# How do you use an insulation jacket's R-value for calculating heat transfer from a hot water tank?

I know how to use thermal resistance circuits to calculate heat transfer in Cartesian and cylindrical coordinates, but I don't know how to deal with real-world insulation values, and I was not able to find much about this online.

For example, I have this hot water tank that has an insulation jacket with 2" R-12.5 insulation. For simplicity, let's assume the insulation is the only thing that separates the hot water from cold outside air (no heat transfer from bottom of tank, top of tank, etc. Just through the insulation jacket). Assume water tank thickness aside from the 2" insulation jacket is negligible.

So it goes hot water -> insulation -> cold air outside. I can look up a free convection coefficient for the tank's hot water and can look up a convection coefficient for air as well. That allows me to use the formula $$R=1/(h*2*pi*r*heightofinsulationjacket)$$ twice (once for water and once for air).

So the problem is how to deal with the insulation jacket. How do I interpret 2" R-12.5 insulation? How do I factor in the thickness of the insulation? The surface area is taken care of in the heat transfer equation.

$$Q''=(Tf-Ti)/R$$ or $$Q=(Tf-Ti)/R*A$$

Updated: How do I use an R-value in cylindrical coordinates?

e.g. $$R/A=1/(hW*2*pi*r1)+Rins/(whatAreaHere)+1/(hA*2*pi*r2)$$

I am missing the area for my R-value insulation

The insulation R-value already accounts for thickness, so you can ignore it. You just have to account for the effect of surface area and temperature difference.

The SI units of R are $\frac{K \cdot m^2}{W}$

Heat transfer increases as the surface area increases, so Q is calculated as:

$$Q_{insulation}=\frac{A (T_f-T_i)}{R}$$ and $$Q''{insulation}=\frac{(T_f-T_i)}{R}$$

• Whoops, can't delete my comment so I'm making a new one. I meant this: How would I implement the R-value in cylindrical coordinates though where you specify the surface area of each of your R-values separately? e.g. R=1/(hWater*2*pi * r1*L)+Rinsulation/(what area here)+1/(hAir*2*pi * r2*L). Do I just divide my insulation R-value by 2*pi * L where L is the height of the tank? Commented Jul 26, 2017 at 18:56
• Are you trying to model the temperature variation inside the tank, or to just figure out the total heat transfer? Normally you'd just sum up the R values, but in the case of cylindrical coordinates, I don't think you can since the area is not constant. Sorry I'm not much help there. Commented Jul 26, 2017 at 19:32
• I am just trying to figure out the total heat transfer through the tank in cylindrical coordinates. I do have a model, but I am not particularly interested in the temperature profile right now. This surely must have been done somewhere since R-values for insulation are common on hot water tanks. How do they use their R-value? Commented Jul 26, 2017 at 22:48
• If it were me, I would assume a constant temperature of the water inside - whatever the thermostat is set on. The temperature gradient probably won't make much of a difference in the calculation, although I'm not sure what kind of accuracy you're looking for. Then I'd create the total R value by summing the insulation conduction with the overall convection you found elsewhere. That seems pretty straightforward. Commented Jul 26, 2017 at 23:20
• I think this explains what I was looking for: physics.stackexchange.com/questions/348770/… Commented Jul 29, 2017 at 16:04

@geekly: agreed. Also, insulation thickness will increase surface area for external convection, as: $$Q_{jacket \to air} = 2\pi (l_{heater})(r_{heater} + t_{jacket})h(T_{jacket} - T_{air})$$ This effect prompted the critical radius of insulation for cyclinders: https://en.wikipedia.org/wiki/Thermal_insulation: $$r_{critical} = \frac{k}{h} = \frac{t_{jacket}}{R_{jacket}h}$$