wrote:
Beartums wrote:

My wife and I are having a discussion about radiator placement...
Currently,
our contractor has placed our hot-water radiators on the shelf under
windows.
My wife is convinced that the radiators will never properly heat the
rooms
since heat goes up...

Warm air rises, and cool air falls near exterior walls. Why fight
nature?
IMO, a radiator near an exterior wall is less efficient, since it's
in
slower air, since it has to fight downgoing air. It also loses more
heat
to outdoors, keeping the wall warmer than a central radiator would
and
making turbulent vs laminar flow near the wall, which increases the
wall's
film conductance. Harry Thomason knew this, but we've mostly
forgotten it.

I believe that the room is actually warmed by being filled with heat

(otherwise, you would need radiators on all walls and even in the
middle of the room to heat the floor).

Warm air rises over hot spots, slides along the ceiling to cooler
spots,
drops to the floor, slides back to hot spots, rewarms and rises. Warm
ceilings radiate heat to the rest of the room with a surprisingly
high
linearized conductance: 4x0.1714x10^-8(460+70)^3 = U1.

While I believe the optimal placement of the radiator would be on
the
floor, the sizing of the radiator is MUCH more important, and the
saving
of floor space in this case is more important that the minimal
increase
in heating efficiency.

Can anyone add any comments or guidance on this issue?

We might keep a room with an 8'x8'/R16 = 4 Btu/h-F exterior wall 70 F
on
a 30 F day with a central radiator that heats C cfm from 70-dT to 70
F,
where CdT = (70-dT/2-30)4 = 10-dT/8. If 70 F air flows down a 8' wide
virtual duct that extends 1' from the wall into the room and 70-dT
air
flows out the bottom and C = 16.6x1x8sqrt(8dT), dT = 0.163 F, the
wall
loses (70-0.163/2-30)4 = 159.67 Btu/h, ignoring the turbulent flow. A
radiator below might make it lose (70+0.163/2-30)4 = 160.33 Btu/h. No
big
deal, altho the difference is larger for windows with less
insulation.

OTOH, warmer walls allow lower room air temps, for the same comfort
level.
A cube with 5 70 F walls and a 70-0.163/2 = 69.919 F wall and a
radiant
temp of 70-0.163/2/6 = 69.986 F would need 70.0104 F room air for
equal
comfort compared to 70 F air and 70 F walls, according to ASHRAE
55-2004,
so the loss from the warmer wall outweighs the gain from cooler air.

With 180 F water in 5 Btu/h-F-ft fin tube, the central radiator might
have
159.67/(5(180-69.84)) = 0.29' of tube in slow-moving air near the
floor.

But the bouyancy force of a column of warm air in some sort of
chimney
above a tube can move air by fins at a higher velocity and raise
their
water-air conductance. If a foot of fin tube has a conductance of 5
Btu/h-F
= A(2+V/2) in V = 0 mph air, its effective area A = 2.5 ft^2. (I
counted 43
2"x2" fins per foot, about 2.4 ft^2, including both sides.) Fin tube
near
the floor in a closet or stairwell or inside wall with an A ft^2 vent
at
the bottom and top and an 8' height diff between them and a dT temp
diff
from room to chimney air should make C = 16.6Asqrt(8dT) = 47Asqrt(dT)
cfm
flow with velocity V = 0.01136C/A = 0.533sqrt(dT) mph, so 1' of 180 F
tube
would lose (180-69.84)2.5(2+0.533sqrt(dT)/2) = 551+73.4sqrt(dT) =
cfmdT
= 47AdT^1.5, ie dT = ((11.73+1.56sqrt(dT)/A)^(2/3). With a 2"x12"
slot,
A = 0.167, so dT = (70.4+9.36sqrt(dT))^(2/3). Plugging in dT = 10 on
the
right makes dT = 21.5 on the left. Repeating makes dT = 23.4, 23.7,
and
23.7, so 2.6 mph air might come out of the top vent at 69.8+23.7 =
93.5 F,
moving 47(0.167)23.7^1.5 = 906 Btu/h of heat, so we only need
159.67/906
= 0.176' of fin tube, ie 40% less than fin tube in free air.

The closet might be a nice place to dry clothes.

Nick

An excellent example of why you should keep calculators away from
people with internet access and too much time on their hands.

All of your calculations mean nothing with respect to heating a room
evenly. Efficiency and comfort are not necessarily the same thing.

R