Question

A liquid of density 1103 kg/m3 flows withspeed 1.37 m/s into a pipe of diameter 0.19 m.The diameter of the pipe decreases to 0.05 mat its exit end. The exit end of the pipe is 6.34 m lower than the entrance of the pipe,
and the pressure at the exit of the pipe is 1.2 atm.Applying Bernoulli’s principle, what is the pressure P1 at the entrance end of the pipe? Assume the viscosity of the fluid is negligible and the fluid is incompressible. The acceleration of gravity is 9.8 m/s2 and Patm= 1.013�105Pa.

Answer 1

Answer:

Pressure at the entrance of pipe is  2.68 x 10⁵ Pa

Explanation:

Given :

Density of liquid, ρ = 1103 kg/m³

Speed of liquid at entrance of pipe, v₁ = 1.37 m/s

Diameter of pipe at entrance, d₁ = 0.19 m

Diameter of pipe at exit, d₂ = 0.05 m

Height difference between pipe, h = 6.34 m

Pressure of liquid at the exit of pipe, P₂ = 1.2 atm

Acceleration due to gravity, g = 9.8 m/s²

Consider A₁ and A₂ be the area of pipe at entrance and exit of pipe respectively, and v₂ be the speed of liquid at exit of pipe.

Applying equation of continuity,

A₁v₁ = A₂v₂

But $A_{1} =\pi \frac{d_{1} ^{2} }{4}$     and $A_{2} =\pi \frac{d_{2} ^{2} }{4}$, so the above equation becomes:

$d_{1} ^{2} v_{1} =d_{2} ^{2} v_{2}$

$\frac{d_{1} ^{2} v_{1}}{d_{2} ^{2} } =v_{2}$

Substitute the suitable values in the above equation.

$\frac{(0.19) ^{2}\times1.37}{(0.05) ^{2} } =v_{2}$

v₂ = 19.78 m/s

Consider P₁ be the pressure of the liquid at the entrance of the pipe.

Applying Bernoulli's equation:

Energy per unit volume before entering = Energy per unit volume after leaving

$P_{1} +\frac{1}{2} \rho v_{1} ^{2} +\rho gh_{1} =P_{2} +\frac{1}{2} \rho v_{2} ^{2} +\rho gh_{2}$

$P_{1} =P_{2} +\frac{1}{2} \rho( v_{2} ^{2}-v_{1} ^{2} ) +\rho g(h_{2}-h_{1})$

Substitute the suitable values in the above equation.

$P_{1} =1.2\times 1.013\times10^{5} +\frac{1}{2} \times1103\times( (19.78) ^{2}-(1.37) ^{2} ) +1103\times9.8\times (-6.34)$

P₁ = 2.68 x 10⁵ Pa

Answer 2

The pressure of the liquid at the entrance of the pipe is $2.68 \times 10^5 \ Pa$.

The exit velocity of the pipe can be determined using continuity equation as shown below;

$A_1V_1 = A_2V_2\\\\\pi(\frac{d_1}{2} )^2V_1 = \pi(\frac{d_2}{2} )^2V_2\\\\\frac{1}{4} \pi d_1^2 V_1 = \frac{1}{4} \pi d_2^2 V_2\\\\d_1^2 V_1 = d_2^2 V_2\\\\V_2 = \frac{d_1^2 V_1}{d_2^2} \\\\\V_2 = \frac{(0.19)^2 \times 1.37}{0.05^2} \\\\V_2 = 19.78 \ m/s$

Apply Bernoulli's equation to determine the pressure of the liquid at the entrance of the pipe;

$P_1 \ + \frac{1}{2} \rho v_1^2 \ + \rho gh_1 = P_2 \ + \frac{1}{2} \rho v_2^2 \ + \rho gh_2\\\\P_1 = \frac{1}{2} \rho(v_2^2 - v_1^2 ) + \rho g(h_2-h_1) \ + \ P_2\\\\P_1 = \frac{1}{2} \times 1103(19.78^2 - 1.37^2) + 1103\times 9.8 (0-6.34) + (1.2 \times 1.013 \times 10^5)\\\\P_1 = 2.68 \times 10^5 \ Pa$

Thus, the pressure of the liquid at the entrance of the pipe is $2.68 \times 10^5 \ Pa$

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