Here is a Btur formula we use for calculating the amount of insulation is

needed.

Btur's = SF x u factor x TD

SF is the exterior surface between the ambient air and the inside

heated area.

u factor is equal to 1/R where R is the R-factor or resistance of the

insulation.

TD is the Temperature difference you want to maintain between the

heated area and outside ambient air.

Btur's is the amount of heat units you lose per hour. 1000 watts is

equal to about 3412 btus or 3.412 btus per watt.

Lets say you do not have a insulated battery box, so the R-factor is 1 and

the u factor is 1/R or 1/1 = 1 u factor

The formula now becomes:

Btur's = 28.31 SF x 1 u x 60 TD

Btur's = 1698.6

Watts = 1698.6/3.412 = 497.8 watts

Therefore it would take about 500 watts of continuous heating per hour to

maintain 60 degrees F. at 0 degrees.

A uninsulated metal container only adds a trace gain, because steel has

about 0.5 R factor per foot thickness.

Lets see what happens when we add 0.5 inch of foam board to the outside of a

battery box which has a R-factor of 2.5 R or (1/2.5) = 0.4 u factor:

Btur's = 28.31 sf x 0.4 u x 60 = 679.44 Btur's

Watts = 679.44 / 3.412 = 199.13 watts

Therefore with 1/2 of foam it will take about 200 watts per hour at 0

degrees F.

Now if you double the thickness of the insulation, the watts will drop by

one half:

No insulation ........... 500 watts per hour

1/2 inch foam ........... 200 watts

1 inch foam ............. 100 watts

2 inch foam ............. 50 watts

Dead air space of 3/4 inch adds about 3 R factor.

A vacuum space using two chamber box where the inside box does not touch the

outside box and apply a vacuum, the R factor goes in to the 1000's.

Example of my insulated battery box, that does not require any additional

heat to maintain 65 degrees at 35 below zero.

My battery box is about 47.5 SF of outside surface area which includes the

bottom, sides and top.

The insulation on the bottom is 2 inches of foam with two layers of marine

nylon carpet, 1/4 inch solid fiberglass which has a R factor of 15 R's

losses 105.6 btuh's

The insulation on the cover is 1/4 inch solid fiberglass, 1/2 inches of foam

and nylon carpet which has about 6 R's losses 265.6 btuh,s.

The insulation on the sides is 1/4 inch solid fiberglass, 2 inches of foam

board, nylon carpet, 4 inches of dead air space, nylon carpet and another

layer of 2 inch foam board for a R-factor of about 40 R factor losses 2.5

btuh,s.

Different R-factors require three calculations for each area. So at

-35 F. the total btuh,s is about 375 btuh which is about 110 watts per hour.

I no longer park the EV outside more than 2 hours in any temperature. The

inside of the battery box temperature is always between 70 and 80 degrees

even when the outside ambient air is at 100 degrees.

When I charge the batteries, it takes in the garage air which maintains a

constant 70 degrees temperature through a exhaust fan circulation that is

exhaust either through a hose going to a garage door port and/or ceiling

fans.

Roland

----- Original Message -----

From: <[email protected]>

To: <[email protected]>

Sent: Wednesday, September 26, 2007 9:43 PM

Subject: [EVDL] Estimate of Battery Heat output when charging?

> I'm working on the math, trying to figure how much heat input will be

> needed to keep the batteries warm this winter. They are in boxes

> insulated to R5.

>

> Dimensions:

> Rear: 12 Batteries US 8VGCHC

> 40.5x27x14 (outside Dimensions)

> Total Surface Area: 28.31 sq.ft.

> Temperature Differential 60 deg. (0 outside, 60 in

> the box, hopefully)

> Calculated Watts needed to keep warm: 339

>

> How much of this heat could come from the charging of the batteries? I

> am hoping to charge from a 120V outlet, and don't think I have 500 Watts

> to spare (including front box)

>

> Andy

> '88 Mazda 323

> Solectria/ElectroAuto Manual Transmission Kit

> 18 US Battery 8VGCHC

>

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