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Question: A Pitot tube (shown in the figure )is mounted along the axis of a gas pipeline whose cross-sectional...

A Pitot tube (shown in the figure )is mounted along the axis of a gas pipeline whose cross-sectional area is equal to S. Assume the viscosity to be negligible, the volume of gas flowing across the section of the pipe per unit time, if the difference in the liquid columns is equal to Δh\Delta h and the densities of the liquid and gas are ρo\rho_o and ρ\rho respectively is Q=SxΔgρogρQ = S\sqrt {\dfrac{x\Delta g \rho_o g}{\rho}} find the value of x.

Explanation

Solution

Apply the Bernoulli principle at the two cross-sections A and B and take the difference of the pressures at A and B equal to ρogΔh\rho_o g \Delta h get the value of the volume of gas flowing per unit time into the section B and thus the value of the x required.

Complete step by step solution:
We have two cross-sections A and B and the gas flows through the cross-section B.
Now the Bernoulli principle can be stated as:
At any cross section the P+12ρv2P + \dfrac{1}{2} \rho v^2 = constant.
Now let us consider the volume of gas flowing across section B per unit time as VBV_B
And the value of the velocity of the gas flowing through section A is zero.
Let the pressure at section A be PAP_A and the pressure at section B is PBP_B
The density of the gas is ρ\rho
Applying the Bernoulli principle in section A we get :
PA+12ρVA2P_A + \dfrac{1}{2} \rho V_A^2 = constant
Similarly applying Bernoulli principle at section B we get:
PB+12ρVB2P_B + \dfrac{1}{2} \rho V_B^2 = constant
Taking VBV_B = V and VAV_A = 0 we get from the about two equations we can get the difference in pressures as
PAPB=12ρv2P_A - P_B = \dfrac{1}{2} \rho v^2
Now we know the difference in pressure at A and B should be equal to
we get PAPB=12ρv2=ρogΔhP_A - P_B = \dfrac{1}{2} \rho v^2 = \rho _o g \Delta h
v=2Δhρogρv = \sqrt {\dfrac{2\Delta h \rho_o g }{\rho}}
The volume of the per unit time given area of the cross-section is: Q=S×vQ = S \times v
So we get
Q=S2ΔhρogρQ = S \sqrt {\dfrac{2\Delta h \rho_o g }{\rho}}
Thus we got the value of using the Bernoulli principle as 2 .

Note: Here we neglected the height difference between the section a and b. The possible mistakes that we can make in this kind of problem are that we may consider the height difference between sections A and B which will lead to mistakes and time-consuming.