Answer:
The answer is below
Explanation:
The complete question is attached.
a) Bernoulli equation is given as:
[tex]P+\frac{1}{2}\rho V^2+ \rho gz=constant\\\\\frac{P}{\rho g} +\frac{V^2}{2g} +z=constant\\[/tex]
Where P = pressure, V = velocity, z = height, g = acceleration due to gravity and ρ = density.
[tex]\frac{P}{\rho g} +\frac{V^2}{2g} +z=constant\\\\\frac{P}{\gamma} +\frac{V^2}{2g} +z=constant\\\\\frac{P_1}{\gamma} +\frac{V_1^2}{2g} +z_1=\frac{P_2}{\gamma} +\frac{V_2^2}{2g} +z_2\\\\but \ z_1=z_2,P_1=0,V_1=0,V_2=60\ mph=88\ ft/s. Hence:\\\\\frac{P_2}{\gamma} =-\frac{V_2^2}{2g} \\\\P_2=\gamma*-\frac{V_2^2}{2g} =\rho g*-\frac{V_2^2}{2g} \\\\P_2=-\frac{V_2^2}{2}*\rho=-\frac{(88.8\ ft/s)^2}{2} * 0.00238\ slug/ft^3=-9.22\ lb/ft^2\\\\P_2+\gamma_{H_2O}h-\gamma_{oil}(1/12 \ ft)=0\\\\[/tex]
[tex]\gamma_{oil}=0.9\gamma_{H_2O}=0.9*62.4\ lb/ft^3=56.2\ lb/ft^3\\\\Therefore:\\\\-9.22\ lb/ft^2+62.4\ lb/ft^3(h)-56.2\ lb/ft^3(1/12\ ft)=0\\\\h=0.223\ ft[/tex]
b)
[tex]\frac{P}{\gamma} +\frac{V^2}{2g} +z=constant\\\\\frac{P_2}{\gamma} +\frac{V_2^2}{2g} +z_2=\frac{P_3}{\gamma} +\frac{V_3^2}{2g} +z_3\\\\but \ z_3=z_2,V_3=0,V_2=60\ mph=88\ ft/s. \\\\\frac{P_2}{\gamma}+\frac{V_2^2}{2g} = \frac{P_3}{\gamma}\\\\\frac{P_3-P_2}{\gamma}=\frac{V_2^2}{2g} \\\\P_3-P_2=\frac{V_2}{2g}*\gamma=\frac{V_2^2}{2g}*\rho g\\\\P_3-P_2=\frac{V_2}{2}*\rho=\frac{(88\ ft/s^2)^2}{2}*0.00238\ slg\ft^3\\\\P_3-P_2=9.22\ lb/ft^2[/tex]
