[2] Repeat Example 8.1 (pages 303-305) but with the following exceptions: steam fed to the turbine isat 9,000 kPa & 600 °C, the turbine exhaust is at 10 kPa, and for parts b & c the efficiencies are both0.8. 304Example 8.1Steam generated in a power plant at a pressure of 8600 kPa and a temperature of500°C is fed to a turbine. Exhaust from the turbine enters a condenser at 10 kPa,where it is condensed to saturated liquid, which is then pumped to the boiler.(a) What is the thermal efficiency of a Rankine cycle operating at these conditions?(b) What is the thermal efficiency of a practical cycle operating at these conditionsif the turbine efficiency and pump efficiency are both 0.75?(c) If the rating of the power cycle of part (b) is 80,000 kW, what is the steam rateand what are the heat-transfer rates in the boiler and condenser?Solution 8.1(a) The turbine operates under the same conditions as the turbine of Ex. 7.6 where,on the basis of 1 kg of steam:ThusMoreover, the enthalpy at the end of isentropic expansion, H₂ in Ex. 7.6, is here:H₂ = 2117.4 kJ-kg-1Subscripts refer to Fig. 8.4. The enthalpy of saturated liquid condensate at 10 kPa(and sat = 45.83°C) is:By Eq. (8.2) applied to the condenser,Q(condenser) = H4 – H₂ = 191.8 – 2117.4 = -1925.6 kJ-kg-¹(AH)s= -1274.2 kJ.kg-¹W, (isentropic) = (AH)s= -1274.2 kJ.kg-¹where the minus sign indicates heat flow out of the system.The pump operates under essentially the same conditions as the pump ofEx. 7.10, where:This result is of course also:and H₁ = H₂ + (AH)s = 191.8 +8.7 = 200.5 kJ-kg-¹The enthalpy of superheated steam at 8,600 kPa and 500°C is:H₂ = 3391.6 kJ.kg-1By Eq. (8.2) applied to the boiler,Q(boiler) = H₂H₁ = 3391.6 - 200.5 = 3191.1 kJ.kg-1The net work of the Rankine cycle is the sum of the turbine work and the pump work:W, (Rankine) = -1274.2 +8.7 = -1265.5 kJ-kg-¹W, (isentropic) = (AH)s = 8.7 kJ-kg-1By Ex. 7.10 for the pump,andH4 191.8 kJ-kgThenThe thermal efficiency of the cycle is:- W, (Rankine)Q(b) With a turbine efficiency of 0.75, then also from Ex. 7.6:W, (turbine) = AH = -955.6 kJ-kg-1and H3 = H₂ + AH = 3391.6-955.6 = 2436.0 kJ.kg-1For the condenser,n =CHAPTER 8. Production of Power from HeatW, (Rankine) = -Q(boiler) - Q(condenser)orThe net work of the cycle is therefore:Q(condenser) = H4 H3 = 191.8 - 2436.0= -2244.2 kJ.kg-18.1. The Steam Power PlantNote that-1= -3191.1 + 1925.6= -1265.5 kJ.kg-1m =W, (pump) = AH = 11.6 kJ.kg-1Then by Eq. (8.1),W, (net) = -955.6 +11.6= -944.0 kJ.kg-¹The thermal efficiency of the cycle is therefore:1265.53191.1H₁ = H4 + AH = 191.8 +11.6 = 203.4 kJ.kg-¹Q(boiler) = H₂ - H₁ = 3391.6 — 203.4 = 3188.2 kJ-kg-1n == 0.3966which may be compared with the result of part (a).(c) For a power rating of 80,000 kW:=W- W,(net) 944.0= 0.2961Q(boiler) 3188.2=W, (net)-80,000 kJ-s-1W, (net) -944.0 kJ-kg-,(net) = mW, (net)-1= 84.75 kg.sQ (boiler) = (84.75)(3188.2) = 270.2 x 10³ kJ.s-¹(condenser) = (84.75)(-2244.2) = -190.2 x 10³ kJ-s-¹Q (boiler) + Q (condenser) = — W, (net)305

Given-Information:- Pressure, p = 9000 kPaTemperature, T = 600°CExhaust Pressure, p = 10…

[2] Repeat Example 8.1 (pages 303-305) but with the following exceptions: steam fed to the turbine is at 9,000 kPa & 600 °C, the turbine exhaust is at 10 kPa, and for parts b & c the efficiencies are both 0.8.

