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 6, Problem 162RQ
To determine

Plot the following graphs for the pressure varies from 10kPa to 50kPa and the exit area varies from 1000cm2 to 3000cm2.

  • Exit velocity versus exit pressure power output of turbine
  • Turbine power output versus exit pressure power output of turbine

Expert Solution & Answer
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Explanation of Solution

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

  m˙1=m˙2=m˙

Write the formula for inlet mass flow rate.

  m˙=A1V1v1        (I)

Here, the cross-sectional area is A, the velocity is V and the specific volume is v; the subscript 1 indicates the inlet condition.

At inlet:

The steam is at the state of superheated condition.

Refer Table A-6, “Superheated water”.

Obtain the inlet enthalpy (h1) and specific volume (v1) corresponding to the pressure of 7MPa and the temperature of 600°C.

  h1=3650.6kJ/kgv1=0.055665m3/kg

The turbine operates steadily. Hence, the inlet and exit mass flow rates are equal.

  m˙1=m˙2=m˙

Write the formula for exit mass flow rate.

  m˙=A2V2v2        (II)

Here, the cross-sectional area is A, the velocity is V and the specific volume is v; the subscript 2 indicates the exit condition.

Rearrange the Equation (II) to obtain exit velocity (V2).

  V2=m˙v2A2        (III)

At exit:

Consider the exit pressure (P2) is 25kPa.

The steam is with the quality of 95%.

Write the formula for exit enthalpy (h2) of the steam with the consideration of its quality.

  h2=hf+xhfg        (IV)

Write the formula for exit specific volume (v2) of the steam with the consideration of its quality.

  v2=vf+x(vgvf)        (V)

Here, the enthalpy is h, the quality of steam is x, and subscript f indicates that fluid state, g indicates the gaseous state, fg indicates the mixed state (vaporization).

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

Obtain the following corresponding to the pressure of 25kPa.

  vf=0.001020m3/kgvg=6.2034m3/kghf=271.96kJ/kghfg=2345.5kJ/kg

Consider the steam flows at steady state. Hence, the inlet and exit mass flow rates are equal.

  m˙1=m˙2=m˙

Write the energy rate balance equation for one inlet and one outlet system.

  [Q˙1+W˙1+m˙(h1+V122+gz1)][Q˙2+W˙2+m˙(h2+V222+gz2)]=ΔE˙system        (VI)

Here, the rate of heat transfer is Q˙, the rate of work transfer is W˙, the enthalpy is h and the velocity is V, the gravitational acceleration is g, the elevation from the datum is z and the rate of change in net energy of the system is ΔE˙system; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The steam flows at steady state through the turbine. Hence, the rate of change in net energy of the system becomes zero.

  ΔE˙system=0

Heat loss occurs at the rate of 20kJ/kg. Neglect the potential energy changes. Here, the work done is by the system (turbine) and the work done on the system is zero i.e. W˙1=0.

The Equations (VI) reduced as follows to obtain the work output (W˙2).

  [0+0+m˙(h1+V122+0)][Q˙2+W˙2+m˙(h2+V222+0)]=0m˙(h1+V122)[Q˙2+W˙2+m˙(h2+V222)]=0m˙(h1+V122)=Q˙2+W˙2+m˙(h2+V222)W˙2=m˙(h1+V122)m˙(h2+V222)Q˙2

  W˙2=m˙(h1h2+V12V222)Q˙2W˙2=m˙(h2h1+V22V122)Q˙2        (VII)

Here, Q˙2=m˙Q2.

Rewrite the Equation (VII) as follows.

