Question

A new residential complex in Starkville will be supplied with 2.674 ft3/s at a pressure of 3600
lb/ft2. The piping system will be connected to the main city supply line (9360 lb/ft2), which has
10‐ft of elevation difference with the city line. Determine the corresponding pipe size to connect
the apartment piping system to the city line. The design considers the use of PVC (ks=5*10‐7 ft).
The system consists of 100‐ft of pipe, 4 threaded 90 elbows (kL=1.5), 10 threaded unions
(kL=0.08) and a swing check valve (kL=2.0). Consider water density to be 62.42lb/ft3 and the
dynamic viscosity to be 2.34*10‐5 lb*s/ft2.

Answer 1

Ans) Apply Bernoulli equation between point 1 and 2 located at main line and supply line respectively,

P1/ + /2g + Z1 = P2/ + /2g + Z2 + Hf +Hm

Given, P1 = 9360 psf and P2 = 3600 psf  

Since, pipe diameter id constant throughout, V1 = V2

Elevation, Z1 = 10 ft and Z2 = 0

Putting given values,

=> (9360/62.4) + 10 = (3600/62.4) + Hf + Hm

=> Hf + Hm = 102.30 ............................................(1)

Now,

Frictional head loss,Hf = 8 f L / ( g )

Since, both friction factor and diameter are unknown, solution is iterative so assume initial friction factor as 0.02 to began iteration.

Iteration 1 :

=> Hf = 8(0.02)(100)() / ( x 32.2 x)

=> Hf = (0.36 / )

Also,

Hm = K / 2 g

K = 4(1.5) + 10(0.08) + 2 = 8.8

A = (/4) = 0.785

=> Hm = 8.8 () / (2 x 32.2 x )

=> Hm = (1.585 / )

Putting values in equation 1,

  => (0.36 / ) + (1.585 / ) = 102.30

On solving above equation , we get D 0.395 ft

.

Iteration 2 :

Use above calculated diameter to fine Reynold number (Re),

Re = V D /

Velocity (V) = Q / A = 2.674 / (/4) = 21.94 ft/s

is dynamic viscosity = 2.34 x lb s/ft2

Densitty of water () = 1.94 slugs/ft3

=> Re = 1.94 x 21.94 x 0.395/( 2.34 x )

=> Re = 718488

Roughness of pipe (ks) = 5 x ft

=> Relative roughness (ks/D) = 5 x / 0.395 = 1.2 x

According to Moody diagram, for Re = 718488 and ks/D = 1.2 x , friction factor (f) = 0.0123

=> Hf = 8(0.0123)(100)() / ( x 32.2 x)

=> Hf = (0.222/ )

Minor loss remain same sa before so,again putting vales in equation 1,

  => (0.222 / ) + (1.585 / ) = 102.30

On solving above equation , we get D 0.38 ft

.

Iteration 3 :

Use above calculated diameter to fine Reynold number (Re),

Re = V D /

Velocity (V) = Q / A = 2.674 / (/4) = 23.59 ft/s

=> Re = 1.94 x 23.59 x 0.38/( 2.34 x )

=> Re = 743186

=> Relative roughness (ks/D) = 5 x / 0.38 = 1.3 x

According to Moody diagram, for Re = 743186 and ks/D = 1.3 x , friction factor (f) = 0.0123

Since, friction factor comes out to be same as above iteration, diameter will remain same as before.

Hence, required pipe diameter (D)= 0.38 ft