Answer:
a) 180 m³/s
b) 213.4 kg/s
Explanation:
[tex]A_1[/tex] = 1 m²
[tex]P_1[/tex] = 100 kPa
[tex]V_1[/tex] = 180 m/s
Flow rate
[tex]Q=A_1V_1\\\Rightarrow Q=1\times 180\\\Rightarrow Q=180\ m^3/s[/tex]
Volumetric flow rate = 180 m³/s
Mass flow rate
[tex]\dot{m}=\rho Q\\\Rightarrow \dot m=\frac{P_1}{RT} Q\\\Rightarrow \dot m=\frac{100000}{287\times 293.15}\times 180\\\Rightarrow \dotm=213.94\ kg/s[/tex]
Mass flow rate = 213.4 kg/s
The volumetric flow rate and mass flow rate are respectively; Q₁ = 180 m³/s and m' = 213.4 kg/s
We are given;
Initial cross sectional area; A₁ = 1 m²
The inlet pressure of the air; P₁ = 100 kPa = 100000 Pa
The inlet velocity; V₁ = 180 m/s
Inlet Temperature; T₁ = 20 °C = 293.15 K
A) Formula for Volumetric rate is;
Q₁ = A₁ * V₁
Q₁ = 1 * 180
Q₁ = 180 m³/s
B) Formula for mass flow rate is;
m' = PQ/RT
where R = 287 J/kg.k
Thus;
m' = (100000 * 180)/(287 * 293.15)
m' = 213.4 kg/s
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