Thermodynamics: An Engineering Approach
Thermodynamics: An Engineering Approach
9th Edition
ISBN: 9781259822674
Author: Yunus A. Cengel Dr., Michael A. Boles
Publisher: McGraw-Hill Education
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Question
Chapter 7.13, Problem 197RP
To determine

The power output produced from the turbine, and the overall isentropic efficiency of turbine.

Expert Solution & Answer
Check Mark

Answer to Problem 197RP

The power output produced from the turbine is 810.1 kJ/kg.

The overall isentropic efficiency of turbine is 95.80%.

Explanation of Solution

Write the formula to calculate the specific entropy of steam from tables (s).

s=sf+x(sfg) (I)

Here, specific entropy of saturated liquid is sf, and the specific entropy of saturated liquid vapor mixture is sfg.

Write the formula to calculate the specific enthalpy of steam from tables (h).

h=hf+x(hfg) (II)

Here, specific entropy of saturated liquid is sf, and the specific entropy of saturated liquid vapor mixture is sfg.

Draw the Ts diagram for the turbine process as in Figure (1).

Thermodynamics: An Engineering Approach, Chapter 7.13, Problem 197RP

Write the formula for isentropic efficiency of the turbine (ηT,1) between the inlet and bleed point.

ηT,1=h1h2h1h2s (III)

Here, specific enthalpy at the turbine inlet is h1, specific enthalpy at the bleed point is h2, and the specific enthalpy at the isentropic exit of bleed point is h2s.

Write the formula for isentropic efficiency of the turbine (ηT,2) between the bleed point and turbine exit.

ηT,2=h2h3h2h3s (IV)

Here, specific enthalpy at the actual turbine exit is h3, and the specific enthalpy at the isentropic exit of turbine is h3s.

Write the expression for the energy balance Equation for a closed system.

E˙inE˙out=ΔE˙system (V)

Here, net energy rate transfer into the control volume is E˙in, net energy rate transfer exit from the control volume is E˙out and rate of change in energy of the system is ΔE˙system.

Write the general expression to calculate the isentropic efficiency of turbine (ηiso).

ηiso=wactualwiso (VI)

Here, actual work output is wactual and the isentropic work output is wiso.

Conclusion:

The rate of change in energy of the system is zero at steady state.

Substitute 0 for ΔE˙system in Equation (V).

E˙inE˙out=0E˙in=E˙out (VII)

Substitute m˙1h1 for E˙in, m˙2h2+m˙3h+W˙actual for E˙out in Equation (VII).

m˙1h1=m˙2h2+m˙3h+W˙actual

W˙actual=m˙1h1m˙2h2m˙3h3wactual=h10.06h20.94h3

wactual=(h1h2)+0.94(h2h3) (VIII)

Re-write the Equation (VIII) for the isentropic work output (wiso).

wactual=(h1h2s)+0.94(h2h3s) (IX)

Here, mass flow rate of steam at inlet is m˙1, mass flow rate of steam at bleed point is m˙2, mass flow rate of steam at the exit of turbine is m˙3, actual work output from turbine is W˙actual, and the actual work output per unit mass is wactual.

From the Table A-6 “Superheated water”, obtain the specific enthalpy (h1) and specific entropy (s1) of superheated steam at pressure (P1) of 4MPa and temperature (T1) of 350°C as 3093.3kJ/kg and 6.5843kJ/kgK.

From the Table A-5 “Saturated water - Pressure”, obtain the following properties of water at pressure of 800 kPa.

sf=2.0457 kJ/kgKsfg=4.6160 kJ/kgKhf=720.87 kJ/kghfg=2047.5 kJ/kg

Substitute 2.0457 kJ/kgK for sf, 4.6160 kJ/kgK for sfg, and 6.5843kJ/kgK for s2s in Equation (I).

6.5843kJ/kgK=2.0457 kJ/kgK+x2s(4.6160 kJ/kgK)x2s=0.9832

Substitute 720.87 kJ/kg for hf, 2047.5 kJ/kg for hfg, and 0.9832 for x2s in

Equation (II).

h2s=720.87 kJ/kg+0.9832(2047.5 kJ/kg)=2734.0 kJ/kg

Substitute 0.97 for ηT,1, 3093.3kJ/kg for h1, and 2734.0 kJ/kg for h2s in

Equation (III).

0.97=3093.3kJ/kgh23093.3kJ/kg2734.0 kJ/kgh2=2744.8kJ/kg

Substitute 720.87 kJ/kg for hf, 2047.5 kJ/kg for hfg, and 2744.8kJ/kg for h2 in Equation (II).

2744.8kJ/kg=720.87 kJ/kg+x2(2047.5 kJ/kg)x2=0.9885

Substitute 2.0457 kJ/kgK for sf, 4.6160 kJ/kgK for sfg, and 0.9885 for x2 in Equation (I).

s2=2.0457 kJ/kgK+0.9885(4.6160 kJ/kgK)s2=6.6086kJ/kgK

From the Table A-5 “Saturated water - Pressure”, obtain the following properties of water at pressure of 30 kPa.

sf=0.9441 kJ/kgKsfg=6.8234 kJ/kgKhf=289.27 kJ/kghfg=2335.3 kJ/kg

Substitute 0.9441 kJ/kgK for sf, 6.8234 kJ/kgK for sfg, and 6.6086kJ/kgK for s3s in Equation (I).

6.6086kJ/kgK=0.9441 kJ/kgK+x3s(6.8234 kJ/kgK)x3s=0.8302

Substitute 289.27 kJ/kg for hf, 2335.3 kJ/kg for hfg, and 0.8302 for x3s in Equation (II).

h3s=289.27 kJ/kg+0.8302(2335.3 kJ/kg)=2227.9 kJ/kg

Substitute 2227.9 kJ/kg for h3s, 0.95 for ηT,2, and 2744.8kJ/kg for h2 in Equation (IV).

0.95=2744.8kJ/kgh32744.8kJ/kg2227.9 kJ/kgh3=2253.7kJ/kg

Substitute 3093.3kJ/kg for h1, 2744.8kJ/kg for h2, and 2253.7kJ/kg for h3 in Equation (IX).

wactual=(3093.3kJ/kg2744.8kJ/kg)+0.94(2744.8kJ/kg2253.7kJ/kg)=810.1kJ/kg

Thus, the power output produced from the turbine is 810.1 kJ/kg,

Substitute 3093.3kJ/kg for h1, 2744.8kJ/kg for h2, 2734.0 kJ/kg for h2s, and 2227.9kJ/kg for h3s in Equation (VIII).

wiso=(3093.3kJ/kg2734.0 kJ/kg)+0.94(2744.8kJ/kg2227.9kJ/kg)=845.2kJ/kg

Substitute 810.1kJ/kg for wactual, and 845.2kJ/kg for wiso in Equation (VI).

ηiso=810.1kJ/kg845.2kJ/kg=0.958×100%=95.80%

Thus, the overall isentropic efficiency of turbine is 95.80%.

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Chapter 7 Solutions

Thermodynamics: An Engineering Approach

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