A liquid of density 1296 km/m^3 with speed 1.07 m/s into a pipe of diameter 0.24

Question

A liquid of density 1296 km/m^3 with speed 1.07 m/s into a pipe of diameter 0.24m. A liquid of density 1296 km/m^3 with speed 1.07 m/s into a pipe of diameter 0.24m. 019 10.0 points A liquid of density 1296 kg/m3 flows with speed 1.07 m/s into a pipe of diameter 0.24 m The diameter of the pipe decreases to 0.05 m at its exit end. The exit end of the pipe is 6.29 m lower than the entrance of the pipe, and the pressure at the exit of the pipe is 1.3 atm 1.3 atm6.29 m 1.07 m/s( 0.24 m 0.05 m Applying Bernoulli's principle, what is the pressure P at the entrance end of the pipe? Assume the viscosity of the fluid

Explanation / Answer

P = pressure
d = density
V = velocity
h = elevation
g = gravitation acceleration

P + (1/2)dV^2 + dgh = constant value = Bernoulli's equation.

Given :
P1 = ?
V1 = 1.07 m/s
A1 = (.24m/2)^2 * pi
A2 = (.05m/2)^2 * pi
P2 = 1.3 atm = 131723 Pascals
V2 = ?

So, to use bernoulli's equation, first solve for V2
Use the formula, A1V1 = A2V2
(.24/2)^2 * pi * 1.07 = (.05/2)^2 * pi * V2
=> V2 = 24.65 m/s

Now, we have all variables needed for bernoulli's equation
x + (1/2)(1296)(1.07)^2 + (1296)*(9.8)*(6.29) = 131723 + (1/2)(1296)(24.65)^2 + (1296)*(9.8)*(0)

=> P1 = 444832 Pa

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