Question

A 31.5 % hydrochloric acid solution is pumped from one storage tank to another. The power input to the pump is 2 kW and is 50% efficient. The pipe is plastic PVC pipe with an internal diameter of 50 mm. At a certain time, the liquid level in the first tank is 4.1 m above the pipe outlet. Because of an accident, the pipe is severed between the pump and the second tank, at a point 2.1 m below the pipe outlet of the first tank. This point is 27 m in equivalent pipe length from the first tank.

Compute the flow rate (in kg/s) from the leak. The viscosity of the solution is 1.8x10^(-3) kg/m.s, and the density is 1600 kg/m^3.

Answer 1

The mass discharge rate is 15.2 kg/s

Step-by-step explanation:

The mechanical energy balance:

`(deltaP)/(\rho ) +(deltau^(2) )/(2\alpha g_(c) ) +(g*deltaz)/(g_(c) ) +F=-(W)/(m) , deltaP=0(net pressure arised),\alpha =1\\(deltau^(2) )/(2\alpha g_(c) ) +(g*deltaz)/(g_(c) ) +F=-(W)/(m) ,deltaz=4.1+2.1=6.2m\\W=2*0.5*1000*1=1000J`

`m=\rho uA=1600u^(2) ((\pi 0.5^(2) )/(4) )=3.14u^(2)`

`F=(K_(ent)+K_(pipe)+K_(exit) )(u^(2) )/(2g_(c) ) \\K_(ent)=(160)/(Re) +0.5\\K_(exit)=1\\F=(1.5+(160)/(Re) +2160f )(u^(2) )/(2g_(c) )`

From balance equation:

`-(u^(2) )/(2) +(9.8*(-6.2))+F=(1.5+(160)/(Re)+2160f )(u^(2) )/(2g_(c) ) =(1000)/(3.14u^(2) )`

The Reynolds number:

`Re=(du\rho )/(u) =(0.05u*1600)/(1.8x10^(-3) ) =4.44x10^(4) u^(2) =4.44x10^(4) *4.8=214889`

`f=0.079Re^(-1/4) =0.0039`

The discharge velocity is 4.83 m/s and the mass discharge rate is 15.2 kg/s

Answer 2

The flow rate (in kg/s) from the leak is 130.973 kg /s

Step-by-step explanation:

Here we have

Power of the pump given by

P = (ρ·g·Q·H)/η

Where:

ρ = Fluid density = 1600 kg/m³

q = Acceleration due to gravity = 9.81 m/s²

Q = Pump flow rate

H = Pump head =

η = Efficiency of the pump = 0.5

Therefore Q·H = P×η/(ρ·g)

Q·H = 6.37 ×10⁻⁵

By Bernoulli equation, we have

P₁ +0.5ρv₁²+ρgh₁ = P₂ +0.5ρv₂²+ρgh₂

The suction head = 4.1 + 2.1 = 6.2 m

The roughness for PVC is 0.00425×10⁻³ m Average

Relative roughness = k/d = 0.00425×10⁻³/0.05 = 8.5×10⁻⁵

From the Moody chart, the friction factor is approximately 0.0125

Therefore the head loss is

hf = f (L/D) × (v²/2g)

The velocity of flow from the tank is

`z_A -z_B = (v^(2) )/(2g)(1+4f(l)/(d) )`

`6.2 = (v^(2) )/(2\cdot9.81)(1+4\cdot0.0125(27)/(0.05) )`

v = 2.084 m/s

Therefore head loss in pipe = 5.979 m

Suction head = 6.2 -5.979 = 0.2214 m

Pump head = Discharge head - Suction head = -0.2214 m

Since the pump has a power capacity of 1 kW = 1 kJ/s

The pump is able to pump m*g*d/t = 1 kJ/s

Here d = distance or head = 1 - 0.2214 = 0.77857

t = 1 s

g = 9.81 m/s²

Therefore the mass of flow rate of the liquid from the leakage is

130.973 kg /s.