When a fluid is flowing past and immersed body, and at a point if the resultant velocity of the body becomes zero, the values of temperature, pressure and density at that point are known as Stagnation Properties. That point is called as stagnation point.

They are denoted by (pressure = $P_S$), (temperature = $T_S$) and (density = $\rho_S$).

Applying Bernoulli's equation for adiabatic process flow,

$\left(\dfrac K{K-1}\right)\dfrac {P_1}{\rho_1 g}+\dfrac {V_1^2}{2g}+z_1=\left(\dfrac K{K-1}\right)\dfrac {P_2}{\rho_2 g}+\dfrac {V_2^2}{2g}+z_2$

But $z_1=z_2$ and cancelling $\dfrac 1g$

$\left(\dfrac K{K-1}\right)\dfrac {P_1}{\rho_1}+\dfrac {V_1^2}{2}=\left(\dfrac K{K-1}\right)\dfrac {P_2}{\rho_2}+\dfrac {V_2^2}{2}$

Point 2 is a stagnation point. Hence velocity will become zero at stagnation point and pressure density will be denoted by $P_S$ and $\rho_S$.

$\therefore V_2=0, P_2=P_S, \rho_2=\rho_S$

Put these values in above equation, we get

$\left(\dfrac K{K-1}\right)\dfrac {P_1}{\rho_1}+\dfrac {V_1^2}{2}=\left(\dfrac K{K-1}\right)\dfrac {P_S}{\rho_S}+0$

$\left(\dfrac K{K-1}\right)\left[\dfrac {P_1}{\rho_1}-\dfrac {P_S}{\rho_S}\right]=-\dfrac {V_1^2}{2}$

$\left(\dfrac K{K-1}\right)\dfrac {P_1}{\rho_1}\left[1-\dfrac {P_S}{\rho_S}\times \dfrac {\rho_1}{P_1}\right]=-\dfrac {V_1^2}{2}$

$\left(\dfrac K{K-1}\right)\dfrac {P_1}{\rho_1}\left[1-\dfrac {P_S}{P_1}\times \dfrac {\rho_1}{\rho_S}\right]=-\dfrac {V_1^2}{2}..................(1)$

But for adiabatic process,

$\dfrac P{\rho^K}=\text{constant}$

or, $\dfrac {P_1}{\rho_1^K}=\dfrac {P_S}{\rho_S^K}$

or, $\left(\dfrac {\rho_1}{\rho_S}\right)=\left(\dfrac {P_1}{P_S}\right)^{1/K}.......................(2)$

Putting the value of $\dfrac {\rho_1}{\rho_S}$ in eqaution (1),

$\left(\dfrac K{K-1}\right)\dfrac {P_1}{\rho_1}\left[1-\dfrac {P_S}{P_1}\times\left(\dfrac {P_1}{P_S}\right)^{1/K}\right]=-\dfrac {V_1^2}{2}$

$\left(\dfrac K{K-1}\right)\dfrac {P_1}{\rho_1}\left[1-\dfrac {P_S}{P_1}\times\left(\dfrac {P_S}{P_1}\right)^{-1/K}\right]=-\dfrac {V_1^2}{2}$

$\left(\dfrac K{K-1}\right)\dfrac {P_1}{\rho_1}\left[1-\left(\dfrac {P_S}{P_1}\right)^{1-1/K}\right]=-\dfrac {V_1^2}{2}$

$\left[1-\left(\dfrac {P_S}{P_1}\right)^{\dfrac {K-1}K}\right]=-\dfrac {V_1^2}{2}\times \left(\dfrac {K-1}K\right)\dfrac {\rho_1}{P_1}$

$1+\dfrac {V_1^2}{2}\times \left(\dfrac {K-1}K\right)\dfrac {\rho_1}{P_1}=\left(\dfrac {P_S}{P_1}\right)^{\dfrac {K-1}K}.........................(3)$

Now for adiabatic process the velocity of sound,

$c=\sqrt{KRT}$

$c=\sqrt{K\dfrac P\rho}$

From point 1,

$c_1=\sqrt{K\dfrac {P_1}{\rho_1}}$

or, $c_1^2={K\dfrac {P_1}{\rho_1}}$

Substituting the value $K\dfrac {P_1}{\rho_1}=c_1^2$ in equation(3), we get

$1+\dfrac {V_1^2}{2}(K-1)\times \dfrac 1{c_1^2}=\left(\dfrac {P_S}{P_1}\right)^{\dfrac {K-1}K}$

$1+\dfrac {V_1^2}{2c_1^2}(K-1)=\left(\dfrac {P_S}{P_1}\right)^{\dfrac {K-1}K}$

$\therefore P_S=P_1\left[1+\dfrac {(K-1)}1M_1^2\right]^{\left(\dfrac K{K-1}\right)}$

Equation gives the values of stagnation pressure.

Expression for stagnation density($\rho_S$)

$\rho_S=\rho_1\left[\left(1+\dfrac {(K-1)}2M_1^2\right)^{\left(\dfrac K{K-1}\right)}\right]^{1/K}$

$\left[\rho_S=\rho_1\left(1+\dfrac {(K-1)}2M_1^2\right)^{\left(\dfrac 1{K-1}\right)}\right]$

Expression for stagnation temperature($T_S$)

$T_S=T_1\left[1+\dfrac {(K-1)}2M_1^2\right]$

Numericals

Q1) Find the Mach number when an aeroplane is flying at 110 zerokm per hour through still having a pressure of $7N/cm^2$ and temperature $-5 ^\circ C$. Wind velocity may be taken as zero. Take $R=287.14J/kg-K$. Calculate the pressure, temperature and density of air at stagnation point on the nose of the plane. (K = 1.4)

Solution:- Given:-

Speed of aeroplane

V=1100km/hr

$V=\dfrac {1100\times1000}{60\times60}=305.55m/s$

Pressure of air,

$P_1=7N/cm^2=7\times10^4N/m^2$

Temperature,

$t_1=-5^\circ C$

Using relation

$c=\sqrt{KRT}$

$c_1=\sqrt{1.4\times287.14\times268}$

$c_1=328.2m/s$

Therefore Mach Number

$M_1=\dfrac {V_1}{c_1}$

$M_1=\dfrac {305.55}{328.20}=0.9309$

$M_1\approx0.931$

Step 1: Stagnation Pressure ($P_S$)

$P_S=P_1\left[1+\dfrac {(K-1)}2M_1^2\right]^{\left(\dfrac K{K-1}\right)}$

$P_S=7\times 10^4\left[1+\dfrac {(1.4-1)}2(0.931)^2\right]^{\left(\dfrac {1.4}{1.4-1}\right)}$

$P_S=7\times 10^4\left[1+0.1733\right]^{\left(\dfrac {1.4}{0.4}\right)}$

$P_S=7\times 10^4\left[1.1733\right]^{\left(3.5\right)}$

$P_S=12.24\times10^{-4}N/m^2$

Step 2: Stagnation Temperature ($T_S$)

$T_S=T_1\left[1+\dfrac {(K-1)}2M_1^2\right]$

$T_S=268\left[1+\dfrac {(1.4-1)}2(0.931)^2\right]$

$T_S=268\left[1.1733\right]$

$T_S=314.44K$

Step 3: Stagnation Density ($\rho_S$)

$\rho_S=\dfrac {P_S}{RT_S}$

$\rho_S=\dfrac {12.24\times10^4}{287.14\times 314.44}$

$\rho_S=1.355 \ kg/m^3$