2.6454 kJ/kg⋅Ksfg@3 MPa=3.5402 kJ/kg⋅Khf@3 MPa=1008.3 kJ/kghfg@3 MPa=1794.9 kJ/kg

Substitute 6.1428 kJ/kg⋅K for s4s, 2.6454 kJ/kg⋅K for sf@3 MPa and 3.5402 kJ/kg⋅K for sfg@3 MPa in Equation (IX).

x4s=6.1428 kJ/kg⋅K−2.6454 kJ/kg⋅K3.5402 kJ/kg⋅K=0.9880

Substitute 0.9880 for x4s, 1008.3 kJ/kg for hf@3 MPa and 1794.9 kJ/kg for hfg@3 MPa in Equation (X).

h4s=1008.3 kJ/kg+(0.9880)(1794.9 kJ/kg)=2781.7 kJ/kg

Substitute 3157.9 kJ/kg for h3, 0.90 for ηT and 2781.7 kJ/kg for h4s in Equation (XI).

h4=3157.9 kJ/kg−(0.90)(3157.9 kJ/kg−2781.7 kJ/kg)=2819.4 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 5 of 3 MPa and temperature at state 5 of 500°C.

h5=3457.2 kJ/kgs5=7.2359 kJ/kg⋅K

Here, enthalpy and entropy at state 5 is h5 and s5 respectively.

Due to throttling process, entropy at state 5 is equal to entropy at state 6s.

s5=s6s=7.2359 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 10 kPa.

sf@10 kPa=0.6492 kJ/kg⋅Ksfg@10 kPa=7.4996 kJ/kg⋅Khf@10 kPa=191.81 kJ/kghfg@10 kPa=2392.1 kJ/kg

Substitute 7.2359 kJ/kg⋅K for s6s, 0.6492 kJ/kg⋅K for sf@10 kPa and 7.4996 kJ/kg⋅K for sfg@10 kPa in Equation (XII).

x6s=7.2359 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.8783

Substitute 0.8783 for x6s, 191.81 kJ/kg for hf@10 kPa and 2392.1 kJ/kg for hfg@10 kPa in Equation (XIII).

h6s=191.81 kJ/kg+(0.8783)(2392.1 kJ/kg)=2292.7 kJ/kg

Substitute 3457.2 kJ/kg for h5, 0.90 for ηT and 2292.7 kJ/kg for h6s in Equation (XIV).

h6=3457.2 kJ/kg−(0.90)(3457.2 kJ/kg−2292.7 kJ/kg)=2409.1 kJ/kg

Substitute 3157.9 kJ/kg for h3, 206.95 kJ/kg for h2, 925.862 kJ/kg for h10, 523.63 kJ/kg for h11, and 30 kg/s for m˙s in Equation (V).

m˙air=3157.9 kJ/kg−206.95 kJ/kg925.862 kJ/kg−523.63 kJ/kg(30 kg/s)=220.1 kg/s

Hence, the mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

(b)

To determine

The rate of total heat input.

(b)

Expert Solution

Answer to Problem 109RP

The rate of total heat input is 227,700 kW.

Explanation of Solution

Express the rate of total heat input.

Q˙in=m˙air(h9−h8)+m˙s(h5−h4) (XVII)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 1515.42 kJ/kg for h9, 567.76 kJ/kg for h8, 3457.2 kJ/kg for h5 and 2819.4 kJ/kg for h4 in Equation (XVII).

Q˙in={(220.1 kg/s)[(1515.42−567.76) kJ/kg]+(30 kg/s)[(3457.2−2819.4) kJ/kg]}=227,700 kJ/s[kWkJ/s]=227,700 kW

Hence, the rate of total heat input is 227,700 kW.

(c)

To determine

The thermal efficiency of the combined cycle.

(c)

Expert Solution

Answer to Problem 109RP

The thermal efficiency of the combined cycle is 48.2%.

Explanation of Solution

Express the rate of total heat output.

Q˙out=m˙air(h11−h7)+m˙s(h6−h1) (XVIII)

Express the thermal efficiency of the combined cycle.

ηth=1−Q˙outQ˙in (XIX)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 523.63 kJ/kg for h11, 290.16 kJ/kg for h7, 2409.1 kJ/kg for h6 and 191.81 kJ/kg for h1 in Equation (XVIII).

Q˙out={(220.1 kg/s)[(523.63−290.16) kJ/kg]+(30 kg/s)[(2409.1−191.81) kJ/kg]}=117,900 kJ/s[kWkJ/s]=117,900 kW

Substitute 117,900 kW for Q˙out and 227,700 kW for Q˙in in Equation (XIX).

ηth=1−117,900 kW227,700 kW=0.482=48.2%

Hence, the thermal efficiency of the combined cycle is 48.2%.

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Answer 5

Chapter 10.9, Problem 109RP

(a)

To determine

The mass flow rate of air in the gas-turbine cycle.

(a)

Expert Solution

Answer to Problem 109RP

The mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

Explanation of Solution

Show the T-s diagram as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 10.9, Problem 109RP

Express Prandtl number at state 8s.

Pr8s=P8sP7Pr7 (I)

Here, pressure at state 8s is P8s, pressure at state 7 is P7 and Prandtl number at state 7 is Pr7.

