
475Power Converter Cooling
where watts = equipment power loss transferred to water; L = total length of cooling
tube in one series path (m); θ = mean temperature difference, θ
tube(constant)
− θ
water(average)
(°C); V = velocity of water in tube (m/s); and D = inside diameter of cooling tube(cm).
In terms of the water quantity ow rate Q in liters/minute,
Q
DV
V
Q
=×=
π
(/)
2 100
1000
60
2
2
or
. (17.17)
Using Equation 17.17 in 17.16, we obtain
Watts
148 2
15
42 9
8
L
Q
D
Q
D×
=
××
=
θπ
.
.
.
.0
0.8
Therefore,
liters
minute
watts
42.9
Q= =D
L ×
θ
1.225
(17.18)
watervelocity 0.212
liters/minute
m/s
cm
2
V=
D
. (17.19)
The left-hand side of Equation 17.18 suggests that ...