Heating and cooling of electric machine

In an iron core (or any part where losses are produced inside the material), heat must flow through the material to reach the cooling surface. The inside is therefore hotter than the surface. The highest temperature, at the centre, is the hot spot, and it decides the safety of nearby insulation.

Step 1: heat flow in one direction

Assume:

• $p$ = heat (loss) produced per unit volume, W/m³, uniform
• $\rho$ = thermal resistivity of the material, °C·m/W ($\rho = 1/k$)
• heat flows only in the $x$ direction (the other dimensions are large), and both faces are at the same surface temperature $\theta_s$.

By symmetry the hottest point is at the centre plane, $x = 0$, where no heat crosses. Consider a plane at distance $x$ from the centre. All heat produced between the centre and this plane must cross it. Per unit area:

$$Q_x = p\,x \quad \text{W/m}^2$$

By Fourier's law, the temperature gradient is

$$\frac{d\theta}{dx} = -\rho\, Q_x = -\rho\, p\, x$$

Integrating from the centre (temperature $\theta_h$):

$$\begin{aligned} \theta(x) &= \theta_h - \frac{\rho\, p\, x^2}{2} \end{aligned}$$

So the temperature is parabolic across the thickness. At the surface, $x = t/2$, $\theta = \theta_s$:

$$\theta_h - \theta_s = \frac{\rho\, p}{2}\left(\frac{t}{2}\right)^2 = \frac{p\,\rho\, t^2}{8}$$

 temperature
     ^        theta_h (hot spot)
     |         _..--.._
     |      .-'        '-.
     |    .'              '.
     |   /                  \
     |  theta_s        theta_s
     +--|---------+---------|--> x
      -t/2        0        +t/2
        <------ thickness t ------>

Step 2: core with heat flow in two directions

A laminated core block can lose heat in two directions:

• along the laminations (direction $x$, length $l_x$ between cooling surfaces or ducts), where thermal resistivity $\rho_x$ is low (through iron);
• across the laminations (direction $y$, length $l_y$), where thermal resistivity $\rho_y$ is much higher, because heat must cross the insulation (varnish, oxide) between sheets.

        cooling surface
   +---------------------+  ^
   | ==================  |  |
   | ==================  |  | l_y (across lam.)
   | ==================  |  |
   +---------------------+  v
   <-------- l_x -------->   (along laminations)

Let $p_x$ and $p_y$ be the parts of $p$ carried away in each direction, with $p_x + p_y = p$. Both paths start at the same hot spot and end at the same surface temperature, so by Step 1:

$$\theta_h - \theta_s = \frac{p_x \rho_x l_x^2}{8} = \frac{p_y \rho_y l_y^2}{8}$$

Hence $p_x = \dfrac{8(\theta_h-\theta_s)}{\rho_x l_x^2}$ and $p_y = \dfrac{8(\theta_h-\theta_s)}{\rho_y l_y^2}$. Adding:

$$p = 8(\theta_h-\theta_s)\left[\frac{1}{\rho_x l_x^2} + \frac{1}{\rho_y l_y^2}\right]$$ $$\theta_h - \theta_s = \frac{p}{8\left[\dfrac{1}{\rho_x l_x^2} + \dfrac{1}{\rho_y l_y^2}\right]}$$

This is the expression for the internal (hot-spot) temperature rise of the core above its surface. The hot-spot temperature itself is $\theta_h = \theta_s + (\theta_h - \theta_s)$, where $\theta_s$ is found from the surface dissipation, $\theta_s - \theta_a = \dfrac{\text{loss}}{S\lambda}$.

Conclusions

• The temperature difference is proportional to the square of the length of the heat path. Splitting the core into packets with ventilating ducts reduces it sharply.
• Since $\rho_y \gg \rho_x$, most heat flows along the laminations. Ducts placed across the laminations, which cut $l_x$, are therefore the most effective.

The heating time constant is the time a machine would take to reach its final steady temperature rise if heat were never lost to the surroundings, i.e. if the initial rate of rise continued. Equivalently, it is the time to reach 63.2 % of the final rise. The cooling time constant is the corresponding time during cooling: the time for the temperature rise to fall to 36.8 % of its initial value.

Heating

Let $P$ = losses (W), $G$ = mass (kg), $h$ = specific heat (J/kg°C), $S$ = cooling surface (m²), $\lambda$ = specific heat dissipation (W/m²°C), $\theta$ = temperature rise. In time $dt$:

$$P\,dt = G h\,d\theta + S\lambda\theta\,dt$$

Solving with $\theta = \theta_i$ at $t=0$:

$$\theta = \theta_m\left(1 - e^{-t/\tau_h}\right) + \theta_i\, e^{-t/\tau_h}$$ $$\theta_m = \frac{P}{S\lambda}, \qquad \tau_h = \frac{G h}{S\lambda}$$

Starting from cold ($\theta_i = 0$): $\theta = \theta_m(1 - e^{-t/\tau_h})$. At $t = \tau_h$, $\theta = 0.632\,\theta_m$. In practice the final temperature is reached after about $4\tau_h$–$5\tau_h$.

