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The ideal gas model is a simplification used in thermodynamics to describe the behavior of gases. In this model, gases are assumed to follow the Ideal Gas Law, which states that the product of pressure (P) and volume (V) is directly proportional to the product of the number of moles (n), the gas constant (R), and the temperature (T). Mathematically, it is represented as:
\[ PV = nRT \]
In the ideal gas model, gas molecules are assumed to occupy negligible volume and exert no intermolecular forces on each other. This model simplifies calculations by providing a straightforward relationship between the state variables of a gas.
In the context of the exercise, the ideal gas model allows us to use the specific heat capacity at constant pressure (\[ c_p \]) to calculate the exergy transfer. Remember, specific heat capacity (\[ c_p \]) is the amount of heat required to raise the temperature of 1 pound of gas by 1-degree Rankine (R).

Exergy transfer is a measure of the usable energy that accompanies a process, such as heat transfer, within a thermodynamic system. It accounts for both the energy carried by the heat and the temperature at which this energy is transferred.
The formula for exergy transfer (\[Q_{ex}\]) associated with heat transfer is:
\[ Q_{ex} = c_p * \big( T_0 \times ln \big( \frac{T_{final}}{T_0} \big) + (T_{initial} - T_{final} ) \big) \]
In this equation:

• \[c_p \] is the specific heat capacity.
• T_0 is the dead state temperature (the reference temperature of the environment).
• T_{initial} and T_{final} are the initial and final temperatures of the gas, respectively.

By incorporating the natural logarithm of the temperature ratio and the difference in temperatures, this formula helps us calculate the amount of exergy, in Btu per lb of gas flowing, transferred during the cooling process.

In steady state heat transfer, the temperatures and other properties of the system remain constant over time, even though heat is being transferred. This means any heat entering or leaving the system does so at a constant rate.
In the exercise, the hot gaseous products of combustion are at steady state as they cool from 3000°F to 250°F.
Ignoring fluid friction and assuming constant pressure, we can simplify our calculations significantly. Under these conditions, the first law of thermodynamics dictates that the rate of heat loss by the gas must equal the rate of heat gain by the surroundings.

To work with temperatures in thermodynamic equations, it's often necessary to convert between different temperature units. The Rankine (°R) scale is the absolute scale associated with Fahrenheit (°F). The conversion between Fahrenheit and Rankine is straightforward:
\[ T_R = T_F + 459.67 \]
In this equation,

• \[T_R \] is the temperature in Rankine
• \[T_F \] is the temperature in Fahrenheit

For example,

• The initial temperature in the exercise is 3000°F, which converts to 3459.67°R.
• The final temperature is 250°F, converting to 709.67°R.
• The dead state temperature of the environment, 80°F, converts to 539.67°R.

These conversions are vital, as thermodynamic equations require absolute temperatures (e.g., in Rankine) to ensure accurate calculations.