## Chapter 6

## 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?

## Step-by-Step

## 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.