Re: Time required to heat half kg of water using thermal solar
- From: DB <abc@xxxxxxxx>
- Date: Sun, 07 Sep 2008 23:25:24 -0700
Jasvinder wrote:
The ambient temp is 110 F.
At 20 mph the air side heat transfer coefficient is 20 watts/m^2-C.
Heat Input = Coefficient * Area * Temp. Diff. * Time
= 20 X 0.055742 X 43.32 X T
= 48.2948 X T
Heat Required
Latent Heat of Ice= 80 Cal/g
1 Pint of water= 473.176 g
Heat required= 80 X 473.176 Cal
= 44.058 Watt-hr
Time= 44.058/48.2948
0.9121 Hr
or 54 Minutes.
Hope we stop at this.
BTW Ben that was good concept elaboration. Thx.
Even if the observation is clear?
http://www.soda-is.com/eng/map/
.
- References:
- Time required to heat half kg of water using thermal solar
- From: Jasvinder
- Re: Time required to heat half kg of water using thermal solar
- From: Uncle Ben
- Re: Time required to heat half kg of water using thermal solar
- From: Jasvinder
- Time required to heat half kg of water using thermal solar
- Prev by Date: Re: Time required to heat half kg of water using thermal solar
- Next by Date: JAPAN'S NEW SOLAR SHIP
- Previous by thread: Re: Time required to heat half kg of water using thermal solar
- Next by thread: Home Solar Hydrogen Production
- Index(es):
Relevant Pages
|
