# Heat Rise in a rack mounted chassis....

M

#### [email protected]

Jan 1, 1970
0
I'm looking for some advise on heat rise. If I have a chassis (10 x 17
x 19) and I mount two components that dissipate 385W and 256W
respectively, and then have a power supply that supplies those
components power to operate, then I'm assuming (if the power supply is
80% efficient) that my total heat load is somewhere in the 1400W range
(641W + 770W (supply input pwr)). First off, is that a correct
assumption of the heat loading or am I making this way too simple?

Next, based on this heat load, the heat rise is a function of the total
surface area of the chassis (calculated to be about 9.5 sqft). Based
on that, the Watts/square ft = 147W/sqft (1400/9.5 = 147). Curves show
that the heat rise will be off the charts since the charts only go up
to 20W/sqft for something close to 80F temp difference (heat rise as a
function of the ambient air temp)

Does anyone see errors in my calculations or is this the expected heat
rise for this scenario?

J

#### John Larkin

Jan 1, 1970
0
I'm looking for some advise on heat rise. If I have a chassis (10 x 17
x 19) and I mount two components that dissipate 385W and 256W
respectively, and then have a power supply that supplies those
components power to operate, then I'm assuming (if the power supply is
80% efficient) that my total heat load is somewhere in the 1400W range
(641W + 770W (supply input pwr)). First off, is that a correct
assumption of the heat loading or am I making this way too simple?

Next, based on this heat load, the heat rise is a function of the total
surface area of the chassis (calculated to be about 9.5 sqft). Based
on that, the Watts/square ft = 147W/sqft (1400/9.5 = 147). Curves show
that the heat rise will be off the charts since the charts only go up
to 20W/sqft for something close to 80F temp difference (heat rise as a
function of the ambient air temp)

Does anyone see errors in my calculations or is this the expected heat
rise for this scenario?

Your box is roughly half the size of a toaster oven, and if you set it
to "broil" the power is in the 1/2 ballpark, too. They get pretty hot
on the outside, as I recall.

John

R

#### Rich Grise

Jan 1, 1970
0
Your box is roughly half the size of a toaster oven, and if you set it
to "broil" the power is in the 1/2 ballpark, too. They get pretty hot
on the outside, as I recall.

Well, 10 x 17 x 19 is about twice the size of any toaster oven I've ever
seen - maybe you're thinking of an ordinary oven?

In either case, unless the OP is actually _making_ an oven (in which
case he'd be better off to insulate it) he'd better poke some holes
in that cabinet and add a bank of fans. With 1400 Watts inside a sealed,
10 x 17 x 19 aluminum rack-panel cabinet, he could easily cook dinner
on top of it.

Cheers!
Rich

J

#### John Larkin

Jan 1, 1970
0
Well, 10 x 17 x 19 is about twice the size of any toaster oven I've ever
seen - maybe you're thinking of an ordinary oven?

Yeah, I meant twice.
In either case, unless the OP is actually _making_ an oven (in which
case he'd be better off to insulate it) he'd better poke some holes
in that cabinet and add a bank of fans. With 1400 Watts inside a sealed,
10 x 17 x 19 aluminum rack-panel cabinet, he could easily cook dinner
on top of it.

My point.

Something like 100 CFM air flow would carry off 1400 watts without too
much exaust air temp rise. But the devil's in the details.

John

J

#### John - KD5YI

Jan 1, 1970
0
I'm looking for some advice on heat rise. If I have a chassis (10 x 17
x 19) and I mount two components that dissipate 385W and 256W
respectively, and then have a power supply that supplies those
components power to operate, then I'm assuming (if the power supply is
80% efficient) that my total heat load is somewhere in the 1400W range
(641W + 770W (supply input pwr)). First off, is that a correct
assumption of the heat loading or am I making this way too simple?