Express enthalpy at state 8.

h8=h7+h8s−h7ηC (II)

Here, enthalpy at state 7 is h7, enthalpy at state 8s is h8s and the compressor efficiency is ηC.

Express Prandtl number at state 10s.

Pr10s=P10sP9Pr9 (III)

Here, pressure at state 10s is P10s, pressure at state 9 is P9 and Prandtl number at state 9 is Pr9.

Express enthalpy at state 10.

h10=h9−ηT(h9−h10s) (IV)

Here, enthalpy at state 9 is h9, compressor of the gas and steam turbine is ηT and enthalpy at state 10s is h10s.

Express enthalpy at state 1.

h1=hf@10 kPa (V)

Here, enthalpy of saturation liquid at pressure of 10 kPa is hf@10 kPa.

Express specific volume at state 1.

v1=vf@10 kPa (VI)

Here, specific volume of saturation liquid at pressure of 10 kPa is vf@10 kPa.

Express initial work input.

wpI,in=v1(P2−P1) (VII)

Here, pressure at state 2 and 1 is P2 and P1 respectively.

Express enthalpy at state 2.

h2=h1+wpI,in (VIII)

Express quality at state 4s.

x4s=s4s−sf@3 MPasfg@3 MPa (IX)

Here, entropy at state 4s is s4s, entropy at saturation liquid and evaporation at pressure of 3 MPa is sf@3 MPa and sfg@3 MPa respectively.

Express enthalpy at state 4s.

h4s=hf@3 MPa+x4shfg@3 MPa (X)

Here, enthalpy at saturation liquid and evaporation at pressure of 3 MPa is hf@3 MPa and hfg@3 MPa respectively.

Express enthalpy at state 4.

h4=h3−ηT(h3−h4s) (XI)

Here, enthalpy at state 3 is h3 and enthalpy at state 4s is h4s.

Express quality at state 6s.

x6s=s6s−sf@10 kPasfg@10 kPa (XII)

Here, entropy at state 6s is s6s, entropy at saturation liquid and evaporation at pressure of 10 kPa is sf@10 kPa and sfg@10 kPa respectively.

Express enthalpy at state 6s.

h6s=hf@10 MPa+x6shfg@10 MPa (XIII)

Here, enthalpy at saturation liquid and evaporation at pressure of 10 MPa is hf@10 MPa and hfg@10 MPa respectively.

Express enthalpy at state 6.

h6=h5−ηT(h5−h6s) (XIV)

Here, enthalpy at state 5 is h5 and enthalpy at state 6s is h6s.

Express the mass flow rate of air in the gas-turbine cycle from energy balance equation.

m˙air=h3−h2h10−h11m˙s (XV)

Here, enthalpy at state 10 is h10, enthalpy at state 11 is h11, enthalpy at state 3 and 2 is h3 and h2 respectively, and mass flow rate of steam is m˙s.

Conclusion:

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 7 and Prandtl number at state 7 corresponding to temperature at state 7 of 290 K.

h7=290.16 kJ/kgPr7=1.2311

Substitute 8 for P8sP7 and 1.2311 for Pr7 in Equation (I).

Pr8s=8(1.2311)=9.849

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XVI)

Here, the variables denote by x and y is Prandtl number at state8s and enthalpy at state 8s respectively.

Show the enthalpy at state 8s corresponding to Prandtl number as in Table (1).

Prandtl number at state 8s Pr8s	Enthalpy at state 8s h8s(kJ/kg)
9.684 (x1)	523.63 (y1)
9.849 (x2)	(y2=?)
10.37 (x3)	533.98 (y3)

Substitute 9.684, 9.849 and 10.37 for x1,x2 and x3 respectively, 523.63 kJ/kg for y1 and 533.98 kJ/kg for y3 in Equation (XVI).

y2=(9.849−9.684)(533.98 kJ/kg−523.63 kJ/kg)(10.37−9.684)+523.63 kJ/kg=526.12 kJ/kg=h8s

Thus, enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 is,

h8s=526.12 kJ/kg

Substitute 290.16 kJ/kg for h7, 526.12 kJ/kg for h8s and 0.85 for ηC in Equation (II).

h8=290.16 kJ/kg+[526.12 kJ/kg−290.16 kJ/kg0.85]=567.76 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 9 and Prandtl number at state 9 corresponding to temperature at state 9 of 1400 K.

h9=1515.42 kJ/kgPr9=450.5

Here, enthalpy at state 9 is h9 and Prandtl number at state 9 is Pr9.

Substitute 18 for P10sP9 and 450.5 for Pr9 in Equation (III).

Pr10s=18(450.5)=56.3

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 10s corresponding to Prandtl number at state 10s of 56.3 using an interpolation method.

Show the enthalpy at state 10s corresponding to Prandtl number as in Table (2).

Prandtl number at state 10s Pr10s	Enthalpy at state 10s h10s(kJ/kg)
52.59 (x1)	843.98 (y1)
56.3 (x2)	(y2=?)
57.60 (x3)	866.08 (y3)

Use excels and substitutes the values from Table (II) in Equation (XVI) to get,

h10s=860.35 kJ/kg

Here, enthalpy at state 10s is h10s.

