##### Power and Efficiency Definitions

• Define Shaft Power, fluid power, volumetric efficiency and overall efficiency.
• Calculate values in a fluid power system utilizing Pascal’s law.
• Calculate flow rate, flow velocity and mechanical advantage in a hydraulic system.
• Learn and understand principles of fluid mechanics and power with applications.
• Provide the student the necessary analytic skills to solve and analyze a variety of fluid mechanics and fluid power related problems.

###### Solved Example: 66-1-01

A hydraulic motor is required to drive a load at 500 rpm with 1000 Nm of torque. What is the output power?

Solution:

N = 500 rpm = 52.36 rad/s

T$_A$ = 1000N.m

$\mathrm{Power} = T_A \times N = 1000 \times 52.36 = 52360\ W$

The output power is 52.360 kW.

###### Solved Example: 66-1-02

In a hydraulic press, the plunger is 40 mm in diameter and the ram is 320 mm in diameter. What effort force must be applied to the plunger to raise a load of 1600 N, assuming no losses?

Solution:
The movement ratio = $\left(\dfrac{D_2}{D_1}\right)^2 = \left(\dfrac{320}{40}\right)^2 =64 \textbf{(NO UNITS)}$

Assuming 100% efficiency,

Force ratio = movement ratio =64

$\dfrac{F_2}{F_1} = 64$ $F_1 = 25N$

###### Solved Example: 66-1-03

A hydraulic jack has a plunger 25 mm. diameter which generates fluid in the system. This pressure acts on a load ram piston of 75 mm diameter. If the effort of the operating pedal is 700 N, with a mechanical advantage of 7 to 1, what is the maximum load the jack will lift if the overall efficiency of the jack is 70%?

Solution:
$\mathrm{Ideal\ output} = 7 \times 700 = 4900N$ \begin{align*} \mathrm{Actual\ output} &= \mathrm{efficiency \times Ideal\ Output}\\ &= 0.70 \times 4900\\ &= 3430\ N \end{align*}

###### Solved Example: 66-1-04

A hydraulic press consist of a 75 mm diameter cylinder fitted with a plunger to which the effort is applied and a larger cylinder 450 mm diameter with a ram which support the load. If the stroke of the plunger is 300 mm, how many strokes must be made to lift the load through 150 mm?

Solution:
\begin{equation*} \begin{split} \mathrm{No.\ of\ strokes} &= \dfrac{\mathrm{Volume\ of\ Output}}{\mathrm{Volume\ of\ Input}}\\ &= \dfrac{\left(\dfrac{\pi}{4}\right)D_2^2 H_2}{\left(\dfrac{\pi}{4}\right) D_1^2 H_1}\\ &= \left(\dfrac{D_2}{D_1}\right)^2 \dfrac{H_2}{H_1}\\ &= \left(\dfrac{450}{75}\right)^2 \dfrac{150}{300} \\ &= 18 \textbf{(NO UNITS)} \end{split} \end{equation*}

The overall efficiency of a centrifugal pump when head is 25 m, discharge = 0.04 m$^3$/s and output power P = 16 kW (take g = 10 $m/s^2$? and $\rho$ = 1000 kg/m$^3$) is: (SSC JE CE Oct 2020 Evening)
Pump power = $\rho g h Q = 1000 \times 10 \times 0.4 \times 25 = 10000\ W$ $\eta = \dfrac{\rho g h Q}{P}= \dfrac{10000}{16000}= 0.625 = 62.5\%$