To solve the exercise, it is necessary to apply the concepts related to the conservation of the energy flow given by Bernoulli and the equation of head loss in the pipe for laminar flow. Through them and the calculation of the flow we can identify the flow rate of oil.
Bernoulli is defined by,
[tex]\frac{p_1}{\rho g}+\frac{V_1^2}{2g}+z_1=\frac{p_2}{\rho g}+\frac{V_2^2}{2g}+z_2+h_f[/tex]
Where,
P = Pressure
[tex]\rho[/tex]= Density
z= Datum height
V = Velocity
g = Gravitaty constant
[tex]h_f[/tex] = frictional head loss in the pipe
There is no change in the final height datum nor initial speed. So,
[tex]\frac{p_{atm}}{850*9.81}+\frac{0^2}{2(9.81)}+1.45=\frac{p_{atm}}{850*9.81}+\frac{V_2^2}{2(9.81)}+0+h_f[/tex]
Re-arrange to find the head loss,
[tex]h_f = 1.45-\frac{V^2}{19.62}[/tex]
In the case of the head loss in the pipe for laminar flow we have that
[tex]h_f = \frac{32\mu VL}{\rho gd^2}[/tex]
Equation we have,
[tex]1.45-\frac{V^2}{19.62}=\frac{32\mu VL}{\rho gd^2}[/tex]
[tex]\frac{V^2}{19.62}+\frac{32\mu VL}{\rho gd^2}-1.45=0[/tex]
At this point our values are given as,
[tex]L=55*10^{-2}m[/tex]
[tex]g = 9.81m/s^2[/tex]
[tex]\rho = 850kg/m^3[/tex]
[tex]\mu = 0.11 kg/m.s[/tex]
[tex]d = 4*10^{-3}m[/tex]
Therefore,
[tex]\frac{V^2}{19.62}+\frac{32(0.11)V(55*10^{-2})}{850(9.81)(4*10^{-3})^2}-1.45=0[/tex]
[tex]0.051V^2+14.51V-1.45=0[/tex]
[tex]V=0.1m/s[/tex]
Finally the discharge is given as
[tex]Q = AV[/tex]
Where,
A= Area
V = Velocity
[tex]Q = \frac{\pi}{4}d^2*V[/tex]
[tex]Q=\frac{\pi}{4}*0.004^2*0.1[/tex]
[tex]Q= 1.256*10^{-6}m^3s[/tex]
That is equal in cm^3 per hour as,
[tex]Q= 1.256*10^{-6}m^3s *(\frac{10^6cm^3}{1m^3})(\frac{3600s}{1h})[/tex]
[tex]Q = 4521cm^3/h[/tex]
Therefore the flow rate of oil is 4521cm^3/h
