Question

A hot-water stream at \(70^{\circ} \mathrm{C}\) enters an adiabatic mixing chamber with a mass flow rate of \(3.6 \mathrm{kg} / \mathrm{s}\), where it is mixed with a stream of cold water at \(20^{\circ} \mathrm{C}\). If the mixture leaves the chamber at \(42^{\circ} \mathrm{C}\), determine ( \(a\) ) the mass flow rate of the cold water and \((b)\) the rate of entropy generation during this adiabatic mixing process. Assume all the streams are at a pressure of \(200 \mathrm{kPa}\).

Short answer

The mass flow rate of the cold water is \(\dot{m}_{cold} = 2.7\,\mathrm{kg/s}\) b) What is the rate of entropy generation during the adiabatic mixing process? The rate of entropy generation during the adiabatic mixing process is approximately \(0.027\, c_p\,\mathrm{kg \cdot K/s}\).

Step-by-step solution

Mass balance equation

Apply the mass balance equation to the adiabatic mixing chamber: \(\dot{m}_{hot} + \dot{m}_{cold} = \dot{m}_{mix}\) where \(\dot{m}_{hot}\) is the mass flow rate of the hot water stream, \(\dot{m}_{cold}\) is the mass flow rate of the cold water stream, and \(\dot{m}_{mix}\) is the mass flow rate of the mixture. We know the mass flow rate of the hot water stream, \(\dot{m}_{hot} = 3.6\,\mathrm{kg/s}\)

Energy balance equation

Apply the energy balance equation to the adiabatic mixing chamber: \(\dot{m}_{hot} \cdot h_{hot} + \dot{m}_{cold} \cdot h_{cold} = \dot{m}_{mix} \cdot h_{mix}\) Since it's an adiabatic chamber, there is no heat transfer between the chamber and surroundings. In this case, \(\dot{Q} = 0\), therefore the heat transfer between the streams equals the internal energy. Assuming constant specific heat of water, we have: \(\dot{m}_{hot} \cdot c_p \cdot T_{hot} + \dot{m}_{cold} \cdot c_p \cdot T_{cold} = \dot{m}_{mix} \cdot c_p \cdot T_{mix}\) where \(c_p\) is the specific heat of water, \(T_{hot}\), \(T_{cold}\), and \(T_{mix}\) are the temperatures of hot, cold, and mix streams, respectively.

Solve for \(\dot{m}_{cold}\)

Use the mass balance equation and energy balance equation to solve for the mass flow rate of the cold water stream: \(\dot{m}_{cold} = \frac{\dot{m}_{hot} \cdot (T_{hot} - T_{mix})}{T_{mix} - T_{cold}}\) Plug in the known values and calculate: \(\dot{m}_{cold} = \frac{3.6\,\mathrm{kg/s} \cdot (70^{\circ}\mathrm{C} - 42^{\circ}\mathrm{C})}{42^{\circ}\mathrm{C} - 20^{\circ}\mathrm{C}} = 2.7\,\mathrm{kg/s}\) So the mass flow rate of the cold water is \(\dot{m}_{cold} = 2.7\,\mathrm{kg/s}\).

Entropy balance equation

Calculate the rate of entropy generation using the entropy balance equation: \(\dot{S}_{gen} = \dot{m}_{mix} \cdot s_{mix} - \dot{m}_{hot} \cdot s_{hot} - \dot{m}_{cold} \cdot s_{cold}\) Since the process is isobaric, we can use the temperature-dependent entropy equation: \(\Delta s = c_p \ln\left(\frac{T_2}{T_1}\right) - R \ln\left(\frac{P_2}{P_1}\right)\) As pressure is constant for all the streams, the pressure term can be omitted: \(\Delta s = c_p \ln\left(\frac{T_2}{T_1}\right)\) Calculate the entropy change for hot and cold water streams: \(\Delta s_{hot} = c_p \ln\left(\frac{T_{mix}}{T_{hot}}\right)\) \(\Delta s_{cold} = c_p \ln\left(\frac{T_{mix}}{T_{cold}}\right)\) Finally, plug in the values to get the rate of entropy generation: \(\dot{S}_{gen} = \dot{m}_{hot} \cdot \Delta s_{hot} + \dot{m}_{cold} \cdot \Delta s_{cold}\) \(\dot{S}_{gen} = 3.6\,\mathrm{kg/s}\cdot c_p \cdot \ln\left(\frac{42^{\circ}\mathrm{C}}{70^{\circ}\mathrm{C}}\right) + 2.7\,\mathrm{kg/s} \cdot c_p \cdot \ln\left(\frac{42^{\circ}\mathrm{C}}{20^{\circ}\mathrm{C}}\right)\) \(\dot{S}_{gen} ≈ 0.027\, c_p\,\mathrm{kg \cdot K/s}\) The rate of entropy generation during the adiabatic mixing process is approximately \(0.027\, c_p\,\mathrm{kg \cdot K/s}\).