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Chapter 5: Problem 38

Ten kmol per hour of air is throttled from upstream conditions of \(298.15 \mathrm{K}\left(25^{\circ} \mathrm{C}\right)\) and 10 bar to a downstream pressure of 1.2 bar. Assume air to be an ideal gas with \(C_{P}=(7 / 2) R\) (a) What is the downstream temperature? (b) What is the entropy change of the air in \(\mathrm{J} \mathrm{mol}^{-1} \mathrm{K}^{-1} ?\) (c) What is the rate of entropy generation in \(\mathrm{W} \mathrm{K}^{-1}\) ? (d) If the surroundings are at \(293.15 \mathrm{K}\left(20^{\circ} \mathrm{C}\right),\) what is the lost work?

Short Answer

Expert verified
(a) 298.15 K (b) 19.15 J/mol路K (c) 191.5 W/K (d) 56.1 kW

Step by step solution

01

Throttling Process

In a throttling process, the enthalpy remains constant: \( h_1 = h_2 \). For an ideal gas with constant \( C_P \), the enthalpy \( h = C_P T \). Therefore, the temperature remains the same across the throttle. The downstream temperature after the process remains \(298.15 \mathrm{K}\).
02

Entropy Change Calculation

For an ideal gas throttling process, the entropy change \( \Delta S \) can be calculated using the equation: \[ \Delta S = C_P \ln{\frac{T_2}{T_1}} - R \ln{\frac{P_2}{P_1}} \]. Since the temperature doesn't change (\(T_1 = T_2\)), the first term is zero and the equation simplifies to: \[ \Delta S = - R \ln{\frac{P_2}{P_1}} \]. Substituting the values, \[ \Delta S = - 8.314 \ln{\frac{1.2}{10}} \approx 19.15\ \mathrm{J/mol\cdot K} \].
03

Rate of Entropy Generation

The rate of entropy generation \(\dot{S}_{gen}\) is the difference between the entropy change of the air flow and the external surroundings. Since the process is adiabatic, \(\dot{n} \cdot \Delta S = \dot{S}_{gen} \). With \(\dot{n}=10\ \mathrm{mol/s}\), \(\dot{S}_{gen} = 10 \times 19.15 = 191.5\ \mathrm{W/K} \).
04

Calculate the Lost Work

Lost work is calculated as \(\dot{W}_{lost} = T_0 \cdot \dot{S}_{gen}\) where \(T_0 = 293.15\ \mathrm{K}\). The lost work is: \(\dot{W}_{lost} = 293.15 \times 191.5 \approx 56131.725\ \mathrm{W} \approx 56.1 \ \mathrm{kW}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ideal Gas
Air is treated as an ideal gas in this problem, which simplifies calculations in thermodynamics. An ideal gas is a theoretical model where gas particles:
  • Do not interact with each other
  • Have elastic collisions
  • Obey the ideal gas law: \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is amount of substance, \( R \) is the gas constant, and \( T \) is temperature.
In this problem, the specific heat capacity at constant pressure, \( C_P \), is given as \( \frac{7}{2} R \). This means the energy required to raise the temperature is proportional to this constant. Recognizing air as an ideal gas allows us to apply these simplified formulas effectively, ensuring the temperature doesn鈥檛 change during a throttling process when there鈥檚 constant enthalpy.
Entropy Change
Entropy is a measure of disorder or randomness in a system. In thermodynamics, especially during processes like throttling, entropy can tell us about the irreversibility of the process. For an ideal gas undergoing a throttling process:
  • The formula to find entropy change is: \( \Delta S = C_P \ln\frac{T_2}{T_1} - R \ln\frac{P_2}{P_1} \).
  • Since the temperature doesn鈥檛 change during throttling, the equation simplifies to \( \Delta S = - R \ln\frac{P_2}{P_1} \).
  • Using given values, calculate \( \Delta S \) as approximately 19.15 J/mol路K.
This simplified calculation of entropy change shows how pressure variations can affect a system when enthalpy is constant.
Throttling Process
A throttling process involves a fluid passing through a valve, rendering minimal resistance to its flow. This process is noted for constant enthalpy (\( h_1 = h_2 \)). In this scenario:
  • An ideal gas assumption signifies the enthalpy can be expressed as \( h = C_P T \).
  • The consequence is that temperature remains the same across the throttle, ensuring the downstream temperature remains constant.
  • Throttling also results in a pressure drop, depicted by the change from 10 bar to 1.2 bar in this problem.
Understanding the throttling process is crucial for industries using systems like refrigeration cycles, where energy conservation is key.
Lost Work
Lost work refers to the work that could have been converted into useful energy but is instead dissipated due to inefficiencies in the system. For this problem, it's a measure of the energy loss owing to entropy generation. It is calculated by:
  • \( \dot{W}_{lost} = T_0 \cdot \dot{S}_{gen} \), with \( T_0 \) as the temperature of the surroundings.
  • From the calculated rate of entropy generation, \( \dot{S}_{gen} = 191.5 \ \mathrm{W/K} \).
  • The resulting lost work amounts to roughly \( 56.1 \ \mathrm{kW} \).
Lost work helps us recognize potential improvements in system efficiency, allowing for adjustments to minimize energy waste.

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