Question

Water with a density of 1.000 kg/m³ flows into the smaller diameter (0,15 m) end of a 10,5 m tapered pipe, with a velocity of 20 m/s under 15 karo ter (0,15m) end of flows out of the bigger diameter (0,3 m) end of the pipe. If the inlet side of the pipe is 3,5 m higher than the outlet side, calculate the pressure exerted at the outlet end of the tapered pipe

Answer 1

To calculate the pressure exerted at the outlet end of the tapered pipe, we will use the Bernoulli's equation:

P1 + ρgh1 + 0.5ρv1^2 = P2 + ρgh2 + 0.5ρv2^2

Where:
P1 = pressure at the inlet end of the pipe
P2 = pressure at the outlet end of the pipe
ρ = density of water = 1000 kg/m³
g = acceleration due to gravity = 9.81 m/s²
h1 = height difference between the inlet and outlet sides = 3.5 m
h2 = 0
v1 = velocity at the inlet end of the pipe = 20 m/s
v2 = velocity at the outlet end of the pipe
The cross-sectional areas at the inlet and outlet ends of the pipe are:
A1 = π(0.15)^2 /4 = 0.01767 m²
A2 = π(0.3)^2 /4 = 0.07069 m²

Since the volume flow rate is constant, we can use the equation of continuity:

A1v1 = A2v2

0.01767 * 20 = 0.07069 * v2

v2 = 5 m/s

Now we can plug in the values into the Bernoulli's equation:

P1 + ρgh1 + 0.5ρv1^2 = P2 + ρgh2 + 0.5ρv2^2

P1 + ρgh1 + 0.5ρv1^2 = P2 + ρgh2 + 0.5ρv2^2

P1 + 1000*9.81*3.5 + 0.5*1000*20^2 = P2

P1 + 34350 + 20000 = P2

P2 = P1 + 54350

Therefore, the pressure exerted at the outlet end of the tapered pipe is 54,350 Pa higher than the pressure at the inlet end of the pipe.