# How to calculate temperature rise on AC induction motor in an evaporative cooler centrifugal blower?

A neighbor recently asked me if his evaporative air conditioner was running correctly because he felt the blower drive motor and it was too hot to leave his hand on it for long.

I was surprised to find nothing helpful to make a back of the envelope approximation of the operating temperature (the motor casing) of an AC induction motor commonly used as the centrifugal blower drive motor (via belt to pulley on the squirrel cage blower) inside the "swamp coolers" commonly used in the US Southwest states (with dry enough weather to make evaporative cooling effective).

So, I read some thermodynamics and looked at some thermal modeling of AC induction motors and came up with the following rough approach (my question will be whether anyone has a better idea or other comment on this approach):

In a simplified thermal circuit representation of a motor the various internal motor loss nodes are lumped circuit representations (with as much breakdown as you have data and computational capacity, e.g., Motor-Cad or the like). Because I have only typical values or nameplate ratings for the target blower motor I tried a rudimentary model where I ignore the internal motor thermal resistance and concentrate on the motor frame/housing/case thermal resistance. This seems reasonable since whatever the complexity internal to the motor, all of that heat is in series with the casing, so the casing can be considered as a single thermal resistance equal to the delta T in C across the motor casing divided by the motor heat loss (known from nameplate hp specified at a given rpm for a two speed motor and from general knowledge of typical efficiency for these typically split phase AC induction motors, the latter 60% at operating hp and proper load).

The cooled air inside the AC enclosure was ambient for the problem and equal to 15 degrees C at the time of measurement. The motor casing was 37.78 degrees C on this occasion while producing 1/6 hp output at 60% efficiency, i.e., 82 watts heat loss. So $\dfrac{37.78 - 15}{82} = 0.2778 \text{ °C}/\text{W}$ thermal resistance calculated for the motor housing. This was a 2-speed motor, i.e. two winding sets, one developing 1/6 hp at 1140 rpm the other 1/2 hp at 1725 rpm. We are operating at 1/6 hp and 1140 rpm in the instant problem. 115 VAC single phase.

This thermal resistance calculated would be a combined convective and radiation resistance. It also includes the likely internal fins on the rotor (this being an ODP open drip proof housing and manufacturers typically include small fanning fins inside on the ends of the rotor shaft to produce an internal forced convection cool utilizing the air flow from ambient through the openings in the case). I looked at a typical rpm vs thermal resistance curve that researchers had developed empirically and derived a cubic polynomial regression fit that allowed me to estimate the 0 speed thermal resistance of the motor case (not true locked rotor, but just the static thermal resistance at a controlled current by the researchers). With that equation I estimated the thermal resistance of the case at 0 rpm without the internal fanning effect would be approx. 0.96626 deg. C/watt. For a motor casing estimated area of 0.1567626 square meters (cylinder) this gives a combined convective and radiation transfer coefficient of $\dfrac{1}{0.96626\cdot0.1567626} = 6.60182\text{ W}/\text{m}^2\text{°C} = h$. Typical combined heat transfer coefficients in this context for simple geometric shapes are in the range of $12 - 14\text{ W}/\text{m}^2\text{°C}$, so this is a bit low, but in the ballpark of expectations in rough initial estimate.

I used the thermal resistance already calculated above and temperature at the occasion my neighbor believed his motor was too hot. That calculation gave me 44 °C for the motor plus ambient (it was 21 °C ambient within the AC enclosure on that occasion). A thermodynamic text opined that 44 °C was too hot to be tolerated on direct contact with skin to surface (depends on one's capacity for pain; theoretically it would require about 5 minutes at 50 °C to produce a major skin burn).

The cubic polynomial modifying the measured casing thermal resistance dependent on rpm was of form $f(x) = ax^3 + bx^2 + cx + d$ where the coefficients are $a = -0.00000124809$, $b = 0.0067928076$, $c = -12.2709059880$, $d = 9884.78844117$ and within 1 - 5% fit within the range of 0 to 2500 rpm. The $f(x)$ result should be divided by 10000 (1e4) to renormalize. Then you multiply that corrective factor times the 0 rpm thermal resistance given (0.96626 °C/W). $x$ is the rpm, e.g, 1140 in the present case with the motor running at "low speed" 1/6 hp. For example, $f(1140) = 2874.7921$. Divide that by 1e4 to get 0.2875. Multiply $0.2875$ by the 0 rpm base thermal resistance of 0.96626 °C/watt to get 0.2779 °C/W case thermal resistance at 1140 rpm (as was experimentally measured).

I should add that the motor is NEMA class F insulation so the motor is permitted to rise up to 105 °C casing temp in up to 40 °C ambient (combined ambient and motor 155 C permitted NEMA specifying the motor rise lower to allow for hot spots internal).

• Thanks to hazzey for the nice demonstration of how to add math symbols. I will study that and try to do better in the future. – Dalton Bentley Mar 14 '16 at 23:11
• On 10/2/16 I posted a detailed analysis as an answer, but your system regards a long answer as "spam." The original text (I stripped it of images and edited down to fit here within rep and length limits) is at Why is my motor hot – Dalton Bentley Oct 2 '16 at 19:17

$$\dot{Q} = hA\Delta T$$
$\dot{Q}$ is the rate of heat transfer, $h$ is the heat transfer coefficient, $A$ is the surface area, and $\Delta T$ is the temperature difference between the free stream air and the surface of the motor. For your case, you already calculated that the rate of heat dissipation is $82 \,W$. For convection, the heat transfer coefficient could be anywhere in a wide range, but $50 \frac{W}{m^2 K}$ is probably a good initial guess. I'm assuming the motor housing is a cylinder $10\, cm$ long and $10\, cm$ diameter, so the surface area would be $0.047\, m^2$. Solving for $\Delta T$ we get a $35^\circ C$ difference between air and motor, which would put the motor surface at $50^\circ C$ (assuming the air is at $15^\circ$). This seems plausible based on your calculation and your neighbor's personal experience.