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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.

Respuesta :

Answer:

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

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

[tex]z_A -z_B = \frac{v^{2} }{2g}(1+4f\frac{l}{d} )[/tex]

[tex]6.2 = \frac{v^{2} }{2\cdot9.81}(1+4\cdot0.0125\frac{27}{0.05} )[/tex]

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.

Answer:

The mass discharge rate is 15.2 kg/s

Explanation:

The mechanical energy balance:

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

[tex]m=\rho uA=1600u^{2} (\frac{\pi 0.5^{2} }{4} )=3.14u^{2}[/tex]

[tex]F=(K_{ent}+K_{pipe}+K_{exit} )\frac{u^{2} }{2g_{c} } \\K_{ent}=\frac{160}{Re} +0.5\\K_{exit}=1\\F=(1.5+\frac{160}{Re} +2160f )\frac{u^{2} }{2g_{c} }[/tex]

From balance equation:

[tex]-\frac{u^{2} }{2} +(9.8*(-6.2))+F=(1.5+\frac{160}{Re}+2160f )\frac{u^{2} }{2g_{c} } =\frac{1000}{3.14u^{2} }[/tex]

The Reynolds number:

[tex]Re=\frac{du\rho }{u} =\frac{0.05u*1600}{1.8x10^{-3} } =4.44x10^{4} u^{2} =4.44x10^{4} *4.8=214889[/tex]

[tex]f=0.079Re^{-1/4} =0.0039[/tex]

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

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