How do you Calculate Ampacity of a Conductor Geometry?

[BuddyBoy]

So what is the cross-sectional area, in metres? Have you looked up the resistivity of nickel? If so, what is the resistance of, say, a strip 20cm long? What is the maximum current that you will draw from the battery pack? And what therefore will be the power that the strip needs to dissipate?


You have already been told that orientation of the strip is only relevant if heat loss by convection is significant. Do you think there is any convection inside the battery pack?


No, you cannot measure the internal resistance of a battery with a multimeter.

I think you should stop thinking about designing battery packs and learn some electronics.

Finding the power dissipated in a strip is easy. But I am wondering how many Watts dissipated by the strip is considered "to much" because of the amount of heat?

Yea I do think there is some convection in the battery pack. There's not much for an air gap, but it is still present. I wanted as high fidelity as possible before making simplifications and assuming something is about zero.

There are different BMSs available, which have different continuous discharge currents supported, assuming the selected cells can support it. Some as low as 20 A, some nearly double that. Regardless finding cells to meet these needs is easy. But I'm struggling with trying to figure out what dimensions of nickel is considered insignificant?

I'm ok with considering it insignificant. But how many watts crossed the line from insignificant to significant?

So let's say at 20 A, 63 mm x 10 mm x 0.1 mm, this would give me, this would give me 1.76 W. At how many watts dissipated in a single strip should I be concerned that my strip is to small? This is just what the numbers gave, and part of my question if it is to much watts. So I'm not suggesting anyone actually send this much current through a strip this size. Theoretical only, and is the whole point of my question, and trying to understand the concepts here of how much power dissipated in a strip is considered to much.

Thanks for the info on internal resistance and cells heating up.

Thanks for all the help, greatly appreciated!!

 

[BuddyBoy]

UPDATE: Like others have mentioned, I completely disregard the heat transfer from to/from the cell. I may been overthinking this problem.

So for a randomly chosen 21700 cell. 50S, in a 3P5S pack, continuous discharge from the pack of $30 A$.

The temperature of the strip $T_{S}$ is greater than the temperature of the cell $T_{C}$, and the maximum operating temperature of the cell of $60\;C$.

The surface temperature of the cell at $10 A$ (30/3, it's a 3P pack) from the graph is $51\;C$. We do not want the surface temperature of the strip to get to $60\;C$ because then heat will transfer from the strip to the cells, eventually warming up the cells to their maximum operating temperature.

Assume that the length of the strip is $63 mm$ to allow for three cells to be connected.

The thermal conductivity of most 21700 cells is generically speaking between $12.6 \frac{W}{m\;C}$ and $16.7 \frac{W}{m\;C}$

The resistivity of nickel is approximately $6.99*10^{-8} \Omega\;m$ or $6.99*10^{-8} \Omega\;m*\frac{1}{10^{3}}\;\frac{m}{mm} = 6.99*10^{-11} \Omega\;mm$

$I^{2} \frac{\rho\;L_{S}}{Wt} = 3\frac{K_{C}}{L_{C}}\pi\;r^{2}(T_{S} - T_{C})$
$\frac{1}{Wt} = \frac{1}{\rho\;L_{S}I^{2}}3\frac{K_{C}}{L_{C}}\pi\;r^{2}(T_{S} - T_{C})$
$Wt = \frac{\rho\;L_{S}I^{2}L_{C}}{3K_{C}\pi\;r^{2}(T_{S}-T_{C})}$
$Wt = \frac{(6.99*10^{-11} \Omega\;mm)\;(63 mm)(30 A)^{2}(70 mm)}{3(12.6 \frac{W}{m\;C})\pi\;(21 mm)^{2}(60 C- 51 C)} \frac{W}{A^{2}\Omega}$
$Wt = \frac{(6.99*10^{-11} mm)\;(63)(30)^{2}(70)}{3(12.6 \frac{1}{m})\pi\;(21)^{2}(60 - 51)}$
$Wt = \frac{(6.99*10^{-11})\;(63)(30)^{2}(70)}{3(12.6)\pi\;(21)^{2}(60 - 51)} m*mm*(10^{3} \frac{mm}{m})$
$Wt = \frac{(6.99*10^{-11})\;(63)(30)^{2}(70)}{3(12.6)\pi\;(21)^{2}(60 - 51)}(10^{3}) mm^{2}$

So if I assume 0.1 mm for thickness, I get a width of 5.88621 * 10 ^(-6) mm

Seems like what others suggest that "any" dimension would work. But I'm just trying to confirm at this point that this approach is correct? I don't see what's wrong with it at the moment.

Again just theoretical, it's just what the numbers gave me. Trying to confirm if my approach based on the concepts is correct. The number is in question, and could be wrong.

 

[pbuk]

I may been overthinking this problem.

Ya think?

 

[BuddyBoy]

Ya think?

Perhaps, but I'm not sure that my logic in post number 62 holds water? I'm not sure that I can assume the temperature of the strip would be 60 deg C, while the temperature of the cell is 51 deg C? Because this would result in heat transfer occurring between the strip and cell. If we want the cells to be the limitation, than we want the strip to be a colder temperature than the cell. So I have to find a way to find the resistance of the strip, such that heat transfer occurs between the cell and strip, as opposed to the other way around.

