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:
Temperature drop of air = -52.5 K
Given Information:
Initial velocity of air = negligible
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:
Since air is assumed to be an ideal gas, the enthalpy change is related to the temperature difference as:
Calculation:
Step 1:
Calculate the enthalpy change
Based on equation (2) we have:
Step 2:
Calculate the temperature drop (ΔT)
Based on equation (3) we have
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
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:
Since air is assumed to be an ideal gas, the enthalpy change is related to the temperature difference as:
Calculation:
Step 1:
Calculate the enthalpy change
Based on equation (2) we have:
Step 2:
Calculate the temperature drop (ΔT)
Based on equation (3) we have
The temperature drop of air is -52.5 K
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Chapter 7 Solutions
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