Question

Problem No. 1 (25 Points) Steady water flow at 400 gpm takes place between the two reservoirs shown below. Water is at 68 deg. F with kinematic viscosity of 1.08E-5 feet square/second. The piping system involves commercial steel, schedule 40, a sharp edged entrance, 5 standard 90 degree flanged elbows, fully open flanged globe valve and a sharp exit. Both reservoirs are open to the atmosphere. a) Determine the Head Gain required across the pump. b) Find the corresponding power delivered to the water. c) Find the power delivered to the shaft if the pump motor is 55% efficient. 1 Ele. 1901 120 ft 4" sch. 40 1 Ele. O T = 60°F VE 80 ft 4" sch. 40 globe valve

Answer 1

Given:

Q = 400 gpm = 0.891204 ft3/s

$\nu$ = 1.08 * 10-5 ft2/s

Solution:

1) Diameter of 4" schedule 40 pipe, d = 4.026 in = 0.3355 ft { from property table }

Loss coefficient for sharpe edge entrance, Ke = 0.5

Loss coefficient for standard 90o flanged elbow, Kel = 0.3 ( 5 numbers )

Loss coefficient for fully open flanged globe valve, Kgv = 10

Loss coefficient for sharpe exit, Kex = 1 { from table }

Velocity of water in the pipe, V = Q / A = Q / ( $\pi$ / 4) * d2

$\therefore$ V = 0.891204 / ( $\pi$ / 4) * 0.33552

V = 10.08 ft/s

Thereforer, Reynolds number of the flow will be,

Re = ( V * d ) / $\nu$

Re = ( 10.08 * 0.3355 ) / 1.08 * 10-5

Re = 313163.3756

Surface roughness of commercial steel pipe, $\varepsilon$ = 0.0002296588 ft { from data table }

Therefore, Relative roughness is, $\varepsilon$ / d = 0.0002296588 / 0.3355 = 6.845 * 10-4

Using Moody's chart for $\varepsilon$ / d = 6.845 * 10-4 and Re = 313163.3756,

friction factor is, f = 0.0191358

Applying Bernouli's equation at point 1 and 2,

P1/$\rho$g + z1 + V12/2g + Hp = P2/$\rho$g + z2 + V22/2g + hL

where, P1/$\rho$g = P2/$\rho$g = Atmospheric pressure

z1 = 0

V12/2g = V22/2g = 0,{ As water is stationary in the tank }

z2 = 190ft

Hp = pump head

hL = total head loss = major loss + minor loss = hmj + hmn

Therefore,

hmj = ( f * L * V2 ) / ( 2 * g * d ) { L = 80 + 120 = 200 ft }

hmj = ( 0.0191358 * 200*10.082 ) / ( 2 * 32.174* 0.3355 )

hmj = 18.0123 ft

And, minor loss will be,

hmn = $\sum$ K * V2 / 2g

hmn = [ 0.5 + ( 5 * 0.3 ) + 10 + 1 ] * 10.082 / ( 2 * 32.174 )

hmn = 20.5353 ft

Therefore, above equation reduces to,

Hp = z2 + hL

Hp = 190 + 18.0123 + 20.5353

Hp = 228.5476 ft

2) Power delivered to the water, $\dot{W}$

$\dot{W}$ = $\rho$ * g * Q* Hp

$\dot{W}$ = 62.4 * 32.174 * 0.891204 * 228.5476

$\dot{W}$ = = 408924.79 lbf.ft/s

3) Power delivered to the shaft if pump motor is 55% efficient,

Pump efficiency,

$\eta _{pump}$ = $\dot{W}$ / P

$\therefore$ P = $\dot{W}$ / $\eta _{pump}$

P = 408924.79 / 0.55

P = 743499.62 lbf.ft/s