Question

An ammonia-water absorption refrigeration cycle is used to keep a space at \(25^{\circ} \mathrm{F}\) when the ambient temperature is \(70^{\circ} \mathrm{F}\). Pure ammonia enters the condenser at 300 psia and \(140^{\circ} \mathrm{F}\) at a rate of \(0.04 \mathrm{lbm} / \mathrm{s}\). Ammonia leaves the condenser as a saturated liquid and is expanded to 30 psia. Ammonia leaves the evaporator as a saturated vapor. Heat is supplied to the generator by geothermal liquid water that enters at \(240^{\circ} \mathrm{F}\) at a rate of \(0.55 \mathrm{lbm} / \mathrm{s}\) and leaves at \(200^{\circ} \mathrm{F}\). Determine ( \(a\) ) the rate of cooling provided by the system, in \(\mathrm{Btu} / \mathrm{h}\), the \(\mathrm{COP}\), and (b) the second-law efficiency of the system. The enthalpies of ammonia at various states of the system are: condenser inlet \(h_{2}=665.7 \mathrm{Btu} / \mathrm{lbm},\) evaporator inlet \(h_{4}=190.9 \mathrm{Btu} / \mathrm{lbm}\) evaporator exit \(h_{1}=619.2 \mathrm{Btu} / \mathrm{lbm} .\) Also, take the specific heat of geothermal water to be \(1.0 \mathrm{Btu} / \mathrm{lbm} \cdot^{\circ} \mathrm{F}\).

Short answer

Based on the given information about the ammonia-water absorption refrigeration cycle, we calculated the following: 1. The rate of cooling provided by the system is 61675.2 Btu/h. 2. The Coefficient of Performance (COP) of the cycle is 9.2. 3. The second-law efficiency of the system is 0.64.

Step-by-step solution

Calculate the rate of cooling provided by the system

First, we find the rate of heat absorption in the evaporator (Q_evap): $$Q_{evap} = \dot{m}_{ammonia}*(h_1 - h_4)$$ Using the given values, we plug in $$ Q_{evap} = 0.04\ lbm/s*(619.2\ Btu/lbm - 190.9\ Btu/lbm)$$ Now calculate the value of Q_evap: $$ Q_{evap} = 17.132\ Btu/s $$ To find the rate of cooling in Btu/h, multiply Q_evap by the conversion factor from seconds to hours: $$ Q_{cooling} = Q_{evap} * (3600\ s / 1\ h) $$ Calculate the value of Q_cooling: $$ Q_{cooling} = 61675.2\ Btu/h $$

Calculate the Coefficient of Performance (COP)

Use the formula for COP of refrigeration cycle $$ COP = \frac{Q_{evap}}{W} $$ To find the work (W) in Btu/s, use the difference in enthalpy during the condensation: $$ W = \dot{m}_{ammonia}*(h_2 - h_1) $$ Plug in the given values: $$ W = 0.04\ lbm/s*(665.7\ Btu/lbm - 619.2\ Btu/lbm) $$ Now calculate the value of W: $$ W = 1.86\ Btu/s $$ Now, plug in the values of Q_evap and W to calculate the COP: $$ COP = \frac{17.132\ Btu/s}{1.86\ Btu/s} $$ Calculate the value of COP: $$ COP = 9.2 $$

