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Refrigerant \(134 a\) enters a counterflow heat exchanger operating at steady state at \(-32^{\circ} \mathrm{C}\) and a quality of \(40 \%\) and exits as saturated vapor at \(-32^{\circ} \mathrm{C}\). Air enters as a separate sticam with a unass fluw 1 ate of \(5 \mathrm{~kg} / \mathrm{s}\) and is couleal al a constant pressure of 1 bar from 300 to \(250 \mathrm{~K}\). Heat transfer between the heat exchanger and its surroundings can be ignored, as can the effects of motion and gravity. (a) As in Fig. E7.6, sketch the variation with position of the temperature of each stream. Locate \(T_{0}\) on the sketch. (b) Determine the rate of exergy destruction within the heat exchanger, in \(\mathrm{kW}\). (c) Devise and evaluate an exergetic efficiency for the heat exchanger. Let \(T_{0}=300 \mathrm{~K}, p_{0}=1\) bar.

Short Answer

Expert verified
a) Sketch the temperature profile. b) Use conservation of energy and entropy principles to find exergy destruction. c) Calculate the exergetic efficiency using the input and destroyed exergy.

Step by step solution

01

Identify Given Data

The given data includes: -Refrigerant 134a enters at inguinal's State: -32°C, quality of 40%, exits as saturated vapor at -32°C.-Air: mass flow rate 5 kg/s, cooled at 1 bar from 300 K to 250 K.-The ambient temperature and pressure are 300 K and 1 bar respectively.
02

Sketch the Temperature Profile

Create a plot with the x-axis representing the position within the heat exchanger, and the y-axis representing the temperature (in Kelvin). Plot two separate lines, one for the Refrigerant 134a (constant at -32°C or 241 K) and one for Air (declining from 300 K to 250 K).
03

Determine the Enthalpy Changes

Use Refrigerant tables to find the specific enthalpies: Initial: Quality = 0.4,inguinal's h1 = hf + x(hfg). Find hf and hfg for -32°C from the tables. Final: Saturated vapor h2 = hg. Air: Assume c_p (specific heat at constant pressure) is 1.005 kJ/kg*K. Then, inguinal's h_air = c_p(T_{in} - T_{out}).
04

Calculate Mass Flow Rates

Mass flow rate of air is given: 5 kg/s. For Refrigerant, use energy balance between input and output states to find it. inguinal's inguinal's Q_ref = m_ref(h2 - h1).
05

Energy Balance

Using the principle of conservation of energy, the energy lost by the air must equal to the energy gained by the refrigerant: inguinal's m_air inguinal's c_p inguinal's (T_out - T_in) = m_ref inguinal's (h2 - h1).
06

Exergy Destruction Calculation

Calculate the exergy destruction using: inguinal's inguinal's () E_destruction = T_0(ΔS_air + ΔS_ref).Where, ΔS_is = m(O_pTinx - T_{exit})/T_{in).
07

Determine Exergetic Efficiency

Define exergetic efficiency as the ratio of useful exergy output to the exergy input. Use the calculated exergy destruction and input to find efficiency: inguinal's E_efficiency = 1 - E_destruction / (E_input).
08

Evaluate Results

Perform all necessary calculations following the formulas and substitution of values.

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Key Concepts

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

Refrigerant Thermodynamics
Refrigerants are substances used in heat exchange systems, like refrigerators and air conditioners, to absorb and transfer heat. In this exercise, we focus on the thermodynamics of Refrigerant 134a, a common refrigerant. The refrigerant enters the heat exchanger at -32°C with a quality of 40%, meaning it's a two-phase mixture of liquid and vapor (40% vapor and 60% liquid). It exits the heat exchanger as saturated vapor at the same temperature of -32°C. This means the refrigerant has absorbed enough heat to convert all its liquid part into vapor, thus exiting as 100% vapor. To fully understand and analyze refrigerant behavior, you often refer to refrigerant tables which provide specific enthalpies (energy content) and other properties at different temperatures and pressures. These properties help calculate the energy and mass flow in the system.
Enthalpy Changes
Enthalpy, denoted as 'h,' is a measure of total energy in a thermodynamic system, including internal energy and the energy needed to make room for it by displacing its environment. The enthalpy changes in the refrigerant and air streams are crucial in this analysis. You calculate the initial and final enthalpy values for the refrigerant. Initially, for Refrigerant 134a at -32°C with 40% quality, you use the formula: h1 = hf + x(hfg). Here, hf is the enthalpy of the fluid, hfg is the enthalpy of vaporization, and x is the quality. For the final state, the refrigerant exits as saturated vapor at -32°C, so h2 = hg, where hg is the enthalpy of the vapor. For the air stream, you utilize the specific heat capacity at constant pressure (cp, typically 1.005 kJ/kg*K). The enthalpy change for the air is calculated using h_air = cp (T_in - T_out).
Energy Balance
Energy balance is the principle that energy entering a system must equal the energy leaving it, plus any changes within the system. In the context of this exercise, the energy lost by the cooling air stream must equal the energy gained by the heating refrigerant stream. We start by considering the given mass flow rate of the air stream, which is 5 kg/s. The air cools from 300 K to 250 K. The energy loss by the air stream can be expressed as: Q_air = m_air * cp * (T_in - T_out). We then equate this to the energy gained by the refrigerant: Q_ref = m_ref * (h2 - h1). Rearranging these equations allows us to solve for the mass flow rate of the refrigerant, m_ref, ensuring that the energy exchange between the air and refrigerant is balanced.
Exergetic Efficiency
Exergetic efficiency is a measure of how well a system converts available energy into useful work, taking into account energy losses. The exergy destruction in a system results from irreversibility due to factors like friction, unrestrained expansion, mixing of different substances, etc. To determine exergetic efficiency, we must first calculate the rate of exergy destruction. This can be found using the formula: E_destruction = T0 * (ΔS_air + ΔS_ref), where ΔS is the change in entropy. For the air: ΔS_air = m_air * cp * ln(T_out / T_in). For the refrigerant: ΔS_ref requires finding the entropy change from the refrigerant tables. Once the exergy destruction is known, the exergetic efficiency (η_ex) can be determined by the formula: η_ex = 1 - (E_destruction / E_input). This ratio provides insight into how effectively the heat exchanger operates, taking into account both the useful energy transfer and the energy losses due to inherent inefficiencies.

