Refrigerant- 22 absorbs heat from a cooled space at 50^∘ F as it flows through an evaporator of

Refrigerant- 22 absorbs heat from a cooled space at 50^∘ F...

Refrigerant- 22 absorbs heat from a cooled space at $50^{\circ} \mathrm{F}$ as it flows through an evaporator of a refrigeration system. R-22 enters the evaporator at $10^{\circ} \mathrm{F}$ at a rate of $0.08 \mathrm{lbm} / \mathrm{s}$ with a quality of 0.3 and leaves as a saturated vapor at the same pressure. Determine $(a)$ the rate of cooling provided, in $\mathrm{Btu} / \mathrm{h}$, (b) the rate of exergy destruction in the evaporator, and (c) the second-law efficiency of the evaporator. Take $T_{0}=77^{\circ} \mathrm{F}$. The properties of $\mathrm{R}-22$ at the inlet and exit of the evaporator are: $h_{1}=107.5 \mathrm{Btu} / \mathrm{lbm}, s_{1}=0.2851 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}, h_{2}=172.1 \mathrm{Btu} /$ Ibm, $s_{2}=0.4225 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}$

Refrigerant- 22 absorbs heat from a cooled space at $50^{\circ} \mathrm{F}$ as it flows through an evaporator of a refrigeration system. R-22 enters the evaporator at $10^{\circ} \mathrm{F}$ at a rate of $0.08 \mathrm{lbm} / \mathrm{s}$ with a quality of 0.3 and leaves as a saturated vapor at the same pressure. Determine $(a)$ the rate of cooling provided, in $\mathrm{Btu} / \mathrm{h}$, (b) the rate of exergy destruction in the evaporator, and (c) the second-law efficiency of the evaporator. Take $T_{0}=77^{\circ} \mathrm{F}$. The properties of $\mathrm{R}-22$ at the inlet and exit of the evaporator are:
$h_{1}=107.5 \mathrm{Btu} / \mathrm{lbm}, s_{1}=0.2851 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}, h_{2}=172.1 \mathrm{Btu} /$ Ibm, $s_{2}=0.4225 \mathrm{Btu} / \mathrm{lbm} \cdot \mathrm{R}$

Key Concepts

Exergy Analysis and Exergy Destruction

Exergy analysis involves evaluating the useful work potential of a system as it interacts with its surroundings. It focuses on identifying the energy that is effectively converted into work versus that which is lost due to irreversibilities such as friction, non-ideal heat transfer, or other inefficiencies. Exergy destruction quantifies the lost work potential and is a crucial concept for improving system performance.

Thermodynamic Properties: Enthalpy, Entropy, and Quality

Enthalpy is a key property representing the total energy content of a substance, while entropy measures the disorder within the system. The specific qualities of the refrigerant, such as its quality (the fraction of the vapor phase in a liquid-vapor mixture), aid in understanding the state of the fluid during phase transitions. These properties are essential for accurately assessing heat and mass transfer in the evaporator.

Second-Law Efficiency

Second-law efficiency measures the performance of a thermodynamic process by comparing the actual work or cooling achieved with the maximum possible work or cooling that could be theoretically obtained according to the second law of thermodynamics. This concept helps in assessing how well a system minimizes irreversibilities and uses its available energy.

Refrigeration Cycle and Evaporator Function

The evaporator is a key component in a refrigeration cycle, where it absorbs heat from a low-temperature environment to cause the refrigerant to evaporate. This phase change enables the system to remove heat from a cooled space, thereby achieving the desired cooling effect. Understanding the role of the evaporator is crucial in analyzing how thermal energy is transferred and used within refrigeration systems.

Energy Balance and Heat Transfer

In thermal systems, the energy balance is used to track the exchange of energy, particularly in the form of heat, through components such as evaporators. The change in the refrigerant's enthalpy, which is a measure of its energy content, quantifies the amount of heat absorbed during the phase change process. This concept is fundamental when calculating the cooling rate provided by the system.

Transcript

00:01 This question our given data is t not is given which is 77 in fahrenheit scale h1 is given that is 107 .5 btu by ibm its units s1 is given that is 0 .2851 btu per ibmr while h2 is given which is equals to 172 2 .1 btu per ibm while s 2 is given that is 0 .4 225 btu per ibm .r.

00:45 While the tl is given that is 50 degree centigrade m dot is given that is 0 .08 ibn per second.

00:56 Now what we have to find in this question we have to find in part a the rate of cooling provided so q .l is find finding formula is m dot h2 minus each one now putting the values in this so 0 .08 and their required units so h2 is 172 .1 minus 107 .5 so their required units is so our required answer is is 5 .162 btu per second.

01:44 Now we change the units into per hour.

01:49 So it's equals to 18580 btu per hour.

01:56 Now in part b, what we have to find, we have to find the anaeropathy generation and the exerge destruction.

02:05 So first of all, the anotropy generation, which formula is, this one now putting the values in this formula which equals to 0 .08 and their units 0 .4 -225 is s2 and s1 is 0 .2851 and their units now the ql that is 5 .162 and tl which we convert into r so plus 460 r so our required answer is 0 .008691 pt u per second r while in thus part b we also have to find exurgy destruction so our exergy destruction is t dot t not into an atropi generation.

03:38 So putting these values, which is equals to these and the units, so the required answer is 4 .0 .467 ptu per second.

03:58 Now in part c, what we have to find, we have to find the energy supplied or expended during this cooling process is the energy decrease of the refrigerator as it evaporates in the evapurator...

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