Question

A 6.8 cm diameter pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 33.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Answer 1

Answer :0.00568358 m³/s

the Bernoulli equation for a horizontal tube is
p1 + 1/2*ρ*v1^2 = p2 + 1/2*ρ*v2^2

v2/v1 = r1^2/r2^2 = 6.8^2/4^2 = 2.89
---> v1 = 2.89v1 ---> substitute in Bernoulli equation:

p1 + 1/2*ρv1^2 = p2 + 1/2*ρ(2.89v1)^2 whith

p1 = 33 000 Pa
p2 = 24 000 Pa
ρ = 1000 kg/m^3

33 000 + 1/2*1000*v1^2 = 24 000 + 1/2*1000*(2.89v1)^2
solution for v1 = 1.565 m/s

the volume/s is therefore
V = (pi*r^2/4)*L = pi*6.8^2*156.5 /4

= 5683.58 cm^3/s

= 0.00568358 m^3/s

Answer 2

density of water = 1000 kg/m³
D1 = 0.068 m A1 = πD1²/4 = 3.632E-3 m²
D2 = 0.040 m A2 = πD2²/4 = 1.26E-3 m²

P/1000 = fluid energy due to pressure
E1 = 33E3/E3 = 33 J/kg
E2 = 24E3/E3 = 24 J/kg
E1 - E2 = 9 J/kg

assuming pipe is level (doesn't change height):
(V2² - V1²)/2 = 9
V2² - V1² = 18
V1(A1) = V2(A2)
V1 = V2(A2/A1) = (1.26/3.632)V2 = 0.347V2
V2² - (0.347V2)² = 18
V2² - 0.120V2² = 18
0.88V2² = 18
V2² = 20.5
V2 =4.52 m/s
V2A2 = (4.52)(1.26E-3) = 5.7E-3 m³/s = ANS

answered by: asen