Substitute 1515.42 kJ/kg for h9, 0.90 for ηT and 860.35 kJ/kg for h10s in Equation (IV).

h10=1515.42 kJ/kg−(0.90)(1515.42 kJ/kg−860.35 kJ/kg)=925.86 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 11 corresponding to temperature at state 11 of 520 K.

h11=523.63 kJ/kg

Here, enthalpy at state 11 is h11.

Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 10 kPa.

hf@10 kPa=191.81 kJ/kgvf@10 kPa=0.00101 m3/kg

Substitute 191.81 kJ/kg for hf@10 kPa in Equation (V).

h1=191.81 kJ/kg

Substitute 0.00101 m3/kg for vf@10 kPa in Equation (VI).

v1=0.00101 m3/kg

Substitute 0.00101 m3/kg for v1, 15 MPa for P2 and 10 kPa for P1 in Equation (VII).

wpI,in=(0.00101 m3/kg)(15 MPa−10 kPa)=(0.00101 m3/kg)(15,000 kPa−10 kPa)kJkPa⋅m3=15.14 kJ/kg

Substitute 191.81 kJ/kg for h1 and 15.14 kJ/kg for wpI,in in Equation (VIII).

h2=191.81 kJ/kg+15.14 kJ/kg=206.95 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 3 of 15 MPa and temperature at state 3 of 450°C.

h3=3157.9 kJ/kgs3=6.1428 kJ/kg⋅K

Here, enthalpy and entropy at state 3 is h3 and s3 respectively.

Due to throttling process, entropy at state 3 is equal to entropy at state 4s.

s3=s4s=6.1428 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 3 MPa(3000 kPa).

sf@3 MPa=2.6454 kJ/kg⋅Ksfg@3 MPa=3.5402 kJ/kg⋅Khf@3 MPa=1008.3 kJ/kghfg@3 MPa=1794.9 kJ/kg

Substitute 6.1428 kJ/kg⋅K for s4s, 2.6454 kJ/kg⋅K for sf@3 MPa and 3.5402 kJ/kg⋅K for sfg@3 MPa in Equation (IX).

x4s=6.1428 kJ/kg⋅K−2.6454 kJ/kg⋅K3.5402 kJ/kg⋅K=0.9880

Substitute 0.9880 for x4s, 1008.3 kJ/kg for hf@3 MPa and 1794.9 kJ/kg for hfg@3 MPa in Equation (X).

h4s=1008.3 kJ/kg+(0.9880)(1794.9 kJ/kg)=2781.7 kJ/kg

Substitute 3157.9 kJ/kg for h3, 0.90 for ηT and 2781.7 kJ/kg for h4s in Equation (XI).

h4=3157.9 kJ/kg−(0.90)(3157.9 kJ/kg−2781.7 kJ/kg)=2819.4 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 5 of 3 MPa and temperature at state 5 of 500°C.

h5=3457.2 kJ/kgs5=7.2359 kJ/kg⋅K

Here, enthalpy and entropy at state 5 is h5 and s5 respectively.

Due to throttling process, entropy at state 5 is equal to entropy at state 6s.

s5=s6s=7.2359 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 10 kPa.

sf@10 kPa=0.6492 kJ/kg⋅Ksfg@10 kPa=7.4996 kJ/kg⋅Khf@10 kPa=191.81 kJ/kghfg@10 kPa=2392.1 kJ/kg

Substitute 7.2359 kJ/kg⋅K for s6s, 0.6492 kJ/kg⋅K for sf@10 kPa and 7.4996 kJ/kg⋅K for sfg@10 kPa in Equation (XII).

x6s=7.2359 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.8783

Substitute 0.8783 for x6s, 191.81 kJ/kg for hf@10 kPa and 2392.1 kJ/kg for hfg@10 kPa in Equation (XIII).

h6s=191.81 kJ/kg+(0.8783)(2392.1 kJ/kg)=2292.7 kJ/kg

Substitute 3457.2 kJ/kg for h5, 0.90 for ηT and 2292.7 kJ/kg for h6s in Equation (XIV).

h6=3457.2 kJ/kg−(0.90)(3457.2 kJ/kg−2292.7 kJ/kg)=2409.1 kJ/kg

Substitute 3157.9 kJ/kg for h3, 206.95 kJ/kg for h2, 925.862 kJ/kg for h10, 523.63 kJ/kg for h11, and 30 kg/s for m˙s in Equation (V).

m˙air=3157.9 kJ/kg−206.95 kJ/kg925.862 kJ/kg−523.63 kJ/kg(30 kg/s)=220.1 kg/s

Hence, the mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

(b)

To determine

The rate of total heat input.

(b)

Expert Solution

Answer to Problem 109RP

The rate of total heat input is 227,700 kW.

Explanation of Solution

Express the rate of total heat input.

Q˙in=m˙air(h9−h8)+m˙s(h5−h4) (XVII)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 1515.42 kJ/kg for h9, 567.76 kJ/kg for h8, 3457.2 kJ/kg for h5 and 2819.4 kJ/kg for h4 in Equation (XVII).

