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## Q. 6.P.21

Glycerol, of density 1260 kg/m³ and viscosity 50 mNs/m², is flowing through a 50 mm pipe and the flowrate is measured using an orifice meter with a 38 mm orifice. The pressure differential is 150 mm as indicated on a manometer filled with a liquid of the same density as the glycerol. There is reason to suppose that the orifice meter may have become partially blocked and that the meter is giving an erroneous reading. A check is therefore made by inserting a pitot tube at the centre of the pipe. It gives a reading of 100 mm on a water manometer. What does this suggest?

## Verified Solution

From the reading taken from the pitot tube, the velocity in the pipe, and hence the mass flowrate, can be calculated. From the orifice meter, the mass flowrate can also be calculated and compared with the accurate value.

For the pitot tube, $u=\sqrt{2 g h}$          (equation 6.10)

where $u=u_{\max }$ at the pipe axis, and the head loss h is in m of the liquid flowing.
Now:                $h=(100 / 1000) \times(1000 / 1260)=0.0794$ m of glycerol

∴                    $u_{\max }=\sqrt{2 \times 9.81 \times 0.0794}=1.25$ m/s
Reynolds number $=(1260 \times 1.25 \times 0.05 / 0.05)=1575$

∴                 $u_{ av }=0.5 u_{ max }=0.63$ m/s          (equation 3.36)

Mass flowrate $=\left(0.63 \times 1260 \times(\pi / 4) \times 0.05^2\right)=1.56$ kg/s

For the orifice meter:

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

$=C_D A_0 \rho \sqrt{\frac{2 g h}{1-\left(d_0 / d\right)^4}}$

$=\left(C_D \times(\pi / 4) \times 0.038^2 \times 1260\right) \sqrt{\frac{2 \times 9.81 \times(150 / 1000)}{1-(0.038 / 0.05)^4}}=2.99 C_D$

$C_D=(1.56 / 2.99)=0.53$ which confirms that the $\underline{\underline{\text { meter is faulty.}}}$