86 CHEMICAL ENGINEERING VOLUME 1 SOLUTIONS
hD 0 .3mofwateror 0. 3 / 0. 9 D 0 .333 m of petroleum product
CDD 0 .62 (assumed)∴ 45. 0 D 0. 62 ðA 0 ð 900
p
2 ð 9. 81 ð 0. 333 Thus: A 0 D 0 .3155 m^2 andd 0 D 0 .2m.
This orifice diameter is larger than the pipe size so that it was clearly wrong to use the
simpler equation.
Thus: GDCDA 0
√
[2gh/ 1 A 0 /A 1 ^2 ] (equation 6.19)A 1 D/ 4 0. 15 ^2 D 0 .0177 m^2∴ 45. 0 D 0. 62 ðA 0 ð 900
√
[2ð 9. 81 ð 0. 33 / 1 A 0 / 0. 0177 ^2 ]Thus: A 0 D 0 .154 m^2 andd 0 D 0 .14 m
PROBLEM 6.11
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-under-
water 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,!DR/u^2 D 0 .0025. Density of mercuryD
13 ,600 kg/m^3. Viscosity of waterD1mNs/m^2.
What type of pump would be used, how would it be driven and what material of
construction would be suitable?
Solution
Area of pipe,A 1 D/ 4 0. 05 ^2 D 0 .00197 m^2.
Area of orifice,A 0 D/ 4 0. 04 ^2 D 0 .00126 m^2.
hD150 mmHg under waterD 0. 15 ð 13600 1000 / 1000 1 .88 m of water.
1 A 0 /A^2 D 0 .591, and hence:
GDCDA 0
√
[2gh/ 1 A 0 /A^2 ] (equation 6.19)D 0. 6 ð 0. 00126 ð 1000 √
2 ð 9. 81 ð 1. 88 / 0. 591 D 5 .97 kg/sReynolds number,ud/ DdG/A 1 / D 0. 05 6. 22 / 0. 00197 / 1 ð 10 ^3 D 1. 52 ð 105
The pressure drop is given by:
P/vD 4 R/u^2 l/dG/A^2
D 4 0. 0025 30 / 0. 05 5. 97 / 0. 00197 ^2 D 5. 74 ð 107 kg^2 /m^4 s^2