Q˙in={(220.1 kg/s)[(1515.42−567.76) kJ/kg]+(30 kg/s)[(3457.2−2819.4) kJ/kg]}=227,700 kJ/s[kWkJ/s]=227,700 kW

Hence, the rate of total heat input is 227,700 kW.

(c)

To determine

The thermal efficiency of the combined cycle.

(c)

Expert Solution

Answer to Problem 109RP

The thermal efficiency of the combined cycle is 48.2%.

Explanation of Solution

Express the rate of total heat output.

Q˙out=m˙air(h11−h7)+m˙s(h6−h1) (XVIII)

Express the thermal efficiency of the combined cycle.

ηth=1−Q˙outQ˙in (XIX)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 523.63 kJ/kg for h11, 290.16 kJ/kg for h7, 2409.1 kJ/kg for h6 and 191.81 kJ/kg for h1 in Equation (XVIII).

Q˙out={(220.1 kg/s)[(523.63−290.16) kJ/kg]+(30 kg/s)[(2409.1−191.81) kJ/kg]}=117,900 kJ/s[kWkJ/s]=117,900 kW

Substitute 117,900 kW for Q˙out and 227,700 kW for Q˙in in Equation (XIX).

ηth=1−117,900 kW227,700 kW=0.482=48.2%

Hence, the thermal efficiency of the combined cycle is 48.2%.

Answer 6

Chapter 10.9, Problem 109RP

(a)

To determine

The mass flow rate of air in the gas-turbine cycle.

(a)

Expert Solution

Answer to Problem 109RP

The mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

Explanation of Solution

Show the T-s diagram as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 10.9, Problem 109RP

Express Prandtl number at state 8s.

Pr8s=P8sP7Pr7 (I)

Here, pressure at state 8s is P8s, pressure at state 7 is P7 and Prandtl number at state 7 is Pr7.

Express enthalpy at state 8.

h8=h7+h8s−h7ηC (II)

Here, enthalpy at state 7 is h7, enthalpy at state 8s is h8s and the compressor efficiency is ηC.

Express Prandtl number at state 10s.

Pr10s=P10sP9Pr9 (III)

Here, pressure at state 10s is P10s, pressure at state 9 is P9 and Prandtl number at state 9 is Pr9.

Express enthalpy at state 10.

h10=h9−ηT(h9−h10s) (IV)

Here, enthalpy at state 9 is h9, compressor of the gas and steam turbine is ηT and enthalpy at state 10s is h10s.

Express enthalpy at state 1.

h1=hf@10 kPa (V)

Here, enthalpy of saturation liquid at pressure of 10 kPa is hf@10 kPa.

Express specific volume at state 1.

v1=vf@10 kPa (VI)

Here, specific volume of saturation liquid at pressure of 10 kPa is vf@10 kPa.

Express initial work input.

wpI,in=v1(P2−P1) (VII)

Here, pressure at state 2 and 1 is P2 and P1 respectively.

Express enthalpy at state 2.

h2=h1+wpI,in (VIII)

Express quality at state 4s.

x4s=s4s−sf@3 MPasfg@3 MPa (IX)

Here, entropy at state 4s is s4s, entropy at saturation liquid and evaporation at pressure of 3 MPa is sf@3 MPa and sfg@3 MPa respectively.

Express enthalpy at state 4s.

h4s=hf@3 MPa+x4shfg@3 MPa (X)

Here, enthalpy at saturation liquid and evaporation at pressure of 3 MPa is hf@3 MPa and hfg@3 MPa respectively.

Express enthalpy at state 4.

h4=h3−ηT(h3−h4s) (XI)

Here, enthalpy at state 3 is h3 and enthalpy at state 4s is h4s.

Express quality at state 6s.

x6s=s6s−sf@10 kPasfg@10 kPa (XII)

Here, entropy at state 6s is s6s, entropy at saturation liquid and evaporation at pressure of 10 kPa is sf@10 kPa and sfg@10 kPa respectively.

Express enthalpy at state 6s.

h6s=hf@10 MPa+x6shfg@10 MPa (XIII)

Here, enthalpy at saturation liquid and evaporation at pressure of 10 MPa is hf@10 MPa and hfg@10 MPa respectively.

Express enthalpy at state 6.

h6=h5−ηT(h5−h6s) (XIV)

Here, enthalpy at state 5 is h5 and enthalpy at state 6s is h6s.

Express the mass flow rate of air in the gas-turbine cycle from energy balance equation.

m˙air=h3−h2h10−h11m˙s (XV)

Here, enthalpy at state 10 is h10, enthalpy at state 11 is h11, enthalpy at state 3 and 2 is h3 and h2 respectively, and mass flow rate of steam is m˙s.

Conclusion:

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 7 and Prandtl number at state 7 corresponding to temperature at state 7 of 290 K.

h7=290.16 kJ/kgPr7=1.2311

Substitute 8 for P8sP7 and 1.2311 for Pr7 in Equation (I).

Pr8s=8(1.2311)=9.849

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XVI)

Here, the variables denote by x and y is Prandtl number at state8s and enthalpy at state 8s respectively.

Show the enthalpy at state 8s corresponding to Prandtl number as in Table (1).

