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
Step-by-step 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:

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

The flow rate

b) Applying the Bernoulli's equation between the points 2 and 4

z₄ - z₂ = 0 (same height)
P₄ = 0 (open at atmosphere)

Solving for v₂
v₂ = 36.5 m/s
To get the height you need to use the continuity expression


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

Here
z₄-z₁ = 0, same height
v₄-v₁ = 0, same speeds
P₄ = 0, open atmosphere
Substituting values:
