For the control volume depicted with dashed-lines on the left (a) and right side
ID: 1862362 • Letter: F
Question
For the control volume depicted with dashed-lines on the left (a) and right side (b) of a heat exchanger on the Figure, derive (i) the mass-(ii) energy- and (iii) entropy-balance equations. List and label all quantity needed for the derivations.
Explanation / Answer
Let subscript 1 as inlet and 2 as outlet
(a) Heat is transferred from fluid B to fluid A
(i)
ma1=ma2=ma
mb1=mb2=mb
(ii)
heat lost by fluid B = heat gained by fluid A
mbCpb(Tb1-Tb2)= maCpa(Ta2-Ta1)
(iii)
entopy change by the fluid B= mb*Cpb* ln(Tb2/Tb1)
entropy change by the fluid A=ma*Cpa*ln(Ta2/Ta2)
So net entropy change=mb*Cpb* ln(Tb2/Tb1) + ma*Cpa*ln(Ta2/Ta2)
Where;
ma= mass flow rate of fluid A
mb= mass flow rate of fluid B
Ta1=Inlet temperature of fluid A
Ta2=Outlet temperature of fluid A
Ta1<Ta2 {since heat is being gained by fluid A}
Tb1=Inlet temperature of fluid B
Tb2=Outlet temperature of fluid B
Tb1>Tb2 {since heat is being lost by B}
Cpa= specific heat constant of fluid A
Cpb= specific heat constant of fluid B
(b)
Here heat is being transferred from fluid A to fluid B
(i)
ma1=ma2=ma
mb1=mb2=mb
(ii)
heat gained by fluid B = heat lost by fluid A
mbCpb(Tb2-Tb1)= maCpa(Ta1-Ta2)
(iii)
entopy change by the fluid B= mb*Cpb* ln(Tb2/Tb1)
entropy change by the fluid A=ma*Cpa*ln(Ta2/Ta2)
So net entropy change=mb*Cpb* ln(Tb2/Tb1) + ma*Cpa*ln(Ta2/Ta2)
Where;
ma= mass flow rate of fluid A
mb= mass flow rate of fluid B
Ta1=Inlet temperature of fluid A
Ta2=Outlet temperature of fluid A
Ta1>Ta2 {since heat is being lost by fluid A}
Tb1=Inlet temperature of fluid B
Tb2=Outlet temperature of fluid B
Tb1<Tb2 {since heat is being gained by fluid B}
Cpa= specific heat constant of fluid A
Cpb= specific heat constant of fluid B