A commercial refrigerator with refrigerant- 134 a as the...

A commercial refrigerator with refrigerant- 134 a as the working fluid is used to keep the refrigerated space at $-35^{\circ} \mathrm{C}$ by rejecting waste heat to cooling water that enters the condenser at $18^{\circ} \mathrm{C}$ at a rate of $0.25 \mathrm{kg} / \mathrm{s}$ and leaves at $26^{\circ} \mathrm{C}$. The refrigerant enters the condenser at $1.2 \mathrm{MPa}$ and $50^{\circ} \mathrm{C}$ and leaves at the same pressure subcooled by $5^{\circ} \mathrm{C}$. If the compressor consumes $3.3 \mathrm{kW}$ of power, determine $(a)$ the mass flow rate of the refrigerant, $(b)$ the refrigeration load, $(c)$ the $C O P,$ and $(d)$ the minimum power input to the compressor for the same refrigeration load.

Key Concepts

Vapor Compression Refrigeration Cycle

This is a thermodynamic cycle used for cooling wherein the refrigerant undergoes phase changes and pressure variation as it circulates through components such as the evaporator, compressor, condenser, and expansion device. It involves absorbing heat in the evaporator at low pressure and rejecting it in the condenser at high pressure, forming the basis of most cooling systems in practical applications.

Refrigerant Property Analysis

Understanding the thermodynamic properties of the refrigerant (such as R-134a) is crucial because its behavior in terms of pressure, temperature, and phase change under different operating conditions determines the performance of the refrigeration cycle. This analysis allows engineers to determine specific states of the refrigerant at various stages of the cycle.

Energy and Mass Balances

This concept involves applying the principles of conservation of mass and energy to the refrigeration cycle. In practical applications, these calculations are used to determine quantities such as the mass flow rates of the refrigerant and the cooling water, the refrigeration load, and the distribution of energy among different components of the system.

Coefficient of Performance (COP)

The COP is a performance measure of refrigeration systems, defined as the ratio of the refrigeration effect (or load) to the power input to the compressor. It is an essential parameter that indicates the energy efficiency of the refrigerator, with higher COP values reflecting a more efficient system.

Carnot Efficiency and Minimum Work

This concept is used to determine the theoretical limit of system performance based on the Carnot cycle. It pertains to calculating the minimum possible work required for a given refrigeration load by relating the operating temperatures to the highest achievable efficiency, serving as an ideal benchmark against which real systems are compared.

Heat Exchanger Analysis

Heat exchangers, such as condensers and evaporators, play a critical role in transferring heat between fluids. Analyzing these components includes understanding how changes in inlet and outlet temperatures and flow rates affect the overall heat transfer process, which is vital for assessing the system's thermal performance and for ensuring efficient heat rejection or absorption.

Transcript

00:01 So here in this problem they are saying that commercial refrigerator is being used to maintain the refrigerated space at minus 35 degrees celsius and the condenser is rejecting heat by using the cooling water which enters at 18 degrees celsius and it exit at 26 degrees celsius and the mass flow rate of water we are given as 0 .25 kg per second and they are saying that the refrigerator enters the condenser let this be as condition one let this be as second condition this is third and this is fourth and they are saying that at condition one pressure and temperature we are given as 1 .2 megapa pascal and 50 degrees celsius and then the refrigerator is sub cooled by 5 degrees celsius after the in the condenser and then the and they are saying the power consumed by the compression it is given given as 3 .3 kilowatt and we have to determine four things the first one is mass flow rate of the refrigerant then the refrigeration load the cop and finally the minimum power input to the compression for the same refrigeration load so first of all we'll be doing one thing we'll be making we'll be finding the properties of pressure and temperature at point one and at point two so we are given that pressure is 1 .2 megapascal and the temperature is 50 degrees celsius at point 1 so now we'll be using the properties of the refrigerant 134a as given in the book from the table given the book we can say that the enthalpy h1 at point 1 it would be nearly 278 .28 .28 kilojoule per kg now next we'll be writing the properties at point 2 but before that we have to calculate the temperature at 0 .2 so temperature at point 2 would be equals to the saturation temperature at 1 .2 megapascal minus delta t of subcooled.

02:17 So we are given that the temperature is sub cooled in the condenser it is 5 degrees celsius and the saturation temperature at corresponding to 1 .2 megapaascal it would be 46 .3.

02:30 It have obtained this value from the table.

02:33 We have to see the value corresponding to 1 .2 megapascal.

02:36 The temperature would be this minus this thing we are given in the equation as 5 degrees celsius so it comes out to be as 41 .3 degrees celsius so we have pressure at point 2 as 1 .2 megapascal and the temperature is 41 point this t2 comes out to be as 41 .3 degrees celsius so corresponding to this value we'll get the enthalpy as 11 .19 .19 .10 .19 kilojoule per kg.

03:11 Now next we'll be writing the properties of water.

03:14 So water enters at 18 degrees celsius and it exit at 26 degrees celsius.

03:22 That means if we see the inlet condition of water it is 18 degrees celsius and the dryness fraction of water is zero...

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