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A manometer of mercury indicates a reading of 1 m when the fluid pressure, water, (P_1) at point A is 0.065 m Hg vacuum.  Determine the fluid pressure, oil, (P_2) at point B.

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P_1

P_2

0.065 \ \si\m

0.035 \ \si\m

0.065 \ \si{m \ Hg}

Given:

h_{water} = (0.35 + 1)m = 1.35m

h_{oil}=0.35m

h_{Hg}=1.00m

\gamma_{Hg}=133 kN/m^3

\gamma_{oil}=8.95 kN/m^3

\gamma_{water}=9.80 kN/m^3

Find:

The fluid pressure (P_2) at point B.

Solution:

P_2+\gamma_{oil}h_{oil}-\gamma_{water}h_{water}=P_1

P_1=-\gamma_{Hg}(0.065m)

P_2=-\gamma_{Hg}(0.065m)-\gamma_{oil}1.35m+\gamma_{Hg}(1.00m)+\gamma_{water}0.35m

P_2=115.706kPa

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Thermodynamics Copyright © by Diana Bairaktarova (Adapted from Engineering Thermodynamics - A Graphical Approach by Israel Urieli and Licensed CC BY NC-SA 3.0) is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.

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