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An air-conditioning system involves the mixing of cold air and warm outdoor air before the mixture is routed to the conditioned room in steady operation. Cold air enters the mixing chamber at \(7^{\circ} \mathrm{C}\) and \(105 \mathrm{kPa}\) at a rate of \(0.55 \mathrm{m}^{3} / \mathrm{s}\) while warm air enters at \(34^{\circ} \mathrm{C}\) and 105 kPa. The air leaves the room at \(24^{\circ} \mathrm{C}\). The ratio of the mass flow rates of the hot to cold air streams is \(1.6 .\) Using variable specific heats, determine \((a)\) the mixture temperature at the inlet of the room and (b) the rate of heat gain of the room.

Short Answer

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Question: Determine the mixture temperature at the inlet of the room and the rate of heat gain of the room, given the following parameters: - Cold air temperature: 7°C - Hot air temperature: 34°C - Cold air volumetric flow rate: 0.55 m³/s - Hot air volumetric flow rate: 1.6 times the cold air flow rate - Cold and hot air pressures: 105 kPa Answer: To determine the mixture temperature at the inlet of the room, follow these steps: 1. Calculate the mass flow rates of the hot and cold air streams using the ideal gas equation and given parameters. 2. Calculate the enthalpies of the incoming and outgoing air streams at the given temperatures using the variable specific heats. 3. Apply the steady flow energy equation (SFEE) to find the mixture temperature (T_out) corresponding to the calculated enthalpy value (h_out). To determine the rate of heat gain of the room, use the energy balance for the room and the enthalpy values calculated in the previous steps: Rate of heat gain (Q_gain) = (total mass flow rate * mixture enthalpy) - (cold air mass flow rate * cold air enthalpy) - (hot air mass flow rate * hot air enthalpy)

Step by step solution

01

Calculate mass flow rates of hot and cold air streams

To calculate the mass flow rates, we first need to determine the specific volume of the air streams using the ideal gas equation as follows: \(v_{c} = \frac{R_{air}T_{c}}{P_{c}}\) and \(v_{h} = \frac{R_{air}T_{h}}{P_{h}}\) where \(v_c\) is the specific volume of cold air, \(v_h\) is the specific volume of hot air, \(R_{air}\) is the specific gas constant for air (approximately 0.287 kJ/kgK), \(T_{c}=7+273.15 \thinspace K\) and \(T_{h}=34+273.15 \thinspace K\) are the absolute temperatures of cold and hot air in Kelvin, and \(P_{c}=P_{h}=105 \thinspace kPa\) are the pressures of cold and hot air. Now we can calculate the mass flow rates using the volumetric flow rates and specific volumes: \(\dot{m}_{c} = \frac{\dot{V}_{c}}{v_{c}}\) and \(\dot{m}_{h} = \frac{1.6\dot{V}_{c}}{v_{h}}\) where \(\dot{m}_{c}\) and \(\dot{m}_{h}\) are the mass flow rates of cold and hot air, and \(\dot{V}_{c}=0.55 \thinspace m^3/s\) is the volumetric flow rate of cold air.
02

Calculate enthalpies of incoming and outgoing air streams

Using variable specific heats, we can obtain the enthalpy values from air property tables (or software) at the given temperatures: \(h_{c} = h(T_{c})\), \(h_{h} = h(T_{h})\), and \(h_{out} = h(T_{out})\) where \(h_{c}\), \(h_{h}\), and \(h_{out}\) are the specific enthalpies of cold, hot, and outgoing air, and \(T_{out}=24+273.15 \thinspace K\) is the temperature of the outgoing air.
03

Apply the steady flow energy equation (SFEE)

SFEE states that the sum of energy input to the mixing chamber equals the sum of energy output. In this case, the only form of energy involved is enthalpy since the kinetic and potential energies can be assumed negligible, and no heat is added or work is done: \(\dot{m}_{c}h_{c} + \dot{m}_{h}h_{h} = \left(\dot{m}_{c}+\dot{m}_{h}\right)h_{out}\) (a) To determine the mixture temperature at the inlet of the room, which is \(T_{out}\), we can solve the above equation for \(h_{out}\) and then find the corresponding temperature value: \(h_{out} = \frac{\dot{m}_{c}h_{c} + \dot{m}_{h}h_{h}}{\dot{m}_{c}+\dot{m}_{h}}\) Find \(T_{out}\) corresponding to \(h_{out}\) from the air property table (or software). (b) To determine the rate of heat gain of the room, we will use the energy balance for the room itself. The incoming enthalpy is the mixture enthalpy at the inlet, while the outgoing enthalpy is for the air leaving the room: \(\dot{Q}_{gain} = \left(\dot{m}_{c}+\dot{m}_{h}\right)h_{out} - \dot{m}_{c}h_{c} - \dot{m}_{h}h_{h}\) where \(\dot{Q}_{gain}\) is the rate of heat gain by the room.

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Key Concepts

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

Steady Flow Energy Equation
Understanding the steady flow energy equation (SFEE) is crucial when analyzing systems like air-conditioning, where fluid flows continuously through a control volume. In essence, the SFEE states that for a steady-flow process, the sum of all energy entering a control volume must equal the sum of all energy leaving that volume. This balance ensures that energy is conserved.

When applying the SFEE to air-conditioning, we focus mainly on the enthalpy associated with the mass flow of air, assuming that changes in kinetic and potential energy are negligible. Thus, the heat gained or lost by the room is directly related to the difference in enthalpy between the incoming and outgoing air streams.
Enthalpy
Enthalpy is a thermodynamic property that represents the total heat content of a system. It is particularly useful for processes occurring at constant pressure, which is common in HVAC applications like air-conditioning systems. Mathematically, enthalpy is the sum of a substance's internal energy and the product of its pressure and volume. In practical terms, as seen in our exercise, enthalpy helps in calculating the heat transfer during the mixing of two air streams.