Prandtl number at state 8s Pr8s	Enthalpy at state 8s h8s(kJ/kg)
9.684 (x1)	523.63 (y1)
9.849 (x2)	(y2=?)
10.37 (x3)	533.98 (y3)

Substitute 9.684, 9.849 and 10.37 for x1,x2 and x3 respectively, 523.63 kJ/kg for y1 and 533.98 kJ/kg for y3 in Equation (XVI).

y2=(9.849−9.684)(533.98 kJ/kg−523.63 kJ/kg)(10.37−9.684)+523.63 kJ/kg=526.12 kJ/kg=h8s

Thus, enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 is,

h8s=526.12 kJ/kg

Substitute 290.16 kJ/kg for h7, 526.12 kJ/kg for h8s and 0.85 for ηC in Equation (II).

h8=290.16 kJ/kg+[526.12 kJ/kg−290.16 kJ/kg0.85]=567.76 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 9 and Prandtl number at state 9 corresponding to temperature at state 9 of 1400 K.

h9=1515.42 kJ/kgPr9=450.5

Here, enthalpy at state 9 is h9 and Prandtl number at state 9 is Pr9.

Substitute 18 for P10sP9 and 450.5 for Pr9 in Equation (III).

Pr10s=18(450.5)=56.3

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 10s corresponding to Prandtl number at state 10s of 56.3 using an interpolation method.

Show the enthalpy at state 10s corresponding to Prandtl number as in Table (2).

Prandtl number at state 10s Pr10s	Enthalpy at state 10s h10s(kJ/kg)
52.59 (x1)	843.98 (y1)
56.3 (x2)	(y2=?)
57.60 (x3)	866.08 (y3)

Use excels and substitutes the values from Table (II) in Equation (XVI) to get,

h10s=860.35 kJ/kg

Here, enthalpy at state 10s is h10s.

Substitute 1515.42 kJ/kg for h9, 0.90 for ηT and 860.35 kJ/kg for h10s in Equation (IV).

h10=1515.42 kJ/kg−(0.90)(1515.42 kJ/kg−860.35 kJ/kg)=925.86 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 11 corresponding to temperature at state 11 of 520 K.

h11=523.63 kJ/kg

Here, enthalpy at state 11 is h11.

Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 10 kPa.

hf@10 kPa=191.81 kJ/kgvf@10 kPa=0.00101 m3/kg

Substitute 191.81 kJ/kg for hf@10 kPa in Equation (V).

h1=191.81 kJ/kg

Substitute 0.00101 m3/kg for vf@10 kPa in Equation (VI).

v1=0.00101 m3/kg

Substitute 0.00101 m3/kg for v1, 15 MPa for P2 and 10 kPa for P1 in Equation (VII).

wpI,in=(0.00101 m3/kg)(15 MPa−10 kPa)=(0.00101 m3/kg)(15,000 kPa−10 kPa)kJkPa⋅m3=15.14 kJ/kg

Substitute 191.81 kJ/kg for h1 and 15.14 kJ/kg for wpI,in in Equation (VIII).

h2=191.81 kJ/kg+15.14 kJ/kg=206.95 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 3 of 15 MPa and temperature at state 3 of 450°C.

h3=3157.9 kJ/kgs3=6.1428 kJ/kg⋅K

Here, enthalpy and entropy at state 3 is h3 and s3 respectively.

Due to throttling process, entropy at state 3 is equal to entropy at state 4s.

s3=s4s=6.1428 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 3 MPa(3000 kPa).

sf@3 MPa=2.6454 kJ/kg⋅Ksfg@3 MPa=3.5402 kJ/kg⋅Khf@3 MPa=1008.3 kJ/kghfg@3 MPa=1794.9 kJ/kg

Substitute 6.1428 kJ/kg⋅K for s4s, 2.6454 kJ/kg⋅K for sf@3 MPa and 3.5402 kJ/kg⋅K for sfg@3 MPa in Equation (IX).

x4s=6.1428 kJ/kg⋅K−2.6454 kJ/kg⋅K3.5402 kJ/kg⋅K=0.9880

Substitute 0.9880 for x4s, 1008.3 kJ/kg for hf@3 MPa and 1794.9 kJ/kg for hfg@3 MPa in Equation (X).

h4s=1008.3 kJ/kg+(0.9880)(1794.9 kJ/kg)=2781.7 kJ/kg

Substitute 3157.9 kJ/kg for h3, 0.90 for ηT and 2781.7 kJ/kg for h4s in Equation (XI).

h4=3157.9 kJ/kg−(0.90)(3157.9 kJ/kg−2781.7 kJ/kg)=2819.4 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 5 of 3 MPa and temperature at state 5 of 500°C.

h5=3457.2 kJ/kgs5=7.2359 kJ/kg⋅K

Here, enthalpy and entropy at state 5 is h5 and s5 respectively.