  W˙2=m˙(h2h1+V22V122)m˙Q2        (VIII)

Calculation:

Substitute 271.96kJ/kg for hf, 95% for x and 2345.5kJ/kg for hfg in Equation (IV).

  h2=271.96kJ/kg+95%(2345.5kJ/kg)=271.96kJ/kg+95100(2345.5kJ/kg)=271.96kJ/kg+2228.225kJ/kg=2500.185kJ/kg

Substitute 0.001020m3/kg for vf, 95% for x and 6.2034m3/kg for vg in

Equation (V).

  v2=0.001020m3/kg+95%(6.2034m3/kg0.001020m3/kg)=0.001020m3/kg+95100(6.20238m3/kg)=0.001020m3/kg+5.892261m3/kg=5.893281m3/kg

Substitute 150cm2 for A1, 60m/s for V1 ,and 0.055665m3/kg for v1 in Equation (I).

  m˙=(150cm2)(60m/s)0.055665m3/kg=(150cm2×1m2104cm2)(60m/s)0.055665m3/kg=16.1681kg/s16.17kg/s

Consider the exit area (A2) of the turbine is 1400cm3.

Substitute 16.17kg/s for m˙, 5.893281m3/kg for v2 , and 1400cm2 for A2 in

Equation (III).

  V2=(16.17kg/s)(5.893281m3/kg)1400cm2=(16.17kg/s)(5.893281m3/kg)1400cm2×1m2104cm2=680.6739m/s

Substitute 16.17kg/s for m˙, 2500.185kJ/kg for h2, 3650.6kJ/kg for h1, 680.6739m/s for V2, 60m/s for V1, and 20kJ/kg for Q2 in Equation (VIII).

  W˙2=16.17kg/s[2500.185kJ/kg3650.6kJ/kg+(680.6739m/s)2(60m/s)22]16.17kg/s(20kJ/kg)=[16.17kg/s(1150.415kJ/kg+229858.4791m2/s2×1kJ/kg1000m2/s2)323.4kJ/s]=16.17kg/s(920.5565kJ/kg)323.4kJ/s=14885.3989kJ/s323.4kJ/s

  =14561.9989kJ/s×1kW1kJ/s14562kW

The exit velocity (V2) and the power output (W˙2) for exit pressure (P2) of 25kPa and the exit area (A2) of 1400cm2 is as follows.

  V2=680.6739m/sW˙2=14562kW

Using excel spread sheet, the exit velocity (V2) and the power output (W˙2) is calculated for the exit pressure varies from 10kPa to 50kPa, for the exit area of 1000cm2 and shown in Table 1.

S.No.P2(kPa)V2(m/s)W˙2(kW)
1102253.540216–22171.1196
2151539.230498–514.857057
3201174.8711047295.806083
425952.943537710965.91684
530803.215013412968.62817
640613.439074714943.44488
750497.767012115822.49054

Table 1

Similarly, the exit velocity (V2) and the power output (W˙2) is calculated for the exit pressure varies from 10kPa to 50kPa, for the exit area of 2000cm2 and shown in Table 2.

S.No.P2(kPa)V2(m/s)W˙2(kW)
1101126.7701088623.292217
215769.615249113851.56455
320587.43555215665.73428
425476.471768916472.41662
530401.607506716880.6829
640306.719537417225.27947
750248.88350617324.918

Table 2

Similarly, the exit velocity (V2) and the power output (W˙2) is calculated for the exit pressure varies from 10kPa to 50kPa, for the exit area of 3000cm2 and shown in Table 3.

S.No.P2(kPa)V2(m/s)W˙2(kW)
110751.18007214325.96107
215513.076832716512.01299
320391.623701317215.72099
425317.647845917492.1388
530267.738337817605.13749
640204.479691617647.84143
750165.922337417603.1453

Table 3

Refer Table 1, 2, and 3.

Plot the graph for the exit pressure (P2) against power output of turbine (W˙2) as shown in Figure 1.

Fundamentals of Thermal-Fluid Sciences, Chapter 6, Problem 162RQ , additional homework tip  1

Refer Table 1, 2, and 3.

Plot the graph for the exit pressure (P2) against exit velocity (V2) as shown in Figure 2.

Fundamentals of Thermal-Fluid Sciences, Chapter 6, Problem 162RQ , additional homework tip  2

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Chapter 6 Solutions

Fundamentals of Thermal-Fluid Sciences

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