Since air-conditioning often uses air as the working fluid, and air's specific heat capacity varies with temperature, we look up the specific enthalpies for the given temperatures in property tables or calculate them using the air's variable specific heat.
Mass Flow Rate
The mass flow rate measures how much mass moves through a given surface per unit time, typically expressed in kilograms per second (kg/s). It's a key component in our exercise, where we calculate the amount of cold and hot air entering the mixing chamber. This rate directly affects the energy balance and therefore the overall efficiency of the system. The mass flow rate can be computed by dividing the volumetric flow rate by the specific volume (which is the inverse of density for ideal gases) of the fluid, a method applied in this exercise to find the rates for hot and cold air streams.
Ideal Gas Equation
The ideal gas equation, which states that the pressure, volume, and temperature of an ideal gas are related as PV = nRT, where P is pressure, V is volume, T is temperature, R is the specific gas constant, and n is the amount of substance (in moles), provides a simple means to determine various state properties of gases under certain assumptions. While real gases may deviate from ideal gas behavior under high pressure or low temperature, air can often be treated as an ideal gas at conditions typical of HVAC applications. An understanding of this equation is fundamental to calculating specific volumes, as done in the provided exercise for both the cold and hot air streams, enabling further calculations of mass flow rates.
Specific Heat
Specific heat, or specific heat capacity, is the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). Air, like any other substance, has a specific heat that varies with temperature -- a detail that's crucial when working with air-conditioning systems. For precise thermodynamic calculations, especially those using the SFEE, we use variable specific heats to determine the exact enthalpy values for the air at different temperatures. This approach increases the accuracy of the results in our exercise, affecting the mixing temperature and the heat gain calculations.

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Most popular questions from this chapter

In large steam power plants, the feedwater is frequently heated in a closed feedwater heater by using steam extracted from the turbine at some stage. Steam enters the feedwater heater at \(1 \mathrm{MPa}\) and \(200^{\circ} \mathrm{C}\) and leaves as saturated liquid at the same pressure. Feedwater enters the heater at \(2.5 \mathrm{MPa}\) and \(50^{\circ} \mathrm{C}\) and leaves at \(10^{\circ} \mathrm{C}\) below the exit temperature of the steam. Determine the ratio of the mass flow rates of the extracted steam and the feedwater.

A heat exchanger is to heat water \(\left(c_{p}=4.18 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) from 25 to \(60^{\circ} \mathrm{C}\) at a rate of \(0.2 \mathrm{kg} / \mathrm{s}\). The heating is to be accomplished by geothermal water \(\left(c_{p}=4.31 \mathrm{kJ} / \mathrm{kg} \cdot^{\circ} \mathrm{C}\right)\) available at \(140^{\circ} \mathrm{C}\) at a mass flow rate of \(0.3 \mathrm{kg} / \mathrm{s}\). Determine the rate of heat transfer in the heat exchanger and the exit temperature of geothermal water.

A tank with an internal volume of \(1 \mathrm{m}^{3}\) contains air at \(800 \mathrm{kPa}\) and \(25^{\circ} \mathrm{C}\). A valve on the tank is opened allowing air to escape and the pressure inside quickly drops to \(150 \mathrm{kPa}\), at which point the valve is closed. Assume there is negligible heat transfer from the tank to the air left in the tank. (a) Using the approximation \(h_{e} \approx\) constant \(=h_{e, \mathrm{avg}}=\) \(0.5\left(h_{1}+h_{2}\right),\) calculate the mass withdrawn during the process. (b) Consider the same process but broken into two parts. That is, consider an intermediate state at \(P_{2}=400 \mathrm{kPa}\), calculate the mass removed during the process from \(P_{1}=800 \mathrm{kPa}\) to \(P_{2}\) and then the mass removed during the process from \(P_{2}\) to \(P_{3}=150 \mathrm{kPa},\) using the type of approximation used in part \((a),\) and add the two to get the total mass removed. (c) Calculate the mass removed if the variation of \(h_{e}\) is accounted for.

Water flows through a shower head steadily at a rate of \(10 \mathrm{L} / \mathrm{min}\). An electric resistance heater placed in the water pipe heats the water from 16 to \(43^{\circ} \mathrm{C}\). Taking the density of water to be \(1 \mathrm{kg} / \mathrm{L},\) determine the electric power input to the heater, in \(\mathrm{kW}\). In an effort to conserve energy, it is proposed to pass the drained warm water at a temperature of \(39^{\circ} \mathrm{C}\) through a heat exchanger to preheat the incoming cold water. If the heat exchanger has an effectiveness of 0.50 (that is, it recovers only half of the energy that can possibly be transferred from the drained water to incoming cold water), determine the electric power input required in this case. If the price of the electric energy is 11.5 e \(/ \mathrm{kWh}\), determine how much money is saved during a 10 -min shower as a result of installing this heat exchanger.

Steam at 80 psia and \(400^{\circ} \mathrm{F}\) is mixed with water at \(60^{\circ} \mathrm{F}\) and 80 psia steadily in an adiabatic device. Steam enters the device at a rate of \(0.05 \mathrm{lbm} / \mathrm{s}\), while the water enters at \(1 \mathrm{lbm} / \mathrm{s}\). Determine the temperature of the mixture leaving this device when the outlet pressure is 80 psia.

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