How do people use a provided insulation R-value to calculate heat transfer? This seems so strange to me because it must have been done before, but I think I'm missing something. So let's say there's 1-D heat transfer through a tank in cylindrical coordinates. The equation for the 1-D heat transfer is
$$
Q=(Tf-Ti)/R
$$
I know the temperature difference, but calculating the entire R-value is tricky to me. I have a hot water storage tank whose 2 inch insulation has an R value of 12.5. So when I go to calculate the total R-value, I get this:
$$
R=1/(hA*2*pi*r1*L)+Rins/(What area do I put here???)+1/(hW*2*pi*r2*L)
$$
The problem is since I have a company-provided R-value for insulation, so I have no idea what area I divide by to work this work in cylindrical coordinates where you need an area for each individual R-value.
I asked this on the engineering beta forums, but it seems like they are not certain either, so I'm wondering if any physicists know.
 A: In cylindrical coordinates, the rate of heat transfer equation through the insulation is described by the equation 
$$Q=-(2\pi r L)k\frac{dT}{dr}$$where k is the thermal conductivity.  Since Q is independent of r, we can integrate this equation to obtain:$$Q=k(2\pi L)\frac{(T_1-T_2)}{\ln{(r_2/r_1)}}=\frac{(2\pi \bar{r} L)}{R}(T_1-T_2)\tag{1}$$where $T_1$ is the temperature at the inner surface of the insulation, T_2 is the temperature at the outer surface of the insulation,  R is the "R value" of the insulation $(r_2-r_1)/k$ and $\bar{r}$ is the "log-mean" radius of the insulation $(r_2-r_1)/\ln(r_2/r_1)$.  This is typically very close in value to the arithmetic mean radius $(r_2+r_1)/2$.
We also have that, at the inner wall, $$Q=2\pi r_1Lh_W(T_W-T_1)\tag{2}$$where $T_W$ is the water temperature and $h_W$ is the corresponding heat transfer coefficient, while, and the outer wall, $$Q=2\pi r_1Lh_A(T_2-T_A)\tag{3}$$If we solve these three equations for the temperature differences, we obtain:$$T_1-T_2=\frac{QR}{2\pi\bar{r}L}\tag{4}$$$$T_W-T_1=\frac{Q}{2\pi r_1Lh_W}\tag{5}$$$$T_2-T_A=\frac{Q}{2\pi r_2Lh_A}\tag{6}$$Adding these three equations together yields:$$T_W-T_A=\frac{Q}{2\pi\bar{r}L}\left(R+\frac{1}{h_W(r_1/\bar{r})}+\frac{1}{h_A(r_2/\bar{r})}\right)$$So finally, $$R_{overall}=R+\frac{1}{h_W(r_1/\bar{r})}+\frac{1}{h_A(r_2/\bar{r})}$$
A: R value is the rate of heat loss per area per temperature difference.  In SI units $m^2 K /Watt$. 
So if you have a 2m tall tank with 0.5m diameter = 3.5$m^2$ surface area and assuming the water inside is at 90C and the room outside is 20C you will lose $3.5 * (90-20)/R$ Watts
If you live in a country with bizarre medieval units of measurement then the same principle applies but the numbers are different.
