NON-CONVENTIONAL ENERGY RESOURCES AND UTILISATION 81

Each electron also carries away its K.E. which is equal to 2KTc, i.e.,

Q 2 c = Jc

2 KTc

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 = Jc

(^2) c
c
KT
V
e
+
 – Ja^
^2
+

c
a
KT
V
e
where e = 1.602 × 10–19 coulomb.
The power output of the generator
W = Vo (Jc – Ja)
The thermal efficiency of the thermionic generator
η =
()
22
ac a
cc ceaa
VJ J
KT KT
JV JV
ee
−
+−+


Now Vo = Vc – Va substituting the following values.
c
c
V
KT
= βc ; a
a
V
KT
= βa and a
c
T
T
= θ.
η =
()()
22
cc aa c a
ce
cc c aa a
KT KT J J
KT KT
JKT J KT
ee
β−β −
 β+ −β +
 
 

2 ()
2 ()
2
()[1 ]
(2) (2)
ca
ce
ca
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
βa = βc
If βa= βc
ηmax = [1 – θ]
2
β
β+
2
2
1
(2)
1
2

−θ

θβ+θ
−
β+
2
2
1
(2)
1
2
−θ
θβ+θ
−
β+
= 1
ηmax = (1 – θ)
2
β
β+