Answer:
the percent increase in the velocity of air is 25.65%
Explanation:
Hello!
The first thing we must consider to solve this problem is the continuity equation that states that the amount of mass flow that enters a system is the same as what should come out.
m1=m2
Now remember that mass flow is given by the product of density, cross-sectional area and velocity
(α1)(V1)(A1)=(α2)(V2)(A2)
where
α=density
V=velocity
A=area
Now we can assume that the input and output areas are equal
(α1)(V1)=(α2)(V2)

Now we can use the equation that defines the percentage of increase, in this case for speed

Now we use the equation obtained in the previous step, and replace values

the percent increase in the velocity of air is 25.65%
Answer:
hL = 0.9627 m
Explanation:
Given
Q = 0.040 m³/s (constant value)
D₁ = 15 cm = 0.15 m ⇒ R₁ = D₁/2 = 0.15 m/2 = 0.075 m
D₂ = 8 cm = 0.08 m ⇒ R₂ = D₂/2 = 0.08 m/2 = 0.04 m
P₁ = 480 kPa = 480*10³Pa
P₂ = 440 kPa = 440*10³Pa
α = 1.05
ρ = 1000 Kg/m³
g = 9.81 m/s²
h₁ = h₂
hL = ? (the irreversible head loss in the reducer)
Using the formula Q = v*A ⇒ v = Q/A
we can find the velocities v₁ and v₂ as follows
v₁ = Q/A₁ = Q/(π*R₁²) = (0.040 m³/s)/(π*(0.075 m)²) = 2.2635 m/s
v₂ = Q/A₂ = Q/(π*R₂²) = (0.040 m³/s)/(π*(0.04 m)²) = 7.9577 m/s
Then we apply the Bernoulli law (for an incompressible flow)
(P₂/(ρ*g)) + (α*v₂²/(2*g)) + h₂ = (P₁/(ρ*g)) + (α*v₁²/(2*g)) + h₁ - hL
Since h₁ = h₂ we obtain
(P₂/(ρ*g)) + (α*v₂²/(2*g)) = (P₁/(ρ*g)) + (α*v₁²/(2*g)) - hL
⇒ hL = ((P₁-P₂)/(ρ*g)) + (α/(2*g))*(v₁²-v₂²)
⇒ hL = ((480*10³Pa-440*10³Pa)/(1000 Kg/m³*9.81 m/s²)) + (1.05/(2*9.81 m/s²))*((2.2635 m/s)²-(7.9577 m/s)²)
⇒ hL = 0.9627 m