Due to throttling process, entropy at state 5 is equal to entropy at state 6s.

s5=s6s=7.2359 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 10 kPa.

sf@10 kPa=0.6492 kJ/kg⋅Ksfg@10 kPa=7.4996 kJ/kg⋅Khf@10 kPa=191.81 kJ/kghfg@10 kPa=2392.1 kJ/kg

Substitute 7.2359 kJ/kg⋅K for s6s, 0.6492 kJ/kg⋅K for sf@10 kPa and 7.4996 kJ/kg⋅K for sfg@10 kPa in Equation (XII).

x6s=7.2359 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.8783

Substitute 0.8783 for x6s, 191.81 kJ/kg for hf@10 kPa and 2392.1 kJ/kg for hfg@10 kPa in Equation (XIII).

h6s=191.81 kJ/kg+(0.8783)(2392.1 kJ/kg)=2292.7 kJ/kg

Substitute 3457.2 kJ/kg for h5, 0.90 for ηT and 2292.7 kJ/kg for h6s in Equation (XIV).

h6=3457.2 kJ/kg−(0.90)(3457.2 kJ/kg−2292.7 kJ/kg)=2409.1 kJ/kg

Substitute 3157.9 kJ/kg for h3, 206.95 kJ/kg for h2, 925.862 kJ/kg for h10, 523.63 kJ/kg for h11, and 30 kg/s for m˙s in Equation (V).

m˙air=3157.9 kJ/kg−206.95 kJ/kg925.862 kJ/kg−523.63 kJ/kg(30 kg/s)=220.1 kg/s

Hence, the mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

Answer 7

Chapter 10.9, Problem 109RP

(a)

To determine

The mass flow rate of air in the gas-turbine cycle.

Answer 8

(a)

Expert Solution

Answer to Problem 109RP

The mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Show the T-s diagram as in Figure (1).

CONNECT FOR THERMODYNAMICS: AN ENGINEERI, Chapter 10.9, Problem 109RP

Express Prandtl number at state 8s.

Pr8s=P8sP7Pr7 (I)

Here, pressure at state 8s is P8s, pressure at state 7 is P7 and Prandtl number at state 7 is Pr7.

Express enthalpy at state 8.

h8=h7+h8s−h7ηC (II)

Here, enthalpy at state 7 is h7, enthalpy at state 8s is h8s and the compressor efficiency is ηC.

Express Prandtl number at state 10s.

Pr10s=P10sP9Pr9 (III)

Here, pressure at state 10s is P10s, pressure at state 9 is P9 and Prandtl number at state 9 is Pr9.

Express enthalpy at state 10.

h10=h9−ηT(h9−h10s) (IV)

Here, enthalpy at state 9 is h9, compressor of the gas and steam turbine is ηT and enthalpy at state 10s is h10s.

Express enthalpy at state 1.

h1=hf@10 kPa (V)

Here, enthalpy of saturation liquid at pressure of 10 kPa is hf@10 kPa.

Express specific volume at state 1.

v1=vf@10 kPa (VI)

Here, specific volume of saturation liquid at pressure of 10 kPa is vf@10 kPa.

Express initial work input.

wpI,in=v1(P2−P1) (VII)

Here, pressure at state 2 and 1 is P2 and P1 respectively.

Express enthalpy at state 2.

h2=h1+wpI,in (VIII)

Express quality at state 4s.

x4s=s4s−sf@3 MPasfg@3 MPa (IX)

Here, entropy at state 4s is s4s, entropy at saturation liquid and evaporation at pressure of 3 MPa is sf@3 MPa and sfg@3 MPa respectively.

Express enthalpy at state 4s.

h4s=hf@3 MPa+x4shfg@3 MPa (X)

Here, enthalpy at saturation liquid and evaporation at pressure of 3 MPa is hf@3 MPa and hfg@3 MPa respectively.

Express enthalpy at state 4.

h4=h3−ηT(h3−h4s) (XI)

Here, enthalpy at state 3 is h3 and enthalpy at state 4s is h4s.

Express quality at state 6s.

x6s=s6s−sf@10 kPasfg@10 kPa (XII)

Here, entropy at state 6s is s6s, entropy at saturation liquid and evaporation at pressure of 10 kPa is sf@10 kPa and sfg@10 kPa respectively.

Express enthalpy at state 6s.

h6s=hf@10 MPa+x6shfg@10 MPa (XIII)

Here, enthalpy at saturation liquid and evaporation at pressure of 10 MPa is hf@10 MPa and hfg@10 MPa respectively.

Express enthalpy at state 6.

h6=h5−ηT(h5−h6s) (XIV)

Here, enthalpy at state 5 is h5 and enthalpy at state 6s is h6s.

Express the mass flow rate of air in the gas-turbine cycle from energy balance equation.

m˙air=h3−h2h10−h11m˙s (XV)

Here, enthalpy at state 10 is h10, enthalpy at state 11 is h11, enthalpy at state 3 and 2 is h3 and h2 respectively, and mass flow rate of steam is m˙s.

Conclusion:

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 7 and Prandtl number at state 7 corresponding to temperature at state 7 of 290 K.

h7=290.16 kJ/kgPr7=1.2311

Substitute 8 for P8sP7 and 1.2311 for Pr7 in Equation (I).

Pr8s=8(1.2311)=9.849

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 using an interpolation method.

Write the formula of interpolation method of two variables.

y2=(x2−x1)(y3−y1)(x3−x1)+y1 (XVI)

Here, the variables denote by x and y is Prandtl number at state8s and enthalpy at state 8s respectively.

Show the enthalpy at state 8s corresponding to Prandtl number as in Table (1).

