alt.coffee here
When you say " . . . because by examination i know that C is 3BTU/F.", I
assume you are referring to the specific heat of water (C sub p in
engineering terms). That value is, for your purposes and for most
practical
purposes, 1.0 btu's/lb/degree F.
The first step in solving your problem is to find the total heat lost
(Btu's) by the coffee in the time you stated (6 hours). Q = (mass of
liquid)(specific heat of liquid)(delta T experienced by the liquid).
1.5 quarts of coffee (assume water) weighs very close to 3.128 lbs., so Q
=
(3.128)(1.0)(203 - 144) = 184.55 Btu's.
The AVERAGE (emphasis intentional) rate of heat loss is 184.55/6 = 30.76
Btu's/hour
Q is also equal to (U)(A)(LMTD) where U is the overall heat transfer
coefficient (Btu's per hour per square foot per degree F), A is the area
the
heat is being lost through (square feet), and LMTD is the log mean
temperature differential (in degrees F). So,
30.76 = (U)(A)(LMTD)
I'll let you go from here.
Eric S. (35 years removed from mechanical engineering school)
<phlegmatico@[EMAIL PROTECTED]
> wrote in message
news:1123132415.488391.212720@[EMAIL PROTECTED]
>i wanted to figure out the equivalent R-value of the vacuum insulation
> in a Nissan brand vacuum-insulated bottle, and see how it compares with
> foam-in-place (I've seen numbers of 7/inch for that).
>
> i chose to play with the model JLN-1400, because coffeeandkitchen.com's
> site, mentioned an actual (claimed) thermal performance;
>
> 1.5 Quarts of liquid capacity, starting at 203 F, sitting in an
> ambient of 68, went to 144 in hours. A cylindrical form factor, 4.5"
> diameter by 9.5" height external.
>
> I started by calculating the internal cubic volume of 1.5 quarts
> (assuming the it's a cylinder of same ratio). I got 86.3"^3 of
> diameter 3.45" and height 7.3", which yields a thermal-leakage surface
> of (I hope i'm copying everything from my notes correctly!) 98"^2.
>
> that means, if i can find R or U, I can calculate R-value.
>
> Well, if I can calculate RC, that will give me R, because by
> examination i know that C is 3BTU/F. yes, i am presuming that coffee
> is just water.
>
> Ok, I know that at Time(0), delta of contents-ambient temp was 135. And
> six hours later, the same delta was 59.
>
> I formed
>
> 144 = 68 +(203-68)e^-[6/RC]
>
> 144 = 68 + 135e^-[6/RC]
>
> 76 = 135e^-[6/RC]
>
> 0.562963 = e^-[6/RC]
>
> ln (0.562963) = -0.5745413
>
> -6/RC = -0.5745413
>
> RC = 10.443 hours
>
>
> is this correct so far?
>
>
> if yes, then RC/C = R, so 10.443/3= 3.481 = R
>
> now, i do (144"^2)/(98"^2) to see that the fluid has 0.68 feet^2
> leakage surface,
>
> so i form 3.481/0.68 = R-value = 5.119
>
>
>
>
>
> So, how am I doing? can you please mention, which group you're
> replying from.
>
>
> by the way, do people really drink coffee at 203F ?!? Likewise, on
> group origin.
>
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