Answer:
a) The flow rate if the water is 0.0553 m³/s
b) The height is 0.0253 m
c) The pressure at the inlet 1 is 0
Explanation:
Given:
Width of the channel is 0.06 m
Height at 1 and 4 is 0.04 m
Height at 3 is 0.02 m
ρwater = 1000 kg/m³
ρair = 1.23 kg/m³
Questions:
a) The flow rate if the water in the small tube connected to the static pressure tap, Q = ?
b) Calculate the height in section 2, h₂ = ?
c) Calculate the pressure at the inlet (1) for the fluid to flow, P₁ = ?
a) To solve this question, you need to apply the Bernoulli's equation between the points 3 and 4:
[tex]\frac{P_{4}-P_{3} }{\rho _{air} } +\frac{v_{4}^{2} -v_{3}^{2} }{2 } +g(z_{4} -z_{3} )=0[/tex]
Here,
P₄ = 0 (open at atmosphere)
z₄-z₃ = 0 (same height)
P₃ = ρgh₃ = 1000*9.8*0.1 = 980
Substituting values and solving for v₄
[tex]v_{4} =\sqrt{\frac{2P_{3} }{3\rho _{air} } } =\sqrt{\frac{2*980}{3*1.23} } =23.047m/s[/tex]
The flow rate
[tex]Q=A_{4} v_{4} =0.06*0.04*23.047=0.0553m^{3} /s[/tex]
b) Applying the Bernoulli's equation between the points 2 and 4
[tex]g(z_{4} -z_{2} )+\frac{P_{4}-P_{2} }{\rho } +\frac{v_{4}^{2}-v_{2}^{2} }{2} =0[/tex]
z₄ - z₂ = 0 (same height)
P₄ = 0 (open at atmosphere)
[tex]0+\frac{1000*9.8*0.05}{1.23} +\frac{23.047^{2}-v_{2}^{2} }{2} =0[/tex]
Solving for v₂
v₂ = 36.5 m/s
To get the height you need to use the continuity expression
[tex]v_{4} A_{4} =v_{2} A_{2}[/tex]
[tex]23.047*0.06*0.04=36.5*0.06*h_{2}[/tex]
Solving for h₂
h₂ = 0.0253 m
c) In this part, you need to apply the Bernoulli's expression between the points 1 and 4
[tex]g(z_{4} -z_{1} )+\frac{P_{4}-P_{1} }{\rho } +\frac{v_{4}^{2}-v_{1}^{2} }{2} =0[/tex]
Here
z₄-z₁ = 0, same height
v₄-v₁ = 0, same speeds
P₄ = 0, open atmosphere
Substituting values:
[tex]0+\frac{0-P_{1} }{\rho } +0=0\\P_{1} =0[/tex]