Prandtl number at state 8s Pr8s	Enthalpy at state 8s h8s(kJ/kg)
9.684 (x1)	523.63 (y1)
9.849 (x2)	(y2=?)
10.37 (x3)	533.98 (y3)

Substitute 9.684, 9.849 and 10.37 for x1,x2 and x3 respectively, 523.63 kJ/kg for y1 and 533.98 kJ/kg for y3 in Equation (XVI).

y2=(9.849−9.684)(533.98 kJ/kg−523.63 kJ/kg)(10.37−9.684)+523.63 kJ/kg=526.12 kJ/kg=h8s

Thus, enthalpy at state 8s corresponding to Prandtl number at state 8s of 9.849 is,

h8s=526.12 kJ/kg

Substitute 290.16 kJ/kg for h7, 526.12 kJ/kg for h8s and 0.85 for ηC in Equation (II).

h8=290.16 kJ/kg+[526.12 kJ/kg−290.16 kJ/kg0.85]=567.76 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 9 and Prandtl number at state 9 corresponding to temperature at state 9 of 1400 K.

h9=1515.42 kJ/kgPr9=450.5

Here, enthalpy at state 9 is h9 and Prandtl number at state 9 is Pr9.

Substitute 18 for P10sP9 and 450.5 for Pr9 in Equation (III).

Pr10s=18(450.5)=56.3

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 10s corresponding to Prandtl number at state 10s of 56.3 using an interpolation method.

Show the enthalpy at state 10s corresponding to Prandtl number as in Table (2).

Prandtl number at state 10s Pr10s	Enthalpy at state 10s h10s(kJ/kg)
52.59 (x1)	843.98 (y1)
56.3 (x2)	(y2=?)
57.60 (x3)	866.08 (y3)

Use excels and substitutes the values from Table (II) in Equation (XVI) to get,

h10s=860.35 kJ/kg

Here, enthalpy at state 10s is h10s.

Substitute 1515.42 kJ/kg for h9, 0.90 for ηT and 860.35 kJ/kg for h10s in Equation (IV).

h10=1515.42 kJ/kg−(0.90)(1515.42 kJ/kg−860.35 kJ/kg)=925.86 kJ/kg

Refer Table A-17, “ideal gas properties of air”, and write the enthalpy at state 11 corresponding to temperature at state 11 of 520 K.

h11=523.63 kJ/kg

Here, enthalpy at state 11 is h11.

Refer Table A-5, “saturated water-pressure table”, and write the properties at pressure of 10 kPa.

hf@10 kPa=191.81 kJ/kgvf@10 kPa=0.00101 m3/kg

Substitute 191.81 kJ/kg for hf@10 kPa in Equation (V).

h1=191.81 kJ/kg

Substitute 0.00101 m3/kg for vf@10 kPa in Equation (VI).

v1=0.00101 m3/kg

Substitute 0.00101 m3/kg for v1, 15 MPa for P2 and 10 kPa for P1 in Equation (VII).

wpI,in=(0.00101 m3/kg)(15 MPa−10 kPa)=(0.00101 m3/kg)(15,000 kPa−10 kPa)kJkPa⋅m3=15.14 kJ/kg

Substitute 191.81 kJ/kg for h1 and 15.14 kJ/kg for wpI,in in Equation (VIII).

h2=191.81 kJ/kg+15.14 kJ/kg=206.95 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 3 of 15 MPa and temperature at state 3 of 450°C.

h3=3157.9 kJ/kgs3=6.1428 kJ/kg⋅K

Here, enthalpy and entropy at state 3 is h3 and s3 respectively.

Due to throttling process, entropy at state 3 is equal to entropy at state 4s.

s3=s4s=6.1428 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 3 MPa(3000 kPa).

sf@3 MPa=2.6454 kJ/kg⋅Ksfg@3 MPa=3.5402 kJ/kg⋅Khf@3 MPa=1008.3 kJ/kghfg@3 MPa=1794.9 kJ/kg

Substitute 6.1428 kJ/kg⋅K for s4s, 2.6454 kJ/kg⋅K for sf@3 MPa and 3.5402 kJ/kg⋅K for sfg@3 MPa in Equation (IX).

x4s=6.1428 kJ/kg⋅K−2.6454 kJ/kg⋅K3.5402 kJ/kg⋅K=0.9880

Substitute 0.9880 for x4s, 1008.3 kJ/kg for hf@3 MPa and 1794.9 kJ/kg for hfg@3 MPa in Equation (X).

h4s=1008.3 kJ/kg+(0.9880)(1794.9 kJ/kg)=2781.7 kJ/kg

Substitute 3157.9 kJ/kg for h3, 0.90 for ηT and 2781.7 kJ/kg for h4s in Equation (XI).

h4=3157.9 kJ/kg−(0.90)(3157.9 kJ/kg−2781.7 kJ/kg)=2819.4 kJ/kg

Refer Table A-6, “superheated water”, and write the properties corresponding to pressure at state 5 of 3 MPa and temperature at state 5 of 500°C.

h5=3457.2 kJ/kgs5=7.2359 kJ/kg⋅K

Here, enthalpy and entropy at state 5 is h5 and s5 respectively.

