Thermodynamics -- total entropy generation during cooling of tank of water....

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SUMMARY

The discussion focuses on calculating the total entropy generation during the cooling process of a 0.2 m³ steel container filled with liquid water, initially at 50°C, cooling to an ambient temperature of 25°C. The methodology involves treating the system as isolated and applying the entropy change equations for the liquid water, steel container, and surrounding air. The calculated total entropy generation (Sgen) is 2.8435 kJ/K, derived from the individual entropy changes of the components involved. The approach and calculations presented are confirmed to be correct by a peer in the discussion.

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  • Understanding of thermodynamic principles, specifically entropy generation.
  • Familiarity with heat transfer concepts and calculations.
  • Knowledge of specific heat capacities for water and steel.
  • Ability to perform logarithmic calculations for temperature ratios.
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  • Learn about heat transfer mechanisms in thermodynamics.
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Homework Statement


A 0.2 m^3 steel container that has a mass of 30 kg is filled with liquid
water. Initially both the steel tank and the water are at 50°C. Now heat is transferred, and the entire system (i.e., steel and water) cools to the surrounding air temperature of 25°C. Determine the total entropy generation during this process.
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Homework Equations

The Attempt at a Solution


I think that take the steel container and liquid water and air(at 25 °C) as an isolated system so from the equation Sin-Sout(Net entropy transfer by heat or mass out the isolated boundary)+Sgen=ΔSsystem so no net entropy changes across isolated system boundaries so Sgen=ΔSsystem and ΔSsystem=ΔSliquid water+ΔSsteel+ΔSair surrounding
then find the mass of water by find the average density of water at 25°C and 50°C I got Density=992.5 kg/m^3 and also mass of water=992.5*0.2=198.5 kg
ΔSliquid water=(198.5 kg)(S@25 °C - S@50 °C)=198.5*(0.3672-0.7038)=-66.8151 kJ/k
and ΔSsteel(incompressible substances)=m*C*ln(T2/T1)=30*0.5*ln((25+273)/(50+273))=-1.2084 kJ/k
and total heat transfer to an air surroundings is Qair=-Qliquid water-Qsteel
find Qliquid=m*c*(T2-T1)=198.5*4.18*(25-50)=-20743.25 kJ
Qsteel=m*c*(T2-T1)=30*0.5*(25-50)=-375 kJ
Qair=-(-375-20743.25)=21118.25 kJ
from ΔSair=Qair/T=21118.25/(25+273)=70.867 kJ/k
then from ΔSliquid water and ΔSsteel and ΔSair I can find Sgen=ΔSsystem=ΔSliquid water+ΔSsteel+ΔSair surrounding =-66.8151 kJ/k-1.2084 kJ/k+70.867 kJ/k=2.8435 kJ/k
Is my solution correct or not? please help me for my thermodynamics homework
thankyou
 
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I didn't check your arithmetic, but your methodology is correct.

Chet
 

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