Answer:
(a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]
(b). Â The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
Explanation:
Given that,
Gauge pressure at bottom = pâ
Suppose, an arrangement with a horizontal pipe carrying fluid of density p . The fluid rises to heights h1 and h2 in the two open-ended tubes (see figure). The cross-sectional area of the pipe is A1 at the position of tube 1, and A2 at the position of tube 2.
Find the speed of the fluid in the left end of the main pipe.
(a). We need to calculate the gauge pressure at the bottom of tube 1
Using bernoulli equation
[tex]P_{1}=\rho g h_{1}[/tex]
(b). We need to calculate the speed of the fluid in the left end of the main pipe
Using bernoulli equation
Pressure for first pipe,
[tex]P_{1}=\rho gh_{1}[/tex].....(I)
Pressure for second pipe,
[tex]P_{2}=\rho gh_{2}[/tex].....(II)
From equation (I) and (II)
[tex]P_{2}-P_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]
Put the value of Pâ and Pâ
[tex]\rho g h_{2}-\rho g h_{1}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)[/tex]
[tex]gh_{2}-gh_{1}=\dfrac{1}{2}(v_{1}^2-v_{2}^2)[/tex]
[tex]2g(h_{2}-h_{1})=v_{1}^2-v_{2}^2[/tex]....(III)
We know that,
The continuity equation
[tex]v_{1}A_{1}=v_{2}A_{2}[/tex]
[tex]v_{2}=v_{1}(\dfrac{A_{1}}{A_{2}})[/tex]
Put the value of vâ in equation (III)
[tex]2g(h_{2}-h_{1})=v_{1}^2-(v_{1}(\dfrac{A_{1}}{A_{2}}))^2[/tex]
[tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\dfrac{A_{1}}{A_{2}}))^2[/tex]
Here, [tex]\dfrac{A_{1}}{A_{2}}=\gamma[/tex]
So, [tex]2g(h_{2}-h_{1})=v_{1}^2(1-(\gamma)^2)[/tex]
[tex]v_{1}=\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
Hence, (a). The gauge pressure at the bottom of tube 1 is [tex]P_{1}=\rho g h_{1}[/tex]
(b). Â The speed of the fluid in the left end of the main pipe [tex]\sqrt{\dfrac{2g(h_{2}-h_{1})}{(1-(\gamma)^2)}}[/tex]
