Question

A reversible absorption refrigerator consists of a reversible heat engine and a reversible refrigerator. The system removes heat from a cooled space at \(-15^{\circ} \mathrm{C}\) at a rate of \(70 \mathrm{kW}\) The refrigerator operates in an environment at \(25^{\circ} \mathrm{C}\). If the heat is supplied to the cycle by condensing saturated steam at \(150^{\circ} \mathrm{C},\) determine \((a)\) the rate at which the steam condenses, and (b) the power input to the reversible refrigerator. (c) If the COP of an actual absorption chiller at the same temperature limits has a COP of \(0.8,\) determine the second-law efficiency of this chiller.

Short answer

Answer: (a) The rate at which steam condenses is approximately \(0.03675 \ \text{kg/s}\). (b) The power input to the reversible refrigerator is approximately \(10.85 \ \text{kW}\). (c) The second-law efficiency of the chiller is approximately \(12.4\%\).

Step-by-step solution

Conversion of temperatures to Kelvin

To make the calculations easier, let's first convert the given temperatures from Celsius to Kelvin:$$ T_\text{low} = -15^\circ\text{C} + 273.15 = 258.15\ \text{K} \\ T_\text{high} = 25^\circ\text{C} + 273.15 = 298.15\ \text{K} \\ T_\text{steam} = 150^\circ\text{C} + 273.15 = 423.15\ \text{K}. $$

Find the reversible COP of the refrigerator

Now we need to find the coefficient of performance (COP) for the refrigeration cycle operating between \(T_\text{low}\) and \(T_\text{high}\). Since it is a reversible refrigerator, we can use the formula for the reversible COP:$$ \text{COP}_\text{rev} = \frac{T_\text{low}}{T_\text{high} - T_\text{low}} = \frac{258.15 \ \text{K}}{298.15\ \text{K} - 258.15\ \text{K}} = \frac{258.15 \ \text{K}}{40 \ \text{K}} = 6.45. $$

Determine the required work input for the reversible refrigerator

Now that we have the COP, we can find the required work input (\(W_\text{in}\)) for the reversible refrigerator:$$ W_\text{in} = \frac{Q_\text{removed}}{\text{COP}_\text{rev}} = \frac{70 \ \text{kW}}{6.45} \approx 10.85 \ \text{kW}. $$Here, \(Q_\text{removed}\) is the heat removed from the cooled space.

Determine the required heat input from the steam

Next, we'll determine the heat input from the steam (\(Q_\text{steam}\)) by using the energy balance for the refrigerator:$$ Q_\text{steam} = Q_\text{removed} + W_\text{in} = 70\ \text{kW} + 10.85\ \text{kW} \approx 80.85\ \text{kW}. $$

Find the rate of steam condensation

Now we can find the required steam condensation rate (\(m_\text{steam}\)) by using the given heat input from the steam:$$ m_\text{steam} = \frac{Q_\text{steam}}{h_\text{fg}} , $$where \(h_\text{fg}\) is the enthalpy of vaporization for steam at \(150^{\circ} \mathrm{C}\). From steam tables, we find that \(h_\text{fg} \approx 2200 \ \text{kJ/kg}\). Thus, we have:$$ m_\text{steam} = \frac{80.85\ \text{kW}}{2200 \ \text{kJ/kg}} = \frac{80.85 \cdot10^3\ \text{W}}{2200 \cdot10^3\ \text{J/kg}} \approx 0.03675 \ \text{kg/s}. $$

Calculate the second-law efficiency of the chiller

Finally, we will find the second-law efficiency of the chiller. For this, we need to compare the real chiller's COP to the COP of the reversible refrigerator:$$ \text{Second-law efficiency} = \frac{\text{COP}_\text{actual}}{\text{COP}_\text{rev}} = \frac{0.8}{6.45} \approx 0.124. $$ To summarize the results: (a) The rate at which the steam condenses is approximately \(0.03675 \ \text{kg/s}\). (b) The power input to the reversible refrigerator is approximately \(10.85 \ \text{kW}\). (c) The second-law efficiency of the chiller is approximately \(12.4\%\).