Applied Fluid Mechanics (7th Edition)
Applied Fluid Mechanics (7th Edition)
7th Edition
ISBN: 9780132558921
Author: Robert L. Mott, Joseph A. Untener
Publisher: PEARSON
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Chapter 4, Problem 4.1PP

figure 4.2 shows a vacuum tank with a flat circular observation window in one end. If the pressure in the tank is 0.12psia when the barometer reads 30.5in of mercury, calculate the total force on the window.

Expert Solution & Answer
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To determine

Total force on the window.

Answer to Problem 4.1PP

The required value of the force is 1672.8lb

Explanation of Solution

Given information:

Applied Fluid Mechanics (7th Edition), Chapter 4, Problem 4.1PP

   Pressure (P) = 0.12 psiaPressure head (h) = 30.5 in of hgDiameter (D) = 12 in

Formula used:

   pressure (P) = force(F)area(A)pressure (P) =γhforce (F) = P×A

Calculation:

We know the standard formula to calculate pressure, i.e.: P = force(F)area(A)

Then, F = P×A (a)

   Area (A) = π4(D)2A=π4(12)2A=113.097 in2

Let us assume that the pressure in the tank is Patm=γh

   Patm=(0.4889)(30.5) (specific weight of mercury (γ)=0.4889lb/in3)Patm=14.911lb/in2

Net pressure acting in the tank is, P=PatmPinside;

   P=PatmPinsideP=14.9110.12P=14.791lb/in2

From equation (a);

   F=P×AF=(14.791)(113.097)F=1672.8lb

Then, the net force acting on the window is;

   F=1672.8lb

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