Respuesta :
Answer:
A. 50.93 m/s
B. 201 kW
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
The required shaft power input to the pump and the water discharge velocity are;
Input Power = 199 Kw
Discharge velocity = 50.93 m/s
We are given;
Nozzle exit diameter; d = 5 cm = 0.05 m
Nozzle exit Radius; r = d/2 = 0.05/2 = 0.025 m
Formula for the cross sectional area of the nozzle is;
A = πr²
A = π × 0.025²
A = 1.9635 × 10⁻³ m²
We are given rate of flow; Q = 0.1 m³/s
To get the speed that the water will pass through the nozzle, we will use the formula;
v = Q/A
v = 0.1/(1.9635 × 10⁻³)
v = 50.93 m/s
From conservation of energy, we know that;
Potential energy = Kinetic energy
Thus;
mgh = ¹/₂mv²
m will cancel out and making h the subject gives us;
h = v²/2g
where h is the required head of water
Plugging in the relevant values gives;
h = 50.93²/(2 × 9.8)
h = 132.34 m
We are given;
Total irreversible head loss of the system = 3 m
Position of nozzle above sea level = 3 m
Now, the formula for total head the pump required is;
Total head required by the pump = Total irreversible head loss of the system + the position of the nozzle above sea level + head of water required)
Thus;
Total head required by the pump = (3 + 3 + 132.34)
Total head required by the pump = 138.34 m
mass of water pumped can be gotten from the formula;
m = ρV
where ρ is density of water = 1030 kg/m
Volume per second; V = 0.1m³/
Thus;
mass of water per second;
m = 1030 × 0.1
m = 103 kg/s
Thus;
Potential energy is;
PE = mgh
But the mass is in mass rate and as such the result will be the power. Thus;
Power = (103 × 9.8 × 138.34)
Power = 139.297 KW
We are given pump efficiency = 70% = 0.7
Thus, the shaft power input to the pump and the water discharge velocity is; P = 139.297/0.7
P = 199 Kw
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