Question

A manometer is used to measure the air pressure in a tank. The fluid used has a specific gravity of 1.25, and the differential height between the two arms of the manometer is 34 in. If the local atmospheric pressure is 12.7 psia, determine the absolute pressure in the tank for the cases of the manometer arm with the a) higher b) lower fluid level being attached to the tank .

Answer 1

Answer:

ΔP = 1.53 psia

a) Higher arm Ph = 11.16 psia

b) Lower fluid level Pl = 14.23 psia

Explanation:

The fluid used has a specific gravity of 1.25. We can calculate the density of the fluid taking into account the definition of specific gravity (GE) as the relation between Fluid density and Water Density.

$GE =\frac{\rho_f }{\rho_w}\\ \rho_f = GE*\rho_w\\\rho_f = 1.25 * 62.4 \frac{lb}{ft^3} = 78 \frac{lb}{ft^3}$

Height of the manometer is 34 in, we can convert this value to feet using 1ft = 12 in. Using the hydrostatic equation to calculate ΔP, replacing and converting to the same units we can calculate the manometric pressure. Atmospheric pressure (Pa) is 12.7 psia  

$\Delta P = \rho g h = (78 \frac{lb}{ft^3} )(32.174 \frac{ft}{s^2} ) (\frac{34 in *1ft}{12 in} )(\frac{1lbf}{32.174 lbm ft/s^2} )(\frac{1ft^2}{144 in^2} )\\ \Delta P=1.53 \frac{lbf}{in^2} =1.53 psi$

So, ΔP = 1.53 psi

a) Using the fluid level higher attached to the tank (Vacuum pressure). We subtract the manometric pressure of atmospheric pressure

$Ph = 12.7 psi - 1.53 psi = 11.16 psi$

So, pressure using the higher level is Ph= 11.16 psi.  

b) Using the fluid level lower attached to the tank. We add the manometric pressure to the atmospheric value

$Pl = 12.7 psi + 1.53 psi = 14.23 psi$

Finally, pressure using the lower level is Pl= 14.23 psi.