Question

A horizontal pipe of diameter 1.09 m has a
smooth constriction to a section of diameter
0.654 m . The density of oil flowing in the pipe
is 821 kg/m3.
If the pressure in the pipe is 7190 N/m2
and in the constricted section is 5392.5 N/m2,
what is the rate at which oil is flowing?
Answer in units of m3/s.
Don't round answer

Answer 1

For an in-compressible liquid:

$$A_1v_1=A_2v_2$$

$$P_1+0.5\rho v_1^2=P_2+0.5\rho v_2^2$$

$$\frac{v_2}{v_1}=\frac{d_1^2}{d_2^2}=\frac{1.09^2}{0.654^2}=2.778$$

$$7190+0.5(821) v_1^2=5392.5+0.5(821)(2.778)^2 v_1^2$$

$$v_1=0.8074\frac{m}{s}$$

So,

$$Q=0.25\pi d_1^2v_1$$

$$Q=0.25\pi (1.09)^2(0.8074)=0.753\frac{m^3}{s}$$