Water at 20 degree C is pumped at a steady rate from a well

Water at 20 degree C is pumped at a steady rate from a well to a reservoir elevated 5 m above the water level in the well. Determine the flow rate if it is known that the pump labeled P supplies a head increase of 7m.

Solution

For threaded elbow, minor loss coeff = 1.5

Total number of elbows = 3 (1 in inlet, 2 in outlet)

Pipe entrance loss coeff = 0.78

Pipe exit loss coeff = 1

For 50 mm pipe, K₁ = 0.78 + 1.5 = 2.28

For 100 mm pipe, K₂ = 2*1.5 + 1 = 4

Total inlet pipe length L₁ = 3+5 = 8 m

Total outlet pipe length L₂ = 5+2+15 = 22 m

Inlet Pipe\'s cross-section area A₁ = pi/4 * d² = 3.14/4 *50² = 1962.5 mm²

Outlet Pipe\'s cross-section area A₂ = pi/4 * d² = 3.14/4 *100² = 7850 mm²

Flow rate Q = Ain*V₁ = Aout*Vout

V_out = (1962.5 / 7850)*V₁

V_out = 0.25*V₁

Taking station 1 as free surface of well inside the 3 m pipe and station 2 at the free surface of reservoir and datum at free surface of well, we have

P₁ = P₂ = Patm

V₂ = 0 since very large surface area of reservoir

P₁ / (rho*g) + V₁² / (2g) + z₁ + H_p = P₂ / (rho*g) + V₂² / (2g) + z₂ + (K₁ + f₁L₁/d₁) V₁² / (2g) + (K₂ + f₂L₂/d₂) V_out² / (2g)

V₁² / (2g) + 0 + 7 = 0 + 5 + (2.28 + f₁*8/0.05) V₁² / (2g) + (4 + f₂*22/0.1) V_out² / (2g)

2 = (1.28 + 160*f₁) V₁² / (2g) + (4 + 220*f₂) (0.25*V₁)² / (2g)

2 = (1.28 + 160*f₁) V₁² / (2g) + (0.25 + 13.75*f₂) V₁² / (2g)

2 = (1.53 + 160*f₁ + 13.75*f₂) V₁² / (2g)

2 = (1.53 + 160*f₁ + 13.75*f₂) V₁² / (2*9.81)

(1.53 + 160*f₁ + 13.75*f₂) V₁² = 39.24...............eqn1

Now, assume in 50 mm dia pipe, Re = 10⁶,

From Moody diagram, at Re = 10⁶ for smooth pipe, we have friction factor f₁ = 0.0118

Re = V₁*d₁ / neu

10⁶ = V₁ * 0.05 / (1.004*10^-6)

V₁ = 20.08 m/s

V_out = 0.25*20.08 = 5.02 m/s

In 100 mm dia pipe, Re = V_out * d₂ / neu

Re = 5.02 * 0.1 / (1.004*10^-6)

Re = 5*10⁵

From Moody diagram, at Re = 5*10⁵ for smooth pipe, we have friction factor f₂ = 0.013

Now putting it in left hand side of eqn 1, we get

(1.53 + 160*0.0118 + 13.75*0.013) * 20.08² = 1450.2

Since 1450.2 is not close to 39.24 our guess of Re was not close.

Assume in 50 mm pipe, Re = 1.5*10⁵

From Moody diagram, at Re = 1.5*10⁵ for smooth pipe, we have friction factor f₁ = 0.016

Re = V₁*d₁ / neu

1.5*10⁵ = V₁ * 0.05 / (1.004*10^-6)

V₁ = 3.01 m/s

V_out = 0.25*3.01 = 0.753 m/s

In 100 mm dia pipe, Re = V_out * d₂ / neu

Re = 0.753 * 0.1 / (1.004*10^-6)

Re = 0.75*10⁵

From Moody diagram, at Re = 0.75*10⁵ for smooth pipe, we have friction factor f₂ = 0.028

Now putting it in left hand side of eqn 1, we get

(1.53 + 160*0.016 + 13.75*0.028) * 3.01² = 40.54

Since 40.54 is close to 39.24 our guess of Re is close.

Hence, we have V1 = 3.01 m/s

Flow rate Q = V1*A1

= 3.01*(1962.5*10^-6)

= 0.0059 m3/s