Like you had mentioned, I should consider the cells being the limitation factor, and not the strips. Meaning that the cells should be hotter than the strips.

Problem Statement:
Determine the dimensions of a nickel strip required to maintain $30 A$, such that the randomly selected 21700 cells, Samsung 50S, within a battery box. Size the strip such that the cells are the limitation, and not the strip. Meaning that the cells maintain a higher temperature than the strip. Assume that cells are identical, with identical internal resistances $r_{int}$. Each individual cell is at the same temperature $T_{C}$ when drawing the same current, therefore there is no heat transfer in between each cell. Assume that the ambient air conditions $T_{\infty} = 100\;deg\;F$. 21700 cells have a diameter of $21 mm$. Assume that insulation is placed around the nickel strip, such that it is in contact with the side walls of the battery box. This assumption limits convection of the stirp to near zero, and can be considered insignificant. Test data shows that the temperature of this particular cell model is $51\;deg\;C$ at $10 A$. Find a generic formula with variables.

Heat transfer from cells to strip:
$$(\frac{I}{3})^{2}\;r_{int} + (\frac{I}{3})^{2}\;r_{int} + (\frac{I}{3})^{2}\;r_{int} = \frac{k_{S}}{t_{S}}A_{C}(T_{C} - T_{S}) + \frac{k_{S}}{t_{S}}A_{C}(T_{C} - T_{S}) + \frac{k_{S}}{t_{S}}A_{C}(T_{C} - T_{S})$$
$$3(\frac{I^{2}}{9})r_{int} =3\frac{k_{S}}{t_{S}}A_{C}(T_{C} - T_{S})$$
$$\frac{1}{3}I^{2}r_{int} =3\frac{k_{S}}{t_{S}}A_{C}(T_{C} - T_{S})$$
Solving for $T_{C} - T_{S}$:
$$T_{C} - T_{S} = \frac{1}{3}I^{2}r_{int}\frac{1}{3}\frac{t_{S}}{k_{S}}\frac{1}{A_{C}}$$
$$T_{C} - T_{S} = \frac{I^{2}\;r_{int}\;t_{S}}{9k_{S}\;A_{C}}$$

Heat transfer from strip to insulation:
$$\frac{1}{3}I^{2}r_{int} + I^{2}R = \frac{k_{I}}{t_{I}}A_{S}(T_{S} - T_{I})$$
$$I^{2}(\frac{1}{3}r_{int} + R) = \frac{k_{I}}{t_{I}}A_{S}(T_{S} - T_{I})$$
Solving for $T_{S} - T_{I}$:
$$T_{S} - T_{I} = (\frac{1}{3}r_{int} + R)\frac{t_{I}I^{2}}{k_{I}A_{S}}$$

Heat transfer from insulation to box:
$$I^{2}(\frac{1}{3}r_{int} + R) = \frac{k_{B}}{t_{B}}A_{I}(T_{I} - T_{B})$$
Solving for $T_{I} - T_{B}$:
$$T_{I} - T_{B} = (\frac{1}{3}r_{int} + R)\frac{t_{B}I^{2}}{k_{B}A_{I}}$$

Heat transfer from box to ambient air:
$$I^{2}(\frac{1}{3}r_{int} + R) =(T_{B} - T_{\infty})\sum_{i=1}^n h_{i}*A_{i}$$
Solve for $T_{B} - T_{\infty}$:
$$(T_{B} - T_{\infty}) = \frac{I^{2}(\frac{1}{3}r_{int} + R)}{\sum_{i=1}^n h_{i}*A_{i}}$$

Solve for $T_{C} - T_{\infty}$:
$$T_{C} - T_{\infty} = T_{C} - T_{S} + T_{S} - T_{I} + T_{I} - T_{B} + T_{B} - T_{\infty}$$
$$T_{C} - T_{\infty} = \frac{I^{2}\;r_{int}\;t_{S}}{9k_{S}\;A_{C}} + (\frac{1}{3}r_{int} + R)\frac{t_{I}I^{2}}{k_{I}A_{S}} + (\frac{1}{3}r_{int} + R)\frac{t_{B}I^{2}}{k_{B}A_{I}} + \frac{I^{2}(\frac{1}{3}r_{int} + R)}{\sum_{i=1}^n h_{i}*A_{i}}$$
$$T_{C} - T_{\infty} = I^{2}(\frac{\;r_{int}\;t_{S}}{9k_{S}\;A_{C}} + (\frac{1}{3}r_{int} + R)(\frac{t_{I}}{k_{I}A_{S}} + \frac{t_{B}}{k_{B}A_{I}} + \frac{1}{\sum_{i=1}^n h_{i}*A_{i}}))$$
Now solve for $R$:
$$\frac{T_{C} - T_{\infty}}{I^{2}} = \frac{\;r_{int}\;t_{S}}{9k_{S}\;A_{C}} + (\frac{1}{3}r_{int} + R)(\frac{t_{I}}{k_{I}A_{S}} + \frac{t_{B}}{k_{B}A_{I}} + \frac{1}{\sum_{i=1}^n h_{i}*A_{i}})$$
$$\frac{T_{C} - T_{\infty}}{I^{2}} - \frac{\;r_{int}\;t_{S}}{9k_{S}\;A_{C}} = (\frac{1}{3}r_{int} + R)(\frac{t_{I}}{k_{I}A_{S}} + \frac{t_{B}}{k_{B}A_{I}} + \frac{1}{\sum_{i=1}^n h_{i}*A_{i}})$$
$$\frac{\frac{T_{C} - T_{\infty}}{I^{2}} - \frac{\;r_{int}\;t_{S}}{9k_{S}\;A_{C}}}{\frac{t_{I}}{k_{I}A_{S}} + \frac{t_{B}}{k_{B}A_{I}} + \frac{1}{\sum_{i=1}^n h_{i}*A_{i}}} = \frac{1}{3}r_{int} + R$$
$$R = \frac{\frac{T_{C} - T_{\infty}}{I^{2}} - \frac{\;r_{int}\;t_{S}}{9k_{S}\;A_{C}}}{\frac{t_{I}}{k_{I}A_{S}} + \frac{t_{B}}{k_{B}A_{I}} + \frac{1}{\sum_{i=1}^n h_{i}*A_{i}}} - \frac{1}{3}r_{int}$$

