Power Plant Engineering

Fig. 2.26. Characteristic Curve.
E
lc
V
c
φ
c
φ
a
δ
V
a
V
o
E
la
Gap Distance
En
e
rg
y
V
o
lt
s
(
–
)

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NON-CONVENTIONAL ENERGY RESOURCES AND UTILISATION
81
Each electron also carries away its K.E. which is equal to 2KT
c
, i.e.,
Q
2c
= J
c
2
c
KT
e
[W/cm
2
]
The back emission from the anode must similarly carry energy to the cath-ode. The net rate of
energy supply to the cathode,
Q
1
= J
c
2
c
c
KT
V
e


+




– J
a
2


+




c
a
KT
V
e
where
e = 1.602 × 10
–19
coulomb.
The power output of the generator
W = V
o
(J
c
– J
a
)
The thermal efficiency of the thermionic generator
η
= 
(
)
2
2
a
c
a
c
e
c
c
a
a
V J
J
KT
KT
J
V
J
V
e
e
−




+
−
+








Now
V
o
= V
c
– V
a
substituting the following values.
c
c
V
KT
= 
β
c
; 
a
a
V
KT
= 
β
a
and 
a
c
T
T
= 
θ
.
η
= 
(
) (
)
2
2
c
c
a
a
c
a
c
e
c
c
c
a
a
a
KT
KT
J
J
KT
KT
J
KT
J
KT
e
e
β
− β
−




β
+
−
β
+








= 
(
)
2
(
)
2
2
(
) [1
]
(
2)
(
2 )
c
a
c
e
c
a
c
e
e
β − β
β − β
β − θβ
− θ
β +
− θ β + θ
It is found that for all values of 
θ
, the efficiency curve peaks are very near to the value of
β
a
= 
β
c
If
β
a
= 
β
c
η
max
= [1 – 
θ
] 
2
β
β +
2
2
1
(
2 )
1
2




− θ




θ β + θ
−


β +




2
2
1
(
2 )
1
2
− θ
θ β + θ
−
β +
= 1
η
max
= (1 – 
θ
)
2
β
β +

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82
POWER PLANT ENGINEERING
Carnot efficiency
η
c
= 1 – 
a
c
T
T
= (1 – 
θ
)
If 
β
= 18,
η
max
= 0.9(1 – 
θ
)
η
max
occurs when 
β
a
= 
β
c
or
c
c
V
T
= 
a
a
V
T
.
2.19 THERMOELECTRIC SYSTEMS
A loop of two dissimilar metals develops an e.m.f. when the two junctions of the loop are kept
at different temperatures. This is called Seebeck effect. This effect is used in a thermocouple to meas-
ure temperature.
Thermoelectric generator is a device which directly converts heat energy into electrical energy
using the Seebeck thermoelectric effect. The device is very simple but thermal efficiency is very low of
the order of 3%. Efficiency of thermoelectric generator depends upon the temperature of hot and cold
junctions and the material properties. The semiconductor materials have more favourable properties
which can withstand high temperatures and can give reasonable efficiency. The probability of develop-
ing peak load power stations of the order of 100 mW working at 20 percent thermal efficiency is high.
Where cheap fuels are available thermoelectric generators can be developed for base load and standby
power generation also. Another important application is the use of radioactive decay heat to generate
power in space and other remote locations. The use of solar energy to supply heat for generating elec-
tricity can be an attractive application of thermoelectric devices if high efficiency materials can be
developed.
2.19.1 PRINCIPLE OF WORKING
The operation of a thermoelectric generator is shown in
Fig. 2.27. The net useful power output is given by
W = I
2
R [W]
where
I = current [A]
R = External load resistance [
Ω
]
The current in the circuit is given by
I = 
(
)
i
T
R
R
α∆
+
[A]
where
α
= Seeback coefficient (V/K]
∆
T = Temperature difference between hot and cold junctions [K]
R
i
= Internal resistance of thermoelectric generator [
Ω
]
The magnitude of potential difference depends on the pair of conductor materials and on the
temperature difference between the junctions.
R
Q
out
–
+
Q
in
I
Fig. 2.27. Thermoelectric Generator.

