Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 2, Problem 85RQ
To determine

The absolute pressure at the center of the pipe.

Expert Solution & Answer
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Explanation of Solution

Given:

Local atmospheric pressure (Patm) is 14.2psia.

Height of the water in the manometer (hw) is 20in..

Height of the mercury in the manometer (hHg) is 25in..

Height of the oil on left arm of the manometer (hoil) is 60in..

Height of the oil on left arm of the manometer (hoil) is 30in..

Density of the water (ρw) is 62.4lbm/ft3.

Specific gravity of the oil (SGoil) is 0.80.

Specific gravity of the mercury (SGHg) is 13.6.

Acceleration due to gravity (g) is 32.2ft/s2.

Calculation:

Calculate the density of the oil (ρoil).

  ρoil=(ρw)(SGoil)=(62.4lbm/ft3)(0.80)=49.92lbm/ft3

Calculate the density of the mercury (ρHg).

  ρHg=(ρw)(SGHg)=(62.4lbm/ft3)(13.6)=848.64lbm/ft3

Calculate the absolute pressure at the center of the pipe (P).

  Pρwghw+ρoilghoilρHgghHgρoilghoil=Patm

  P=Patm+ρwghwρoilghoil+ρHgghHg+ρoilghoil=Patm+g(ρwhwρoilhoil+ρHghHg+ρoilhoil)=14.2psia+(32.2ft/s2)[(62.4lbm/ft3)(20in.)(49.92lbm/ft3)(60in.)+(848.64lbm/ft3)(25in.)+(49.92lbm/ft3)(30in.)]=14.2psia+(32.2ft/s2)[(62.4lbm/ft3)(2012ft)(49.92lbm/ft3)(6012ft)+(848.64lbm/ft3)(2512ft)+(49.92lbm/ft3)(3012ft)]=14.2psia+(32.2ft/s2)[(62.4lbm/ft3)(2012ft)(49.92lbm/ft3)(6012ft)+(848.64lbm/ft3)(2512ft)+(49.92lbm/ft3)(3012ft)](1lbf32.2lbmft/s2)(1ft2144in.2)=26.4psia

Thus, the absolute pressure at the center of the pipe is 26.4psia_.

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Chapter 2 Solutions

Fundamentals of Thermal-Fluid Sciences

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