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
and
, so the above equation becomes:
![d_{1} ^{2} v_{1} =d_{2} ^{2} v_{2}](https://tex.z-dn.net/?f=d_%7B1%7D%20%5E%7B2%7D%20v_%7B1%7D%20%3Dd_%7B2%7D%20%5E%7B2%7D%20v_%7B2%7D)
![\frac{d_{1} ^{2} v_{1}}{d_{2} ^{2} } =v_{2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd_%7B1%7D%20%5E%7B2%7D%20v_%7B1%7D%7D%7Bd_%7B2%7D%20%5E%7B2%7D%20%7D%20%20%3Dv_%7B2%7D)
Substitute the suitable values in the above equation.
![\frac{(0.19) ^{2}\times1.37}{(0.05) ^{2} } =v_{2}](https://tex.z-dn.net/?f=%5Cfrac%7B%280.19%29%20%5E%7B2%7D%5Ctimes1.37%7D%7B%280.05%29%20%5E%7B2%7D%20%7D%20%20%3Dv_%7B2%7D)
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}](https://tex.z-dn.net/?f=P_%7B1%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20v_%7B1%7D%20%5E%7B2%7D%20%2B%5Crho%20gh_%7B1%7D%20%3DP_%7B2%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Crho%20v_%7B2%7D%20%5E%7B2%7D%20%2B%5Crho%20gh_%7B2%7D)
![P_{1} =P_{2} +\frac{1}{2} \rho( v_{2} ^{2}-v_{1} ^{2} ) +\rho g(h_{2}-h_{1})](https://tex.z-dn.net/?f=P_%7B1%7D%20%20%3DP_%7B2%7D%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Crho%28%20v_%7B2%7D%20%5E%7B2%7D-v_%7B1%7D%20%5E%7B2%7D%20%29%20%2B%5Crho%20g%28h_%7B2%7D-h_%7B1%7D%29)
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)](https://tex.z-dn.net/?f=P_%7B1%7D%20%20%3D1.2%5Ctimes%201.013%5Ctimes10%5E%7B5%7D%20%20%2B%5Cfrac%7B1%7D%7B2%7D%20%5Ctimes1103%5Ctimes%28%20%2819.78%29%20%5E%7B2%7D-%281.37%29%20%5E%7B2%7D%20%29%20%2B1103%5Ctimes9.8%5Ctimes%20%28-6.34%29)
P₁ = 2.68 x 10⁵ Pa