Calculate the second-law efficiency of the system

We are given the mass flow rate and temperatures of the geothermal water, as well as the specific heat of the geothermal water. We can find the rate at which heat is supplied by the geothermal water (Q_geothermal) as: $$ Q_{geothermal} = \dot{m}_{geothermal} * c_p * (T_{inlet} - T_{outlet}) $$ Using the given values: $$ Q_{geothermal} = 0.55\ lbm/s * 1.0\ Btu/lbm\ °F * (240°F - 200°F) $$ Calculate the value of Q_geothermal: $$ Q_{geothermal} = 22\ Btu/s $$ Now find the maximum possible COP (COP_max) using the ambient and desired temperatures: $$COP_{max} = \frac{T_{low}}{T_{high} - T_{low}}$$ , where temperatures should be in Rankine. Convert °F to Rankine: $$ T_{low} = 25°F + 459.67 = 484.67\ R $$ $$ T_{high} = 70°F + 459.67 = 529.67\ R $$ Now calculate the COP_max: $$COP_{max} = \frac{484.67\ R}{529.67\ R - 484.67\ R}$$ $$ COP_{max} = 14.37 $$ Finally, we calculate the second-law efficiency as: $$\eta_{second-law} = \frac{COP}{COP_{max}}$$ Plug in the values for COP and COP_max: $$\eta_{second-law} = \frac{9.2}{14.37}$$ Calculate the value of the second-law efficiency: $$\eta_{second-law} = 0.64 $$ A summary of the results: (a) The rate of cooling provided by the system is 61675.2 Btu/h, and the COP is 9.2. (b) The second-law efficiency of the system is 0.64.

Key concepts

Thermodynamic Cycles

Thermodynamic cycles are foundational to understanding how refrigeration systems, like the ammonia-water absorption refrigeration cycle described in the exercise, work. Essentially, these cycles are a series of processes that a fluid undergoes where it returns to its original state at the end of the cycle. In refrigeration, the main goal is to remove heat from a low-temperature reservoir and transfer it to a higher-temperature one. A key part of a thermodynamic cycle is the working fluid, in this case, ammonia. It changes phase from gas to liquid and back again, absorbing and releasing heat in the process.
In the given exercise, the refrigeration cycle begins with ammonia entering the condenser with certain thermodynamic properties. As it moves through the system, it undergoes phase changes, heat addition, and heat removal, which keeps the space cool. The steps within the cycle involve heat exchange with the environment and the internal work of the system, such as when the ammonia is expanded. Understanding each of these steps is crucial for analyzing the cycle’s efficiency and performance.
An essential aspect of these cycles is that they operate consistently over time, maintaining the desired temperature difference between the space to be cooled and the ambient environment. The reliability and efficiency of a thermodynamic cycle are what make modern refrigeration possible.

Coefficient of Performance (COP)

When examining refrigeration systems, the coefficient of performance (COP) is a vital metric that measures the efficiency of the cycle. The COP is the ratio of the amount of cooling effect produced to the work input required to produce that cooling effect. This concept directly relates to how well the refrigeration system is utilizing energy.
In mathematical terms, the COP for a refrigeration cycle is expressed as
\[\begin{equation}COP = \frac{Q_{evap}}{W}\end{equation}\]
where:

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  For the provided exercise, you calculated the COP to be 9.2. This means that for every unit of work invested in the system, over nine units of cooling effect are achieved. This indicates a highly efficient system, especially compared to conventional refrigeration cycles. By constantly analyzing the COP, one can determine the points of inefficiencies that may exist within the system and address them for enhanced performance.

Second-law Efficiency

Second-law efficiency is a comprehensive measure of how a thermal system's actual performance compares to its maximum theoretical performance based on the Second Law of Thermodynamics. It considers the irreversibilities in a system and gives a realistic view of the system's effectiveness in converting supplied energy to work or useful heat. Simply put, it's a ratio of the real-world efficiency to the ideal efficiency that would be achieved if the process were reversible.
In the context of the absorption refrigeration cycle, the second-law efficiency is calculated by dividing the actual COP by the maximum possible COP (COP_max), which is dependent on the temperature limits of the cycle. The closer this value is to 1, the more effectively the system is utilizing energy, considering the limitations imposed by the Second Law of Thermodynamics.
To calculate the second-law efficiency, the exercise shows:
\[\begin{equation}\eta_{second-law} = \frac{COP}{COP_{max}}\end{equation}\]
In this exercise, you found the second-law efficiency to be 0.64, or 64%. This means the refrigeration system is converting 64% of the ideal energy potential into the actual cooling effect. While it's impossible to reach an efficiency of 100% due to factors such as friction and non-ideal heat transfer, understanding the second-law efficiency can help in making improvements to the system to make it as efficient as possible within the bounds of physical laws.