Water at 60° F flows from the basement to the first floor through the 0.0625 – ft. diameter copper pipe (a drawn tubing) at a rate of 12 gal/min and exits through a faucet of diameter 0.5 – in. as shown in Figure 1. KĻ = 2 based on pipe velocity - 12 ft (4) 0.75-in. diameter (2) copper pipe 10 ft 0.50-in. Q =12.0 gal/min (3) diameter (1) Threaded 15 ft- - 90° elbows Figure 1: Piping system Assuming the flow is incompressible, steady, and with uniform velocity profile, the general energy equation is, P v,? y* 29 P2, V,2 +z = + 2g + zz + h̟ (1) With this /p = 8 x 10-5 and the calculated Reynolds number Re = 45,000, the value of f is obtained from the Moody chart f = 0.0215. For threaded 90° elbow, K̟ = 1.5 As shown in Fig. 1, the fluid exits through the faucet freely, thus, P2 = 0. a) Knowing that the flow rate Q = AV is constant, determine the fluid velocity at point (1) and point (2). Note that, the pipe diameter is 0.0625 – ft., and the faucet diameter is 0.5 – in. Q1 = Q3 = Q4 = Q2 A = Az = A, # A2
Water at 60° F flows from the basement to the first floor through the 0.0625 – ft. diameter copper pipe (a drawn tubing) at a rate of 12 gal/min and exits through a faucet of diameter 0.5 – in. as shown in Figure 1. KĻ = 2 based on pipe velocity - 12 ft (4) 0.75-in. diameter (2) copper pipe 10 ft 0.50-in. Q =12.0 gal/min (3) diameter (1) Threaded 15 ft- - 90° elbows Figure 1: Piping system Assuming the flow is incompressible, steady, and with uniform velocity profile, the general energy equation is, P v,? y* 29 P2, V,2 +z = + 2g + zz + h̟ (1) With this /p = 8 x 10-5 and the calculated Reynolds number Re = 45,000, the value of f is obtained from the Moody chart f = 0.0215. For threaded 90° elbow, K̟ = 1.5 As shown in Fig. 1, the fluid exits through the faucet freely, thus, P2 = 0. a) Knowing that the flow rate Q = AV is constant, determine the fluid velocity at point (1) and point (2). Note that, the pipe diameter is 0.0625 – ft., and the faucet diameter is 0.5 – in. Q1 = Q3 = Q4 = Q2 A = Az = A, # A2
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Transcribed Image Text:Thus,
pKL
c) Plot a graph to show the changes of pressure over the pipe length. A template for the plot is
shown below,
30|
20
10
10
20
30
40
Distance along pipe from point (1), ft
d) Using the same graph in (c), sketch another line (no calculation is needed) to show the possible
pressure changes supposed both the major and minor losses are neglected.
sd 'd
Transcribed Image Text:Water at 60° F flows from the basement to the first floor through the 0.0625 – ft. diameter copper
pipe (a drawn tubing) at a rate of 12 gal/min and exits through a faucet of diameter 0.5 – in. as
shown in Figure 1.
KL = 2 based on
pipe velocity
- 12 ft
(4)
0.75-in. diameter
(2)
copper pipe
10'ft
0.50-in.
Q =12.0
gal/min
(3)
diameter
(1)
Threaded
15 ft-
90° elbows
Figure 1: Piping system
Assuming the flow is incompressible, steady, and with uniform velocity profile, the general energy
equation is,
P2, V2
+z, =
2g
+ z2 + h.
(1)
Y' 2g
With this /p = 8 x 10-5 and the calculated Reynolds number Re = 45,000, the value of f is
obtained from the Moody chart f = 0.0215. For threaded 90° elbow, K, = 1.5
As shown in Fig. 1, the fluid exits through the faucet freely, thus, P2 = 0.
a) Knowing that the flow rate Q = AV is constant, determine the fluid velocity at point (1) and
point (2). Note that, the pipe diameter is 0.0625 – ft., and the faucet diameter is 0.5 – in.
Q1 = Q3 = Qs = Q2
A = Az = A, + A2
b) Determine the pressure changes along this pipe, P1, P2, P3 and P4 (with the major and minor
losses).
P, = P, +(V,² – v,?) +r(z2 – z4) + h̟ (2)
where the total head loss, h,
hi = major loss + minor loss
%3D
hi = ef
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