Answer:
height of the water rise in tank is 10ft
Explanation:
Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)
[tex]\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2[/tex]
[tex]\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}[/tex]-------(1)
substitute [tex]P_a_t_m for P_1, (P_a_t_m +pgh) for P_2[/tex]
0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)
[tex]\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}[/tex]
[tex]\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}[/tex]
Applying bernoulli's equation between tank surface (3) and orifice exit (4)
[tex]\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4[/tex]
substitute
[tex]P_a_t_m for P_3, P_a_t_m for P_4[/tex]
0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g
[tex]\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}[/tex]
At equillibrium Fow rate at point 2 is equal to flow rate at point 4
Q₂ = Q₄
A₂V₂ = A₃V₃
The diameter of the orifice and the siphon are equal , hence there area should be the same
substitute A₂ for A₃
[tex]\sqrt{64.4(20-h)}[/tex] for V₂
[tex]\sqrt{64.4h}[/tex] for V₄
A₂V₂ = A₃V₃
[tex]A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft[/tex]
Therefore ,height of the water rise in tank is 10ft