Cooling

When the machine is switched off ($P = 0$) from a rise $\theta_i$:

$$0 = G h\,d\theta + S\lambda_c\theta\,dt \;\Rightarrow\; \theta = \theta_i\, e^{-t/\tau_c}, \qquad \tau_c = \frac{G h}{S\lambda_c}$$

At $t = \tau_c$, $\theta = 0.368\,\theta_i$.

Curves

 rise                       rise
  ^   theta_m ----------     ^ theta_i
  |        _.-~~~~~          |\
  |     .-'   heating        | \   cooling
  |   /                      |  '.
  |  / 0.632 theta_m at tau  |    '-._ 0.368 at tau_c
  | /                        |        ~~--.____
  +------------------> t     +------------------> t

Comparison

A hot spot is the point of highest temperature inside a body in which heat is produced, such as a core, a coil or a conductor. Heat produced inside must flow outwards to the cooling surfaces, so the interior is hotter than the surface. Insulation near the hot spot ages fastest, so the hot-spot temperature, not the average, must be kept within the limit of the insulation class.

Derivation for a plate of thickness $t$

Consider a plate whose length and width are large compared with its thickness $t$, cooled equally on both faces. Then heat flows only across the thickness. Assume:

• $p$ = heat (loss) produced per unit volume, W/m³, uniform
• $\rho$ = thermal resistivity of the material, °C·m/W ($\rho = 1/k$)
• heat flows only in the $x$ direction (the other dimensions are large), and both faces are at the same surface temperature $\theta_s$.

By symmetry the hottest point is at the centre plane, $x = 0$, where no heat crosses. Consider a plane at distance $x$ from the centre. All heat produced between the centre and this plane must cross it. Per unit area:

$$Q_x = p\,x \quad \text{W/m}^2$$

By Fourier's law, the temperature gradient is

$$\frac{d\theta}{dx} = -\rho\, Q_x = -\rho\, p\, x$$

Integrating from the centre (temperature $\theta_h$):

$$\begin{aligned} \theta(x) &= \theta_h - \frac{\rho\, p\, x^2}{2} \end{aligned}$$

So the temperature is parabolic across the thickness. At the surface, $x = t/2$, $\theta = \theta_s$:

$$\theta_h - \theta_s = \frac{\rho\, p}{2}\left(\frac{t}{2}\right)^2 = \frac{p\,\rho\, t^2}{8}$$

 temperature
     ^        theta_h (hot spot)
     |         _..--.._
     |      .-'        '-.
     |    .'              '.
     |   /                  \
     |  theta_s        theta_s
     +--|---------+---------|--> x
      -t/2        0        +t/2
        <------ thickness t ------>

Result

$$\theta_h = \theta_s + \frac{p\,\rho\,t^2}{8}$$

where $\theta_h$ is the hottest-spot temperature (at the mid-plane) and $\theta_s$ the surface temperature.

Points to note

• The temperature profile across the plate is a parabola with its peak at the centre.
• The hot-spot rise above the surface is proportional to $t^2$. Halving the thickness, for example by a cooling duct in the middle, reduces it to one-quarter.
• It is also proportional to the loss density $p$ and to the thermal resistivity $\rho$. Materials with poor thermal conductivity (insulation, or laminated iron across the sheets) produce large internal gradients.
• If only one face is cooled, the hot spot is at the insulated face, and $\theta_h - \theta_s = p\rho t^2/2$ (the same formula with thickness $2t$).

When the slot insulation is very thick, heat from the conductor cannot pass through it into the core. Heat therefore flows axially along the copper to the overhang, and a temperature gradient appears along the length of the conductor.

