THERMODYNAMICS-SI ED. EBOOK >I<
THERMODYNAMICS-SI ED. EBOOK >I<
9th Edition
ISBN: 9781307573022
Author: CENGEL
Publisher: MCG/CREATE
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Chapter 1.11, Problem 98RP

A vertical piston–cylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of 10 kg and a diameter of 14 cm. Pressure of the gas is to be increased by placing some weights on the piston. Determine the local atmospheric pressure and the mass of the weights that will double the pressure of the gas inside the cylinder Answers: 93.6 kPa, 157 kg

Chapter 1.11, Problem 98RP, A vertical pistoncylinder device contains a gas at a pressure of 100 kPa. The piston has a mass of

Expert Solution & Answer
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To determine

The local atmospheric pressure of the vertical piston-cylinder device.

The mass of the weights that will double the pressure of the vertical piston-cylinder device.

Answer to Problem 98RP

The local atmospheric pressure of the vertical piston-cylinder device is 93.63kPa_.

The mass of the weights that will double the pressure of the vertical piston-cylinder device is 157kg_.

Explanation of Solution

Show the free body diagram of the vertical piston-cylinder device.

THERMODYNAMICS-SI ED. EBOOK >I<, Chapter 1.11, Problem 98RP

Write the expression of vertical force in the piston-cylinder device.

Patm=PmpistongA (I)

Here, the mass of piston is mpiston, the absolute pressure is P, the atmospheric pressure is Patm, the cross-sectional area is A, and acceleration gravity is g.

Write the expression of balance force in the piston-cylinder device.

P=Patm+(mpiston+mweight)gA (II)

Here, the mass of the weights that will double the pressure is mweight.

Determine the area of the piston-cylinder.

A=πD2/4 (III)

Conclusion:

Substitute 14cm for D in Equation (III).

A=π(14cm)2/4=153.9cm2×(104m21cm2)0.0154m2

Substitute 0.0154m2 for A, 10kg for mpiston, 9.81m/s2 for g, and 100kPa for P in Equation (I).

Patm=100kPa(10kg)(9.81m/s2)0.0154m2=100kPa(6370.129kgm/s2)×(1kN1000kgm/s2)(1kPa1kN/m2)=93.63kPa

Thus, the local atmospheric pressure of the vertical piston-cylinder device is 93.63kPa_.

Substitute 0.0154m2 for A, 10kg for mpiston, 9.81m/s2 for g, 93.63kPa for Patm, and 200kPa for P in Equation (II).

200kPa=93.63kPa(10kg+mweight)(9.81m/s2)0.0154m2(10kg+mweight)=(93.63kPa)(9.81m/s20.0154m2)×(1kN1000kgm/s2)mweight=157kg

Thus, the mass of the weights that will double the pressure of the vertical piston-cylinder device is 157kg_.

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Chapter 1 Solutions

THERMODYNAMICS-SI ED. EBOOK >I<

Ch. 1.11 - The value of the gravitational acceleration g...Ch. 1.11 - A 3-kg plastic tank that has a volume of 0.2 m3 is...Ch. 1.11 - A 2-kg rock is thrown upward with a force of 200 N...Ch. 1.11 - Solve Prob. 113 using appropriate software. 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