Chapter 7, Problem 7.1P

Interpretation Introduction

Interpretation:

The temperature drops of air as it expands adiabatically through a nozzle under the given conditions needs to be calculated

Concept Introduction:

• An adiabatic process is one in which there is no transfer of heat (Q) between the system and surroundings.
• Based on the first law of thermodynamics, a steady state, steady flow process between one entrance and one exit (like in the case of a nozzle) can be expressed mathematically as:

  $\begin{array}{l}\text{ΔH + }\frac{{\text{Δu}}^{\text{2}}}{\text{2}}\text{ + gΔz = Q + W ------(1)}\\ \text{where: }\Delta \text{H = change in enthalpy}\\ \Delta \text{u = change in velocity}\\ \text{g = acceleration due to gravity}\\ \Delta \text{z = elevation difference}\\ \text{Q = heat involved}\\ \text{W = work done}\end{array}$

  $$

Temperature drop of air = -52.5 K

Given Information:

Initial velocity of air = negligible $₹{\text{ 0 ms}}^{\text{-1}}$

Final velocity of air = 325 ms^-1

Explanation:

Since the process is adiabatic, Q = 0. If we assume that the elevation difference is negligible, and no shat work is done, then, Δz = W = 0. The equation (1) reduces to:

  $\text{ΔH + }\frac{{\text{Δu}}^{\text{2}}}{\text{2}}\text{ = 0 ----(2)}$

Since air is assumed to be an ideal gas, the enthalpy change is related to the temperature difference as:

  $\begin{array}{l}{\text{ΔH = C}}_{\text{p}}\Delta \text{T----(3)}\\ {\text{where C}}_{\text{p}}\text{ is the heat capacity at constant pressure}\end{array}$

Calculation:

Step 1:

Calculate the enthalpy change

Based on equation (2) we have:

  $\begin{array}{c}\text{ΔH = - }\frac{{\text{u}}_2^2-{\text{u}}_1^2}{\text{2}}\\ \text{= - }\frac{{(325)}^2-{(0)}^2}{2}\left(\frac{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{2}}}\right)\\ {\text{= -52813 m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{Since 1 kJ/kg = 1000 m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{ΔH = -52}\text{.813 kJ/kg}\\ \text{Now, molar mass of air = 0}\text{.02897 kg/mol}\\ \text{Therefore, }\Delta \text{H = -52}\text{.813 kJ/kg }\times \text{ 0}\text{.02897 kg/mol }\\ \text{= -1}\text{.529 kJ/mol}\end{array}$

Step 2:

Calculate the temperature drop (ΔT)

Based on equation (3) we have

  $\begin{array}{c}\text{ΔT = }\frac{\text{ΔH}}{{\text{C}}_{\text{p}}}\\ =\frac{-1.529\text{ kJ}{\text{.mol}}^{-1}}{(7/2)\times 0.008314\text{ kJ}{\text{.K}}^{-1}.{\text{mol}}^{-1}}\\ =-52.5\text{ K}\end{array}$

The temperature drop of air is -52.5 K

Answer to Problem 7.1P

Temperature drop of air = -52.5 K

Explanation of Solution

Given Information:

Initial velocity of air = negligible $₹{\text{ 0 ms}}^{\text{-1}}$

Final velocity of air = 325 ms^-1

Since the process is adiabatic, Q = 0. If we assume that the elevation difference is negligible, and no shat work is done, then, Δz = W = 0. The equation (1) reduces to:

  $\text{ΔH + }\frac{{\text{Δu}}^{\text{2}}}{\text{2}}\text{ = 0 ----(2)}$

Since air is assumed to be an ideal gas, the enthalpy change is related to the temperature difference as:

  $\begin{array}{l}{\text{ΔH = C}}_{\text{p}}\Delta \text{T----(3)}\\ {\text{where C}}_{\text{p}}\text{ is the heat capacity at constant pressure}\end{array}$

Calculation:

Step 1:

Calculate the enthalpy change

Based on equation (2) we have:

  $\begin{array}{c}\text{ΔH = - }\frac{{\text{u}}_2^2-{\text{u}}_1^2}{\text{2}}\\ \text{= - }\frac{{(325)}^2-{(0)}^2}{2}\left(\frac{{\text{m}}^{\text{2}}}{{\text{s}}^{\text{2}}}\right)\\ {\text{= -52813 m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ {\text{Since 1 kJ/kg = 1000 m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \text{ΔH = -52}\text{.813 kJ/kg}\\ \text{Now, molar mass of air = 0}\text{.02897 kg/mol}\\ \text{Therefore, }\Delta \text{H = -52}\text{.813 kJ/kg }\times \text{ 0}\text{.02897 kg/mol }\\ \text{= -1}\text{.529 kJ/mol}\end{array}$

Step 2:

Calculate the temperature drop (ΔT)

Based on equation (3) we have

  $\begin{array}{c}\text{ΔT = }\frac{\text{ΔH}}{{\text{C}}_{\text{p}}}\\ =\frac{-1.529\text{ kJ}{\text{.mol}}^{-1}}{(7/2)\times 0.008314\text{ kJ}{\text{.K}}^{-1}.{\text{mol}}^{-1}}\\ =-52.5\text{ K}\end{array}$

Conclusion

The temperature drop of air is -52.5 K