Question

Water at 20 psia and \(50^{\circ} \mathrm{F}\) enters a mixing chamber at a rate of 300 lbm/min where it is mixed steadily with steam entering at 20 psia and \(240^{\circ} \mathrm{F}\). The mixture leaves the chamber at 20 psia and \(130^{\circ} \mathrm{F}\), and heat is lost to the surrounding air at \(70^{\circ} \mathrm{F}\) at a rate of \(180 \mathrm{Btu} / \mathrm{min}\). Neglecting the changes in kinetic and potential energies, determine the rate of entropy generation during this process?

Short answer

Summarize in one sentence the result or conclusion of this problem. The rate of entropy generation during the steady mixing process of water and steam in a chamber can be calculated using the entropy balance equation and the specific entropies found in thermodynamic property tables.

Step-by-step solution

Find the specific entropies

First, we need to find the specific entropies of the water and steam entering the control volume, and the mixture leaving the chamber. We can use the table of thermodynamic properties for water and steam (available in textbooks or online) to find these values, making sure to use the correct pressure and temperature values for each stream. For example, for water entering the mixing chamber at 20 psia and \(50^{\circ} \mathrm{F}\), we can find its specific entropy value, \(s_{in,1}\) in the table. Similarly, we can find the specific entropy value, \(s_{in,2}\), for steam entering the chamber at 20 psia and \(240^{\circ} \mathrm{F}\). Finally, we can find the specific entropy value, \(s_{out}\), for the mixture leaving the chamber at 20 psia and \(130^{\circ} \mathrm{F}\).

Calculate the rates of entropy entering and leaving the chamber

Next, using the mass flow rates of the water and steam and the specific entropies from step 1, we can calculate the rates of entropy entering and leaving the control volume. \(dS_{in} = \dot{m}_{in,1}\times s_{in,1} + \dot{m}_{in,2}\times s_{in,2}\) \(dS_{out} = \dot{m}_{out}\times s_{out}\) where \(\dot{m}_{in,1}\), \(\dot{m}_{in,2}\), and \(\dot{m}_{out}\) are the mass flow rates of water and steam entering the control volume, and the mixture leaving the chamber, respectively. As this is a steady-state process, the mass flow rates entering and leaving the chamber should be equal: \(\dot{m}_{in,1} + \dot{m}_{in,2} = \dot{m}_{out}\)

Calculate the rate of entropy generation

Now that we have the rates of entropy entering and leaving the control volume, we can use the entropy balance equation to find the rate of entropy generation, \(S_{gen}\): \(S_{gen} = dS_{out} - dS_{in} + \frac{dQ}{dt}\frac{1}{T_{boundary}}\) Substitute all the calculated values from the previous steps into the equation and compute the rate of entropy generation during this mixing process.