Chapter 2, Problem 2.15P

To determine

(a)

The specific weight of the oil.

Answer to Problem 2.15P

The specific weight of the oil is $55.2\text{lbf}/{\text{ft}}^3$.

Explanation of Solution

Concept Used:

Write the hydrostatic formula from gage $B$ to gage $C$.

$p_B+{\gamma }_{\text{oil}}\left(\frac{h_{{}_{\text{oil}}}}{2}\right)+\left({\gamma }_{\text{water}}\right)\left(h_{{}_{\text{water}}}\right)=p_C$ …… (I)

Here, the pressure of gage $B$ is $p_B$, the specific weight of oil is ${\gamma }_{\text{oil}}$, the height of oil column is $h_{{}_{\text{oil}}}$, the specific weight of water is ${\gamma }_{\text{water}}$, the height of water column is $h_{{}_{\text{water}}}$ and pressure of gage $C$ is $p_C$.

The reading of gage $B$ is $1.25\text{lbf}/{\text{in}}^2$ less than the gage reading $C$.

Write the expression for gage pressure $C$.

$\begin{array}{l}{p}_C-\left(1.25\text{lbf}/{\text{in}}^2\right)=p_B\\ p_C=p_B+\left(1.25\text{lbf}/{\text{in}}^2\right)\\ p_C=p_B+\left(1.25\text{lbf}/{\text{in}}^2\right)\left(\frac{144\text{lbf}/{\text{ft}}^2}{1\text{lbf}/{\text{in}}^2}\right)\\ p_C=p_B+\left(180\text{lbf}/{\text{ft}}^2\right)\end{array}$ …… (II)

Substitute $p_B+\left(180\text{lbf}/{\text{ft}}^2\right)$ for $p_C$ in Equation (I).

$\begin{array}{l}{p}_B+{\gamma }_{\text{oil}}\left(\frac{h_{{}_{\text{oil}}}}{2}\right)+\left({\gamma }_{\text{water}}\right)\left(h_{{}_{\text{water}}}\right)=p_B+\left(180\text{lbf}/{\text{ft}}^2\right)\\ {\gamma }_{\text{oil}}\left(\frac{h_{{}_{\text{oil}}}}{2}\right)+\left({\gamma }_{\text{water}}\right)\left(h_{{}_{\text{water}}}\right)=\left(180\text{lbf}/{\text{ft}}^2\right)\end{array}$ …… (III)

Substitute $2\text{ft}$ for $h_{{}_{\text{oil}}}$, $62.4\text{lbf}/{\text{ft}}^3$ for ${\gamma }_{\text{water}}$ and $2\text{ft}$ for $h_{{}_{\text{water}}}$ in Equation (III).

$\begin{array}{l}{\gamma }_{\text{oil}}\left(\frac{2\text{ft}}{2}\right)+\left(62.4\text{lbf}/{\text{ft}}^3\right)\left(2\text{ft}\right)=\left(180\text{lbf}/{\text{ft}}^2\right)\\ {\gamma }_{\text{oil}}\left(\frac{2\text{ft}}{2}\right)=\left(55.2\text{lbf}/{\text{ft}}^2\right)\\ {\gamma }_{\text{oil}}=55.2\text{lbf}/{\text{ft}}^3\end{array}$

Conclusion:

Thus, the specific weight of the oil is $55.2\text{lbf}/{\text{ft}}^3$.

To determine

(b)

The actual gage reading of $C$ in $\text{lbf}/{\text{in}}^2$.

Answer to Problem 2.15P

The actual gage reading of $C$ is $16.6\text{lbf}/{\text{in}}^2$.

Explanation of Solution

Calculate the actual gage reading of $C$.

$p_C=p_a+{\gamma }_{\text{air}}\left(h_{\text{air}}\right)+{\gamma }_{\text{oil}}\left(h_{{}_{\text{oil}}}\right)+{\gamma }_{\text{water}}\left(h_{{}_{\text{water}}}\right)$ …… (IV)

Here, the actual gage reading of $C$ is $p_C$, the pressure gage reading of $A$ is $p_a$, the specific weight of water is ${\gamma }_{\text{water}}$, the height of water column is $h_{{}_{\text{water}}}$, the specific weight of oil is ${\gamma }_{\text{oil}}$, the height of oil column is $h_{{}_{\text{oil}}}$, the specific weight of air is ${\gamma }_{\text{air}}$ and the height of air column is $h_{\text{air}}$.

Substitute $15\text{lbf}/{\text{in}}^2$ for $p_a$, $0.0767\text{lbf}/{\text{ft}}^3$ for ${\gamma }_{\text{air}}$, $2\text{ft}$ for $h_{\text{air}}$, $55.2\text{lbf}/{\text{ft}}^3$ for ${\gamma }_{\text{oil}}$, $2\text{ft}$ for $h_{{}_{\text{oil}}}$, $62.4\text{lbf}/{\text{ft}}^3$ for ${\gamma }_{\text{water}}$ and $2\text{ft}$ for $h_{{}_{\text{water}}}$ in Equation (IV).

$\begin{array}{c}{p}_C=\left[\begin{array}{l}\left(15\text{lbf}/{\text{in}}^2\right)+\left(0.0767\text{lbf}/{\text{ft}}^3\right)\left(2\text{ft}\right)\\ +\left(55.2\text{lbf}/{\text{ft}}^3\right)\left(2\text{ft}\right)+\left(62.4\text{lbf}/{\text{ft}}^3\right)\left(2\text{ft}\right)\end{array}\right]\\ =\left(15\text{lbf}/{\text{in}}^2\right)\left(\frac{144\text{lbf}/{\text{ft}}^2}{1\text{lbf}/{\text{in}}^2}\right)+\left(235.3534\text{lbf}/{\text{ft}}^2\right)\\ =\left(2395.3534\text{lbf}/{\text{ft}}^2\right)\left(\frac{1\text{lbf}/{\text{in}}^2}{144\text{lbf}/{\text{ft}}^2}\right)\\ \approx 16.6\text{lbf}/{\text{in}}^2\end{array}$

Conclusion:

Thus, the actual gage reading of $C$ is $16.6\text{lbf}/{\text{in}}^2$.