Page 48 - Battery Reference Book

Calculation of operating parameters for a lead-acid battery from calorimetric measurements 1/33

Q. As well as heat produced in the electrochemical = 3.237 mol sulphuric acid
reaction, additional heat will be produced due to Joule = molecular weight of sulphuric acid x n
heating:
WA = 98 x
12xRxt (1.86)
cal in t s
4.18 = 317.226g sulphwic acid
150000 x 4.18
(where I is current in amps, and R is the electrical C=
resistance of cell and external contacts). 3600 (2.0100 - 293 x 0.000 01)
Every attempt should be made to minimize Joule = 86.76Ah
heating by using low electrical resistance components Suppose that the initial concentration of sulphuric
in the experimental arrangement. acid at the start of discharge is Cl% by weight, its rel-
Suppose that the ohmic heating component is ative density is S1 and its volume is VI nil, all of which
denoted by Q' calories during the discharge; then are known (Cl = 29, S1 = 1.2104, VI = 1624ml). As
the heat production due to the above electrochemical 3 17.226 g of sulphuric acid have been consumed in the
reaction above is given by electrochemical reaction we can, from Equation 1.84,
(2" = Q - Q' find the final volume concentration (C2) and relative
density (S2) of sulphuric acid at the end of the dis-
If Q" is known, it is then possible (using charge:
Equation 1.20) to calculate n; as heat is evolved Q
has a negative value:
(1.20)
(1.84)
Le.
(1.87)
(1.90)
+-
where Q is the calories evolved during discharge due 196 1.8305 WA
to electrochemical reaction, n the number of moles of Since
sulphuric acid involved in the electrochemical reaction, C1 = 29% by weight
F = 1 F (96 SO0 C), E is the cell e.m.f. (V), T the cell
temperature (K) and dEldT the temperature coefficient Si = 1.2104
of e.m.f. CIS1 = 35.102g sulphuric acid100ml
The weight of sulphuric acid (W,) is given by
VI = 16241111
Q" x 4.18 x 98
WA =98 X I;, = - (1.88) WA = 317.228
Then
Now nF = 3600 C, where C is the Ah capacity of the CzS2 = 16.75 g sulphuric acid100ml
battery. Therefore and, since Sz (the relative density of 100% sulphuric
acid) is 1.8305,
nF 0" x 4.18
(1.89) 16.75 x 100
c2 =
(
100f 16.75 1 - ~ 1.A05)
To take a particular example, if Q" is - 150 000 cal per C2 = 15.56g sulphuric acid100g
discharge,
Therefore
E = 2.010V 16.75
s2 = ~ = 1.076
!E = 0.00001 15.56
dT The final volume (VFml) of electrolyte is then given
T = 293 Km(20"C) by Equation 1.85:
36
I
then VF=vI-(---> 1.8305 196
150000 x 4.18
n= = 0.3627W~ = 15091111
96500(2.0100 - 293 x 0.00001)