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Most popular questions from this chapter

Four kilograms of a two-phase liquid-vapor mixture of water initially at \(300^{\circ} \mathrm{C}\) and \(x_{1}=0.5\) undergo the two different processes described below. In each case, the mixture is brought from the initial state to a saturated vapor state, while the volume remains constant. For each process, determine the change in exergy of the water, the net amounts of exergy transfer by work and heat, and the amount of exergy destruction, each in kJ. Let \(T_{0}=300 \mathrm{~K}, p_{0}=1\) bar, and ignore the effects of motion and gravity. Comment on the difference between the exergy destruction values. (a) The process is brought about adiabatically by stirring the mixture with a paddle wheel. (b) The process is brought about by heat transfer from a thermal reservoir at \(610 \mathrm{~K}\). The temperature of the water at the location where the heat transfer occurs is \(610 \mathrm{~K}\).

Nitrogen \(\left(\mathrm{N}_{2}\right)\) at 25 bar, \(450 \mathrm{~K}\) enters a turbine and expands to \(2 \mathrm{bar}, 250 \mathrm{~K}\) with a mass flow rate of \(0.2 \mathrm{~kg} / \mathrm{s}\) The turbine operates at steady state with negligible heat transfer with its surroundings. Assuming the ideal gas model with \(k=\) \(1.399\) and ignoring the effects of motion and gravity, determine (a) the isentropic turbine efficiency. (b) the exergetic turbine efficiency. Let \(T_{0}=25^{\circ} \mathrm{C}, p_{0}=1 \mathrm{~atm}\).

Air enters an insulated turbine operating at steady state at \(8 \mathrm{bar}, 500 \mathrm{~K}\), and \(150 \mathrm{~m} / \mathrm{s}\). At the exit the conditions are \(1 \mathrm{bar}, 320 \mathrm{~K}\), and \(10 \mathrm{~m} / \mathrm{s}\). There is no significant change in elevation. Determine the work developed and the exergy destruction, each in \(\mathrm{kJ}\) per kg of air flowing. Let \(T_{0}=300 \mathrm{~K}\), \(p_{0}=1\) bar.

A vessel contains \(1 \mathrm{lb}\) of air at pressure \(p\) and \(200^{\circ} \mathrm{F}\). Using the ideal gas model, plot the specific exergy of the air, in Btu/lb, for \(p\) ranging from \(0.5\) to 2 atm. The air is at rest and negligible elevation relative to an exergy reference environment for which \(T_{0}=60^{\circ} \mathrm{F}, p_{0}=1 \mathrm{~atm}\).

Steam enters a turbine operating at steady state at \(4 \mathrm{MPa}\), \(500^{\circ} \mathrm{C}\) with a mass flow rate of \(50 \mathrm{~kg} / \mathrm{s}\). Saturated vapor exits at \(10 \mathrm{kPa}\) and the corresponding power developed is \(42 \mathrm{MW}\). The effects of motion and gravity are negligible. (a) For a control volume enclosing the turbine, determine the rate of heat transfer, in MW, from the turbine to its surroundings Assuming an average turbine outer surface temperature of \(50^{\circ} \mathrm{C}\), determine the rate of exergy destruction, in MW. (b) If the turbine is located in a facility where the ambient temperature is \(27^{\circ} \mathrm{C}\), determine the rate of exergy destruction for an enlarged control volume including the turbine and its immediate surroundings so heat transfer takes place at the ambient temperature. Explain why the exergy destruction values of parts (a) and (b) differ. Let \(T_{0}=300 \mathrm{~K}, p_{0}=100 \mathrm{kPa}\).

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