Calculate the rate of flow of glycerine of density 1.25x 10 kgm through the conical section of a horizontal pipe, if the radii of its ends are 0.1 m and 0.04m and the pressure drop across its length is 10 Nm-2?
Answers
[Q]__?
=>According to the Question we have drawn the following diagram [in attachment]
From Continuity equation,
A1v1 = A2v2
v1/v2 = A2/A1 = π(r2 ^2)/π(r1 ^2)
=[r2/r1]^2 = [0.04 /0.1]^2 = 4/25 _[i]
From Bernoulli's Equation,
P1 + ½ ρv1^2 = P2 + ½ ρv2^2
_[As h1=h2]
•ρ= fluid density
•g = acceleration due to gravity
•P1 = pressure at elevation 1
•v1 = velocity at elevation 1
•P2= pressure at elevation 2
•v2= velocity at elevation 2
∴ v2^2- v1^2 = 2(P1 - P2 )/ρ
∴ v2^2- v1^2 = 2× 10/1. 25 × 10^3
= 1.6×10^-2 m^2s^-2 _[ii]
_________
Solving Equations [i] & [ii] we get:
v2 ≈ 0.128 ms^-1
∴Rate of volume flow through the tube
Q = A2v2 = [π(r2^2)]v2
= π (0.04)^2 (0.128)
= 6.43 × 10^-4 m^3 s^-1
____________________________________
Answer: