91Ó°ÊÓ

Heat transfer is the movement of thermal energy from one object or substance to another due to a temperature difference. In the context of the exercise, we are analyzing how heat is transferred as air flows through a nozzle in a steady state.
When dealing with the first law of thermodynamics for steady-flow systems, heat transfer is a vital component. The equation for steady-flow energy conservation is: \[ \frac{dq}{dN} = \frac{dh}{dN} + \frac{1}{2} (v_2^2 - v_1^2) \] Here,

• \ \( \frac{dq}{dN} \ \) represents heat transfer per unit mass,
• \ \( \frac{dh}{dN} \ \) is the change in specific enthalpy,
• \ \( \frac{1}{2} (v_2^2 - v_1^2) \ \) represents the change in kinetic energy.

Using this equation, we can see how energy is conserved, and how heat transfer affects the internal and kinetic energy of the system. The goal, in this case, is to find the amount of heat transferred per pound of air flowing through the nozzle.

Specific enthalpy is a measure of the total energy of a fluid per unit mass. It includes both the internal energy and the product of pressure and volume. In the given exercise, we are dealing with changes in specific enthalpy due to temperature changes as air flows through the nozzle.
For ideal gases, specific enthalpy can be approximated using the specific heat at constant pressure ( \( c_p \). The formula is: \[ \frac{dh}{dN} = c_p (T_2 - T_1) \] In this case,

• \( T_1 \ \) is the initial temperature of 720°R,
• \( T_2 \ \) is the final temperature of 500°R,
• \( c_p \ \) for air is 0.24 \( \ \frac{Btu}{lb °R} \).

Substituting these values, the change in specific enthalpy is calculated as: \[ \frac{dh}{dN} = 0.24 (500 - 720) = 0.24 (-220) = -52.8 \frac{Btu}{lb} \] This negative value indicates a drop in specific enthalpy as the air cools down while passing through the nozzle.

Kinetic energy in thermodynamics refers to the energy a substance possesses due to its motion. In the context of steady-flow systems, kinetic energy changes are crucial as they affect the overall energy balance of the system.
The change in kinetic energy per unit mass can be calculated using the velocities at the inlet and outlet of the nozzle: \[ \frac{1}{2} (v_2^2 - v_1^2) \] In the exercise,

• \( v_1 \ \) is the initial velocity, given as negligible, thus approximated to 0,
• \( v_2 \ \) is the final velocity, 1450 ft/s.

The change in kinetic energy can be computed as: \[ \frac{1}{2} (1450^2 - 0^2) \times \frac{1}{25032} \frac{Btu}{lb \ \text{ft}^2/s^2} \] Simplifying this gives: \[ 41.98 \frac{Btu}{lb} \] This significant positive change in kinetic energy indicates an increase in the speed of the air as it exits the nozzle, highlighting the conversion of thermal energy into kinetic energy.