Answer:
The mass flow rate is 0.27 kg/s
The exit velocity is 76.1 m/s
The exit pressure is 695 KPa
Explanation:
Assuming the flow to be steady state and the behavior of air as an ideal gas.
The mass flow rate of the air is given as:
Mass Flow Rate = ρ x A1 x V1
where,
ρ = density of air
A1 = inlet area = 3.8 cm² = 3.8 x 10^-4 m²
V1 = inlet velocity = 100 m/s
For density using general gas equation:
PV = nRT
PV = (m/M)RT
PM/RT = ρ
ρ = (680000 N/m²)(0.02897 kg/mol)/(8.314 J/mol.k)(60 + 273)k
ρ = 7.11 kg/m³
Therefore,
Mass Flow Rate = (7.11 kg/m³)(3.8 x 10^-4 m²)(100 m/s)
<u>Mass Flow Rate = 0.27 kg/s = 270 g/s</u>
Now, for steady flow, the mass flow rate remains constant throughout the flow. Hence, flow rate at inlet will be equal to the flow rate at outlet:
Mass Flow Rate = ρ x A2 x V2
where,
ρ = density of air = 7.11 kg/m³ (Assuming in-compressible flow)
A2 = exit area = 5 cm² = 5 x 10^-4 m²
V2 = exit velocity = ?
Therefore:
0.27 kg/s = (7.11 kg/m³)(5 x 10^-4 m²) V2
<u>V2 = 76.1 m/s</u>
Now, for exit pressure, we use Bernoulli's equation between inlet and exit, using subscript 1 for inlet and 2 for exit:
P1 + (1/2) ρ V1² + ρ g h1 = P2 + (1/2) ρ V2² + ρ g h2
Since, both inlet and exit are at same temperature.
Therefore, h1 = h2, and those terms will cancel out.
P1 + (1/2) ρ V1² = P2 + (1/2) ρ V2²
P2 = P1 + (1/2) ρ V1² - (1/2) ρ V2²
P2 = P1 + (1/2) ρ (V1² - V2²)
P2 = 680000 Pa + (0.5)(7.11 kg/m³)[(100m/s)² - (76.1 m/s)²]
P2 = 680000 Pa + 14962.25 Pa
<u>P2 = 694962.25 Pa = 695 KPa</u>
