Answer
a) $\eta=66.88%$
b) $\Delta P=147.14kPa$
Work Step by Step
Here we are working with potential energy:
The potential work is: $W_{p}=mgh$
And the potential power is: $P_{p}=\frac{W_{p}}{t}=\frac{mgh}{t}=\rho Qgh$
The potential power needed is:
$P_{p}=1000\frac{kg}{m^3}*(70\frac{L}{s}*\frac{1m^3}{1000L})*9.81\frac{m}{s^2}*15m=10.3kW$
Then the overall efficiency is:
$\eta=\frac{10.3kW}{15.4kW}=0.6688$
$\eta=66.88%$
For the pressure:
$P_{p}=Q\Delta P$
Then:
$\Delta P=\frac{P_{p}}{Q}=\frac{10.3kW}{70\frac{L}{s}*\frac{1m^3}{1000L}}=147.14kPa$