Due to throttling process, entropy at state 5 is equal to entropy at state 6s.

s5=s6s=7.2359 kJ/kg⋅K

Refer Table A-5, “saturated water-pressure table”, and write the properties corresponding to pressure of 10 kPa.

sf@10 kPa=0.6492 kJ/kg⋅Ksfg@10 kPa=7.4996 kJ/kg⋅Khf@10 kPa=191.81 kJ/kghfg@10 kPa=2392.1 kJ/kg

Substitute 7.2359 kJ/kg⋅K for s6s, 0.6492 kJ/kg⋅K for sf@10 kPa and 7.4996 kJ/kg⋅K for sfg@10 kPa in Equation (XII).

x6s=7.2359 kJ/kg⋅K−0.6492 kJ/kg⋅K7.4996 kJ/kg⋅K=0.8783

Substitute 0.8783 for x6s, 191.81 kJ/kg for hf@10 kPa and 2392.1 kJ/kg for hfg@10 kPa in Equation (XIII).

h6s=191.81 kJ/kg+(0.8783)(2392.1 kJ/kg)=2292.7 kJ/kg

Substitute 3457.2 kJ/kg for h5, 0.90 for ηT and 2292.7 kJ/kg for h6s in Equation (XIV).

h6=3457.2 kJ/kg−(0.90)(3457.2 kJ/kg−2292.7 kJ/kg)=2409.1 kJ/kg

Substitute 3157.9 kJ/kg for h3, 206.95 kJ/kg for h2, 925.862 kJ/kg for h10, 523.63 kJ/kg for h11, and 30 kg/s for m˙s in Equation (V).

m˙air=3157.9 kJ/kg−206.95 kJ/kg925.862 kJ/kg−523.63 kJ/kg(30 kg/s)=220.1 kg/s

Hence, the mass flow rate of air in the gas-turbine cycle is 220.1 kg/s.

Answer 13

(b)

To determine

The rate of total heat input.

(b)

Expert Solution

Answer to Problem 109RP

The rate of total heat input is 227,700 kW.

Explanation of Solution

Express the rate of total heat input.

Q˙in=m˙air(h9−h8)+m˙s(h5−h4) (XVII)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 1515.42 kJ/kg for h9, 567.76 kJ/kg for h8, 3457.2 kJ/kg for h5 and 2819.4 kJ/kg for h4 in Equation (XVII).

Q˙in={(220.1 kg/s)[(1515.42−567.76) kJ/kg]+(30 kg/s)[(3457.2−2819.4) kJ/kg]}=227,700 kJ/s[kWkJ/s]=227,700 kW

Hence, the rate of total heat input is 227,700 kW.

Answer 14

(b)

To determine

The rate of total heat input.

Answer 15

(b)

Expert Solution

Answer to Problem 109RP

The rate of total heat input is 227,700 kW.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Express the rate of total heat input.

Q˙in=m˙air(h9−h8)+m˙s(h5−h4) (XVII)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 1515.42 kJ/kg for h9, 567.76 kJ/kg for h8, 3457.2 kJ/kg for h5 and 2819.4 kJ/kg for h4 in Equation (XVII).

Q˙in={(220.1 kg/s)[(1515.42−567.76) kJ/kg]+(30 kg/s)[(3457.2−2819.4) kJ/kg]}=227,700 kJ/s[kWkJ/s]=227,700 kW

Hence, the rate of total heat input is 227,700 kW.

Answer 20

(c)

To determine

The thermal efficiency of the combined cycle.

(c)

Expert Solution

Answer to Problem 109RP

The thermal efficiency of the combined cycle is 48.2%.

Explanation of Solution

Express the rate of total heat output.

Q˙out=m˙air(h11−h7)+m˙s(h6−h1) (XVIII)

Express the thermal efficiency of the combined cycle.

ηth=1−Q˙outQ˙in (XIX)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 523.63 kJ/kg for h11, 290.16 kJ/kg for h7, 2409.1 kJ/kg for h6 and 191.81 kJ/kg for h1 in Equation (XVIII).

Q˙out={(220.1 kg/s)[(523.63−290.16) kJ/kg]+(30 kg/s)[(2409.1−191.81) kJ/kg]}=117,900 kJ/s[kWkJ/s]=117,900 kW

Substitute 117,900 kW for Q˙out and 227,700 kW for Q˙in in Equation (XIX).

ηth=1−117,900 kW227,700 kW=0.482=48.2%

Hence, the thermal efficiency of the combined cycle is 48.2%.

Answer 21

(c)

To determine

The thermal efficiency of the combined cycle.

Answer 22

(c)

Expert Solution

Answer to Problem 109RP

The thermal efficiency of the combined cycle is 48.2%.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Express the rate of total heat output.

Q˙out=m˙air(h11−h7)+m˙s(h6−h1) (XVIII)

Express the thermal efficiency of the combined cycle.

ηth=1−Q˙outQ˙in (XIX)

Conclusion:

Substitute 220.1 kg/s for m˙air, 30 kg/s for m˙s, 523.63 kJ/kg for h11, 290.16 kJ/kg for h7, 2409.1 kJ/kg for h6 and 191.81 kJ/kg for h1 in Equation (XVIII).