[1] Repeat Example 7.10 (pages 288-289) but with the following exceptions: water is discharged at 9,000 kPa, and n=0.8. Calculate the work of the pump and AT using steam tables and equation 7.25
[1] Repeat Example 7.10 (pages 288-289) but with the following exceptions: water is discharged at 9,000 kPa, and n=0.8. Calculate the work of the pump and AT using steam tables and equation 7.25
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
Related questions
Question
![Example 7.10
Water at 45°C and 10 kPa enters an adiabatic pump and is discharged at a pressure
of 8600 kPa. Assume the pump efficiency to be 0.75. Calculate the work of the pump,
the temperature change of the water, and the entropy change of the water.
Solution 7.10
The following are properties for saturated liquid water at 45°C (318.15 K):
V = 1010 cm³.kg-1
B = 425 x 10-6 K-1 Cp = 4.178 kJ.kg-¹.K-1
7.4. Synopsis
By Eq. (7.24),
W, (isentropic) = (AH)s = (1010)(8600 — 10) = 8.676 × 106 kPa.cm³.kg-1
Because 1 kJ = 106 kPa.cm³,
By Eq. (7.17),
and
W,(isentropic) = (AH)s:
AH =
Solution for AT gives:
AS = 4.178 In
(ΔΗ), 8.676
n
0.75
=
W₁ = AH = 11.57 kJ.kg-¹
The temperature change of the water during pumping, from Eq. (7.25):
8590
11.57 = 4.178 AT + 1010[1 − (425 × 10-6)(318.15)]
106
8.676 kJ-kg-1
= 11.57 kJ-kg-1
or
AT = 0.97 K
The entropy change of the water is given by Eq. (7.26):
319.12
8590
(425 x 10-6)(1010)- = 0.0090 kJ.kg-¹·K-
318.15
106
0.97°C
289](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc21f0c70-127e-4e3a-ac97-62cd90f92e67%2Fcd537c42-b1d5-4a9c-ab1e-040707c31cc2%2Fqejgbw_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Example 7.10
Water at 45°C and 10 kPa enters an adiabatic pump and is discharged at a pressure
of 8600 kPa. Assume the pump efficiency to be 0.75. Calculate the work of the pump,
the temperature change of the water, and the entropy change of the water.
Solution 7.10
The following are properties for saturated liquid water at 45°C (318.15 K):
V = 1010 cm³.kg-1
B = 425 x 10-6 K-1 Cp = 4.178 kJ.kg-¹.K-1
7.4. Synopsis
By Eq. (7.24),
W, (isentropic) = (AH)s = (1010)(8600 — 10) = 8.676 × 106 kPa.cm³.kg-1
Because 1 kJ = 106 kPa.cm³,
By Eq. (7.17),
and
W,(isentropic) = (AH)s:
AH =
Solution for AT gives:
AS = 4.178 In
(ΔΗ), 8.676
n
0.75
=
W₁ = AH = 11.57 kJ.kg-¹
The temperature change of the water during pumping, from Eq. (7.25):
8590
11.57 = 4.178 AT + 1010[1 − (425 × 10-6)(318.15)]
106
8.676 kJ-kg-1
= 11.57 kJ-kg-1
or
AT = 0.97 K
The entropy change of the water is given by Eq. (7.26):
319.12
8590
(425 x 10-6)(1010)- = 0.0090 kJ.kg-¹·K-
318.15
106
0.97°C
289
![[1] Repeat Example 7.10 (pages 288-289) but with the following exceptions: water is discharged at
9,000 kPa, and n=0.8. Calculate the work of the pump and AT using steam tables and equation 7.25](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fc21f0c70-127e-4e3a-ac97-62cd90f92e67%2Fcd537c42-b1d5-4a9c-ab1e-040707c31cc2%2F3a4ikfq_processed.jpeg&w=3840&q=75)
Transcribed Image Text:[1] Repeat Example 7.10 (pages 288-289) but with the following exceptions: water is discharged at
9,000 kPa, and n=0.8. Calculate the work of the pump and AT using steam tables and equation 7.25
Expert Solution
Step 1
Initial temperature of water = 45 0C
Initial pressure = 10 kPa
Efficiency of the pump = 0.8
Final pressure = 9000 kPa
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