Respuesta :
Answer:
1. ρ₂ = 1.76 kg/m³ , P₂ = 168.699 kPa
2. T₀ = 345.75 K, P₀ = 191.742 kPa, ρ₀= 1.93 kg/m³
3. V₁ = 340.26 m/s , V₂ = 365.96 m/s
4. the local airspeed at 0.5 atm point = 308.2 m/s
5. The percentage difference is = 6.98 % ≈ 7 %
Explanation:
Pressure, P = 101.325 kPa
Density = 1.225 kg/m³
Temperature, T = 288.15 K = 518.67 °R
Gas constant of air, Rair = 287.057 J/(kg·K)
1. Thus since the density, ρ₁ at sea level standard, SLS = 1.225 kg/m³
T₁ = 288.15, γ = 1.005, T₂ = 600 °R = 333.33 K
ρ₂ = ρ₁× [tex](\frac{T_{2} }{T_{1} }) ^{\frac{1}{\gamma - 1} }[/tex]
= 1.225× [tex](\frac{333.33}{288.15}) ^{\frac{1}{1.4 - 1} }[/tex]
= 1.76 kg/m³
Similarly the pressure P₂ is given by [tex]P_{1} (\frac{T_{2} }{T_{1} }) ^{\frac{\gamma}{\gamma - 1} }[/tex]
From where P₁ = 101.325 kPa hence P₂ = 168.699 kPa
2). Speed = [tex]\sqrt{kRT}[/tex]
= Where R = 287.057 J/(kg·K)
Hence speed = [tex]\sqrt{1.4*287.057*288.15}[/tex] = 340.26 m/s
Therefore T₀ = T₁ + [tex]\frac{V_{1} ^{2} }{2C_{p} }[/tex]
= 288.15 + 340.26²/(2×1.005))×(1/1000) = 345.75 K
Therefore as we have
[tex]P_{0} = P_{1} (\frac{T_{01} }{T_{1} }) ^{\frac{1}{\gamma - 1} }[/tex]
, P₀ =101325 ×(345.75/288.15)^(1.4/(1.4-1)
= 191.742 kPa
and ρ₀ = ρ₁ ×[tex](\frac{T_{01} }{T_{1} }) ^{\frac{1}{\gamma - 1} }[/tex]
ρ₁ = 1.225 kg/m³, T₀ = 345.75 K, T₁ = 288.15 K
∴ ρ₀= 1.93 kg/m³
3.) The speed is given as Speed = [tex]\sqrt{kRT}[/tex]
Hence at 1 speed as previously calculated = 340.26 m/s
while at 2 speed = [tex]\sqrt{1.4*287.057*333.33}[/tex]
= 365.96 m/s
4.) If the speed drops to 0.5 atm we have
[tex]\frac{P_{2} }{P_{1} } =(\frac{T_{2} }{T_{1} }) ^{\frac{\gamma}{\gamma - 1} }[/tex] from where P₂/P₁ = 101.325 kPa/ 50.6625 kPa = 0.5 hence
0.5 = [tex](\frac{T_{2} }{288.15 }) ^{\frac{1.4}{1.4 - 1} }[/tex]
From where T₂ = 236.4 K
Hence speed = [tex]\sqrt{kRT}[/tex] =
= [tex]\sqrt{1.4*287.057*308.2}[/tex]
= 308.2 m/s
5.) With Bernoulli's equation we have
We have ΔP/ρ -v₁²/2 = -v₂²/2
Hence we have (101.325 kPa -168.699 kPa )/1.225 kg/m3 - (340.26 m/s )²/2 = -v₂²/2
or v₂ = 340.42 m/s
The percentage difference is 100* (365.96 m/s - 340.42 m/s)/365.96 m/s = 6.98 % ≈ 7 %
