Question

A fireboat is to fight fires at coastal areas by drawing seawater with a density of 1030 kg/m through a 20-cm-diameter pipe at a rate of 0.1 m /s and discharging it through a hose nozzle with an exit diameter of 5 cm. The total irreversible head loss of the system is 3 m, and the position of the nozzle is 3 m above sea level. For a pump efficiency of 70 percent, determine the required shaft power input to the pump and the water discharge velocity.

Answer 1

A. 50.93 m/s

B. 201 kW

Step-by-step explanation:

We have that pump is moving 0.1m³ of water every second. The nozzle is 5cm diameter = 0.05m , or 2.5cm radius. 2.5cm = 0.025m. The cross sectional area of the nozzle is therefore, A = πr²

π×0.025²

= 0.001963m²

Therefore, for 0.1m³ of water to pass through this nozzle in 1 second, it must be travelling at

0.1 / 0.001963

= 50.93 m/s

Pump is meant to impart energy to the water to raise it 4m from sea level to the nozzle and accelerate it from zero to 50.93m/s through the nozzle. Now, assuming no losses, the dynamic energy of the water is equal to the potential energy it would have if it was stationary at some height h.

P.E = k.E

mgh = 0.5mv²

h = v² / 2g

h = 50.93² / (2 ×9.81)

h = 132.21m

So the equivalent head of water required due to accelerating the water is 132.21m. The static head is 4m since the nozzle is 4m above the sea and there is a further loss of 3m of head in the system due to friction etc. So the total head the pump must provide is

132.21 + 4 + 3

= 139.21m

Density = mass / volume so mass = density × volume.

The Pump is moving 0.1m³ per second of water. Therefore in one second the mass of water pumped is

0.1 × 1030 = 103kg

The total energy transferred from the pump to the water is therefore

Potential enegry, which is = mgh

= 103 ×9.81 × 139.21

= 140,662.0 Nm/s

= 140,662.0 Watts

But, we are told that the pump is only 70% efficient. Which implies

140,662 = 0.7P where P is the power supplied to the pump

so P = 140,662 / 0.7

= 200,946 Watts

= 201 kW