Question 6.P.23: The flow of liquid in a 25 mm diameter pipe is metered with ...
The flow of liquid in a 25 mm diameter pipe is metered with an orifice meter in which the orifice has a diameter of 19 mm. The aperture becomes partially blocked with dirt from the liquid. What fraction of the area can become blocked before the error in flowrate at a given pressure differential exceeds 15 per cent? Assume that the coefficient of discharge of the meter remains constant when calculated on the basis of the actual free area of the orifice.
If two sections in the pipe are chosen, 1 being upstream and 2 at the orifice, then from an energy balance:
u_1^2 / 2+P_1 v=u_2^2 / 2+P_2 v (from equation 2.55)
and G, the mass flowrate =u_2 A_2 / v=u_1 A_1 / v
∴ \left(u_2^2 / 2\right)\left(A_2 / A_1\right)^2+P_1 v=u_2^2 / 2+P_2 v
or: u_2^2=\frac{2\left(P_1-P_2\right) v}{1-\left(A_2 / A_1\right)^2}
The volumetric flowrate, Q=C_D A_2 u_2
∴ Q^2=C_D^2 A_2^2 \times \frac{2\left(P_1-P_2\right) v}{1-\left(A_2 / A_1\right)^2}
=2 C_D^2\left(P_1-P_2\right) v A_1^2 A_2^2 /\left(A_1^2-A_2^2\right)
or: Q=K \frac{A_1 A_2}{\sqrt{A_1^2-A_2^2}} (1)
If the area of the orifice is reduced by partial blocking, the new orifice area =r A_2 where f is the fraction available for flow. The new flowrate = 0.85 Q when the error is 15 per cent and:
0.85 Q=\frac{K A_1 f A_2}{\sqrt{A_1^2-f^2 A_2^2}} (2)
A_1=(\pi / 4) \times 25^2=491 mm ^2
A_2=(\pi / 4) \times 19^2=284 mm ^2
∴ Dividing equation (2) by equation (1) and substituting gives:
0.85=\frac{f \sqrt{\left(491^2-284^2\right)}}{\sqrt{\left(491^2-r^2 284^2\right)}}
from which f = 0.89 or \underline{\underline{11 \text { per cent }}} of the area is blocked.