Question

Moist air at \(20^{\circ} \mathrm{C}, 1.05\) bar, \(85 \%\) relative humidity and a volumetric flow rate of \(0.3 \mathrm{~m}^{3} / \mathrm{s}\) enters a well- insulated compressor operating at steady state. If moist air exits at \(100^{\circ} \mathrm{C}\), \(2.0\) bar, determine (a) the relative humidity at the exit. (b) the power input, in \(\mathrm{kW}\). (c) the rate of entropy production, in \(\mathrm{kW} / \mathrm{K}\).

Short answer

For part (a), calculate the exit RH. For part (b), apply energy balance to find power input. For part (c), use entropy balance equation.

Step-by-step solution

Identify Initial and Final States

The initial state of the moist air is given as: - Temperature: 20°C - Pressure: 1.05 bar - Relative humidity: 85% - Volumetric flow rate: 0.3 m³/s The final state is: - Temperature: 100°C - Pressure: 2.0 bar

Determine Specific Humidities

Use the given temperature and relative humidity to find the specific humidity at the entrance. For the exit, assume the water vapor is saturated and calculate the specific humidity. Use psychrometric charts or steam tables to find: - Specific humidity at 20°C and 85% RH - Specific humidity at 100°C saturation

Apply the Mass and Energy Balances

Since the system is steady and well-insulated, apply the mass balance for the dry air and the water vapor individually. Then use the energy balance: Energy in = Energy out + Work done Consider the specific enthalpies of dry air and water vapor at the initial and final states.

Calculate the Relative Humidity at the Exit

Using the specific humidity at the exit calculated in Step 2 and the saturation specific humidity at 100°C, determine the relative humidity at 100°C, 2.0 bar: Relative humidity = (Calculated specific humidity / Saturation specific humidity) * 100%

Calculate the Power Input

Utilize the energy balance equation formulated in Step 3 to solve for the power input. Use the enthalpy values extracted for the specific states: Power input (kW) = Mass flow rate * (specific enthalpy change between states)

Calculate the Rate of Entropy Production

Formulate the entropy balance equation incorporating both the inlet and outlet states. Use the mass flow rate and specific entropy values of both the dry air and water vapor: Rate of entropy production = Mass flow rate * (specific entropy change between states)

Key concepts

relative humidity

Relative humidity (RH) is a measure of how much moisture air contains compared to the maximum amount it can hold at a given temperature. It is expressed as a percentage. If the air is holding all the moisture it can at that temperature, the RH is 100%. In the given problem, the initial RH is 85% at 20°C and 1.05 bar. To calculate RH at the exit (100°C, 2.0 bar), we assume the air at the exit is saturated (100% RH) and calculate specific humidity at both states using psychrometric charts or steam tables. Finally, the RH at the exit is recalculated by comparing the specific humidity at the exit with the saturation-specific humidity at 100°C.

compressor work

Compressor work refers to the energy required to compress the air. It is calculated using the energy balance equation. For a well-insulated, steady-state system, apply the energy balance: energy in = energy out + work done. Given the specific enthalpies of dry air and water vapor at the initial and final states, and using the mass flow rate, Power input (kW) is determined by: \(\text{Power input} (kW) = \text{mass flow rate} \times (\text{specific enthalpy change between states})\).

entropy production

Entropy production rate measures the 'irreversibility' of a thermodynamic process. For the given problem, entropy production is determined by formulating the entropy balance equation for the inlet and outlet states. Using the specific entropy values of both the dry air and water vapor, the rate of entropy production is calculated as: \( \text{Rate of entropy production} = \text{mass flow rate} \times (\text{specific entropy change between states}) \).

energy balance

Energy balance in thermodynamics is used to ensure that all forms of energy are accounted for in a system. In this problem, we apply the energy balance principle: \( \text{Energy In} = \text{Energy Out} + \text{Work Done} \). Given the specific enthalpies for the initial and final states, the power input (work done by compressor) is calculated accordingly. As it's a steady-state and well-insulated system, no heat losses are considered.

mass balance

Mass balance ensures that mass is conserved within a system. For the compressor problem, mass balance is applied individually to dry air and water vapor. This means the mass flow rate of the dry air and the mass flow rate of the water vapor are calculated at both the inlet and outlet. By doing so, we ensure the mass flow rates entering the system are equal to those leaving, complying with the conservation of mass principle.