values from steam table for 9000 kPa and 600 CH = 3631.1S = 6.9574 Example 7.6A steam turbine with rated capacity of 56,400 kW (56,400 kJ.s-1) operates with steamat inlet conditions of 8600 kPa and 500°C, and discharges into a condenser at apressure of 10 kPa. Assuming a turbine efficiency of 0.75, determine the state ofthe steam at discharge and the mass rate of flow of the steam.Solution 7.6At the inlet conditions of 8600 kPa and 500°C, the steam tables provide:H₁ = 3391.6 kJ.kg¯1S₁ = 6.6858 kJ.kg-1.K-1If the expansion to 10 kPa is isentropic, then, S₂ = $₁ 6.6858 kJ-kg-¹·K-¹.Steam with this entropy at 10 kPa is wet. Applying the “lever rule” [Eq. (6.96b),with M = S and x = x2], the quality is obtained as follows:S₂ = S2₂2 + x₂ (S₂2 - S¹½)Then, 6.6858 0.6493 + x2(8.1511 - 0.6493)x₂ = 0.8047This is the quality (fraction vapor) of the discharge stream at point 2'. The enthalpyH₂ is also given by Eq. (6.96b), written:Thus,=H₂ = H₂2 + x₂(H²₂ − H ½₂)-H₂ = 191.8+ (0.8047)(2584.8 – 191.8) = 2117.4 kJ.kg¯1(AH)s = H₂ − H₁ = 2117.4 - 3391.6 = -1274.2 kJ-kg-1-and by Eq. (7.16),AH = n(AH)s= (0.75)(-1274.2) = −955.6 kJ.kg-1Whence, H₂ = H₁ + AH = 3391.6 - 955.6 = 2436.0 kJ-kg-¹Thus the steam in its actual final state is also wet, with its quality given by:2436.0 191.8 + x₂(2584.8 - 191.8)Then S₂ = 0.6493 + (0.9378)(8.1511 - 0.6493) = 7.6846 kJ.kg-¹.K-1x2 = 0.9378This value may be compared with the initial value of S₁ = 6.6858.The steam rate in is given by Eq. (7.13). For a work rate of 56,400 kJ-s¯¹,W₁ = -56,400 = m (2436.0 – 3391.6) m = 59.02 kg.s-¹S=Example 7.6 was solved with data from the steam tables. When a comparable set oftables is not available for the working fluid, the generalized correlations of Sec. 6.4 may beused in conjunction with Eqs. (6.74) and (6.75), as illustrated in the following example. [4] Repeat Example 7.6 (pages 279-280) but with the following conditions: inlet conditions are 9,000kPa and 600 °C, and n = 0.8.

Given, Work rate of Steam turbine, Ws˙ = -56400 KW Inlet condition : Pressure, P1 =9000KPa…

Answered: mass rate of flow of the steam.

Introduction to Chemical Engineering Thermodynamics

8th Edition

ISBN:9781259696527

Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart

Chapter1: Introduction

Section: Chapter Questions

Problem 1.1P

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Question

values from steam table for 9000 kPa and 600 C

H = 3631.1

S = 6.9574

Example 7.6
A steam turbine with rated capacity of 56,400 kW (56,400 kJ.s-1) operates with steam
at inlet conditions of 8600 kPa and 500°C, and discharges into a condenser at a
pressure of 10 kPa. Assuming a turbine efficiency of 0.75, determine the state of
the steam at discharge and the mass rate of flow of the steam.
Solution 7.6
At the inlet conditions of 8600 kPa and 500°C, the steam tables provide:
H₁ = 3391.6 kJ.kg¯1
S₁ = 6.6858 kJ.kg-1.K-1
If the expansion to 10 kPa is isentropic, then, S₂ = $₁ 6.6858 kJ-kg-¹·K-¹.
Steam with this entropy at 10 kPa is wet. Applying the “lever rule” [Eq. (6.96b),
with M = S and x = x2], the quality is obtained as follows:
S₂ = S2₂2 + x₂ (S₂2 - S¹½)
Then, 6.6858 0.6493 + x2(8.1511 - 0.6493)
x₂ = 0.8047
This is the quality (fraction vapor) of the discharge stream at point 2'. The enthalpy
H₂ is also given by Eq. (6.96b), written:
Thus,
=
H₂ = H₂2 + x₂(H²₂ − H ½₂)
-
H₂ = 191.8+ (0.8047)(2584.8 – 191.8) = 2117.4 kJ.kg¯1
(AH)s = H₂ − H₁ = 2117.4 - 3391.6 = -1274.2 kJ-kg-1
-
and by Eq. (7.16),
AH = n(AH)s= (0.75)(-1274.2) = −955.6 kJ.kg-1
Whence, H₂ = H₁ + AH = 3391.6 - 955.6 = 2436.0 kJ-kg-¹
Thus the steam in its actual final state is also wet, with its quality given by:
2436.0 191.8 + x₂(2584.8 - 191.8)
Then S₂ = 0.6493 + (0.9378)(8.1511 - 0.6493) = 7.6846 kJ.kg-¹.K-1
x2 = 0.9378
This value may be compared with the initial value of S₁ = 6.6858.
The steam rate in is given by Eq. (7.13). For a work rate of 56,400 kJ-s¯¹,
W₁ = -56,400 = m (2436.0 – 3391.6) m = 59.02 kg.s-¹
S
=
Example 7.6 was solved with data from the steam tables. When a comparable set of
tables is not available for the working fluid, the generalized correlations of Sec. 6.4 may be
used in conjunction with Eqs. (6.74) and (6.75), as illustrated in the following example.

[4] Repeat Example 7.6 (pages 279-280) but with the following conditions: inlet conditions are 9,000
kPa and 600 °C, and n = 0.8.

Expert Solution

Step 1

Given,

Work rate of Steam turbine, $\dot{W_{s}}$ = -56400 KW

Inlet condition : Pressure, P₁ =9000KPa

Temperature, T₁ = $600^{\circ} C$

@ P₁ & T₁: Steam condition given as,

$H_{1} = 3631.1 K J K g^{- 1} S_{1} = 6.9574 K J K g^{- 1} K^{- 1}$

Discharge pressure, P₂ = 10KPa

Turbine efficiency given as $η = 0.8$

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