Answer
$P_{centerpipe}=26.33psia$
Work Step by Step
Solving the manometer:
$P_{centerpipe}-\rho_{water}gh_{water}+\rho_{oil}gh_{1,oil}-\rho_{mercury}gh_{mercury}-\rho_{oil}gh_{2,oil}=P_{atm}$
$P_{centerpipe}=P_{atm}+\rho_{water}gh_{water}-\rho_{oil}gh_{1,oil}+\rho_{mercury}gh_{mercury}+\rho_{oil}gh_{2,oil}$
$P_{centerpipe}=P_{atm}+g\rho_{water}*(h_{water}-0.8h_{1,oil}+13.6h_{mercury}+0.8h_{2,oil})$
$h_{water}=(20in)*(\frac{1ft}{12in})=\frac{20}{12}ft$
$h_{1,oil}=(60in)*(\frac{1ft}{12in})=5ft$
$h_{mercury}=(25in)*(\frac{1ft}{12in})=\frac{25}{12}ft$
$h_{2,oil}=(30in)*(\frac{1ft}{12in})=2.5ft$
Substituting:
$P_{centerpipe}=14.2psia+32.2\frac{ft}{s^2}*62.4\frac{lbm}{ft^3}*(\frac{20}{12}-0.8*5ft+13.6*\frac{25}{12}ft+0.8*2.5ft)*(\frac{1lbf}{32.2\frac{lbm*ft}{s^2}})*(\frac{1ft^2}{144in^2})$
$P_{centerpipe}=26.33psia$
