Question

Problem 2: Water at 10°C is pumped between two reservoirs, with and elevation increase of 15 m, in a pipeline with the following characteristics: D= 300 mm, L = 70 m, f = 0.025 and for minor losses, K_ = 2.5. The pump curve is approximated by the formula AH, = 22.9-1119 where AH, is in meters and Q is in m/s. Determine the volume flow rate and operating head if a single pump is used.

Answer 1

Given data 1

ΔΗ, 22.9 - 111Q

KL 2.5

$D=0.3\textup{ m}$

L= 70 m

$f=0.025$

$z_{2}-z_{1}=15\textup{ m}$

$\textbf{Solution}\rightarrow$

$\textup{Apply Bernoulli's equation to the pump}$

$\frac{P_{1}}{\rho g}+\frac{V_{1}^{2}}{2g}+z_{1}+\Delta H_{p}=\frac{P_{2}}{\rho g}+\frac{V_{2}^{2}}{2g}+z_{2}+h_{f}$

$z_{1}+\Delta H_{p}=\frac{V_{2}^{2}}{2g}+z_{2}+h_{f}$

$\Delta H_{p}=\frac{V_{2}^{2}}{2g}+\left ( z_{2}-z_{1} \right )+h_{f}$--------------------------------1

$h_{f}=\frac{fLV^{2}}{2gD}+\sum K_{L}\frac{V^{2}}{2g}$

$h_{f}=\frac{fL\left ( \frac{Q}{A} \right )^{2}}{2gD}+\sum K_{L}\frac{\left ( \frac{Q}{A} \right )^{2}}{2g}$

$A=\frac{\pi }{4}*0.3^{2}=0.070685\textup{ m}^{2}$

$h_{f}=\frac{fL\left ( \frac{Q}{A} \right )^{2}}{2gD}+\sum K_{L}\frac{\left ( \frac{Q}{A} \right )^{2}}{2g}$

$h_{f}=\frac{0.025*70*\left ( \frac{Q}{0.070685} \right )^{2}}{2*9.81*0.3}+2.5*\frac{\left ( \frac{Q}{0.070685} \right )^{2}}{2*9.81}$

$h_{f}=85Q^{2}$

$\Delta H_{p}=\frac{V_{2}^{2}}{2g}+\left ( z_{2}-z_{1} \right )+h_{f}$

$22.9-111Q^{2}=\frac{\left ( \frac{Q}{0.070685} \right )^{2}}{2*9.81}+15+85Q^{2}$

$22.9-111Q^{2}=10.20Q^{2}+15+85Q^{2}$

$22.9-15=111Q^{2}+10.20Q^{2}+85Q^{2}$

$Q^{2}=0.03831$

$Q=0.1957\textup{ m}^{3}\textup{/s}$------------------------------- Ans

ΔΗ, 22.9 - 111Q

$\Delta H_{p}=22.9-111*0.03831$

$\Delta H_{p}=18.6473\textup{ m}$------------------------------- Ans