A 21 gallon per minute showerhead is operating at full blast

A 2.1 gallon per minute showerhead is operating at full blast. The city provides water
into your house at 90psig, and the shower discharges into atmosphere. The plumbing line
between your house and the city connection has 60ft of 3/4 in copper with two 90 degree elbows
(treat as flanged)
That line connects to 50ft of 1/2 in line feeding your bathroom, which
has three tees in line flow, one tee in branch flow, and four 90 degree elbows (again, treat all as flanged)

You may assume the nominal diameters are the inner diameters of the pipes, and

treat the showerhead as though it were 1/2 in pipe for the velocity in the K-factor. Assume
roughness of 0.05mm for copper pipe.
If the showerhead is the only tap open, what is its K_L coecient?

Solution

solution:

1)here head of water available is

P=density*g*h

90*g=.03613*g*h

h=2491.00 in

h=2491*2.54/30.48

hs=207.6 ft

2)here velocity in .75 in and .5 in pipe are

Q=2.1 gal/min=8.085 in^3/s

Q=A1*V1

V1=18.30 in/s

3)here by k method,head loss is

Hf1=2*k1(V1^2/2*g)

for flange elbow k1=.3

Hf1=.2601 in=.6607 ft

4)major hjead loss is by william hagen method as

p=4.52*Q^1.85/(130^1.85*d^4.8655) psi/ft

Hf2=1.232 ft

5)velocity in smaller pipe

V2=Q/A2=41.176 in/s

where value of k are

line flow tee=k2=.2

branch flow tee=k3=1

elbow=K4=.3

head loss in 3 tee are=Hf3=3*K2(V2^2/2*g)=1.3170 in

head loss in branch tee=Hf4=K3(V2^2/2*g)=2.195 in

head loss in 2 elbow=Hf5=K4(V2^2/2*g)*2=2.634 in

total loss=Hf3+Hf4+Hf5=.512 ft

6)major loss in second pipe by william method is

p=4.52*Q^1.85/(130^1.85*d^4.8655) psi/ft

Hf6=7.38 ft

7)for total haed loss as

Hf=Hf1+Hf2+Hf3+Hf4+Hf5+Hf6=9.7847 ft

8)total head avail in bathroom is

H=Hs-Hf=207.6-9.7847=197.8153 ft