Question

40000 litre of oil of density 0.9 g cc is pumped from oil tanker ship into a storage tank at 10m higher level than the ship in half an hour what should be the power of the pump

Answer 1

Answer:

1726.56 Watts.

Explanation:

In the question,

Volume of the oil = 40,000 Litre

Density of oil, d = 0.9 g/cm³

Height of the storage tank, h = 10 m

Time taken, t = 1/2 hr.

Power can be given by,

$P=h\gamma. Q$

where,

Q is the volume flow rate in m³/s.

So,

$Q=\frac{40,000}{1/2}\times \frac{1}{1000}\times \frac{1}{3600}\\Q=0.022\,\frac{m^{3}}{s}$

Now,

$\gamma =\rho. g\\\gamma = 0.8\times 1000\times 9.8\\\gamma=7848\frac{N}{m^{3}}$

So,

On putting all the values we get,

$Power, P=h.\gamma.Q\\P=10\times 7848\times0.022\\P=1726.56\,Watts$

Therefore, the power needed is 1726.56 Watts.

Answer 2

Answer:

Power is 1960.15 Watts

Explanation:

Power=hAγQ

where:

hA= energy added to the fluid [m]

γ= specific weight of the fluid [Nm3]

Q= volume flow rate [m3s].

The energy added to the fluid (hA) is normally calculated using the energy equation which accounts for velocity, elevation, pressure, friction losses, etc. If we assume that there is no change in fluid velocity and no change in fluid pressure and no friction losses, then hA is simply the change in elevation of the fluid

Q=(80000Lhr)/(m31000L)/(hr3600sec)=0.005556m3sec

= 0.022224 m3/sec

Now 0.9 g cc = 900kg/m3

γ = pg= 900 x 9.8 = 8820

Power=hAγQ

             = 10 x 0.022224 x 8820

             = 1960.15 Watts