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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Chapter 5.5, Problem 179RP

Reconsider Prob. 5–178. Using appropriate software, investigate the effects of turbine exit area and turbine exit pressure on the exit velocity and power output of the turbine. Let the exit pressure vary from 10 to 50 kPa (with the same quality), and let the exit area vary from 1000 to 3000 cm2. Plot the exit velocity and the power outlet against the exit pressure for the exit areas of 1000, 2000, and 3000 cm2, and discuss the

results.

5–178 Steam enters a turbine steadily at 7 MPa and 600°C with a velocity of 60 m/s and leaves at 25 kPa with a quality of 95 percent. A heat loss of 20 kJ/kg occurs during the process. The inlet area of the turbine is 150 cm2, and the exit area is 1400 cm2. Determine (a) the mass flow rate of the steam, (b) the exit velocity, and (c) the power output.

Expert Solution & Answer
Check Mark
To determine

Plot the exit pressure against power output of turbine and exit pressure against exit velocity for varying pressure from 10kPa to 50kPa, for the varying exit area of 1000cm2 to 3000cm2.

Answer to Problem 179RP

The plot for the exit pressure against power output of turbine and exit pressure against exit velocity for varying pressure from 10kPa to 50kPa, for the varying exit area of 1000cm2 to 3000cm2 is plotted and shown in Figure 1 and Figure 2 respectively.

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 refrigerant flows at steady state through the compressor. 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)

Conclusion:

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.

Thermodynamics: An Engineering Approach, Chapter 5.5, Problem 179RP , 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.

Thermodynamics: An Engineering Approach, Chapter 5.5, Problem 179RP , additional homework tip  2

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Chapter 5 Solutions

Thermodynamics: An Engineering Approach

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