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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.

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  • $\begingroup$ I'm not sure if "entire R value" has much meaning here. It appears that the R value is a measure of the thermal resistance of a material per unit area. So it's meant to be a property of a specific insulating material, not a property of an object like a water tank as a whole. From the perspective of analyzing the effectiveness of the thermal insulation on the water tank, you're really more interested in Q, the heat flux density, and you can easily calculate the Q for your water tank by knowing the R value of the insulation, the dimensions of the tank, and the internal and exterior temperatures. $\endgroup$
    – user93237
    Jul 26, 2017 at 23:35
  • $\begingroup$ Does the insulation come pre-shaped to fit the cylinder, or is it just a layer of material you can apply to any geometry? $\endgroup$
    – JMac
    Jul 27, 2017 at 10:37
  • $\begingroup$ I realize that I didn't really understand the question properly until I saw Chester's answer and, yes, it makes sense that the central issue is how to combine the heat transfer co-efficients across the interfaces with the R value. Having seen Chester's reasoning I am almost certain that his answer is what you need. $\endgroup$ Jul 29, 2017 at 3:08

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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, $$R_{overall}=R+\frac{1}{h_W(r_1/\bar{r})}+\frac{1}{h_A(r_2/\bar{r})}$$

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  • $\begingroup$ This I believe is indeed exactly what the OP needs. I realize I didn't really understand the question properly until I saw your answer and, yes, it makes sense that the central issue is how to combine the heat transfer co-efficients across the interfaces with the R value. $\endgroup$ Jul 29, 2017 at 3:06
  • $\begingroup$ Why is Q independent of r? Can you show how you did that integration? $\endgroup$
    – MechE
    Jul 29, 2017 at 15:40
  • $\begingroup$ Because the same amount of heat flows through each cylindrical surface of constant radius. Where else would it go? The integration is done by "separation of variables." You know, dr/r and dT. $\endgroup$ Jul 29, 2017 at 15:46
  • $\begingroup$ So what you're saying is that the radius of the insulation (thickness maybe be clearer here) is constant since it's a parameter of the problem, so even though we may look at different end points (r2 values), heat flow is still constant through each of those regions regardless of what r2 value we choose (since the radius of the insulation is a constant and we just choose r2=rinsulation to solve the problem)? $\endgroup$
    – MechE
    Jul 29, 2017 at 15:53
  • $\begingroup$ Sorry, but I don't follow what you are asking. The heat is flowing radially, and the amount flowing through the r1 surface is equal to the amount flowing through the r2 surface (as well as through all the constant r surfaces in between). $\endgroup$ Jul 29, 2017 at 18:51
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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.

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