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 Posted: Tue Jun 17, 2008 6:37 pm Post subject: A practical question |
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I have been entrusted to ship a certain ammount of botox from New York
to Miami and need to keep it cool. My question is how much water ice
should I use to keep the botox cool while it's in transit?
Specifics:
insulation, styrofoam case 5 cm thick, surface area of .77m2
assumed average ambiant temprature, 30c
target internal temp, 2c
time in transit 24 hrs
mass of styrofoam box .4 kg
mass of botox, .5 kg with packing
I'd really appriciate the help.
Evan
evan@raiainternational.com |
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Brian Whatcott
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| Posted: Wed Jun 18, 2008 6:39 am Post subject: Re: A practical question |
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On Tue, 17 Jun 2008 11:37:17 -0700 (PDT), emcourtney@gmail.com wrote:
| Quote: |
I have been entrusted to ship a certain ammount of botox from New York
to Miami and need to keep it cool. My question is how much water ice
should I use to keep the botox cool while it's in transit?
Specifics:
insulation, styrofoam case 5 cm thick, surface area of .77m2
assumed average ambiant temprature, 30c
target internal temp, 2c
time in transit 24 hrs
mass of styrofoam box .4 kg
mass of botox, .5 kg with packing
I'd really appreciate the help.
Evan
evan@raiainternational.com
|
I only have numbers for the domestic units handy,
which is a nuisance.
R values for styrofoam run R = 4 to 7 usually. Use R = 4
1/R = k which in the domestic system, means
1/4 BTU per hour per sq ft exposed at 1 inch thick for 1 deg F
difference
Let's start with some conversions....
1 BTU = 1055 joules
1055 joules per hour = 1055 / 60X60 = 0.293 watts
1 square foot area = 12 X 12 / 39.37 X 39.37 = 0.0929 sq meters
1 degF = 5/9 deg C = 0.555 degC
So the thermal resistance in question is
0.25 X 0.293W per 0.0929 sq meters per 2.5 cm thickness at 0.555 degC
delta (Phew!)
We have 0.77 m^2 area ,
5 cm = 2 inch thickness and
28 degF = 15.6 degC delta
and time = 24 hours = 86400 seconds.
The heat flow in terms of power transfer - watts is
0.25 X ( 0.77 / 0.0929 ) X 0.293 W X 15.6 degC / (0.555 degC/degF
X 2 in thickness ) = 8.53 watts during 24 hours.
How to provide this heat flow rate?
Thinking about ice to ice water, that's 80 cal/gm
or 80 X 4.2 joules / gm of melt = 336 joules /gm
8.53 watts for 24 hours = 8.53 X 60 X 60 X 24 = 737000 joules
Thats 737000 / 336 gms of melting ice = 2.2 kg of ice.
[If this were my chore, I would double check these figures!!]
You're welcome
Brian W |
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Brian Whatcott
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| Posted: Wed Jun 18, 2008 8:17 am Post subject: Re: A practical question |
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On Tue, 17 Jun 2008 20:39:21 -0500, Brian Whatcott
<betwys1@sbcglobal.net> wrote:
| Quote: |
On Tue, 17 Jun 2008 11:37:17 -0700 (PDT), emcourtney@gmail.com wrote:
I have been entrusted to ship a certain ammount of botox from New York
to Miami and need to keep it cool. My question is how much water ice
should I use to keep the botox cool while it's in transit?
Specifics:
insulation, styrofoam case 5 cm thick, surface area of .77m2
assumed average ambiant temprature, 30c
target internal temp, 2c
time in transit 24 hrs
mass of styrofoam box .4 kg
mass of botox, .5 kg with packing
I'd really appreciate the help.
Evan
evan@raiainternational.com
I only have numbers for the domestic units handy,
which is a nuisance.
R values for styrofoam run R = 4 to 7 usually. Use R = 4
1/R = k which in the domestic system, means
1/4 BTU per hour per sq ft exposed at 1 inch thick for 1 deg F
difference
Let's start with some conversions....
1 BTU = 1055 joules
1055 joules per hour = 1055 / 60X60 = 0.293 watts
1 square foot area = 12 X 12 / 39.37 X 39.37 = 0.0929 sq meters
1 degF = 5/9 deg C = 0.555 degC
So the thermal resistance in question is
0.25 X 0.293W per 0.0929 sq meters per 2.5 cm thickness at 0.555 degC
delta (Phew!)
We have 0.77 m^2 area ,
5 cm = 2 inch thickness and
28 degF = 15.6 degC delta
and time = 24 hours = 86400 seconds.
The heat flow in terms of power transfer - watts is
0.25 X ( 0.77 / 0.0929 ) X 0.293 W X 15.6 degC / (0.555 degC/degF
X 2 in thickness ) = 8.53 watts during 24 hours.
How to provide this heat flow rate?
Thinking about ice to ice water, that's 80 cal/gm
or 80 X 4.2 joules / gm of melt = 336 joules /gm
8.53 watts for 24 hours = 8.53 X 60 X 60 X 24 = 737000 joules
Thats 737000 / 336 gms of melting ice = 2.2 kg of ice.
[If this were my chore, I would double check these figures!!]
You're welcome
Brian W
|
This doesn't count as a recheck even, but I notice I used a delta of
28 degF WRONG! s/b 28degC = 28 X 9/5 degF = 50degF
That's much more ice - unless there are other numeric zits in
there.....
[that's what hapeens when you mess round converting units.]
Brian W |
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