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## Q. 6.P.11

The flow of water through a 50 mm pipe is measured by means of an orifice meter with a 40 mm aperture. The pressure drop recorded is 150 mm on a mercury-underwater manometer and the coefficient of discharge of the meter is 0.6. What is the Reynolds number in the pipe and what would the pressure drop over a 30 m length of the pipe be expected to be? Friction factor, $\phi=R / \rho u^2=0.0025$. Density of mercury = 13,600 kg/m³. Viscosity of water = 1 mN s/m².
What type of pump would be used, how would it be driven and what material of construction would be suitable?

## Verified Solution

Area of pipe, $A_1=(\pi / 4)(0.05)^2=0.00197 m ^2$.
Area of orifice, $A_0=(\pi / 4)(0.04)^2=0.00126 m ^2$.
h = 150 mmHg under water $=0.15 \times(13600-1000) / 1000 \equiv 1.88$ m of water.

$1-\left(A_0 / A\right)^2=0.591$, and hence:

$G=C_D A_0 \rho \sqrt{\left[2 g h /\left(1-\left(A_0 / A\right)^2\right)\right]}$             (equation 6.19)

$=(0.6 \times 0.00126 \times 1000) \sqrt{2 \times 9.81 \times 1.88 / 0.591}=5.97$ kg/s

Reynolds number, $\rho u d / \mu=d\left(G / A_1\right) / \mu=0.05(6.22 / 0.00197) /\left(1 \times 10^{-3}\right)=\underline{\underline{1.52 \times 10^5}}$

The pressure drop is given by:

$-\Delta P / v=4\left(R / \rho u^2\right)(l / d)(G / A)^2$

$=4(0.0025)(30 / 0.05)(5.97 / 0.00197)^2=5.74 \times 10^7 kg ^2 / m ^4 s ^2$

$-\Delta P=5.74 \times 10^7 \times(1 / 1000)$

$=\left(5.74 \times 10^4\right) N / m ^2$ or $\underline{\underline{57.4 kN / m ^2}}$

Power required = head loss (m) × G × g

$=\left(5.74 \times 10^4 / 1000 \times 9.81\right)(5.97 \times 9.81)=343$ W

For a pump efficiency of 60%, the actual power requirement $=(343 / 0.6)=571$ W.

Water velocity $=5.97 /(0.00197 \times 1000)=3.03$ m/s.
For this low-power requirement at a low head and comparatively low flowrate, a centrifugal pump, electrically driven and made of stainless steel, would be suitable.