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In a refrigeration system, would you recommend condensing the refrigerant-134a at a pressure of 0.7 or \(1.0 \mathrm{MPa}\) if heat is to be rejected to a cooling medium at \(15^{\circ} \mathrm{C} ?\) Why?

Short Answer

Expert verified
Answer: It is more energy-efficient to condense the refrigerant-134a at a pressure of 0.7 MPa due to the smaller temperature difference (14.4°C) compared to that of the 1.0 MPa (24.7°C). A smaller temperature difference improves system efficiency.

Step by step solution

01

Determine the saturation temperatures for the given pressures

We first need to find the saturation temperatures corresponding to the given pressures 0.7 MPa and 1.0 MPa for refrigerant-134a. We can do this by consulting a refrigerant-134a saturation table or using a refrigeration software. For refrigerant-134a: At 0.7 MPa, saturation temperature, \(T_{sat_1} = 29.4^{\circ} \mathrm{C}\) At 1.0 MPa, saturation temperature, \(T_{sat_2} = 39.7^{\circ} \mathrm{C}\)
02

Calculate the temperature difference between the condensing refrigerant and the cooling medium for each pressure

Next, we will find the temperature difference between the condensing refrigerant and the cooling medium for each pressure. The cooling medium temperature is \(15^{\circ} \mathrm{C}\). For 0.7 MPa: Temperature difference, \(\Delta T_1 = T_{sat_1} - T_{cool}\) \(\Delta T_1 = 29.4 - 15 = 14.4^{\circ} \mathrm{C}\) For 1.0 MPa: Temperature difference, \(\Delta T_2 = T_{sat_2} - T_{cool}\) \(\Delta T_2 = 39.7 - 15 = 24.7^{\circ} \mathrm{C}\)
03

Compare the temperature differences and decide on an appropriate pressure.

Now we will compare the temperature differences calculated in step 2 to determine which pressure is more efficient for our refrigeration system. Since \(\Delta T_1 = 14.4^{\circ} \mathrm{C}\) (0.7 MPa) is lower than \(\Delta T_2 = 24.7^{\circ} \mathrm{C}\) (1.0 MPa), we can conclude that it is more energy-efficient and better to condense the refrigerant-134a at a pressure of 0.7 MPa. This is because a smaller temperature difference between the condensing refrigerant and the cooling medium will result in less heat being rejected to the cooling medium, which improves the system efficiency.

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Key Concepts

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

Saturation Temperature
Understanding the saturation temperature is essential for evaluating the performance of refrigeration systems. Saturation temperature, often referred to as the boiling point, is the temperature at which a refrigerant changes phase, either from liquid to gas during evaporation or from gas to liquid during condensation, at a given pressure. This is a crucial property because it determines the heat exchange conditions within the system.

In the context of a refrigeration system rejecting heat to a cooling medium, the saturation temperature of the refrigerant must be higher than the temperature of the cooling medium to ensure heat flows from the refrigerant to the medium. Conversely, within the evaporator, the refrigerant's saturation temperature should be lower than the surroundings to effectively absorb heat.

For refrigerant-134a at 0.7 MPa, the saturation temperature is found to be around 29.4°C, highlighting the refrigerant's ability to condense at a reasonably low temperature while still effectively transferring heat to a medium at 15°C, which is common in many practical applications.
Refrigerant-134a Properties
Refrigerant-134a, also known as R-134a, is a hydrofluorocarbon (HFC) that's widely used in air conditioning and refrigeration systems. A key factor in its popularity has been its relatively low environmental impact compared to other refrigerants in terms of ozone depletion potential and global warming potential.

When scrutinizing R-134a properties, engineers consider aspects such as thermal conductivity, specific heat, and density – all of which influence the system's capacity and efficiency. These properties are temperature and pressure-dependent and play critical roles in determining the operational conditions and the design of the refrigeration system components.

The choice of R-134a often balances efficiency with environmental considerations. It's crucial to recognize that finding the proper condensing pressure, as discussed in the exercise, isn't just a question of immediate system efficiency but also influences long-term sustainability and regulatory compliance.
Condensing Pressure
The condensing pressure in a refrigeration system is closely related to the saturation temperature of the refrigerant during the phase-change process in the condenser. Ensuring the correct condensing pressure is vital for maintaining system reliability and achieving optimum efficiency.

Lower condensing pressures generally mean lower saturation temperatures, which can lead to a higher efficiency because the compressor requires less work to reach the condensing pressure. However, pressures must be kept within a range that ensures adequate temperature differences for heat exchange, compatible with the operational temperatures of the system's components.

