Question

Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW of energy to the water and the head loss of the flow is 10 m. Determine the velocity of the water leaving the pump and discharging into tank B.

Answer 1

Complete Question

Complete Question is attached below.

Answer:

$V'=5m/s$

Explanation:

From the question we are told that:

Diameter $d=0.10m$

Power $P=4.0kW$

Head loss $\mu=10m$

 $\frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu$

 $\frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10$

 $H_m=(\frac{200*10^3}{1000*9.8}-10)$

 $H_m=10.39m$

Generally the equation for Power is mathematically given by

 $P=\rho gQH_m$

Therefore

 $Q=\frac{P}{\rho g H_m}$

 $Q=\frac{4*10^4}{1000*9.81*10.9}$

 $Q=0.03935m^3/sec$

Since

 $Q=AV'$

Where

 $A=\pi r^2\\A=3.142 (0.05)^2$

 $A=7.85*10^{-3}$

Therefore

 $V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}$

 $V'=5m/s$