Respuesta :
Answer:
For air
The pressure downstream of normal shock is 447.76 kPa
The temperature downstream of normal shock is 604.26 K
The mach number downstream of normal shock is 0.504
The velocity of air downstream of normal shock is 248.2 m/s
The stagnation pressure downstream of normal shock is 532.498 kPa
For helium
The pressure downstream of normal shock is 475.89 kPa
The temperature downstream of normal shock is 801.36 K
The mach number downstream of normal shock is 0.546
The velocity of air downstream of normal shock is 309.64 m/s
The stagnation pressure downstream of normal shock is 603.258 kPa
the values of the temperature, [tex]{T_x}[/tex], pressure, [tex]{P_x}[/tex], velocity, v and mach number, [tex]{M_x}[/tex] are higher downstream for helium than for air
Change in entropy, Δ[tex]S_{air}[/tex] for air is 7.333 J/K
Change in entropy, [tex]\triangle S_{helium}[/tex] for helium is 5.784 J/K
Explanation:
Here we have the relations and ratios from compressible flow tables for normal shock as follows;
For Mach number, 2.6 the values of the relations are;
[tex]M_y[/tex] = 0.504
[tex]\frac{P_y}{P_x} = 7.720[/tex]
[tex]\frac{T_y}{T_x} = 2.238[/tex]
[tex]\frac{P_{0y}}{P_{0x}} = 0.460[/tex]
[tex]\frac{P_{0y}}{P_{x}} = 9.181[/tex]
Where:
[tex]M_y[/tex] = Mach number downstream of normal shock
[tex]{P_y}[/tex] = Pressure downstream of normal shock
[tex]{T_y}[/tex] = Temperature downstream of normal shock
[tex]{P_{0y}[/tex] = Stagnation pressure downstream of normal shock
[tex]M_x[/tex] = Mach number upstream of normal shock = 2.6
[tex]{P_x}[/tex] = Pressure upstream of normal shock = 58 kPa
[tex]{T_x}[/tex] = Temperature upstream of normal shock = 270 K
[tex]{P_{0x}[/tex] = Stagnation pressure upstream of normal shock
Therefore;
From;
[tex]\frac{P_y}{P_x} = 7.720[/tex]
[tex]{P_y}[/tex] = [tex]{P_x}[/tex] × 7.720 = 58 kPa × 7.720 = 447.76 kPa
From;
[tex]\frac{T_y}{T_x} = 2.238[/tex]
[tex]{T_y}[/tex] = [tex]{T_x}[/tex] × 2.238 = 270 K × 2.238 = 604.26 K
From;
[tex]\frac{P_{0y}}{P_{x}} = 9.181[/tex]
[tex]{P_{0y}[/tex] = [tex]{P_x}[/tex] × 9.181 = 58 kPa × 9.181 = 532.498 kPa
Velocity of sound in air at 604.26 K is given by the following relation;
[tex]V_{Sound \ in \ air} = 331\frac{m}{s} \times \sqrt{\frac{604.26 \, K}{273 \, K} } = 492.45 \, m/s[/tex]
The velocity of the air is then found from the relation;
[tex]Mach \ number, M_y = \frac{Velocity \ of \ air}{ Velocity \ of \ sound}[/tex]
Therefore;
Velocity of air downstream of normal shock = 492.45 m/s × 0.504 = 248.2 m/s
For helium, γ = 1.667 from where we obtain from similar tables the relationship as follows;
Here we have the relations and ratios from compressible flow tables for normal shock as follows;
For Mach number 2.6 the values of the relations are;
[tex]M_y[/tex] = 0.546
[tex]\frac{P_y}{P_x} = 8.205[/tex]
[tex]\frac{P_{0y}}{P_{0x}} = 0.545[/tex]
[tex]\frac{P_{0y}}{P_{x}} = 10.401[/tex]
Solving as before, where:
[tex]{P_x}[/tex] = Pressure upstream of normal shock = 58 kPa
[tex]{T_x}[/tex] = Temperature upstream of normal shock = 270 K
We have;
[tex]{P_y}[/tex] = [tex]{P_x}[/tex] × 8.205 = 58 kPa × 8.205 = 475.89 kPa
[tex]{T_y}[/tex] = [tex]{T_x}[/tex] × 2.968 = 270 K × 2.968 = 801.36 K
[tex]{P_{0y}[/tex] = [tex]{P_x}[/tex] × 10.401 = 58 kPa × 10.401 = 603.258 kPa
[tex]V_{Sound \ in \ air} = 331\frac{m}{s} \times \sqrt{\frac{801.36 \, K}{273 \, K} } = 567.1 \, m/s[/tex]
Velocity of helium downstream of shock = [tex]V_{Sound \ in \ air} \times M_y[/tex]
= 567.1 m/s × 0.546 = 309.64 m/s
Therefore, the values of the temperature, [tex]{T_x}[/tex], pressure, [tex]{P_x}[/tex], velocity, v and mach number, [tex]{M_x}[/tex] are higher downstream for helium than for air.
The entropy change is given by the following relation;
Change in entropy = ΔS
[tex]\frac{\triangle S}{R} = \frac{\gamma }{\gamma -1} ln\left (\frac{2}{\left (\gamma +1 \right )M_{x}^{2}}+\frac{\gamma -1}{\gamma +1} \right )+\frac{1 }{\gamma -1} ln\left (\frac{2\gamma }{\left (\gamma +1 \right )}M_{x}^{2}-\frac{\gamma -1}{\gamma +1} \right )[/tex]
Where:
[tex]M_x[/tex] is given as 2.6
For air, γ = 1.4
For helium γ = 1.67
R = Universal gas constant
Plugging in the values, we have;
[tex]\frac{\triangle S_{air}}{R} = \frac{1.4 }{1.4 -1} ln\left (\frac{2}{\left (1.4+1 \right )2.6^{2}}+\frac{1.4 -1}{1.4 +1} \right )+\frac{1 }{1.4 -1} ln\left (\frac{2 \times 1.4 }{\left (1.4 +1 \right )}2.6^{2}-\frac{1.4 -1}{1.4 +1} \right )[/tex]
[tex]\frac{\triangle S_{air}}{R} =0.882[/tex]
∴ Δ[tex]S_{air}[/tex] = R × 0.882 = 8.3145 × 0.882 = 7.333 J/K
Similarly
[tex]\frac{\triangle S_{air}}{R} = \frac{1.67 }{1.67 -1} ln\left (\frac{2}{\left (1.67+1 \right )2.6^{2}}+\frac{1.67 -1}{1.67 +1} \right )+\frac{1 }{1.67 -1} ln\left (\frac{2 \times 1.67 }{\left (1.67 +1 \right )}2.6^{2}-\frac{1.67 -1}{1.67 +1} \right )[/tex]
[tex]\frac{\triangle S_{helium}}{R} =0.696[/tex]
[tex]\triangle S_{helium}[/tex] = R × 0.696 = 8.3145 × 0.696 = 5.784 J/K.