The choice between maintaining a condensing pressure of 0.7 MPa or 1.0 MPa affects not only the system's performance but also its design. For instance, components like the condenser will be different in size and design depending on the chosen condensing pressure to handle the heat rejection needs effectively.
Temperature Difference Calculation
Calculating the temperature difference between the condensing refrigerant and the cooling medium is a fundamental step in evaluating the performance of a refrigeration system. This difference is a driving factor for heat rejection and ultimately influences the system's energy consumption.

In our exercise, we first calculated the saturation temperatures of refrigerant-134a at two different pressures. Then, by subtracting the cooling medium temperature from these saturation temperatures, we derived the temperature differences for each pressure scenario.

A smaller temperature difference, as seen with the 0.7 MPa pressure, implies that less energy is required to expel heat to the cooling medium, favoring better system efficiency. However, engineers must consider that too small a temperature difference could also mean inadequate heat transfer and potential system inefficiencies or increased costs due to larger heat exchanger sizes.

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Most popular questions from this chapter

A refrigeration unit operates on the ideal vapor compression refrigeration cycle and uses refrigerant-22 as the working fluid. The operating conditions for this unit are evaporator saturation temperature of \(-5^{\circ} \mathrm{C}\) and the condenser saturation temperature of \(45^{\circ} \mathrm{C}\). Selected data for refrigerant- 22 are provided in the table below. $$\begin{array}{lcccc}\hline T,^{\circ} \mathrm{C} & P_{\text {sat }}, \mathrm{kPa} & h_{f}, \mathrm{kJ} / \mathrm{kg} & h_{g}, \mathrm{kJ} / \mathrm{kg} & s_{g}, \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K} \\ \hline-5 & 421.2 & 38.76 & 248.1 & 0.9344 \\\45 & 1728 & 101 & 261.9 & 0.8682 \\\\\hline\end{array}$$ For \(R-22\) at \(P=1728\) kPa and \(s=0.9344 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) \(T=68.15^{\circ} \mathrm{C}\) and \(h=283.7 \quad \mathrm{kJ} / \mathrm{kg} .\) Also, take \(c_{p, \text { air }}=1.005 \mathrm{kJ} / \mathrm{kg} \cdot \mathrm{K}\) (a) Sketch the hardware and the \(T\) -s diagram for this heat pump application. (b) Determine the COP for this refrigeration unit. (c) The evaporator of this unit is located inside the air handler of the building. The air flowing through the air handler enters the air handler at \(27^{\circ} \mathrm{C}\) and is limited to a \(20^{\circ} \mathrm{C}\) temperature drop. Determine the ratio of volume flow rate of air entering the air handler \(\left(m_{\text {ail }}^{3} / \min \right)\) to mass flow rate of \(\mathrm{R}-22\left(\mathrm{kg}_{\mathrm{R}-22} / \mathrm{s}\right)\) through the air handler, in \(\left(m_{\mathrm{air}}^{3} / \mathrm{min}\right) /\left(\mathrm{kg}_{\mathrm{R}-22} / \mathrm{s}\right) .\) Assume the air pressure is \(100 \mathrm{kPa}\).

An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at \(800 \mathrm{kPa}\) and the evaporator at \(-12^{\circ} \mathrm{C}\). Determine this system's COP and the amount of power required to service a \(150 \mathrm{kW}\) cooling load.

A thermoelectric cooler has a COP of 0.18 and the power input to the cooler is 1.8 hp. Determine the rate of heat removed from the refrigerated space, in \(\mathrm{Btu} / \mathrm{min}\).

It is proposed to run a thermoelectric generator in conjunction with a solar pond that can supply heat at a rate of \(7 \times 10^{6} \mathrm{kJ} / \mathrm{h}\) at \(90^{\circ} \mathrm{C}\). The waste heat is to be rejected to the environment at \(22^{\circ} \mathrm{C}\). What is the maximum power this thermoelectric generator can produce?

An ideal gas refrigeration cycle using air as the working fluid operates between the pressure limits of 80 and \(280 \mathrm{kPa} .\) Air is cooled to \(35^{\circ} \mathrm{C}\) before entering the turbine. The lowest temperature of this cycle is \((a)-58^{\circ} \mathrm{C}\) \((b)-26^{\circ} \mathrm{C}\) \((c) 5^{\circ} \mathrm{C}\) \((d) 11^{\circ} \mathrm{C}\) \((e) 24^{\circ} \mathrm{C}\)

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