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Engineering and Tech / Fluid Mechanics 2 / Chapter 3 / Problem 6P

Fluid Mechanics | 2nd Edition | ISBN: 9780071284219 | Authors: Yunus A. Cengel, John M. Cimbala

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QUESTION:

Problem 6P

The piston of a vertical piston–cylinder device containing a gas has a mass of 85 kg and a crosssectional area of 0.04 m2 (Fig P3–7). The local atmospheric pressure is 95 kPa, and the gravitational acceleration is 9.81 m/s2.(a) Determine the pressure inside the cylinder. (b) If some heat is transferred to the gas and its volume is doubled, do you expect the pressure inside the cylinder to change?

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QUESTION:

Problem 6P

The piston of a vertical piston–cylinder device containing a gas has a mass of 85 kg and a crosssectional area of 0.04 m2 (Fig P3–7). The local atmospheric pressure is 95 kPa, and the gravitational acceleration is 9.81 m/s2.(a) Determine the pressure inside the cylinder. (b) If some heat is transferred to the gas and its volume is doubled, do you expect the pressure inside the cylinder to change?

ANSWER:

Solution 6P

To find the pressure inside the cylinder, we shall have to draw a figure to understand the situation better.

From the figure above, the force due to the atmospheric pressure and the weight of the piston balances the force due to the pressure from inside the cylinder.

(a)

Step 1

Given, mass of the piston $m=85\ kg$

Crosssectional area of the piston $A=0.04\ {m}^{2}$

Gravitational acceleration $g=9.81\ m/{s}^{2}$

Atmospheric pressure $P{\ }_{atm}=95\ kPa$

Therefore, the weight of the cylinder $W=mg=85\ kg\times{}9.81\ m/{s}^{2}\approx{}834\ N$

The force due to the atmospheric pressure $F{\ }_{atm}=A\times{}95\ kPa=0.04\ {m}^{2}\times{}95\ kPa=3800\ N$

Let $P$ be the pressure inside the cylinder.

Force due to the pressure inside the cylinder $F=P\times{}A=P\times{}0.04\ {m}^{2}$

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