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## Q. 6.P.9

Water flowing at 1.5 l/s in a 50 mm diameter pipe is metered by means of a simple orifice of diameter 25 mm. If the coefficient of discharge of the meter is 0.62, what will be the reading on a mercury-under-water manometer connected to the meter? What is the Reynolds number for the flow in the pipe? Density of water = 1000 kg/m³. Viscosity of water = 1 mN s/m².

## Verified Solution

Mass flowrate, $G=\left(1500 \times 10^{-6} \times 1000\right)=1.5$ kg/s.
Area of orifice, $A_0=(\pi / 4)(0.025)^2=0.00049$ m².
Area of pipe, $A_1=(\pi / 4)(0.050)^2=0.00196$ m².
Reynolds number $=\rho u d / \mu=d\left(G / A_1\right) / \mu$

$=0.05(1.5 / 0.00196) /\left(1 \times 10^{-3}\right)=\underline{\underline{3.83 \times 10^4}}$

The orifice meter equations are 6.19 and 6.21; the latter being used when $\sqrt{\left[1-\left(A_0 / A_1\right)^2\right]}$ approaches unity.

$G=\frac{C_D A_0}{v} \sqrt{\frac{2 v\left(P_1-P_2\right)}{1-\left(A_0 / A_1\right)^2}}$           (equation 6.19)

Thus:          $\sqrt{\left[1-\left(A_0 / A_1\right)^2\right]}=\sqrt{\left[1-\left(25^2 / 50^2\right)^2\right]}=0.968$

Using equation 6.21, $G=C_D A_0 \rho \sqrt{(2 g h)}$ gives:

$1.5=0.62 \times 0.00049 \times 1000 \sqrt{(2 \times 9.81 h)}$, and h = 1.24 m of water

Using equation 6.19 in terms of h gives:

$1.5=(0.62 \times 0.00049 \times 1000 / 0.968) \sqrt{(2 g h)}$, and h = 1.16 m of water

This latter value of h should be used. The height of a mercury-under-water manometer would then be $1.16 /((13.55-1.00) / 1.00)=0.092$ m or $\underline{\underline{92 mm Hg}}$