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Air at \(30^{\circ} \mathrm{C}, 1.05\) bar, and \(80 \%\) relative humidity enters a dehumidifier operating at steady state. Moist air exits at \(15^{\circ} \mathrm{C}, 1\) bar, and \(95 \%\) relative humidity. Condensate exits in a separate stream at \(15^{\circ} \mathrm{C}\). A refrigerant flows through the cooling coil of the dehumidifier with an increase in its specific enthalpy of \(100 \mathrm{~kJ}\) per \(\mathrm{kg}\) of refrigerant flowing. Heat transfer between the humidifier and its surroundings and kinetic and potential energy effects can be ignored. Determine the refrigerant flow rate, in \(\mathrm{kg}\) per \(\mathrm{kg}\) of dry air.

Short Answer

Expert verified
Use psychrometric charts to determine air properties, apply mass balance for condensate, use energy balance to relate enthalpies, and solve for refrigerant flow rate.

Step by step solution

01

Gather Inputs and Assumptions

Given: - Inlet air: \(30^{\circ} \text{C}, 1.05 \text{ bar}, 80\% \text{ relative humidity}\)- Outlet air: \(15^{\circ} \text{C}, 1 \text{ bar}, 95\% \text{ relative humidity}\)- Condensate temperature: \(15^{\circ} \text{C}\)- Increase in refrigerant specific enthalpy: \(100 \text{ kJ/kg}\)- Neglecting heat transfer with surroundings and kinetic/potential energy effects
02

Determine Inlet and Outlet Saturation Properties

Use psychrometric charts or properties tables to find corresponding enthalpies and humidity ratios for the given temperatures, pressures, and relative humidities:- Inlet air (T=30°C, RH=80%): - Saturation pressure, \(P_s\) - Humidity ratio, \(W_1\) - Specific enthalpy, \(h_1\)- Outlet air (T=15°C, RH=95%): - Saturation pressure, \(P_s\) - Humidity ratio, \(W_2\) - Specific enthalpy, \(h_2\)
03

Calculate the Mass of Condensate

The mass of the water vapor that condenses can be calculated using the difference in humidity ratios:\[ m_{condensate} = W_1 - W_2 \]
04

Apply Energy Balance

Since energy changes in the dehumidifier are due to the air cooling and condensation of water vapor, apply the energy balance assuming no heat transfer with surroundings:\[ \text{Energy removed} = \text{(dry air flow rate)} \times (h_1 - h_2) + \text{(condensate flow rate)} \times h_f \text{(at 15°C)} \]Here, \(h_f\) is the enthalpy of liquid water at 15°C.
05

Relate Refrigerant to Energy Removed

The energy removed by the refrigerant is related to its mass flow rate and enthalpy change:\[ \text{Energy removed} = (\text{Refrigerant flow rate}) \times (100 \text{ kJ/kg}) \]
06

Solve for Refrigerant Flow Rate

By equating the energy removed by the dehumidifier to the energy removed by the refrigerant, solve for the refrigerant flow rate per kg of dry air flow:\[ (\text{dry air flow rate}) \times (h_1 - h_2) + (\text{condensate flow rate}) \times h_f = (\text{Refrigerant flow rate}) \times (100 \text{ kJ/kg}) \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Psychrometric properties
The term 'psychrometric properties' refers to the various parameters used to describe moist air conditions in thermodynamics. These include temperature, pressure, relative humidity, specific enthalpy, and humidity ratio.

In the given exercise, we need to work with these properties to track how air changes as it flows through the dehumidifier. Psychrometric charts or tables can provide the necessary values. Here's a breakdown:
  • Saturation pressure \(\text{P}_s\): The pressure at which air becomes saturated with moisture at a given temperature.

  • Humidity ratio \( \text{W} \): The mass of water vapor per mass of dry air.

  • Specific enthalpy \( \text{h} \): The total heat content of the moist air per unit mass.
By using these properties, we can determine how much energy is transferred and how much water vapor condenses during the dehumidification process.
Energy balance
An energy balance is used to account for all energy entering and leaving a system. In this exercise, the dehumidifier's change in energy is due to cooling the moist air and condensing water vapor.

The energy balance equation can be written as:
\[\text{Energy removed} = (\text{dry air flow rate}) \times (h_1 - h_2) + (\text{condensate flow rate}) \times h_f \]
This equation means that the total energy removed from the air is the sum of:
  • The energy due to cooling the dry air, which is the difference between the specific enthalpy of the inlet and outlet air multiplied by the dry air flow rate.

  • The energy due to condensing water vapor, which is the flow rate of the condensate multiplied by the enthalpy of the liquid water at 15°C \( h_f \).

Ignoring heat transfer between the dehumidifier and surroundings simplifies the calculation and focuses solely on the main energy transactions within the dehumidifier.
Humidity ratio calculation
Calculating the humidity ratio is crucial to understanding the amount of water vapor in the air. This in turn helps in determining the mass of condensate produced.

The initial and final humidity ratios (\text{W}\text{_1} and \text{W}\text{_2}) can be obtained from psychrometric charts or tables.

Using these values, the mass of condensate produced can be found by:
\[\text{m}_{\text{condensate}} = \text{W}_{1} - \text{W}_{2} \]
This equation calculates the difference between the initial and final humidity ratios, effectively giving us the total mass of water vapor that condenses out of the air.

