Calculation and comparison of cooling rates require careful specification of conditions, be-
cause it varies with position and time. Most useful method is to determine the cooling rate on the center line of the weld at the instant the metal is passing through a particular tempera-
ture of interest, Tc. At temperatures well below melting, the cooling rate in the weld and its immediate HAZ is substantially independent of position.
In carbon and low alloy steels the temperature of interest is best taken near the pearlite ‘‘nose’’ temperature on the TTT diagram. The exact temperature is not critical but should be the same for all calculations and comparisons. A value of Tc = 550 is quite satisfactory for most steels.
For thickplates requiring several passes (more than six) to complete the joint. The cooling rate (for the first pass or each pass). R is given by :
r = 2n K(Tc Tq)2
where R = cooling rate at a point on the weld centerline, °C/s at just that moment when point
is cooling past TC.
K = Thermal conductivity of the metal J/mm-s°C.
Tc = temperature at which cooling rate is calculated T0 = initial plate temperature, °C.
The cooling rate is maximum at the weld centreline. The above equation gives this maximum cooling rate. At fusion boundary it is only a few percent lower. Thus this equation applies
to the entire weld and the HAZ. If the plates are thin requiring fewer than four passes :
V Hnet J
R = 2п KpC - H - (Tc - T))3
, Hnet J
where t = thickness of base metal mm p = density of metal, g/mm3 C = sp. heat of base metal, J/g. °C The difference between thick and thin plate.
In thick plates the heat flow is three dimensional. This equation (eq. 2) applies to small boad-on-plate welds on thin plates.
Relative plate thickness factor, т is defined as follows to distinguish between thick and
т = h
т < 0.75 thin plate equation is valid т > 0.75 thick plate equation is valid.
Fig. 6.3 Relative plate thickness factor т for cooling rate calculations
Example. Find the best welding speed to be used for the welding of 6 mm steel plates with an ambient temperature of 30°C with the welding transformer set at 25 V and current passing is 300 A. The arc efficiency is 0.9 and possible travel speeds are 6 to 9 mm/s. The limiting cooling rate for satisfactory performance is 6°C/s at a temperature of 550°C.
Solution. Given T0 _ 30°C, Tc _ 550°C, K _ 0.028 J/mm-s-°C
R _ 6°C/s, V_ 25 V, I _ 300 A, h _ 6 mm, f1 _ 0.9, Pc _ 0.0044 J./mm3oC.
1. Assume a travel speed of 9 mm/s
f1 VI 0.9 x 25 x 300
Heat input _ Hnet _ —v— _---------------- 9---------- _--- 750 J/mm
To check whether it is a thick or thin plate
|Pc(Tc - T>) 6 /.0044(550 - 30)
т _ h
V—HHnet— = *V------------- 750----------- _ 03314
This being less than 0.6, it is thin plate, cooling rate will be calculated by using the thin plate equation
R _ 2n KpC
H Hnet J
(Tc - T.)3 . 2
_ 2n x 0.028 x 0.0044 J (550 - 30)d _ 6.9659oC/s.
This value is higher than the critical cooling rate required, we may reduce the travel speed to 8 mm/s and recalculate the cooling rate.
This cooling rate is higher than the limiting cooling rate of 60C/s (given) at a temperature of 550oC : We, therefore, reduce the travel speed to 8 mm/s and recalculate : v _ 8 mm/s
0. 9 x 25 x 300 Heat input, Hnet _ 8 _ 843.75---- J/mm
, Pc (Tc - T)) 6 0.0044 (550 - 30)
т _ h J 77------------------ =- 6 _ 0.312.
To check whether it is a thick or thin plate :
This being less than 0.6, it is a thin plate. Using thin plate equation for cooling rate.
(Tc - T0)3
R = 2n K pC
= 2п X 0.028 X 0.0044
(550 - 30)3 = 5.5040C/s.
This is a satisfactory cooling rate, the welding speed can be finalised at 8 mm/s.
These equations could also be used to calculate the preheat temperature required to avoid martensitic transformation in the weld zone.