Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The differential height is [tex]h= 1.458cm[/tex]
Explanation:
The schematic diagram of the venturimeter is shown on the second uploaded image
The continuity equation is mathematically given as
[tex]\r V = Av[/tex]
Where A is the area
[tex]\r V[/tex] is the flow rate
[tex]v_1[/tex] is the velocity
At the inlet making v the subject to obtain the inlet velocity we have
[tex]v_1 = \frac{\r V}{A_1}[/tex]
Substituting 124 L/s = [tex]0.124 m^3 /s[/tex] for [tex]\r V[/tex] and [tex]A_1 = \frac{\pi}{4 } *d^2 = \frac{\pi}{4} * 0.22^2[/tex] given that
[tex]d =22cm= \frac{22}{100} = 0.22m[/tex]
So [tex]v_1 =\frac{0.124}{\frac{\pi}{4} * 0.22^2 }[/tex]
[tex]=3.26m/s[/tex]
At the exit point the velocity is
[tex]v_2 = \frac{\r V}{A_2}[/tex]
Where [tex]A_ 2 = \frac{\pi}{4}d^2 = \frac{\pi}{4} * 0.10^2[/tex] given that [tex]d =10cm= \frac{10}{100} = 0.10m[/tex]
So [tex]v_2 = \frac{0.124}{\frac{\pi}{4} *(0.10)^2}[/tex]
[tex]= 15.78m/s[/tex]
The Bernoulli's flow equation between the inlet and exist is mathematically given as
[tex]\frac{P_1}{\rho_o } + \frac{v_1^2}{2} = \frac{P_2}{\rho_0} +\frac{v_2^2}{2} +gz_2 ---(1)[/tex]
And
[tex]P_1 - P_2 = \rho_o [\frac{v_2^2}{2} + \frac{v_1^2}{2} ] = \rho_{water} gh ---(2)[/tex]
Where [tex]P_1[/tex] is the pressure at inlet
[tex]P_2[/tex] is the pressure at exist
[tex]\rho_{water}[/tex] is the density of water with value of [tex]1000kg/m^3[/tex]
g is acceleration due to gravity
h is the height of the water column
making h the subject in the equation 2
[tex]h = \rho _o \frac{v_2^2 - v_1^2}{2g\rho_{water}}[/tex]
where [tex]\rho _o[/tex] is the density of air given in the question
Substituting value
[tex]h = (1.20) \frac{15.78^2 - 3.26^2}{2 (9.81 )(1000)}[/tex]
[tex]h= 1.458cm[/tex]
