Water flows from a reservoir in a horizontal pipe in a power plant as shown. If the pipe diameter is 20 cm, the friction factor is 0.015. and the velocity by the pipe is 1.0 m/sec. What is the pressure in the pipe at point 2. Assume kinetic energy correction factors of 1.0.

Here, E = Kinetic energy + Potential Energy 

$E=\frac{1}{2}m{V}^2+mgh$
$p_w=1000 kg/m^3$

$Q=50 m^3/sec$

V= 3m /sec 

h = 90m 

Total mechanical energy per unit mass, 

$\frac{E}{m}=\frac{1}{2}\times V^2+gh=\frac{1}{2}\times 3^2+9.81\times 90$

$= 887.4J / kg\left(Ans\right)$

Power generation potential, 

$P=\frac{E}{m}\times Q\times {\rho }_w=887.4\times 50\times 1000$

$=44.37\times {10}^{6}W=44.37MW\left(Ans\right)$

Here, 

$A=25k{m}^2=25\times {10}^6{m}^2$
$V=1.5\times {10}^6{m}^3$
i = 10cm = 0.1m

Solution: 

Effective rainfall  $=\frac{V}{A}=\frac{1.5\times {10}^6}{25\times {10}^6}=0.06m=6cm$

Runoff coefficient $= \frac{6}{10} = 0.6\left(Ans\right)$