1 atm = 2116 lb/ft2 = 1.01 × 10^5 N/m^2. Consider the isentropic flow over an airfoil. The freestream conditions are T∞ = 245 K and p∞ = 4.35 × 10^4 N/m^2. At a point on the airfoil, the pressure is 3.6 × 10^4 N/m^2. Calculate the density at this point.
The Correct Answer and Explanation is:
To calculate the density at the point on the airfoil, we need to use the Ideal Gas Law for air, which is:p=ρRTp = \rho R Tp=ρRT
Where:
- ppp is the pressure,
- ρ\rhoρ is the density,
- RRR is the specific gas constant for air (R=287 J/kg\cdotpKR = 287 \, \text{J/kg·K}R=287J/kg\cdotpK),
- TTT is the temperature.
Step 1: Use the isentropic flow assumption
In isentropic flow, the temperature and pressure are related. However, since the temperature at the point on the airfoil isn’t given directly, we will use the isentropic relation to approximate it. First, we can calculate the density in terms of pressure and temperature.
Step 2: Rearranging the Ideal Gas Law
We can rearrange the Ideal Gas Law to solve for density:ρ=pRT\rho = \frac{p}{RT}ρ=RTp
Given:
- p=3.6×104 N/m2p = 3.6 \times 10^4 \, \text{N/m}^2p=3.6×104N/m2,
- R=287 J/kg\cdotpKR = 287 \, \text{J/kg·K}R=287J/kg\cdotpK,
- T∞=245 KT_{\infty} = 245 \, \text{K}T∞=245K (freestream temperature).
Since the flow is isentropic, we assume that the temperature at the point on the airfoil is nearly the same as the freestream temperature, so T≈T∞=245 KT \approx T_{\infty} = 245 \, \text{K}T≈T∞=245K.
Step 3: Calculate the density
Now, substitute the values into the Ideal Gas Law equation:ρ=3.6×104287×245\rho = \frac{3.6 \times 10^4}{287 \times 245}ρ=287×2453.6×104ρ=3.6×10470315\rho = \frac{3.6 \times 10^4}{70315}ρ=703153.6×104ρ≈0.512 kg/m3\rho \approx 0.512 \, \text{kg/m}^3ρ≈0.512kg/m3
Conclusion:
The density at the point on the airfoil is approximately 0.512 kg/m³. This calculation assumes that the temperature at the point on the airfoil is close to the freestream temperature due to the isentropic assumption.
