Fluid Mechanics and Heat Transfer Exercises Reference Answers English1 Water is pumped at a constant velocity 1ms from large reservoir resting on the floor to the open top of an absorption tower The p

1. From the Bernoulli's equation, we have:

P1/ρ + V1^2/2g + Z1 = P2/ρ + V2^2/2g + Z2 + hL

where: P1/ρ = pressure head at the inlet (reservoir) V1 = velocity at the inlet (reservoir) Z1 = elevation head at the inlet (reservoir) P2/ρ = pressure head at the outlet (tower) V2 = velocity at the outlet (tower) Z2 = elevation head at the outlet (tower) hL = head loss due to friction

Assuming that the pressure at the outlet (tower) is atmospheric (zero gauge pressure), we have:

P1/ρ + V1^2/2g + Z1 = Z2 + hL

Solving for Z1, we have:

Z1 = Z2 + hL - V1^2/2g - P1/ρ

Substituting the given values, we have:

Z1 = 4 + 30/9.81 - 1^2/2*9.81 - 60/9.81/1200

Z1 = 4.67 meters

Therefore, the water level in the reservoir must be kept at 4.67 meters above the floor.

2. From the Bernoulli's equation, we have:

P1/ρ + V1^2/2g + Z1 = P2/ρ + V2^2/2g + Z2 + hL

where: P1/ρ = pressure head at the inlet (reservoir) V1 = velocity at the inlet (reservoir) Z1 = elevation head at the inlet (reservoir) P2/ρ = pressure head at the outlet (evaporator) V2 = velocity at the outlet (evaporator) Z2 = elevation head at the outlet (evaporator) hL = head loss due to friction

Assuming that the pressure at the inlet (reservoir) is atmospheric (zero gauge pressure), we have:

P1/ρ + V1^2/2g + Z1 = P2/ρ + V2^2/2g + Z2 + hL

Solving for V1, we have:

V1 = sqrt((P2 - P1)/ρ + 2g(Z2 - Z1 - hL))

Substituting the given values, we have:

V1 = sqrt((200/760 - 1)/1200 + 29.8115 - 120)

V1 = 4.025 m/s

The pump effective work is given by:

Wp = ρQgΔh

where: ρ = fluid density Q = volumetric flow rate g = gravitational acceleration Δh = total head difference

Solving for Δh, we have:

Δh = Wp/(ρQg)

Substituting the given values, we have:

Δh = 2010^3/(12009.81*60/1000)

Δh = 27.1 meters

The pump power is given by:

Pp = ρQgΔh/η

where: η = pump efficiency

Assuming an efficiency of 80%, we have:

Pp = 1200209.81*27.1/0.8/1000

Pp = 63.4 kW

Therefore, the pump effective work is 27.1 meters and the pump power is 63.4 kW.

3. From the given equation, we have:

hf = 6.5U^2

where: hf = head loss due to friction U = velocity in the pipe

Solving for U, we have:

U = sqrt(hf/6.5)

Substituting the given values, we have:

U = sqrt(120/6.5)

U = 6.19 m/s

The water flow rate is given by:

Q = π/4D^2U

where: D = pipe diameter

Substituting the given values, we have:

Q = π/4*(108/1000)^26.193600

Q = 0.268 m3/h

Therefore, the water velocity at section A-A' is 6.19 m/s and the water flow rate is 0.268 m3/h.

4. From the Bernoulli's equation, we have:

P1/ρ + V1^2/2g + Z1 = P2/ρ + V2^2/2g + Z2 + hL

where: P1/ρ = pressure head at the inlet (section A) V1 = velocity at the inlet (section A) Z1 = elevation head at the inlet (section A) P2/ρ = pressure head at the outlet (section B) V2 = velocity at the outlet (section B) Z2 = elevation head at the outlet (section B) hL = head loss due to friction

Assuming that the pressure at the outlet (section B) is atmospheric (zero gauge pressure), we have:

P1/ρ + V1^2/2g + Z1 = Z2 + hL

Solving for Z2 - Z1, we have:

Z2 - Z1 = hL - V1^2/2g

Substituting the given values, we have:

Z2 - Z1 = 15 - 2.5^2/2*9.81

Z2 - Z1 = 7.07 meters

Therefore, the water column difference between two glass tubes is 7.07 meters. The relative liquid column height of two tubes in the Fig is shown below:

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   |___________ |  7.07 m
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   |____________|  0