---

NON-CONVENTIONAL ENERGY RESOURCES AND UTILISATION
83
For a loop made of copper and constant wires, the value of Seebeck coefficient a is 0.04 mV/K.
For a temperature difference of 600 K between the junctions, a voltage of 24 mV will be developed. In
order to achieve higher potential difference many generators have to be connected in parallel.
For increasing the useful power output, parallel and series connections are used.
 
Parallel connection
Series connection
Fig. 2.28. Cascading of Thermoelectric Generators.
2.19.2 PERFORMANCE
The thermo-elements of a thermoelectric generator are made up of semiconduc-tors p and n
type. Heat is supplied to the hot junction and from the cold junction heat is removed. Both the junctions
are made of copper, see Fig. 2.29.
Let T
1
= source temperature [K]
T
o
= sink temperature [K]
L
p
, L
n
= length of semiconductor elements [m]
A
p
, A
n
= cross-sectional area of thermoelectric elements [m
2
]
k
p
, k
n
= thermal conductivity of elements [W/mK]
ρ
p
, 
ρ
n
 = electric resistivity of elements [
Ω
-m]
k
p
, k
n
= thermal conductivity of elements [W/K] = 
kA
L
R
p
, R
n
= electrical resistance of elements [
Ω
] = 
L
A
ρ
α
p
, 
α
n
= Seebeck coefficient [V/K]
π
p
, 
π
n
= Peltier coefficient [V]
Seebeck coefficient, 
α
p, n
= 
0
Lt
T
V
T
∆ →
∆
∆
Peltier heat,
α
p, n
= 
π
p, n
I
When a current (I) flows through the junction of two elements, Peltier heat is produced. This is
called Peltier effect. The Peltier coefficient.
π
p, n
= ap, n . I
From Ist law of thermodynamics as applied to upper plate (as control volume), the temperature
difference (T
1
– T
o
) will generate a Seebeck voltage, a
pn
(T
l
– T
o
). There will be an electrical current I
which will flow through the external load R
L
.
Fig 2.29. Circuit Diagram of Thermo-
electric Power Generator.
I
A
p
p
I
A
n
n
Q
1
T
1
R
L
V
L
I
Q
o
Q
o
L
p
L
n

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84
POWER PLANT ENGINEERING
The heat Q
1
will flow into hot junction and conducted into the two legs, Q
k
. The Peltier heat Q
p
will be produced at the junction due to current flowing through the circuit.
Joule heat Q
1
/2 will flow into the junction. It is assumed that half Joulean heat appears at each
junction. Heat balance at the junction will give,
Q
1
= 
1
2
j
Q
= 
p
Q
+ 
k
Q
where
p
Q
= 
π
p, n
I = 
α
p, n
= I.T
1
j
Q
=I
2
R = l
2
(R
p
+ R
n
)
= I
2
p
p
n n
p
n
L
L
A
A


ρ
ρ
+






k
Q
= k
∆
T = (k
p
+ k
n
) (T
1
– T
0
)
= 
p
p
n
n
p
n
k A
k A
L
L


+






(T
1
– T
0
)
Substituting the values into above equation
1
Q
= 
α
p, n
IT
1
– 
1
2
p
p
n n
p
n
L
L
A
A


ρ
ρ
+






I
2
+ 


+






p
p
n
n
p
n
k A
k A
L
L
(T
1
– T
2
)
The useful power generated,
W
L
= I
2
R
L
= 
2
L
L
V
R
The voltage across the load,
V
L
= 
α
p, n
(T
1
= T
0
) – I(R
p
– R
n
)
By Kirchhoff’s Law
I = 
,
1
0
(
)
α
−
+
+
p n
L
p
n
T
T
R
R
R
Let
m = 
L
p
n
R
R
R
+
= Resistance ratio
∴
m + 1 = 
p
n
L
p
n
R
R
R
R
R
+
+
+
Now
 I = 
,
1
0
(
)
(
) (1
)
p n
p
n
T
T
R
R
m
α
−
+
+

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86
POWER PLANT ENGINEERING
For a given pair of elements,
L
m
dW
d

= 0 which gives m = 1.
∴
L
W
= 
1
4
2
2
,
1
0
(
)
α
−
+
p n
p
n
T
T
R
R
Maximum efficiency corresponding to max power,
η
max, power
= 
1
0
1
T
T
T
−
. 
1
0
1
1
1
1
4
2
2
T
T
T
ZT
−
−
+