Question

Ammonia enters a heat exchanger operating at steady state as a superheated vapor at 14 bar, \(60^{\circ} \mathrm{C}\), where it is cooled and condensed to saturated liquid at 14 bar. The mass flow rate of the refrigerant is \(450 \mathrm{~kg} / \mathrm{h}\). A separate stream of air enters the heat exchanger at \(17^{\circ} \mathrm{C}, 1\) bar and exits at \(42^{\circ} \mathrm{C}, 1\) bar. Ignoring heat transfer from the outside of the heat exchanger and neglecting kinetic and potential energy effects, determine the mass flow rate of the air, in \(\mathrm{kg} / \mathrm{min}\).

Short answer

The mass flow rate of the air is approximately 6.86 kg/min.

Step-by-step solution

- Analyze ammonia energy transfer

Determine the change in enthalpy of ammonia as it is cooled and condensed from a superheated vapor to a saturated liquid at 14 bar. Use enthalpy data from a thermodynamic table for ammonia.

- Calculate ammonia enthalpy values

From the ammonia tables: - For 14 bar and 60°C: Superheated vapor enthalpy, \(h_{1} = 1682.3 \, \text{kJ/kg} \) - For 14 bar and saturated liquid: Saturated liquid enthalpy, \(h_{2} = 300.54 \, \text{kJ/kg} \)

- Find enthalpy change for ammonia

Compute the change in enthalpy: \ \[ \Delta h = h_{2} - h_{1} = 300.54 - 1682.3 = -1381.76 \, \text{kJ/kg}. \]

- Calculate energy transferred for ammonia

Determine the total energy transferred from ammonia, \[ Q_{\text{ammonia}} = \text{mass flow rate of ammonia} \, \times \, \Delta h \] \[ Q_{\text{ammonia}} = \frac{450 \, \text{kg/h}}{3600 \, \text{s/h}} \, \times \, (-1381.76 \, \text{kJ/kg}) \approx -172.72 \text{kW}. \]}

- Analyze energy transfer by air

Since the heat exchanger operates steadily and neglecting heat loss, the energy gained by air is equal to the absolute value of energy lost by ammonia: \[ Q_{\text{air}} = 172.72 \text{kW} \]

- Determine air properties

From air tables, determine the specific heat capacity of air at constant pressure, \(c_p = 1.005 \, \text{kJ/kg}^\bullet\text{C} \). The change in temperature for air is: \[ \Delta T_{air} = T_{exit} - T_{enter} = 42^\bullet\text{C} - 17^\bullet\text{C} = 25^\bullet\text{C}. \]

- Calculate mass flow rate of air

Use the energy balance equation for the air side: \[ Q_{\text{air}} = \dot{m}_{air} \, \times \, c_p \, \times \, \Delta T_{air} \] Re-arrange to solve for \, \dot{m}_{air}: \[ \dot{m}_{air} = \frac{Q_{\text{air}}}{c_p \times \, \Delta T_{air}} = \, \frac{172.72}{1.005 \, \times \, 25} = 6.86 \, \text{kg/min}. \]

Key concepts

enthalpy change

Enthalpy change refers to the total heat content of a system, which can vary when the system undergoes transformations such as heating, cooling, or phase change. In the context of the heat exchanger problem, ammonia enters as a superheated vapor, which means it is above its boiling point. As it is cooled and condensed to a saturated liquid, its enthalpy decreases significantly. This decrease in enthalpy is calculated by subtracting the enthalpy of the saturated liquid state from the superheated vapor state. Mathematically, this is given by \(\ \Delta h = h_2 - h_1 = 300.54 {\text{kJ/kg}} - 1682.3 {\text{kJ/kg}} = -1381.76 {\text{kJ/kg}}.\).Understanding how to read and interpret enthalpy values from thermodynamic tables is crucial for solving problems that involve phase changes.

energy transfer

Energy transfer in a heat exchanger is pivotal, as it represents the amount of heat energy moved from one fluid to another. For ammonia to cool down and condense, it needs to lose energy. Conversely, the air, serving as the cooling medium, absorbs this lost energy. The amount of energy transferred by ammonia can be calculated using the mass flow rate and the change in enthalpy. The formula is \[ Q_{\text{ammonia}} = \frac{450 {\text{kg/h}}}{3600 {\text{s/h}}} * (-1381.76 {\text{kJ/kg}}) \approx -172.72 {\text{kW}}.\]This same energy is transferred to the air, which gains energy, increasing its temperature. For steady-state operations with no heat loss, the amount of energy lost by the hot fluid (ammonia) is equal to the energy gained by the cold fluid (air).

mass flow rate calculation

Calculating the mass flow rate of air in the heat exchanger requires an understanding of the relationship between energy transfer and the properties of the fluid involved. Given the energy gained by air (172.72 kW), we use the specific heat capacity \( c_p \) and the temperature change \( \Delta T \). The formula is: \[ \dot{m}_{air} = \frac{Q_{air}}{c_p * \Delta T_{air}} \approx \frac{172.72 {\text{kW}}}{1.005 {\text{kJ/kg}}^\bullet\text{C} * 25 ^\bullet\text{C}} \approx 6.86 {\text{kg/min}.} \]Here, \( c_p \) is the specific heat capacity of air at constant pressure, and \( \Delta T \) is the temperature difference across the heat exchanger. This computation provides the amount of air required per minute to balance the heat transfer from ammonia.

superheated vapor

Superheated vapor is a state where a fluid is at a temperature higher than its boiling point at a given pressure. In the provided exercise, ammonia enters the heat exchanger as a superheated vapor at 14 bar and 60°C. This means the ammonia is beyond its boiling point at this pressure, containing extra energy compared to its saturated state. This excess energy is part of what’s released during the cooldown to a saturated liquid. Understanding the state of the fluid is essential because it dictates the amount of heat exchanged and influences the enthalpy values used in calculations.

saturated liquid

A saturated liquid is a state where the fluid is at its boiling point and ready to evaporate but has not yet started doing so. In the heat exchanger problem, ammonia leaves the heat exchanger as a saturated liquid at 14 bar after being cooled. The transition from superheated vapor to saturated liquid involves releasing a significant amount of energy, evident in the large change in enthalpy. This concept is crucial in thermodynamics as it marks the end of the cooling or condensing process and signifies that the fluid has released all its excess heat.