Question

A pump increases the water pressure from 70 kPa at the inlet to 700 kPa at the outlet. Water enters this pump at 15 °C through a 1-cm-diameter opening and exits through a 1.5-cm-diameter opening. Determine the velocity of the water at the inlet and outlet when the mass flow rate through the pump is 0.5 kg/s. How much work is required to operate the pump?

Answer 1

Answer:

$v_{in} =6.373 \frac{m}{s}$

$v_{out} =2.832 \frac{m}{s}$

$W=0.63063*0.5=0.3153kW$

Explanation:

First, we are going to need the water specific volume at 15ºC: v=0.001001 $\frac{m^{3}}{kg}$. The density '$p$' of the water is the inverse of the specific volume: $p=999 \frac{kg}{m^{3} }$

First, consider the mass flow, which is related to the volumetric flow (density and velocity) and the area:

$m=pvA$

The area of each cross-section is:

$A_{in} =\frac{\pi*0.01^{2} }{4}=7.854*10^{-5}$ (in square meters). Here, the radius was not used but the diameter, which means a division by 4 (2 squared).

$A_{out} =\frac{\pi*0.015^{2} }{4}=1.767*10^{-4}m^{2}$

From mass flow isolate the velocity and calculate it:

$v=\frac{m}{pA}$

$v_{in} =\frac{0.5}{999*7.854*10^{-5} }=6.373 \frac{m}{s}$

$v_{out} =\frac{0.5}{999*1.767*10^{-4} }=2.832 \frac{m}{s}$

The work of the pump is calculated considering an energy balance on the pump:

$w=h_{in}-h_{out}$

Considering the isentropic process may give us the relation:

$dh=vdP\\ h_{2} -h_{1}=v*(P_{2} -P_{1})$

Applying that to the pump,

$w=-0.001001*(700-70)=-0.63063 \frac{kJ}{kg}$

Multiplying it by the mass flow:

$W=-0.63063*0.5=-0.3153kW$

The work is negative because it is entering to the system, but the required is positive. (It is just a standard rule)