The main flaw in this may be that lightbulbs only hit 220 C when
confined in fixtures. For a fair comparison with the exposed hose, we'd
need to look at an exposed lightbulb. My impression is that they don't
go a whole lot above boiling in that case. (Water dripped on a free
standing lightbulb doesn't typically fizz, as far as I can recall.)
If the bulb envelope were only, say, 100C above ambient in that case,
the conclusion here would be very different: I'd get a loss rate for
the hose of 1920 watts, or 77% of the device's output power, leaving
just 580 watts to be carried off by the plume. And that probably
wouldn't be enough heat to support GG's conclusion.
So, all in all, this plausibility check may be too squishy to be of much
use.
On 11-06-21 11:30 AM, Stephen A. Lawrence wrote:
On 11-06-21 10:35 AM, Stephen A. Lawrence wrote:
Jed sez a 150 watt hose wouldn't be very hot. GoatGuy says it
would. Well, let's do a little plausibility calculation, and see
whose ballpark we wind up in.
Now, let's just slow down here. There's another plausibility check
we can do, and it comes out a little differently. Let's try to
calculate how much heat the surface of the hose might be dumping into
the air, using a lightbulb for a "standard", and assuming the hose is
really quite hot.
Assume a 100 watt bulb is a sphere 2" in diameter, or 5 cm. (That
ignores the stem, of course, but it's not too far off, I think. Note
that making this value too small will be "conservative", which is the
direction in which we want to err here.)
Assume it dumps ALL of its energy to convection with the air. (That's
an overestimate, which is conservative, as you will see.)
Assume, further, that its envelope temperature is 200 C above ambient
(which is in the ballpark, according to Wiki, and according to what
I'd guess based on how they feel).
Bulb area is then 4 * pi * 2.5^2 = 79 cm^2.
Now, let's assume the hose has a surface temp of 100C. The rubber is
an insulator, so it'll really be a bit cooler than that; consequently
this is a "conservative" assumption.
So, hose temp is about 80 C above ambient (which I'm assuming is 20 C).
Hose radius is 1 cm.
The 3 meter hose has area = 2 * radius * pi * length = 2 * 1 * pi *
300 = 1885 cm^2.
Convective loss rate of the hose, relative to the bulb, will be
(area(hose) / area(bulb)) * (temp above ambient(hose) / temp above
ambient(bulb)
or
(1885/79) * (80/200) = 9.6
The bulb is assumed to be losing 100 watts, so that's a loss rate of
960 watts for the hose.
That's a loss of 38% of the output power, which was stated to be 2.5 kW.
Furthermore, that's (probably) a liberal estimate: An incandescent
bulb actually radiates away something like 20% of its energy (if I
recall correctly), while the hose, being a lot cooler, will radiate
away quite a bit less. Furthermore, the hose is made of an insulator
(rubber) so its outside surface temp will certainly be cooler than its
inside temp; consequently, its outside temperature must actually be
*less* than 80C above ambient. Furthermore, room temperature is often
higher than 20, rarely lower, which would reduce the relative temp of
the hose further. These effects combine to reduce the power being
dumped by the hose to something less than 38% of the output power of
the device.
Consequently, it seems to me that the end of the hose should still be
putting out a steam plume which carries away on the order of 1.5 kW.
That may still be enough energy to carry GoatGuy's main conclusion,
which is that the hose should have been a lot livelier than it
appeared on the video.
