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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: McGraw-Hill Education
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Water at 60°F flows from the basement to the second floor through the 0.75‐in. (0.0625‐ft)‐diameter
copper pipe (a drawn tubing) at a rate of Q=12.0 gal/min=0.0267 ft3/s and exits through a faucet of
diameter 0.50 in. as shown in the Fig. Determine the pressure at point (1) if all losses are included.
Its already answered however, I dont know how he got the numbers for example 1248, 299 , 1321 ..
Also could you explain why z1 is 0, can you answer the question with steps thank you

Transcribed Image Text:Example
Water at 60°F flows from the basement to the second floor through the 0.75-in. (0.0625-ft)-diameter
copper pipe (a drawn tubing) at a rate of Q=12.0 gal/min=0.0267 ft3/s and exits through a faucet of
diameter 0.50 in. as shown in the Fig. Determine the pressure at point (1) if all losses are included.
2
P₁
8
1
29
- 7 = 1/2 + 2 ₂ ² 2 2
V.
+
2
28
29
P₁ = 8 Z₂ + + p (V₂²_V ²) + Pfl V² + { PKLY ²
2
= 1248+299 + 1515 +1321=4383
=45000
Re = PVD
u
2=0.000005
K6=1,5
2
+ Z₂ +h
Królo
0.75-in.-diameter
copper pipe
16/ft²
Q =
12.0
gal/min
(1)
(4),
10 ft
15 ft.
K ₁₂ = (1.5 * 4) +10 +2
major
(6)
5 ft
(3)
the
(5)
-10 ft
10 ft
(7) (8)
8
'L minor
Threaded
90° elbows
10 ft
K₁ = 2 based on
pipe
velocity
Wide open
globe valve
V₁= Q/A
=Q1A₂.
√2
(2)
0.50-in.
diameter
3625
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