The exact values of \text{W}_{1} and \text{W}_{2} are derived from the given conditions of the inlet and outlet air, which ensures accurate calculation of the condensate mass.
Refrigerant flow rate
The refrigerant flow rate is key in determining the efficiency of the dehumidifier. The refrigerant absorbs heat as it passes through the cooling coil, indicated by an increase in specific enthalpy.

According to the exercise, the specific enthalpy of the refrigerant increases by 100 kJ/kg. Using this information along with the energy balance, we can find the refrigerant flow rate.

The formula is:
\[\text{Energy removed} = (\text{Refrigerant flow rate}) \times (100 \text{kJ/kg}) \]

Setting the energy removed from the air equal to the energy absorbed by the refrigerant and solving for the refrigerant flow rate:

\[( \text{dry air flow rate}) \times (h_1 - h_2) + ( \text{condensate flow rate}) \times h_f = (\text{Refrigerant flow rate}) \times (100 \text{kJ/kg}) \]
Doing this will give you the refrigerant flow rate per kg of dry air.

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Most popular questions from this chapter

Outside air at \(50^{\circ} \mathrm{F}, 1\) atm, and \(40 \%\) relative humidity enters an air-conditioning device operating at steady state. Liquid water is injected at \(45^{\circ} \mathrm{F}\) and a moist air stream exits with a volumetric flow rate of \(1000 \mathrm{ft}^{3} / \mathrm{min}\) at \(90^{\circ} \mathrm{F}, 1 \mathrm{~atm}\) and a relative humidity of \(40 \%\). Neglecting kinetic and potential energy effects, determine (a) the rate water is injected, in \(\mathrm{lb} / \mathrm{min}\). (b) the rate of heat transfer to the moist air, in Btu/h.

Figure P12.82 shows a compressor followed by an aftercooler. Atmospheric air at \(14.7 \mathrm{lbf} / \mathrm{in}^{2}, 900^{\circ} \mathrm{F}\), and \(75 \%\) relative humidity enters the compressor with a volumetric flow rate of \(100 \mathrm{ft}^{3} / \mathrm{min}\). The compressor power input is \(15 \mathrm{hp}\). The moist air exiting the compressor at \(100 \mathrm{lbf} / \mathrm{in}^{2}, 400^{\circ} \mathrm{F}\) flows through the aftercooler, where it is cooled at constant pressure, exiting saturated at \(100^{\circ} \mathrm{F}\). Condensate also exits the aftercooler at \(100^{\circ} \mathrm{F}\). For steady-state operation and negligible kinetic and potential energy effects, determine (a) the rate of heat transfer from the compressor to its surroundings, in Btu/min. (b) the mass flow rate of the condensate, in \(\mathrm{lb} / \mathrm{min}\). (c) the rate of heat transfer from the moist air to the refrigerant circulating in the cooling coil, in tons of refrigeration.

A stream of air (stream 1 ) at \(60^{\circ} \mathrm{F}, 1 \mathrm{~atm}, 30 \%\) relative humidity is mixed adiabatically with a stream of air (stream 2) at \(90^{\circ} \mathrm{F}, 1 \mathrm{~atm}, 80 \%\) relative humidity. A single stream (stream 3 ) exits the mixing chamber at temperature \(T_{3}\) and \(1 \mathrm{~atm}\). Assume steady state and ignore kinetic and potential energy effects Letting \(r\) denote the ratio of dry air mass flow rates \(\dot{m}_{\mathrm{a} 1} / \dot{m}_{\mathrm{a} 2}\) (a) determine \(T_{3}\), in \({ }^{\circ} \mathrm{F}\), for \(r=2\). (b) plot \(T_{3}\), in \({ }^{\circ} \mathrm{F}\), versus \(r\) ranging from 0 to 10 .

At steady state, moist air at \(29^{\circ} \mathrm{C}, 1\) bar, and \(50 \%\) relative humidity enters a device with a volumetric flow rate of \(13 \mathrm{~m}^{3} / \mathrm{s}\) Liquid water at \(40^{\circ} \mathrm{C}\) is sprayed into the moist air with a mass flow rate of \(22 \mathrm{~kg} / \mathrm{s}\). The liquid water that does not evaporate into the moist air stream is drained and flows to another device at \(26^{\circ} \mathrm{C}\) with a mass flow rate of \(21.55 \mathrm{~kg} / \mathrm{s}\). A single moist air stream exits at 1 bar. Determine the temperature and relative humidity of the moist air stream exiting. Ignore heat transfer between the device and its surroundings and kinetic and potential energy effects.

Atmospheric air having dry-bulb and wet-bulb temperatures of 33 and \(29^{\circ} \mathrm{C}\), respectively, enters a wellinsulated chamber operating at steady state and mixes with air entering with dry-bulb and wet- bulb temperatures of 16 and \(12^{\circ} \mathrm{C}\), respectively. The volumetric flow rate of the lower temperature stream is twice that of the other stream. A single mixed stream exits. Determine for the exiting stream (a) the relative humidity. (b) the temperature, in \({ }^{\circ} \mathrm{C}\). Pressure is uniform throughout at \(1 \mathrm{~atm}\). Neglect kinetic and potential energy effects.

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