Answer
$P_{gage}=0.2274psi$
Work Step by Step
Solving the manometer:
$P_{air}+\rho_{oil}gh_{oil}-\rho_{water}gh_{water}=P_{atm}$
$P_{air}-P_{atm}=-\rho_{oil}gh_{oil}+\rho_{water}gh_{water}$
$P_{gage}=g*(\rho_{water}h_{water}-\rho_{oil}h_{oil})$
$30in*\frac{1ft}{12in}=2.5ft$
Substituting:
$P_{gage}=32.2\frac{ft}{s^2}*(62.4\frac{lbm}{ft^3}*2.5ft-49.3\frac{lbm}{ft^3}*2.5ft)*(\frac{1lbf}{32.2\frac{lbm*ft}{s^2}})*(\frac{1ft^2}{144in^2})$
$P_{gage}=0.2274psi$
