C11: In the previous HW set, you evaluated a transient (time dependent) process using a numerical multi-step technique until steady-state is reached (see figure 1 on the right). Use the same technique, and modify your solution for to evaluate entropy production and exergy destruction for the entire process. Since, temperature of the hot and cold liquids will change during the process, entropy production for each step will be different. Suppose j is the step number. At stepj, temperature of hot fluid goes from TH, -)TH,j+1 and temperature of the cold fluid goes from Tm 913,141. FIGURE 1 We expect THU,+1 THJ 0 since Tc is increasing during this process. You evaluated heat passing from hot to cold liquid for each step [6%) in the prior assignment. Having all these parameters, you can compute entropy production in hot the liquid, in the cold liquid, and in between where heat passes from one side to other using entropy balance equation for each step. Please note that small entropy changes in the hot and cold liquids, assuming incompressible substance would be TH, -+ _ _ Tc, '+ ASHij+1 : 5H,j+1 _ 5H,} : mHC in 7'1:le and A5Cj,j+1 5C,j+1 SCJ mcc in 7':le You should also consider incremental entropy transfer by heat term for incremental each step, for example _ CLO __'5QI' _ ('53) _ - 65%, f(r )1) ThH and 65% f r b TM. Boundary temperature here Will change at each step as well as during the step. However, if you keep incremental temperature changes in the hot and cold liquids small, approximation errors here will be small. You can either set Tb to the temperature at the beginning of the step or set Tb to the average of temperatures at the beginning and end of the step. For example, ifyou want to set it to the temperature at the beginning of the step Tb for hot side becomes Tb = THJ- or if you prefer to use average temperature Tb = (THJ- + TH,j+1)/2- Both of them are approximations with different approximation errors. The latter will give you smaller error than the former. If temperature change is small for each step approximation error will be small in both options. Once you have entropy change and entropy transfer by heat, you can compute small incremental entropy production for each step using entropy balance equation. Let's call them 6030,11,}. for hot side (Song. and for cold side. If you add all incremental entropy changes of every step you can find entropy production for the entire process, i.e. p,H = 311:1 6on5\" assuming we have 11 steps to reach equilibrium. You have to do all these calculations for the interface between hot liquid and cold liquid where heat passes through and temperature changes along the path, which is indicated by red dashed line in Figure 1. For the interface, assume that there is no mass, and no entropy [or no entropy change] but there is entropy transfer by heat and hence there is entropy production. Then find entropy production in the interface O'pJ. You can find exergy destruction directly by multiplying entropy production with the dead state temperature [To]. You can assume To = 275 K. Compare entropy production, and exergy destruction for the entire process to the values you have found in Q2 of previous assignment