Answer:
50.93 m/s
199.5 kW
Explanation:
From the question, the nozzle exit diameter = 5 cm, Radius= diameter/2= 5cm/2= 2.5cm. we can convert it to metre for unit consistency= (2.5×0.01)=
0.025m
We can calculate the The cross sectional area of the nozzle as
A= πr^2
A= π ×0.025^2
= 1.9635 ×10^- ³ m²
From the question, the water is moving through the pipe at a rate of 0.1 m /s , then for the water to move through it at a seconds, it must move at
(0.1 / 1.9635 ×10^- ³ m²)
= 50.93 m/s
During the Operation of the pump, the Dynamic energy of the water= potential energy provided there is no loss during the Operation
mgh = 1/2mv²
We can make "h" subject of the formula, which is the height of required head of water
h = (1/2mv²)/mg
h= v² / 2g
h = 50.93² / (2 ×9.81)
h = 132.21m
From the question;
The total irreversible head loss of the system = 3 m,
the given position of nozzle = 3 m
the total head the pump needed=(The total irreversible head loss of the system + the position of the nozzle + required head of water )
=(3 + 3 + 132.21m)
=138.21m
mass of water pumped in a seconds can be calculated since we know that mass is a product of volume and density
Volume= 0.1m³
Density of sea water=1030 kg/m
(0.1 m^3× 1030)
= 103kg
We can calculate the Potential enegry, which is = mgh
= (103 ×9.81 × 138.21)
= 139651.5 Watts
= 139.65kW
To determine required shaft power input to the pump and the water discharge velocity
Energy= efficiency × power
But we are given efficiency of 70 percent, then
139651.5 Watts = 0.7P
=199502.18 Watts
P=199.5 kW
Therefore, the required shaft power input to the pump and the water discharge velocity is 199.5 kW
