Question

A venture tube is used to measure the flow rate of a liquid in a pipe (liquid density is 800 kg/m3). The pipe has a diameter of 10 cm and the smallest diameter of the venture has a diameter of 4 cm. A manometer with a manometer fluid of mercury (specific weight of 133 kN/m3) is used to calculate the flow rate which is connected to the venture section such that one leg is far upstream and the second leg is at the minimum diameter of the venture tube. If the flow rate is 0.05 m3/s determine the elevation change in the manometer fluid.
a. 14.6 m
b. 9.28 m
c. 4.64 m
d. 2.32 m

Answer 1

Answer:

$\triangle h=4.935m$

Explanation:

From the question we are told that:

Liquid density $\rho=800$

Diameter of pipe $d=4cm \approx 0.004m$

Diameter of venture $d=10cm \approx 0.010m$

Specific weight of mercury P_mg $133 kN/m^3$

Flow rate $r=0.05 m^3/s$

Area A:

            $A_1=\frac{\pi}{4}0.1^2\\A_1=0.00785m^2\\A_2=\frac{\pi}{4}0.04^2\\A_2=0.001256m^2$

Generally the Bernoulli's equation is mathematically given by

$\frac{P_1}{\rho_1g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}$

Where

$V_1=\frac{r}{A_1} \\\\ &V_1=\frac{r}{A_2}$

Therefore

$P_1-P_2=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})$

Generally the equation for pressure difference b/w manometer fluid is given as

$P_1-P_2=(p_mg-pg)\triangle h$

Therefore

$(p_mg-pg)\triangle h=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})$

$\triangle h=\frac{\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})}{(p_mg-pg)}$

$\triangle h=\frac{\frac{(800)(0.05)^2}{2}(\frac{(0.1)^2-(0.4)^2}{(0.1)^2(0.04)^2})}{(1.33*10^3-800*9.81)}$

$\triangle h=4.935m$

Therefore elevation change is mathematically given by

$\triangle h=4.935m$