Question

(b) Water flows under gravity between two reservoirs through a pipe of length 5000m. The diameter of the pipe is 0.2 m and the roughness size is 0.04 mm. The water levels in the two reservoirs are maintained with a difference of 50 m. Determine the discharge through the pipe. Neglect all minor losses. Use the attached Moody diagram for estimation of friction factor. (10 marks) (c) In (b), now include entry loss at the upper reservoir with loss coefficient at 0.5, exit loss at the lower reservoir, and loss due to a slightly closed valve along the pipe with loss coefficient at 80.0. Determine the discharge through the pipe, taking into account all these minor losses. (5 marks)

Answer 1

Ans) Apply Bernoulli equation between water surface of two reservoirs as :

P₁/$\rho$g + V1²/2g + Z1 = P₂/$\rho$g + V2²/2g + Z2 + H_f + H_m

where, P1 and P2 are pressures at point 1 and 2 respectively. Since , the reservoirs are open to atmosphere , pressue is only atmospheric pressure, hence gauge pressure P1 = P2 = 0

Also, velocities V1 and V2 at surface = 0

Z1 and Z2 are datum head. Let point 2 be datum , then

Z1 = 50 m and Z2 = 0

H_f = head loss due to friction

a) Since minor loss has to neglected, therefore, above equation reduces to,

0 + 0 + 50 = 0 + 0 + 0 + H_f

=> H_f =f L V²/ D(2g) = 50

where, f = Darcy friction factor

L = length of pipe

V = velocity of flow

D =diameter of pipe

To determine 'f' , let us find Reynold number (Re)

Re = $\rho$ V D/ $\mu$

where, $\mu$ = dynamic viscosity of water = 0.001 Ns/m²

Re = 1000 x V x 0.2 / 10^-3

= 2V x 10⁵

According to Moody diagram, approx value of friction factor can be determined by 'Re' and relative roughness = e/D

= 0.00004 / 0.2 = 0.0002

For Re = 2 x 10⁵ and e/D = 0.0002 , 'f' = 0.016

=> 0.016 x 5000 x V² / (0.2 x 2 x 9.81) = 50

=> V = 1.566 m/s

Discharge = Area x velocity

= ($\pi$/4) x 0.2 x 0.2 x 1.566

= 0.049 m³/s

Ans b) Now, if minor loss also be considered then,

H_f + H_m = 50 [from part (a)]

Friction loss + loss due to entry + loss due to exit + loss due to valve = 50

=> f L V²/ D(2g) + Ke V²/2g + Kv V² /2g = 50

=> V²/2g ( 0016 x 5000/0.2 + 0.50 + 80) = 50

=> V² = 0.104 x 2 x 9.81

=> V= 1.428 m/s

Discharge = area x velocity

= ($\pi$/4) x 0.2 x 0.2 x 1.428

= 0.0448 m³/s