Hot combustion gases at 1500 K loses 3000 W to the steel walls at 700 K which in
ID: 707570 • Letter: H
Question
Hot combustion gases at 1500 K loses 3000 W to the steel walls at 700 K which in turn delivers the 3000 W to a flow of glycol at 100°C and the glycol has heat transfer to atmospheric air at 20°C. Assume steady state and find the rates of entropy generation and where it is generated. O at the combustion gases/steel wall: Sgen 2.286 WIK at the steel wall/glycol: Sgen 3.754 WIK at the glycol/atmosphere: Sgen 2.194 W/K O at the combustion gases/steel wall: Sgen -2.0 W/K at the steel wall/glycol: Sgen-3.0 WIK at the glycol/atmosphere: Sgen -10.0 WIK at the combustion gases/steel wall: Sgen -2.286 W/K at the steel walliglycol: Sgen- -3.754 W/K at the glycol/atmosphere: Sgen -2.194 WIK at the combustion gases/steel wall: Sgen -2 286 W/K at the steel wall/glycol: Sgen 3.0 WIK at the glycol/atmosphere: Sgen 2.194 W/KExplanation / Answer
Entropy balance
Entropy in - Entropy out + Entropy generation = Accumulation
At steady state, Accumulation = 0
Entropy generation = - Entropy in + Entropy out
From combustion gas to steel (1500 K to 700 K)
S gen = - Q/T1 + Q/T2
= - (3000W)/(1500K) + (3000W)/(700K)
= 2.286 W/K
From steel ball to glycol (700 K to 373.15 K)
S gen = - Q/T1 + Q/T2
= - (3000W)/(700K) + (3000W)/(373.15K)
= 3.754 W/K
From glycol to air (373.15 K to 293.15 K)
S gen = - Q/T1 + Q/T2
= - (3000W)/(373.15K) + (3000W)/(293.15K)
= 2.194 W/K
Largest entropy generated From steel ball to glycol
S = Q/T
Lower the temperature, higher the entropy
At 293.15 K (lowest), entropy is largest
Option A is the correct answer
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