fluid dynamics power by pump and pressure at outlet of hose

fluid dynamics.... power by pump and pressure at outlet of hose.

Consider a hose that can supply a chemical propellant to an elevated platform. The nozzle at the end of the hose will not operate unless a minimum pressure of 140 kPa (above atmospheric) is available. The hose consists of a smooth plastic tubing with an internal diameter of 25 nun. The chemical propellant has a specific gravity (S.G.) of 1.10 and a dynamic viscosity of 2 times 10^-3 Pas. If the length of the hose is L = 85 m, determine: The power delivered by the pump to the propellant. The pressure at the outlet of the pump. You may ignore all minor losses and assume that the volumetric flow rate is 95 L/min.

Solution

Power delivered by pump to propellant P_Hydraulic(W) = q g h - Watt

Where

P_hydraulic(W) = hydraulic power (W)

q = flow capacity (m³/s)

Given q = 95L/min = 95*0.001m³/60s =1.58*10^-3 m³/s

= density of fluid (kg/m³) = Sg *1000 = 1.10 *1000 = 1100 kg/m³

g = gravity (9.81 m/s²)

h = differential head (m)

d = diameter of hose = 0.025m

A = Area of hose = (pi d²/4) =(3.14 * 0.025²/4) = 4.90 *10^-4 m²

Now first find ignoring minor losses by Bernoulli’s equation

h = differential head = (pressure head + datum head + kinetic head + head losses) _exit -----equation 1

Pressure head at exit of hose= (P₂/ g) = (140 *1000 /1100*9.81) = 12.97m

Datum head at exit of hose= Z₂ - Z₁= 10- 1.5-1.2m = 7.3m

Velocity of fluid in hose = q/A = 1.58 *10^-3 / 4.90 *10^-4 = 3.22m/s

Kinetic head = (V²/2g) = (3.22²/2*9.81) = 0.528 m

Kinetic head will not change from entry to exit as there is no variation of area.

Head losses H_f = (fl V²/2gd)

Friction factor f = 64/Reynolds number

Dynamic viscosity coefficient = 2*10^-3 Pa s

Reynolds No = ( V d/ dynamic viscosity coefficient) = (1100 *3.22 * 0.025/2*10^-3) = 44275

f = 64/44275 = 0.001445

l = length of hose = 85 m

H_f = (fl V²/2gd) = (0.111445 *85* 3.22²/2*9.81*0.025) =2.59m

Substitute the above values in the equation 1.

h = pressure head + datum head + velocity head + head losses

   = 12.97 + 7.3 + 0.528 + 2.59

    = 23.388 m

Power delivered by pump to propellant P_Hydraulic(W) = 1.58*10^-3 * 1100 *9.81* 23.388

                                                                                      = 397.90W

Pressure at the outlet of pump P₂ =?

Use Bernoulli’s equation between outlet of pump and outlet of hose.

(Pressure head + datum head + velocity head) _{inlet =} h

Pressure head at inlet = (P₂/ g) = (P₂ /1100*9.81) = P₂ /10791 m

Velocity head = (V²/2g) = (3.22²/2*9.81) = 0.528 m²

Datum head at inlet= Z₁ = 1.5m

(Pressure head + datum head + velocity head) _{inlet =} h

(P₁ /10791) +0.528 +1.5 = 23.388

P₁ = 230495 Pa = 230 kPa

Pressure at outlet of pump = 230 kPa