Where:
$R$ is the resistance of the strip
$T_{C}$ is the temperature of the conductor
$T_{\infty}$ is the temperature of the ambient air conditions
$I$ is the current going through the nickel strip
$r_{int}$ is the internal resistance of the cell
$t_{S}$ is the thickness of the nickel strip
$k_{S}$ is the thermal conductivity of the nickel strip
$A_{C}$ is the surface area that the cell is in contact with the strip
$t_{I}$ is the thickness of the insulation
$k_{I}$ is the thermal conductivity of the insulation
$A_{S}$ is the surface area that the strip is in contact with the insulation
$t_{B}$ is the thickness of the box wall
$k_{B}$ is the thermal conductivity of the box
$A_{I}$ is the surface area that the insulation is contact with the box
$h_{i}$ is the heat transfer coefficient for surface area $i$ for the box to the ambient air
$A_{i}$ is the surface area $i$ of the box in contact with ambient air

Now that I know $R$ I can select an appropriately dimensioned strip that has this resistance.

Does this look better? I'm not making any assumptions about the temperature of the strip $T_{S}$, other than that it is less than the temperature of the cell $T_{C}$, which is what, to indicate that the cells are the limitation and not the strip. I don't need to find $T_{S}$ in this method. I'm not sure if my heat transfer for the strip to insulation is correct?

Thanks for all the help! I'm ok accepting that "any" dimensioned strip will work, but I just want to understand why it's insignificant. If it is insignificant, what is the exact point in which it does become significant? Certainly there is a boundary point where I most consider the dimensions. I wonder what this point is and how to find it out.

 

[BuddyBoy]

This may have been a whole simpler that I had thought! The only problem is that the answer I'm getting below seems to be independent of the current, and only dependent on the internal resistance. This is very confusing to me. See below.

So if the cells are supposed to be the limiting factor, than the at the very worse case scenario, the cells are to be the same temperature as the strip. There would be no heat transfer between the cells and the strip.

The heat transfer from the cells to the strip is
$$\frac{1}{3}I^{2}r_{int}$$
See the post above for where I derived it.

The heat transferred from the strip to the cells is
$$I^{2}R$$

So if the cells are at the same temperature as the strip, than the heat transferred between the objects should be identical.
$$\frac{1}{3}I^{2}r_{int} = I^{2}R$$
Solve for $R$:
$$R = \frac{1}{3}r_{int}$$
$$\frac{\rho\;L}{Wt} = \frac{1}{3}r_{int}$$
$$\frac{Wt}{\rho\;L} = \frac{3}{r_{int}}$$
$$Wt = \frac{3\rho\;L}{\;r_{int}}$$

So for Samsung 50S, $r_{int} <= 14 m\Omega$. So I will just take the value of $14 m\Omega$ for now. But understanding that if $r_{int}$ gets smaller, than my $Wt$ quantity needs to be larger. Assuming a nickel strip that is $0.1 mm$ thick and $63 mm$ long. How wide of a strip do I need?

$$W = \frac{3\rho\;L}{\;r_{int}t}$$
$$W = \frac{3(6.99*10^{-8}\;\Omega\;m)(10^{3}\;\frac{mm}{m})(63\;mm)}{(14\;m\Omega)(\frac{1}{10^{3}}\;\frac{\Omega}{m\Omega})(0.1\;mm)} = 9.4365\;mm$$

So per this a 0.1 mm x 9.4365 mm x 63 mm long nickel strip can handle 30 A when it's connected to three cells that have an internal resistance of 14 mOhm.

All theoretical here, this is just what the numbers gave me, and seems incorrect to me. The formula I'm getting is independent of the current, which seems wrong to me. But I'm not seeing what's wrong with this approach though.

Thanks for any help!