Answer:
[tex]\triangle h=4.935m[/tex]
Explanation:
From the question we are told that:
Liquid density [tex]\rho=800[/tex]
Diameter of pipe [tex]d=4cm \approx 0.004m[/tex]
Diameter of venture [tex]d=10cm \approx 0.010m[/tex]
Specific weight of mercury P_mg [tex]133 kN/m^3[/tex]
Flow rate [tex]r=0.05 m^3/s[/tex]
Area A:
[tex]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\\[/tex]
Generally the Bernoulli's equation is mathematically given by
[tex]\frac{P_1}{\rho_1g}+\frac{V_1^2}{2g}=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}\\[/tex]
Where
[tex]V_1=\frac{r}{A_1} \\\\ &V_1=\frac{r}{A_2}[/tex]
Therefore
[tex]P_1-P_2=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})[/tex]
Generally the equation for pressure difference b/w manometer fluid is given as
[tex]P_1-P_2=(p_mg-pg)\triangle h[/tex]
Therefore
[tex](p_mg-pg)\triangle h=\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})[/tex]
[tex]\triangle h=\frac{\frac{Pr^2}{2}(\frac{A_1^2-A_2^2}{A_1^2A_2^2})}{(p_mg-pg)}[/tex]
[tex]\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)}[/tex]
[tex]\triangle h=4.935m[/tex]
Therefore elevation change is mathematically given by
[tex]\triangle h=4.935m[/tex]
