Question #50086

Pressure and force
A pneumatic ram has a chamber piston diameter of 150 mm and is subjected to a variable source of pressure for actuation. What pressure will be required to produce an actuation force of 2 kN (consider the cylinder to be friction less)
1

Expert's answer

2014-12-25T01:31:51-0500

Answer on Question #50086 – Physics – Mechanics | Kinematics | Dynamics

Pressure and force

A pneumatic ram has a chamber piston diameter of 150 mm and is subjected to a variable source of pressure for actuation. What pressure will be required to produce an actuation force of 2 kN (consider the cylinder to be friction less)

Solution:


F=2kN=2000Nactuation force;d=150mm=0.15mpiston diameter;\begin{array}{l} F = 2 k N = 2000 N - \text{actuation force}; \\ d = 150 \, \text{mm} = 0.15 \, \text{m} - \text{piston diameter}; \end{array}


Pressure is defined as force per unit area.


p=ForceArea=FA=5kg9.8Nkg104m2=490kPap = \frac{\text{Force}}{\text{Area}} = \frac{F}{A} = \frac{5kg \cdot 9.8 \frac{N}{kg}}{10^{-4} \, \text{m}^2} = 490 \, \text{kPa}A=πd24A = \frac{\pi d^2}{4}(2)in(1):(2) \text{in}(1):p=Fπd24=4Fπd2=42000N3.14(0.15m)2=113.2kPa.p = \frac{F}{\frac{\pi d^2}{4}} = \frac{4F}{\pi d^2} = \frac{4 \cdot 2000 \, N}{3.14 \cdot (0.15 \, \text{m})^2} = 113.2 \, \text{kPa}.


Answer: pressure is equal to p=113.2kPap = 113.2 \, \text{kPa}.

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