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.

To determine

Total force on the window.

Answer to Problem 4.1PP

The required value of the force is $1672.8$lb

Explanation of Solution

Given information:

   $\begin{array}{l}Pressure\left(P\right)=0.12psia\hfill \\ Pressurehead\left(h\right)=30.5inofhg\hfill \\ Diameter\left(D\right)=12in\hfill \end{array}$

Formula used:

   $\begin{array}{l}\text{pressure (P) = }\frac{\text{force(F)}}{\text{area(A)}}\\ \text{pressure (P) =}\gamma h\\ \text{force (F) = P}\times \text{A}\end{array}$

Calculation:

We know the standard formula to calculate pressure, i.e.: $\text{P = }\frac{\text{force(F)}}{\text{area(A)}}$

Then, $\text{F = P}\times \text{A}\text{(a)}$

   $\begin{array}{l}\text{Area (A) = }\frac{\pi }{4}{\left(D\right)}^2\\ \Rightarrow A=\frac{\pi }{4}{\left(12\right)}^2\\ \Rightarrow A=113.097{\text{ in}}^2\end{array}$

Let us assume that the pressure in the tank is $P_{atm}=\gamma h$

   $\begin{array}{l}{P}_{atm}=\left(0.4889\right)\left(30.5\right)\text{(specific weight of mercury (γ)=0}{\text{.4889lb/in}}^{\text{3}}\text{)}\\ \Rightarrow P_{atm}=14.911lb/i{n}^2\end{array}$

Net pressure acting in the tank is, $P=P_{atm}-P_{inside}$;

   $\begin{array}{l}P=P_{atm}-P_{inside}\\ \Rightarrow P=14.911-0.12\\ \Rightarrow P=14.791lb/i{n}^2\end{array}$

From equation (a);

   $\begin{array}{l}F=P\times A\\ \Rightarrow F=\left(14.791\right)\left(113.097\right)\\ \Rightarrow F=1672.8lb\end{array}$

Then, the net force acting on the window is;

   $F=1672.8lb$