Question

A horizontal pipe of diameter 0.985 m has a smooth constriction to a section of diameter 0.591 m . the density of oil flowing in the pipe is 821 kg/m3 . if the pressure in the pipe is 8100 n/m2 and in the constricted section is 6075 n/m2 , what is the rate at which oil is flowing

Answer 1

Answer:

For an in-compressible liquid:

$a_1v_1=a_2v_2$ (Equation 1)

and

$P_1+\frac{1}{2}\rhov_1^2=P_2+\frac{1}{2}\rhov_2^2$ (Equation 2)

a is the area of cross-section of the pipe, v is the rate at which fluid flows through the cross-section.

P is the pressure of the fluid and ρ is the density of the fluid.

From equation 1,

$\frac{v_2}{v_1} = \frac{a_1}{a_2}=\frac{r_1^2}{r_2^2}=\frac{0.985^2m^2}{0.591^2m^2}=1.78$

From equation 2:

$P_1-P_2=\frac{1}{2}\rho{v_2^2-v_1^2}$

$\Rightarrow \frac{2(P_1-P_2)}{\rho}=v_1^2{\frac{v_2^2}{v_1^2}-1}$

$\frac{2(8100-6075)N/m^2}{821 kg/m^3}=v_1^2(1.78)$

$\Rightarrow v_1=1.66 m/s$

$v_2= 1.78\times 1.66 m/s = 2.95 m/s$

The rate at which oil is flowing the at the cross-section of diameter 0.985 m is 1.66 m/s. The at which oil is flowing through the cross-section of diameter 0.591 m is 2.95 m/s.

Answer 2

Ρgh is the gravitational component of Bernoulli's equation 
the pipe is horizontal 
so no change in h 
ρgh is equal on both sides 
cancels out 
now 
v (velocity) is inversely proportional to square of radius 
so 
v_2 / v_1 = (r_1 )^2 / (r_2)^2 = 1.1^2 / 0.66^2 = 2.78 
v_2 = 2.78 v_2 
and 
8130 N/m² + ((821 kg/m³) * (v_1)²)/2 = 6097.5 N/m² + ((821 kg/m³) * (2.78v_1)²)/2 
solve for v_1 
then 
R = 2 pi (1.1)^2 * v_1