際際滷

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Absolute thermodynamic temp scale
 = 1 
2
1
For any heat engine. For a reversible cycle  =  1, 2
So,
1
2
=  1, 2
T1
T2
Q1
Q2
Q2
Q3
T3
W1=Q1-Q2
W2=Q2-Q3
Q1
Q3
W3=Q1-Q3
1
2
=  1, 2 ;
2
3
=  2, 3
1
3
=  1, 3
,  1, 2 *  2, 3 =  1, 3
Hence the simplest function is :
1
2
=
1
2
From this functional relationship we can deduce absolute zero can not be reached. Lets us
connect n number of engines in series and let the last engine deliver some work while rejecting
heat to a sink at Tn. According to this relation
 1
 
=
 1
 
and hence     ю, since
  can not be zero. So absolute zero on a Kelvin scale can not be reached
Third law of thermodynamics.
Absolute zero on Kelvin scale
Can not be reached without violating
2nd law of thermodynamics
Engine
Engine
Engine
 = 1 
2
1
= 1 
2
1
Efficiency of reversible heat engine
 ref =
Q2
Q1  Q2
=
T2
T1  T2
 HP =
Q1
Q1  Q2
=
T1
T1  T2
Internally reversible and externally irreversible process
TH
TL
QH
QL
Ta
Tb
Engine W
  =  (   )
  =  (  )
 =     
If     and  =  then  is max
But QH=0 and w=0
On the other hand if    ,    
   0,   0,    0
So there is some optimal set of cycle temp for maximum
power output. Find optimum   for max power.
Known are:  , 駒,  ,   ゐ    , find max W
ch=100; cl=80; th=900; tl=300
qh=ch*(th-ta)
ql=cl*(tb-tl)
w=qh-ql
qh/ta=ql/tb
ta_theo=ch/(ch+cl)*th+cl/(ch+cl)*sqrt(th*tl)
tb_theo=cl/(ch+cl)*tl+ch/(ch+cl)*sqrt(th*tl)
Make a parametric table with Ta, Tb, Qh, QL, and w
Vary Ta from 900 to 630K and get the table solved.
Now you can see W as a function of Ta and Tb. Find
W_max from the table.

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Lect 5

  • 1. Absolute thermodynamic temp scale = 1 2 1 For any heat engine. For a reversible cycle = 1, 2 So, 1 2 = 1, 2 T1 T2 Q1 Q2 Q2 Q3 T3 W1=Q1-Q2 W2=Q2-Q3 Q1 Q3 W3=Q1-Q3 1 2 = 1, 2 ; 2 3 = 2, 3 1 3 = 1, 3 , 1, 2 * 2, 3 = 1, 3 Hence the simplest function is : 1 2 = 1 2 From this functional relationship we can deduce absolute zero can not be reached. Lets us connect n number of engines in series and let the last engine deliver some work while rejecting heat to a sink at Tn. According to this relation 1 = 1 and hence ю, since can not be zero. So absolute zero on a Kelvin scale can not be reached Third law of thermodynamics. Absolute zero on Kelvin scale Can not be reached without violating 2nd law of thermodynamics Engine Engine Engine
  • 2. = 1 2 1 = 1 2 1 Efficiency of reversible heat engine ref = Q2 Q1 Q2 = T2 T1 T2 HP = Q1 Q1 Q2 = T1 T1 T2 Internally reversible and externally irreversible process TH TL QH QL Ta Tb Engine W = ( ) = ( ) = If and = then is max But QH=0 and w=0 On the other hand if , 0, 0, 0 So there is some optimal set of cycle temp for maximum power output. Find optimum for max power. Known are: , 駒, , ゐ , find max W ch=100; cl=80; th=900; tl=300 qh=ch*(th-ta) ql=cl*(tb-tl) w=qh-ql qh/ta=ql/tb ta_theo=ch/(ch+cl)*th+cl/(ch+cl)*sqrt(th*tl) tb_theo=cl/(ch+cl)*tl+ch/(ch+cl)*sqrt(th*tl) Make a parametric table with Ta, Tb, Qh, QL, and w Vary Ta from 900 to 630K and get the table solved. Now you can see W as a function of Ta and Tb. Find W_max from the table.