Answer:
Manometric difference x=142.85 mm.
Explanation:
Given :
Pipe diameter [tex]d_1=40 mm[/tex]
venturi meter [tex]d_2=20 mm[/tex]
We can know that discharge through venturi meter is given as
[tex]Q=C_d\dfrac{A_1A_2\sqrt{2gh}}{\sqrt {A_1^2-A_2^2}}[/tex]
[tex]A_1=1.24\times 10^{-3},A_2=3.12\times 10^{-4}[/tex]
[tex]Q=A_1V_1[/tex]
[tex]Q=1.24\times 10^{-3}\times 1.5=0.00186 m^3/s[/tex]
[tex]0.00186=0.97\dfrac{1.24\times 10^{-3}\times 3.12\times 10^{-4} \sqrt{2gh}}{\sqrt {(1.24\times 10^{-3})^2-(3.12\times 10^{-4})^2}}[/tex]
h=1.8 m
We know that [tex]h=x\left (\dfrac{\rho_{hg}}{\rho_w}-1\right )[/tex]
Where x is the manometric deflection
⇒ [tex]1.8=x\left (\dfrac{13600}{1000}-1\right )[/tex]
So x=14.28 mm
Manometric difference x=142.85 mm.