Question

Moist air at \(20^{\circ} \mathrm{C}, 1 \mathrm{~atm}, 43 \%\) relative humidity and a volumetric flow rate of \(900 \mathrm{~m}^{3} / \mathrm{h}\) enters a control volume at steady state and flows along a surface maintained at \(65^{\circ} \mathrm{C}\), through which heat transfer occurs. Liquid water at \(20^{\circ} \mathrm{C}\) is injected at a rate of \(5 \mathrm{~kg} / \mathrm{h}\) and evaporates into the flowing stream. For the control volume, \(\dot{W}_{\mathrm{cv}}=0\), and kinetic and potential energy effects are negligible. Moist air exits at \(32^{\circ} \mathrm{C}, 1 \mathrm{~atm}\). Determine (a) the rate of heat transfer, in \(\mathrm{kW}\). (b) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\).

Short answer

For part (a), the rate of heat transfer, \( \dot{Q}_{cv} \), is calculated using energy conservation principles and properties of moist air. For part (b), the entropy generation rate, \( \dot{S}_{gen} \), is found based on the entropy balance.

Step-by-step solution

List known values and assumptions

Identify and list all given values and assumptions for the problem: Initial conditions:-Temperature of moist air: \(20^{\circ} \text{C}\)-Pressure: \(1 \text{ atm}\)-Relative Humidity: 43%-Volumetric flow rate: \(900 \text{ m}^{3}/\text{h}\)Additional details:-Liquid water injected: \(5 \text{ kg}/\text{h}\)-Heat transfer surface temperature: \(65^{\circ} \text{C}\)-Moist air exit temperature: \(32^{\circ} \text{C}\)-Kinetic and potential energy effects are negligible.

Calculate the mass flow rate of dry air

Use the volumetric flow rate and the ideal gas law to determine the mass flow rate of dry air. The formula for the ideal gas law is: \[ PV = nRT \rightarrow V = \frac{nRT}{P} \]where:-\( P \) is pressure (1 atm)-\( V \) is volume (900 m³/h)-\( R \) is the specific gas constant for air (287 J/kg·K)-\( T \) is temperature in Kelvin (293 K)\[ n = \dot{m}_a \rightarrow \dot{m}_a = \frac{PV}{RT} \] convert units as necessary to find \( \dot{m}_a \approx 1.247.8 kg/h \)

Calculate the initial and final specific humidity

Use the relative humidity and saturation pressure to calculate the specific humidity: The saturation pressure at 20ºC: \( P_{sat,20ºC} = 2.34 \ kPa \) Relative Humidity (RH) = 0.43 Using the specific humidity formula: \[ \text{Specific Humidity} \left( \omega \right) = \frac{0.622 \cdot \text{RH} \cdot P_{sat}}{P - (RH \cdot P_{sat})} \]

Apply the Mass and Energy Balances

Use the conservation of mass and energy equations for the control volume: Mass balance: \[ \dot{m}_a \cdot \omega_1 + \dot{m}_{water} = \dot{m}_a \cdot \omega_2 \] Energy balance: \[ \dot{m}_a \cdot (h_{a2} - h_{a1}) + \dot{m}_{water}\cdot h_{fg,65ºC} = \dot{Q}_{cv} \] Solve for \( \dot{Q}_{cv} \)

Calculation the entropy production rate

The entropy balance for this control volume looks like: \[ \dot{S}_{gen} = \sum \dot{m} s_2 - \sum \dot{m} s_1 \] where: -\( s_2 \) is the specific entropy of the exiting moist air-\( s_1 \) is the specific entropy of the entering moist airPlug known values into these formulas and solve for \( \dot{Q}_{cv} \) and \( \dot{S}_{gen} \)

Key concepts

entropy production calculation

Entropy production is important in understanding the irreversibility in a thermodynamic process. For this control volume, the entropy balance is given by the formula:\[ \dot{S}_{gen} = \sum \dot{m} s_2 - \sum \dot{m} s_1 \]Here,- \( \dot{S}_{gen} \) is the rate of entropy production- \( s_2 \) and \( s_1 \) are the specific entropies of the moist air exiting and entering, respectivelyBy substituting the known values into this equation, we can determine the system's entropy production rate. This calculation reveals the amount of disorder or energy dispersion occurring within the control volume. Ensuring that students understand this concept is key to grasping the second law of thermodynamics.

specific humidity

Specific humidity is a measure of the water vapor present in the air and is essential in drying and air conditioning processes. Calculating specific humidity involves the relative humidity and the saturation pressure of water vapor.The formula for specific humidity (\( \omega \)) is:\[ \omega = \frac{0.622 \cdot \text{RH} \cdot P_{sat}}{P - (\text{RH} \cdot P_{sat})} \]By using the initial conditions:- Relative Humidity (RH): 43%- Saturation pressure at 20ºC: \( P_{sat,20ºC} = 2.34 \text{ kPa} \)- Atmospheric pressure (P): 1 atmWe can calculate the initial and final specific humidities. Understanding specific humidity and its role in thermodynamic calculations assists in determining the moisture content changes in air processing applications.

ideal gas law

The ideal gas law, represented by the equation \[ PV = nRT \], relates a gas's pressure, volume, temperature, and amount. Here, \( P \) is pressure, \( V \) is volume, \( n \) is the amount of substance, \( R \) is the gas constant, and \( T \) is temperature.In many thermodynamic problems, like the mass flow rate determination, the ideal gas law provides a foundation for connecting these properties. Whether converting volumes or calculating the effects of temperature changes, sound knowledge of this law is pivotal. For instance, we used this law to determine:\[ \dot{m}_a = \frac{PV}{RT} \]The idea is that under standard conditions, air behaves almost like an ideal gas, permitting these calculations.