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## Q. 2.19

A differential gage is attached to two cross sections A and B in a horizontal pipe in which water is flowing. The deflection of the mercury in the gage is 1.92 ft, the level nearer A being the lower one. Calculate the difference in pressure in psi between sections A and B. Refer to Fig. 2-13.

## Verified Solution

Note: A sketch will be of assistance in clarifying the analysis of all problems as well as in reducing mistakes. Even a single-line diagram will help.

pressure head at C = pressure head at D

or, using ft of water units, $p_{A}/\gamma – z=[(p_{B}/\gamma)-(z+1.92)]+(13.57)(1.92)$

Then $p_{A}/\gamma – p_{B}/\gamma = difference\ in\ pressure\ heads = (1.92)(13.57-1)=24.1\ ft\ water$

and $p_{A} – p_{B}=(24.1)(62.4/144)=10.44\ psi$.

If ($p_{A} – p_{B}$) had been negative, the proper interpretation of the sign would be that the pressure at B was greater than the pressure at A by 10.44 psi.

Differential gages should have the air extracted from all tubing before readings are taken.