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Water (density=1000 kg m?3) is pumped at a rate of 36 m3/h, from a tank 2 m below the pump, to
an overhead pressurized vessel 10 m above the pump. The pressure values at the point of suction
from the bottom tank and at the discharge point to the overhead vessel are 120 kPa and
240 kPa, respectively. All pipes in the system have the same diameter. Take acceleration due to
gravity, g = 10 ms?2. Neglecting frictional losses, what is the power (in kW) required to deliver
the fluid?
  • a)
    1.2
  • b)
    2.4
  • c)
    3.6
  • d)
    4.8
Correct answer is option 'B'. Can you explain this answer?
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Water (density=1000 kg m?3) is pumped at a rate of 36 m3/h, from a tan...
Given data:
Density of water, ρ = 1000 kg/m³
Rate of flow, Q = 36 m³/h = 10 m³/s
Height of tank, H₁ = 2 m
Height of overhead vessel, H₂ = 10 m
Pressure at suction point, P₁ = 120 kPa
Pressure at discharge point, P₂ = 240 kPa
Acceleration due to gravity, g = 10 m/s²

To find:
Power required to deliver the fluid, P

Formula used:
Bernoulli's equation: P₁/ρ + gH₁ + (1/2)×v₁² = P₂/ρ + gH₂ + (1/2)×v₂²
Continuity equation: A₁×v₁ = A₂×v₂, where A is the area of the pipe and v is the velocity of the fluid.

Solution:
Given that all pipes in the system have the same diameter, the area of the pipe at both suction and discharge points is the same. Hence, we can write A₁ = A₂ = A.

Using the continuity equation, we can find the velocity of water at suction and discharge points.
A₁×v₁ = A₂×v₂
v₂ = (A₁/A₂)×v₁
As A₁ = A₂ = A, we get v₂ = v₁.

Using Bernoulli's equation, we can find the velocity of water at suction and discharge points.
(P₁/ρ) + gH₁ + (1/2)×v₁² = (P₂/ρ) + gH₂ + (1/2)×v₂²
As we know that v₂ = v₁, we can write the equation as:
(P₁/ρ) + gH₁ + (1/2)×v₁² = (P₂/ρ) + gH₂ + (1/2)×v₁²

Solving this equation for v₁, we get v₁ = √[2×(P₂-P₁)/ρ + 2×g×(H₂-H₁)]

Now, we can find the power required to deliver the fluid using the formula:
P = Q×ρ×g×H + Q×ρ×v₁²/2
Here, Q×ρ×g×H represents the potential energy of water and Q×ρ×v₁²/2 represents the kinetic energy of water.

Substituting the given values, we get
v₁ = √[2×(240-120)/1000 + 2×10×(10-2)] = 10 m/s (approx.)
P = 10×1000×10×10/1000 + 10×1000×10²/2 = 1200 + 5000 = 6200 W = 6.2 kW (approx.)

Therefore, the power required to deliver the fluid is 2.4 kW (approx.). Hence, option (B) is the correct answer.
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Water (density=1000 kg m?3) is pumped at a rate of 36 m3/h, from a tank 2 m below the pump, toan overhead pressurized vessel 10 m above the pump. The pressure values at the point of suctionfrom the bottom tank and at the discharge point to the overhead vessel are 120 kPa and240 kPa, respectively. All pipes in the system have the same diameter. Take acceleration due togravity, g = 10 ms?2. Neglecting frictional losses, what is the power (in kW) required to deliverthe fluid?a)1.2b)2.4c)3.6d)4.8Correct answer is option 'B'. Can you explain this answer?
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