Learning Objectives:

• 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.

Mechanical Power is the rate of doing work. It is equivalent to an amount of energy consumed per unit time. In the MKS system, the unit of power is the joule per second (J/s), known as the Watt in honor of James Watt. Larger units are 1KW = 1000 W or 1 Horsepower (HP) = 746 W.

$\mathrm{Fluid\ Power} = \rho g h Q = \gamma H Q$       Page 192

$\mathrm{Pump\ (brake)\ power\ \dot{W}} = \dfrac{\rho g h Q}{\eta_{\mathrm{pump}}} $       Page 192

$\mathrm{Purchased\ power} = \dfrac{\dot{W}}{\eta_{\mathrm{motor}}} $       Page 192

Electric Power in watts produced by an electric current I consisting of a charge of Q coulombs every t seconds passing through an electric potential (voltage) difference of V is: \[P = V \times I = \dfrac{V^2}{R} = I^2 R\] Where,
P = Electric power in Watts
t = time in seconds
I = electric current in Amperes
V = electric potential or voltage in Volts
Mechanical Power, \(P = \dfrac{dU}{dt} = \bar{F} \cdot \bar{v}\)
Mechanical Efficiency, \(\eta = \dfrac{P_{out}}{P_{in}} = \dfrac{U_{out}}{U_{in}}\)

Zureks, Public domain, via Wikimedia Commons

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?

A. 52.36 kW

B. 78.75 kW

C. 112.86 kW

D. 164.40 kW

N = 500 rpm

T = 1000 N.m

The output power is 52.36 kW.

Correct Answer: A

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?

A. 25 N

B. 12.5 N

C. 50 N

D. 200 N-s

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

Correct Answer: A

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%?

A. 6630 N

B. 1330 N

C. 3430 N

D. 3885 N

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

Correct Answer: C

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?

A. 36

B. 64

C. 9

D. 18

\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*}

Correct Answer: D