Question 10.2.2: A single acting reciprocating pump is to raise a liquid of d......
A single acting reciprocating pump is to raise a liquid of density 1200 kg/m³ through a vertical height of 11.5 m, from 2.5 m below pump axis to 9 m above it. The plunger moves with simple harmonic motion, has diameter 125 mm and stroke 225 mm. The suction and delivery pipes are of 75 mm diameter and 3.5 and 13.5 m long respectively. There is a long vessel placed on the delivery pipe near the pump exit but there is no air vessel on the suction pipe. If separation takes place 0.88 bar below atmospheric pressure, find: (i) maximum speed with which the pump can run without separation taking place, (ii) power required to drive the pump, if f = 0.02. Neglect slip for pump.
Given:
ρ = 1200, H = 11.5 m, H_s = 2.5 m, H_d = 9.0 m, D = 0.125 m, L = 0.225 m
d_d = d_s = 0.075 m, l_s = 3.5 m, l_d = 13.5 m, and f = 0.02
1. Separation of flow is possible when the piston is at θ = 0 for suction stroke and θ = 2π for discharge stroke.
Since air vessel is provided in delivery pipe, separation is likely to occur during suction stroke only. Separation will occur when totalsuction head at entrance to pump is > 0.88 bar = (0.88 × 10^5 )/(1200 × 9.81) = 10.97 m of the liquid. But
H_{as} = (L_s/g) × (A_p/A_s) ω^2 r = (3.5/9.81) × (0.125/.075)^2 × ω^2 × (0.225/2) = 44.49 ω^2
∴ In limiting condition, suction head at θ = 0 = H_s + H_{as} = 44.49 ω^2
⇒ N = 63.39 rpm. This is maximum rpm.
2. Total head to be developed by pump = H + H_{fd} + 2/3 H_{fs}. Factor 2/3 on suction side comes because there is no air vessel on this side and the fictional losses are parabolic, as discussed above. Whereas in delivery pipe, velocity remains constant. For a single acting pump,
Q = π D²L/4 × N = 0.00294 m³/s
→
V_s = V_d = Q/Ad = 0.665 m/s
H_{fd}=\frac{fl_d}{d_d}\frac{V^2_d}{2g} =0.081 m
and
H_{fs}=\frac{fl_s}{2gd_s}\frac{(A\omega r)^2}{A_s}=0.207 m
Maximum friction head occurs at θ = π/2
∴ P = 1200 × 9.81 × 0.00294 (11.5 + 0.081 + 2/3 × 0.207) = 406 kW