1. Absolute thermodynamic temp scale
= 1 −
𝑄2
𝑄1
For any heat engine. For a reversible cycle = 𝑓 𝑇1, 𝑇2
So,
𝑄1
𝑄2
= 𝐹 𝑇1, 𝑇2
T1
T2
Q1
Q2
Q2
Q3
T3
W1=Q1-Q2
W2=Q2-Q3
Q1
Q3
W3=Q1-Q3
𝑄1
𝑄2
= 𝐹 𝑇1, 𝑇2 ;
𝑄2
𝑄3
= 𝐹 𝑇2, 𝑇3
𝑄1
𝑄3
= 𝐹 𝑇1, 𝑇3
𝑠𝑜, 𝐹 𝑇1, 𝑇2 * 𝐹 𝑇2, 𝑇3 = 𝐹 𝑇1, 𝑇3
Hence the simplest function is :
𝑄1
𝑄2
=
𝑇1
𝑇2
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
𝑄 𝑛−1
𝑄 𝑛
=
𝑇 𝑛−1
𝑇 𝑛
and hence 𝑇𝑛 𝑐𝑎𝑛 𝑛𝑜𝑡 𝑏𝑒 𝑧𝑒𝑟𝑜, since
𝑄 𝑛 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 −
𝑄2
𝑄1
= 1 −
𝑇2
𝑇1
Efficiency of reversible heat engine
𝑐𝑜𝑝 ref =
Q2
Q1 − Q2
=
T2
T1 − T2
𝑐𝑜𝑝 HP =
Q1
Q1 − Q2
=
T1
T1 − T2
Internally reversible and externally irreversible process
TH
TL
QH
QL
Ta
Tb
Engine W
𝑄 𝐻 = 𝑐 𝐻(𝑇 𝐻 − 𝑇𝑎)
𝑄 𝐿 = 𝑐 𝐿(𝑇𝑏 − 𝑇𝐿)
𝑊 = 𝑄 𝐻 − 𝑄 𝐿
If 𝑇 𝐻 ≈ 𝑇𝑎 and 𝑇𝑏 = 𝑇𝐿 then is max
But QH=0 and w=0
On the other hand if 𝑇𝑎 ≪ 𝑇 𝐻, 𝑎𝑛𝑑 𝑇𝑏 ≫ 𝑇𝐿
𝑄 𝐻 ≫ 0, ≅ 0, 𝑎𝑛𝑑 𝑤 ≅ 0
So there is some optimal set of cycle temp for maximum
power output. Find optimum 𝑇𝑎 𝑎𝑛𝑑 𝑇𝑏for max power.
Known are: 𝐶 𝐻, 𝐶𝐿, 𝑇 𝐻, 𝑇𝐿 𝑎𝑛𝑑 𝑢𝑛𝑘𝑛𝑜𝑤𝑛 𝑎𝑟𝑒 𝑇𝑎 𝑎𝑛𝑑 𝑇𝑏, 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.
