Question

The pressure (gauge) at a point in water is 150 kPa. The velocity at that point is 10 m/sec and the point is 5 m above the datum. The total energy of the water per Newton weight of the water is $(Takeg=10m/se{c}^2)$

• A. 25

Answer 1

The correct option is A 25
Total head = Total energy/ Newton weight of water

$=\frac{p}{\rho g}+z+\frac{V^2}{2g}$

$=\frac{150000}{1000\times 10}+5+\frac{(10{)}^2}{2\times 10}$

$=15+5+5$

$=25m(J/N=\frac{N-m}{N}=m)$