Question

Air flowing steadily in a nozzle experiences a normal shock at a Mach number of Ma 5 2.6. If the pressure and temperature of air are 58 kPa and 270 K, respectively, upstream of the shock, calculate the pressure, temperature, velocity, Mach number, and stagnation pressure downstream of the shock. Compare these results to those for helium undergoing a normal shock under the same conditions. Calculate the entropy changes.

Answer 1

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,
`{T_x}`, pressure,
`{P_x}`, velocity, v and mach number,
`{M_x}` are higher downstream for helium than for air

Change in entropy, Δ
`S_(air)` for air is 7.333 J/K

Change in entropy,
`\triangle S_(helium)` for helium is 5.784 J/K

Step-by-step 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;

`M_y` = 0.504

`(P_y)/(P_x) = 7.720`

`(T_y)/(T_x) = 2.238`

`(P_(0y))/(P_(0x)) = 0.460`

`(P_(0y))/(P_(x)) = 9.181`

Where:

`M_y` = Mach number downstream of normal shock

`{P_y}` = Pressure downstream of normal shock

`{T_y}` = Temperature downstream of normal shock

`{P_(0y)` = Stagnation pressure downstream of normal shock

`M_x` = Mach number upstream of normal shock = 2.6

`{P_x}` = Pressure upstream of normal shock = 58 kPa

`{T_x}` = Temperature upstream of normal shock = 270 K

`{P_(0x)` = Stagnation pressure upstream of normal shock

Therefore;

From;

`(P_y)/(P_x) = 7.720`

`{P_y}` =
`{P_x}` × 7.720 = 58 kPa × 7.720 = 447.76 kPa

From;

`(T_y)/(T_x) = 2.238`

`{T_y}` =
`{T_x}` × 2.238 = 270 K × 2.238 = 604.26 K

From;

`(P_(0y))/(P_(x)) = 9.181`

`{P_(0y)` =
`{P_x}` × 9.181 = 58 kPa × 9.181 = 532.498 kPa

Velocity of sound in air at 604.26 K is given by the following relation;

`V_(Sound \ in \ air) = 331(m)/(s) * \sqrt{(604.26 \, K)/(273 \, K) } = 492.45 \, m/s`

The velocity of the air is then found from the relation;

`Mach \ number, M_y = (Velocity \ of \ air)/( Velocity \ of \ sound)`

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;

`M_y` = 0.546

`(P_y)/(P_x) = 8.205`

`(P_(0y))/(P_(0x)) = 0.545`

`(P_(0y))/(P_(x)) = 10.401`

Solving as before, where:

`{P_x}` = Pressure upstream of normal shock = 58 kPa

`{T_x}` = Temperature upstream of normal shock = 270 K

We have;

`{P_y}` =
`{P_x}` × 8.205 = 58 kPa × 8.205 = 475.89 kPa

`{T_y}` =
`{T_x}` × 2.968 = 270 K × 2.968 = 801.36 K

`{P_(0y)` =
`{P_x}` × 10.401 = 58 kPa × 10.401 = 603.258 kPa

`V_(Sound \ in \ air) = 331(m)/(s) * \sqrt{(801.36 \, K)/(273 \, K) } = 567.1 \, m/s`

Velocity of helium downstream of shock =
`V_(Sound \ in \ air) * M_y`

= 567.1 m/s × 0.546 = 309.64 m/s

Therefore, the values of the temperature,
`{T_x}`, pressure,
`{P_x}`, velocity, v and mach number,
`{M_x}` are higher downstream for helium than for air.

The entropy change is given by the following relation;

Change in entropy = ΔS

`(\triangle S)/(R) = (\gamma )/(\gamma -1) ln\left ((2)/(\left (\gamma +1 \right )M_(x)^(2))+(\gamma -1)/(\gamma +1) \right )+(1 )/(\gamma -1) ln\left ((2\gamma )/(\left (\gamma +1 \right ))M_(x)^(2)-(\gamma -1)/(\gamma +1) \right )`

Where:

`M_x` 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;

`(\triangle S_(air))/(R) = (1.4 )/(1.4 -1) ln\left ((2)/(\left (1.4+1 \right )2.6^(2))+(1.4 -1)/(1.4 +1) \right )+(1 )/(1.4 -1) ln\left ((2 * 1.4 )/(\left (1.4 +1 \right ))2.6^(2)-(1.4 -1)/(1.4 +1) \right )`

`(\triangle S_(air))/(R) =0.882`

∴ Δ
`S_(air)` = R × 0.882 = 8.3145 × 0.882 = 7.333 J/K

Similarly

`(\triangle S_(air))/(R) = (1.67 )/(1.67 -1) ln\left ((2)/(\left (1.67+1 \right )2.6^(2))+(1.67 -1)/(1.67 +1) \right )+(1 )/(1.67 -1) ln\left ((2 * 1.67 )/(\left (1.67 +1 \right ))2.6^(2)-(1.67 -1)/(1.67 +1) \right )`

`(\triangle S_(helium))/(R) =0.696`

`\triangle S_(helium)` = R × 0.696 = 8.3145 × 0.696 = 5.784 J/K.