Question

For the hydraulic motor shown below, the pump pulls linseed oil from tank A via a 3 ½”

Sch40 steel pipe at a flow rate of 620 L/min. The pump is rated at 30hp and has an efficiency of 80%. If the pipe from the pump outlet to the hydraulic motor is 2” Sch40 steel pipe and the pipe from the motor outlet back to tank A is 3 ½” Sch40 steel pipe and the following data are known:

1.Energy loss from tank A to the pump inlet is 6.2 Nm/N.

2.Energy loss from pump outlet to inlet to hydraulic motor is 1.2 Nm/N.

3.Energy loss from outlet of hydraulic motor to tank A inlet is 2.5 Nm/N.

Ignoring all other losses, compute the following:

1. Pressure at the pump inlet.

2. Pressure at the pump outlet.

3.Pressure at the hydraulic motor inlet.

4.Pressure at the hydraulic motor outlet.

5.Power removed from the fluid by the hydraulic motor

Answer 1

1) pressure at pump inlet=P₂

Apply Bernoulli's equation

$\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_{1-2}$

$p_1=423606N/m^2$

V₁=V₂ hence cancels

z₁=2m, z₂=0(datum)

h_1-2=5.8m

The equation reduces to $\frac{p_1}{\rho g}+z_1=\frac{p_2}{\rho g}+h_{1-2}$

$P_2=[\frac{p_1}{\rho g}+z_1-h_{1-2}]\rho g$

$=[\frac{423606}{1260*9.81}+2-5.8]*1260*9.81$

=376635.62N/m²

2) Pressure at the pump outlet=p₃

By applying Bernoulli's equation;

$p_3=[\frac{p_2}{\rho g}+\frac{V_2^2-V_3^2}{2g}+H_p]\rho g$

$=[\frac{376635.62}{1260*9.81}+\frac{2.097^2-7.596^2}{2*9.81}+138.032]*1260*9.81$

=2049213.921N/m²

3) pressure at the hydraulic motor inlet=p_3'

$p_{3'}=(\frac{p_3}{\rho g}-h_{3-3'})\rho g$ $=(\frac{2049213.921}{1260*9.81}-1.5)*1260*9.81$

=2030673.021N/m²

4) pressure at the hydraulic motor outlet=p₄

$p_4=(\frac{p_{4'}}{\rho g}+h_{4-4'})\rho g$

$p_{4'}=p_B+h_2\rho g$ where p_B=atmospheric pressure

$p_{4'}=101325+3*1260*9.81=138406.8N/m^2$

$p_4=(\frac{p_{4'}}{\rho g}+h_{4-4'})\rho g$

$=(\frac{138406.8}{1260*9.81}+2.2)*1260*9.81$

=165600.12N/m²

5)power removed from liquid=p_3'-p₄

=2030673.021-165600.12

=1865072.901N/m²