3

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.

$P_1$

$P_2$

$0.065 \ \si\m$

$0.035 \ \si\m$

$0.065 \ \si{m \ Hg}$

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:

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

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

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

$P_2=115.706kPa$

License

Icon for the Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License

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.

Share This Book