Question

A liquid of density 1270 kg/m31270 kg/m3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.61 m/s9.61 m/s and the pipe diameter ?1d1 is 10.7 cm10.7 cm. At Location 2, the pipe diameter ?2d2 is 16.1 cm16.1 cm. At Location 1, the pipe is Δ?=8.31 mΔy=8.31 m higher than it is at Location 2. Ignoring viscosity, calculate the difference Δ?ΔPbetween the fluid pressure at Location 2 and the fluid pressure at Location 1.

Answer #1

Gravitational acceleration = g = 9.81 m/s^{2}

Density of liquid =
= 1270 kg/m^{3}

Pressure at location 1 = P_{1}

Speed of flow at location 1 = V_{1} = 9.61 m/s

Diameter of pipe at location 1 = d_{1} = 10.7 cm = 0.107
m

Area of pipe at location 1 = A_{1} =
d_{1}^{2}/4

Pressure at location 2 = P_{2}

Speed of flow at location 2 = V_{2}

Diameter of pipe at location 2 = d_{2} = 16.1 cm = 0.161
m

Area of pipe at location 2 = A_{2} =
d_{2}^{2}/4

At location 1 the pipe is 8.31m higher than at location 2.

y = 8.31 m

Pressure difference at location 2 and 1 =
P = P_{2} - P_{1}

By continuity equation,

A_{1}V_{1} = A_{2}V_{2}

(d_{1}^{2}/4)V_{1}
= (d_{2}^{2}/4)V_{2}

d_{1}^{2}V_{1} =
d_{2}^{2}V_{2}

(0.107)^{2}(9.61) = (0.161)^{2}V_{2}

V_{2} = 4.244 m/s

By applying Bernoulli's equation,

P = 150738.05 Pa

P
= 1.507 x 10^{5} Pa

The difference between the fluid pressure at location 2 and the
fluid pressure at location 1 = 1.507 x 10^{5} Pa

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