[2] Repeat Example 8.1 (pages 303-305) but with the following exceptions: steam fed to the turbine is at 9,000 kPa & 600 °C, the turbine exhaust is at 10 kPa, and for parts b & c the efficiencies are both 0.8.

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Question

[2] Repeat Example 8.1 (pages 303-305) but with the following exceptions: steam fed to the turbine is
at 9,000 kPa & 600 °C, the turbine exhaust is at 10 kPa, and for parts b & c the efficiencies are both
0.8.

304
Example 8.1
Steam generated in a power plant at a pressure of 8600 kPa and a temperature of
500°C is fed to a turbine. Exhaust from the turbine enters a condenser at 10 kPa,
where it is condensed to saturated liquid, which is then pumped to the boiler.
(a) What is the thermal efficiency of a Rankine cycle operating at these conditions?
(b) What is the thermal efficiency of a practical cycle operating at these conditions
if the turbine efficiency and pump efficiency are both 0.75?
(c) If the rating of the power cycle of part (b) is 80,000 kW, what is the steam rate
and what are the heat-transfer rates in the boiler and condenser?
Solution 8.1
(a) The turbine operates under the same conditions as the turbine of Ex. 7.6 where,
on the basis of 1 kg of steam:
Thus
Moreover, the enthalpy at the end of isentropic expansion, H₂ in Ex. 7.6, is here:
H₂ = 2117.4 kJ-kg-1
Subscripts refer to Fig. 8.4. The enthalpy of saturated liquid condensate at 10 kPa
(and sat = 45.83°C) is:
By Eq. (8.2) applied to the condenser,
Q(condenser) = H4 – H₂ = 191.8 – 2117.4 = -1925.6 kJ-kg-¹
(AH)s= -1274.2 kJ.kg-¹
W, (isentropic) = (AH)s= -1274.2 kJ.kg-¹
where the minus sign indicates heat flow out of the system.
The pump operates under essentially the same conditions as the pump of
Ex. 7.10, where:
This result is of course also:
and H₁ = H₂ + (AH)s = 191.8 +8.7 = 200.5 kJ-kg-¹
The enthalpy of superheated steam at 8,600 kPa and 500°C is:
H₂ = 3391.6 kJ.kg-1
By Eq. (8.2) applied to the boiler,
Q(boiler) = H₂H₁ = 3391.6 - 200.5 = 3191.1 kJ.kg-1
The net work of the Rankine cycle is the sum of the turbine work and the pump work:
W, (Rankine) = -1274.2 +8.7 = -1265.5 kJ-kg-¹
W, (isentropic) = (AH)s = 8.7 kJ-kg-1
By Ex. 7.10 for the pump,
and
H4 191.8 kJ-kg
Then
The thermal efficiency of the cycle is:
- W, (Rankine)
Q
(b) With a turbine efficiency of 0.75, then also from Ex. 7.6:
W, (turbine) = AH = -955.6 kJ-kg-1
and H3 = H₂ + AH = 3391.6-955.6 = 2436.0 kJ.kg-1
For the condenser,
n =
CHAPTER 8. Production of Power from Heat
W, (Rankine) = -Q(boiler) - Q(condenser)
or
The net work of the cycle is therefore:
Q(condenser) = H4 H3 = 191.8 - 2436.0= -2244.2 kJ.kg-1
8.1. The Steam Power Plant
Note that
-1
= -3191.1 + 1925.6= -1265.5 kJ.kg-1
m =
W, (pump) = AH = 11.6 kJ.kg-1
Then by Eq. (8.1),
W, (net) = -955.6 +11.6= -944.0 kJ.kg-¹
The thermal efficiency of the cycle is therefore:
1265.5
3191.1
H₁ = H4 + AH = 191.8 +11.6 = 203.4 kJ.kg-¹
Q(boiler) = H₂ - H₁ = 3391.6 — 203.4 = 3188.2 kJ-kg-1
n =
= 0.3966
which may be compared with the result of part (a).
(c) For a power rating of 80,000 kW:
=
W
- W,(net) 944.0
= 0.2961
Q(boiler) 3188.2
=
W, (net)
-80,000 kJ-s-1
W, (net) -944.0 kJ-kg-
,(net) = mW, (net)
-1
= 84.75 kg.s
Q (boiler) = (84.75)(3188.2) = 270.2 x 10³ kJ.s-¹
(condenser) = (84.75)(-2244.2) = -190.2 x 10³ kJ-s-¹
Q (boiler) + Q (condenser) = — W, (net)
305

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