Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 8, Problem 113P

(a)

To determine

The mass flow rate of the steam flowing to the turbine.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given:

The initial pressure of the steam (P1) is 5 MPa.

The initial temperature of the steam (T1) is 650°C.

The inlet velocity of the steam (V1) is 80m/s.

The final pressure of the steam (P2) is 50 kPa.

The final temperature of the steam (T2) is 150°C.

The exit velocity of the steam (V2) is 140m/s.

The power output of the turbine (W˙out) is 8MW.

Calculation:

Refer Table A-6, “Superheated water”, for the pressure of 5 MPa and temperature of 650°C to obtain the following properties using interpolation method.

Temperature, °C

Enthalpy kJ/kg

Entropy kJ/kgK

6003666.92511.5
650?  3783.60?7.387
7003900.32506.2

Write the formula of interpolation method of two variables.

  y2=(x2x1)(y3y1)(x3x1)+y1        (I)

Substitute x1=600°C, x2=650°C, x3=700°C, y1=3666.9 kJ/kg, and y3=3900.3 kJ/kg in Equation (I) to obtain the enthalpy at state 1(h1).

  y2=(650600)(3900.33666.9)(700600)+3666.9=3783.60 kJ/kgK

Substitute x1=600°C, x2=650°C, x3=700°C, y1=2511.5 kJ/kgK, and y3=2506.2 kJ/kgK in Equation (I) to obtain the entropy at state 1(s1).

  y2=(650600)(3900.33666.9)(700600)+3666.9=3783.60 kJ/kgK

Refer Table A-6, “Superheated water”, select the actual enthalpy (h2) at the pressure of 50kPa and temperature of 150°C as 2780.2 kJ/kg.

Write the energy balance equation for closed system.

  E˙inE˙out=ΔE˙system        (I)

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

At steady state, rate of change in internal energy of the system is zero.

Rewrite Equation (I) for energy balance for the closed system.

  E˙inE˙out=0E˙in=E˙outm˙(h1+V122)=W˙a,out+m˙(h2+V222)W˙a,out=m˙(h2h1+V22V122)        (II)

  (8MW)(106kJ/s1MW)=m˙(2,780.2kJ/kg3,783.2kJ/kg+(140m/s)2(80m/s)22(1kJ/kg1,000m2/s2))8,000=m˙(996.4)m˙=8.029kg/s

Thus, the mass flow rate of the steam flowing to the turbine is 8.03 kg/s_.

(b)

To determine

The isentropic efficiency of the turbine

(b)

Expert Solution
Check Mark

Explanation of Solution

Refer to Table A-5, “Saturated water-Pressure table”, obtain the following properties at the pressure of 50kPa.

  hf=340.54kJ/kghfg=2,304.7kJ/kgsf=1.0912kJ/kgKsfg=6.5019kJ/kgK

Calculate the vapor quality is (x2s) using the final entropy equation.

  x2s=s2ssfsfgx2s=7.3901kJ/kgK1.0912kJ/kgK6.5019kJ/kgK=0.9688

Calculate the final enthalpy for the isentropic process.

  h2s=hf+x2shfgh2s=340.54kJ/kg+(0.9688)(2,304.7kJ/kg)=2,573.3kJ/kg

For the reversible adiabatic process, the entropy remains constant.

s1=s2s

Write the expression for the power output of the isentropic turbine (W˙s,out).

  W˙s,out=m˙(h2sh1+V22V122)W˙s,out=(8.029kg/s)(2,573.3kJ/kg3,783.2kJ/kg+(140m/s)2(80m/s)22(1kJ/kg1,000m2/s2))=9,661kJ/s(1kW1kJ/s)=9,661kW

Calculate the isentropic efficiency of the turbine (η).

  η=W˙a,outW˙s,outη=8,000kW9,661kW=0.828=82.8%

Thus, the isentropic efficiency of the turbine is 82.8%_

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Chapter 8 Solutions

Fundamentals of Thermal-Fluid Sciences

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