Assuming your components of 385W and 256W are resistors (or equivalent),
your total heat load caused by the components is 641W. You state that the
device which supplies them power is 80% efficient. Therefore, the power
to 801W. (Total watts = load/efficiency)

Next, based on this heat load, the heat rise is a function of the total
surface area of the chassis (calculated to be about 9.5 sqft). Based
on that, the Watts/square ft = 147W/sqft (1400/9.5 = 147). Curves show
that the heat rise will be off the charts since the charts only go up
to 20W/sqft for something close to 80F temp difference (heat rise as a
function of the ambient air temp)

Does anyone see errors in my calculations or is this the expected heat
rise for this scenario?

I agree with your surface area calculation, but I use a slightly different
rule-of-thumb involving square inches. You have 801W/1366 sqin which gives
about .59w/sq-in. I use the crude rule of about 100 times the w/sqin as the
temperature rise (because it is easy to remember. I think it may be higher
than that). So I would guess at about 60C or so rise in temperature of the
surface of the box.

Re-do your estimate based on the lower heat load (85W/sqft) and see if our
results seem to be reasonable.

John

D

#### Dan Akers

Jan 1, 1970
0
"I'm looking for some advise on heat rise. If I have a chassis (10 x 17
x 19) and I mount two components that dissipate 385W and 256W
respectively, and then have a power supply that supplies those
components power to operate, then I'm assuming (if the power supply is
80% efficient) that my total heat load is somewhere in the 1400W range
(641W + 770W (supply input pwr)). First off, is that a correct
assumption of the heat loading or am I making this way too simple?
Next, based on this heat load, the heat rise is a function of the total
surface area of the chassis (calculated to be about 9.5 sqft). Based on
that, the Watts/square ft = 147W/sqft (1400/9.5 = 147). Curves show that
the heat rise will be off the charts since the charts only go up to
20W/sqft for something close to 80F temp difference (heat rise as a
function of the ambient air temp)
Does anyone see errors in my calculations or is this the expected heat
rise for this scenario?"
____________________________________
Re;
I'm wondering what sort of components these are; are they discrete,
passive, components, like resistors, or are they circuit boards? Is the
chassis vented in any way, or are you relying on the heat to conduct
through the sides of the chassis? What is the max. temperature rise
allowed? In what sort of environment will it operate?
In any event, I don't see how you get 1400W. The total for the
components is 641W and the 20% of the power supply is 128W giving a
total of 769W. Never-the-less, in my opinion, the cabinet is going to
need some forced ventilation to obtain a viable temperature rise if
semi-conductors and such are exposed inside this chassis. If you assume
1000W dissipation inside the chassis, you're going to need to move about
320cfm to obtain a 10F temp. rise; for a 20F rise you need 160cfm; and
so on.
I do think that your assumption for natural convection around the
cabinet based solely on it's total surface area is overly simplistic in
that the heat flux will certainly not be evenly distributed over the
surface area of the chassis if the components are simpling naturally
convecting and radiating within. Most of the heat will convect, and to
a lesser extent, radiate off of the upper portion of the chassis, with
lesser amounts from the sides and ends and even less from the bottom.

-Dan Akers

J

#### Jasen Betts

Jan 1, 1970
0
I'm looking for some advise on heat rise. If I have a chassis (10 x 17
x 19) and I mount two components that dissipate 385W and 256W
respectively, and then have a power supply that supplies those
components power to operate, then I'm assuming (if the power supply is
80% efficient) that my total heat load is somewhere in the 1400W range
(641W + 770W (supply input pwr)). First off, is that a correct
assumption of the heat loading or am I making this way too simple?

Next, based on this heat load, the heat rise is a function of the total
surface area of the chassis (calculated to be about 9.5 sqft). Based
on that, the Watts/square ft = 147W/sqft (1400/9.5 = 147). Curves show
that the heat rise will be off the charts since the charts only go up
to 20W/sqft for something close to 80F temp difference (heat rise as a
function of the ambient air temp)

Does anyone see errors in my calculations or is this the expected heat
rise for this scenario?

Just looking at the figure 1400W tells me you've got a firestarter there.
1400W - that's a fan heater... tring to cool something that powerful
without using a big fan and/or a wheelbarrow load of heatsinking is

So what will it be? a radio transmitter or something?

Bye.
Jasen

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