Answer to Question #239566 in Molecular Physics | Thermodynamics for Unknown346307

Question

An aeroplane is flying at 1000 km/h through still air having a pressure of

78.5 kN/m2 (abs.) and temperature – 8°C. Calculate on the stagnation point on the nose of the

plane :

(i) Stagnation pressure,

(ii) Stagnation temperature, and

(iii) Stagnation density.

Take for air : R = 287 J/kg K and γ = 1.4

Answer 1

The sonic velocity for adiabatic flow is given by $C_0=\sqrt{\gamma RT}$ and then the Mach number as $M_0=\cfrac{V_0}{C_0}$. These values need to be calculated because the stagnation pressure is given by

$p_s=p_0 {\Big[ 1+\Big( \dfrac{\gamma-1}{2} \Big){M_0}^2 \Big]}^{\dfrac{\gamma}{\gamma-1}}$

If we substitute the prior equation we have the following relation between all the variables we have and the stagnation pressure:

$p_s=p_0 {\Big[ 1+\Big( \dfrac{\gamma-1}{2\gamma RT} \Big){V_0}^2 \Big]}^{\dfrac{\gamma}{\gamma-1}}$

We proceed to substitute and calculate with

$p_0=78.5\,{kN/m^2} \\T_0=-8+273=265 \,K \\R=287\,J/kgK \\ \gamma=1.4$

$V_0=\cfrac{1000\,km}{h}\times\cfrac{1000\,m}{1\,km}\times\cfrac{1\,h}{3600\,s} \\ \implies V_0=277.77\,{m/s}$

and we find:

$p_s=(78.5\,kN/m^2) {\bigg[ 1+\Big( \dfrac{1.4-1}{2(1.4)(287\,{J/kgK})(265\,K)} \Big){(277.77\,{m/s})}^2 \bigg]}^{\cfrac{1.4}{1.4-1}} \\ \text{ } \\ \implies p_s=(78.5\,kN/m^2) {\Big[ 1.14492552 \Big]}^{3.5} \\ \text{ } \\ p_s=126.064\,{kN/m^2}$

(b) The stagnation temperature is given by

$T_s=T_0 {\Big[ 1+( \frac{\gamma-1}{2} ){M_0}^2 \Big]} \\ \implies T_s=T_0{\Big[ 1+\Big( \dfrac{\gamma-1}{2\gamma RT} \Big){V_0}^2 \Big]}$

After the same substitution we can find the temperature:

$T_s=(265\,K) {\Big[ 1+\Big( \dfrac{1.4-1}{2(1.4)(287\,{J/kgK})(265\,K)} \Big){(277.77\,{m/s})}^2 \Big]} \\ \text { } \\ T_s=(265\,K) {\Big[ 1.14492552 \Big]} \\ \text { } \\ \implies T_s=303.4\,K =30.4°C$

(c) At last, the stagnation density (ρ_s) is given by

$\cfrac{p_s}{\rho_s}=RT_s \implies \rho_s = \cfrac{p_s}{RT_s }$

We substitute the prior values and we find

$\rho_s = \cfrac{126064\,{N/m^2}}{(287\,{J/kgK})(303.4\,K)} \\ \text { } \\ \implies \rho_s =1.448\,{kg/m^3}$

In conclusion, (i) the stagnation pressure is $p_s=126.064\,{kN/m^2}$ , while the (ii) stagnation temperature is $T_s=303.4\,K =30.4°C$, and (iii) the stagnation density is $\rho_s =1.448\,{kg/m^3}$.

Reference:

• Rajput, R. K. (2005). A textbook of engineering thermodynamics. Laxmi Publications.