Solution :Sq =C,dTT?313dT(1 -Pxq473T3133131.004×[(313 -473) - 298lnJ473473=-37kJ/kg
Solution ： q c dT = p 313 0 473 313 473 (1 ) 313 1.004 [(313 473) 298ln ] 473 37 kJ/kg xq p T e c dT T = − = − − = −
5.2.3LowtemperatureheatexergyPotential work of heat at a temperature belowtheenvironmentaltemperatureTToSE)80OxT.Q. = W+maxZ0-1)80SE,QWoTEro =」 (-10Q =T,AS-QTAno =」SQ' = T -△S
5.2.3 Low temperature heat exergy Potential work of heat at a temperature below the environmental temperature T0 T W0 0 0 ' (1 ) Q T T ExQ = − ' Q0 = Wmax +Q ' 0 ' ( 1) Q T T Ex Q = − ' 0 ' ' 0 ( 1) x Q T E Q T S Q T = − = − ' 0 ' nQ 0 T A Q T S T = =
TTo43Exo'2Q'56SFor constant temperatureT(%-1)ExQ2QAnQT
' 0 ' ( 1) xQ T E Q T = − ' 0 ' nQ T A Q T = For constant temperature
TTo34Exq'2T1Q'56SFor finite heat resourceExo =[(-1)60=(-1)mcpdTT=Z-TTT11 1n12(T, -T))= mcp[TInT, -T""T
For finite heat resource 2 1 0 0 2 1 2 0 2 1 2 1 1 0 ( 1) ( 1) [ ln ( )] ( 1) T XQ P T P T T E Q mc dT T T T T T mc T T T T T T T Q T = − = − − = − − − = − 1 2 2 1 ln T T T T T − = T
Influential factors:Heat quantityHeat resource temperatureT, → Ex↑, ^ST个 → Ex, △S个T.[Ex| >[0]when T2Environmental temperature TaT=T。。Ex=0If the system absorbs heat, it discharges exergy;If the system discharges heat, it absorbs exergy;
Influential factors: ⚫Heat quantity ⚫Heat resource temperature 0 , , , when , 2 xQ xQ xQ T E S T E S T T E Q ⚫Environmental temperature T0 0 0 T T E = = xQ If the system absorbs heat, it discharges exergy; If the system discharges heat, it absorbs exergy;