Example 10.5. A centrifugal pump has the following dimensions: inlet pipe 2-in pipe size (diameter = 2.067 in), outlet pipe 1.5-in pipe size (diameter = 1.61 in), impeller inner diameter = 2.067 in, outer diameter = 6.75 in, rotational velocity = 1750 rpm. What is the pump head, according to Eq. 10.20, assuming zero friction, for a volumetric flow rate of 100 gpm? Here we can write 1750 min @= 2π -= 183.26 1 (10.F) min 60 s S and 100 gal/min 231 in³ min ft m V₁ = 9.56 -= 2.91 (/4) (2.067 in)² gal 60 s 12 in S S = and similarly, V2 15.76 ft / s = 4.81 m/s. The impeller inner and outer radii are 0.086 ft and 0.336 ft, respectively. Substituting these values in Eq. 10.20, with zero friction, we find h = (183.26/s)² 32.2 ft/s² [(0.336 ft) (0.086 ft)²] + = 74.76 ft - 2.44 ft = 72.3 ft = 30.4 m (10.G) [(15.75 ft/s)² - (9.56 ft/s)²] 2.32 ft/s²
Example 10.5. A centrifugal pump has the following dimensions: inlet pipe 2-in pipe size (diameter = 2.067 in), outlet pipe 1.5-in pipe size (diameter = 1.61 in), impeller inner diameter = 2.067 in, outer diameter = 6.75 in, rotational velocity = 1750 rpm. What is the pump head, according to Eq. 10.20, assuming zero friction, for a volumetric flow rate of 100 gpm? Here we can write 1750 min @= 2π -= 183.26 1 (10.F) min 60 s S and 100 gal/min 231 in³ min ft m V₁ = 9.56 -= 2.91 (/4) (2.067 in)² gal 60 s 12 in S S = and similarly, V2 15.76 ft / s = 4.81 m/s. The impeller inner and outer radii are 0.086 ft and 0.336 ft, respectively. Substituting these values in Eq. 10.20, with zero friction, we find h = (183.26/s)² 32.2 ft/s² [(0.336 ft) (0.086 ft)²] + = 74.76 ft - 2.44 ft = 72.3 ft = 30.4 m (10.G) [(15.75 ft/s)² - (9.56 ft/s)²] 2.32 ft/s²
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
For the flow in Example 10.5, the experimental values are h ≈ 32 ft, Po = 1.5 hp, and η = 71%. Those are for pumping water. If we use this pump to pump gasoline at Q = 100 gpm, what are the predicted values of h, Po and η?
![Example 10.5. A centrifugal pump has the following dimensions:
inlet pipe 2-in pipe size (diameter = 2.067 in), outlet pipe 1.5-in pipe
size (diameter = 1.61 in), impeller inner diameter = 2.067 in, outer
diameter = 6.75 in, rotational velocity = 1750 rpm. What is the pump
head, according to Eq. 10.20, assuming zero friction, for a volumetric
flow rate of 100 gpm?
Here we can write
1750
min
@=
2π
-= 183.26 1
(10.F)
min
60 s
S
and
100 gal/min
231 in³
min
ft
m
V₁ =
9.56 -= 2.91
(/4) (2.067 in)²
gal
60 s 12 in
S
S
=
and similarly, V2
15.76 ft / s = 4.81 m/s. The impeller
inner and outer radii are 0.086 ft and 0.336 ft, respectively.
Substituting these values in Eq. 10.20, with zero friction, we find
h =
(183.26/s)²
32.2 ft/s²
[(0.336 ft) (0.086 ft)²] +
= 74.76 ft - 2.44 ft = 72.3 ft = 30.4 m
(10.G)
[(15.75 ft/s)² - (9.56 ft/s)²]
2.32 ft/s²](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb957c537-60c7-415f-b6cc-9e02c08b1c01%2F1011ec73-aa6c-42b3-8023-fb758fe1dda4%2Flm6z81t_processed.png&w=3840&q=75)
Transcribed Image Text:Example 10.5. A centrifugal pump has the following dimensions:
inlet pipe 2-in pipe size (diameter = 2.067 in), outlet pipe 1.5-in pipe
size (diameter = 1.61 in), impeller inner diameter = 2.067 in, outer
diameter = 6.75 in, rotational velocity = 1750 rpm. What is the pump
head, according to Eq. 10.20, assuming zero friction, for a volumetric
flow rate of 100 gpm?
Here we can write
1750
min
@=
2π
-= 183.26 1
(10.F)
min
60 s
S
and
100 gal/min
231 in³
min
ft
m
V₁ =
9.56 -= 2.91
(/4) (2.067 in)²
gal
60 s 12 in
S
S
=
and similarly, V2
15.76 ft / s = 4.81 m/s. The impeller
inner and outer radii are 0.086 ft and 0.336 ft, respectively.
Substituting these values in Eq. 10.20, with zero friction, we find
h =
(183.26/s)²
32.2 ft/s²
[(0.336 ft) (0.086 ft)²] +
= 74.76 ft - 2.44 ft = 72.3 ft = 30.4 m
(10.G)
[(15.75 ft/s)² - (9.56 ft/s)²]
2.32 ft/s²
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