Assumptions

• The conductor carries current density $J$. Loss per unit volume is $p = J^2\rho_e$, where $\rho_e$ is the electrical resistivity.
• The slot insulation is very thick (a poor conductor of heat). No heat flows sideways into the iron; all heat produced in the slot portion flows along the conductor to the overhang (end connections), where it is dissipated.
• $\rho_t$ = thermal resistivity of the conductor (copper), $A$ = conductor cross-section, $L$ = embedded (slot) length, $\theta_o$ = temperature of the conductor at the slot ends (overhang).

   overhang      slot portion (length L)      overhang
  ~~~~~~~~~ |============================| ~~~~~~~~~
            |<--- heat    hottest  heat --->|
           x=-L/2         x = 0          x=+L/2
               temperature: parabola, max at centre

Derivation

By symmetry the hottest point is at the centre of the slot length ($x = 0$). Heat produced between the centre and a section at distance $x$ flows through that section:

$$Q_x = p\,A\,x$$

Fourier's law along the conductor gives

$$\frac{d\theta}{dx} = -\frac{\rho_t\,Q_x}{A} = -\rho_t\,p\,x$$

Integrating from the centre ($\theta = \theta_m$ at $x = 0$):

$$\theta = \theta_m - \frac{\rho_t\,p\,x^2}{2}$$

At the slot ends, $x = L/2$, $\theta = \theta_o$:

$$\theta_m - \theta_o = \frac{p\,\rho_t\,L^2}{8} = \frac{J^2\rho_e\,\rho_t\,L^2}{8}$$

So the temperature varies parabolically along the conductor. The gradient at any point is $\rho_t p x$, which is zero at the centre and largest at the slot ends.

Conclusions

• The temperature rise of the centre over the ends is proportional to $J^2$ and to $L^2$. Long cores with high current density (for example turbo-alternators) have large axial gradients.
• Copper has a very low thermal resistivity, so in normal machines this rise is small. Its importance is that it shows heat cannot pass through thick insulation easily, so large machines need ventilating ducts and direct conductor cooling.

A conductor in a slot produces copper loss $p = J^2\rho_e$ per unit volume. This heat can leave by two paths: across the slot insulation into the iron (then to the cooling air), or along the conductor to the overhang, where it is dissipated to the air. Which path dominates depends on the thickness of the slot insulation and the length of the overhang. This gives two limiting cases.

Case 1: slot insulation very thick (heat flows along the conductor)

Assumptions

• The conductor carries current density $J$. Loss per unit volume is $p = J^2\rho_e$, where $\rho_e$ is the electrical resistivity.
• The slot insulation is very thick (a poor conductor of heat). No heat flows sideways into the iron; all heat produced in the slot portion flows along the conductor to the overhang (end connections), where it is dissipated.
• $\rho_t$ = thermal resistivity of the conductor (copper), $A$ = conductor cross-section, $L$ = embedded (slot) length, $\theta_o$ = temperature of the conductor at the slot ends (overhang).

   overhang      slot portion (length L)      overhang
  ~~~~~~~~~ |============================| ~~~~~~~~~
            |<--- heat    hottest  heat --->|
           x=-L/2         x = 0          x=+L/2
               temperature: parabola, max at centre

Derivation By symmetry the hottest point is at the centre of the slot length ($x = 0$). Heat produced between the centre and a section at distance $x$ flows through that section:

$$Q_x = p\,A\,x$$

Fourier's law along the conductor gives

$$\frac{d\theta}{dx} = -\frac{\rho_t\,Q_x}{A} = -\rho_t\,p\,x$$

Integrating from the centre ($\theta = \theta_m$ at $x = 0$):

$$\theta = \theta_m - \frac{\rho_t\,p\,x^2}{2}$$

At the slot ends, $x = L/2$, $\theta = \theta_o$:

$$\theta_m - \theta_o = \frac{p\,\rho_t\,L^2}{8} = \frac{J^2\rho_e\,\rho_t\,L^2}{8}$$

So the temperature varies parabolically along the conductor. The gradient at any point is $\rho_t p x$, which is zero at the centre and largest at the slot ends.

Case 2: slot insulation thin (heat flows across the insulation)

If the insulation is thin, almost all the heat of the embedded part flows radially through it to the teeth and core. The conductor temperature is then nearly uniform along its length, and the temperature drop occurs across the insulation.

  iron | insulation | copper (theta_c)
       |  t_i       |
  theta_i <-- heat --

Heat produced in length $L$: $Q = p A L$. It flows through insulation of thickness $t_i$, thermal resistivity $\rho_i$ and area $S_p L$, where $S_p$ is the perimeter of the slot insulation:

$$\theta_c - \theta_{iron} = Q \times \frac{\rho_i\,t_i}{S_p L} = \frac{p\,A\,\rho_i\,t_i}{S_p}$$

The gradient is now linear across the insulation: $\dfrac{d\theta}{dy} = \dfrac{p A \rho_i}{S_p}$.

Summary

In practice both paths act together. Thin, high-conductivity insulation (mica-epoxy) and ventilating ducts keep the conductor hot spot low.

The one-hour rating is the output the motor can deliver for one hour, starting from cold, without its temperature rise exceeding the permitted value (here 40 °C, the same as the continuous rating).

Given

• Rated output = 20 hp, final steady rise at rated output $\theta_m = 40$ °C
• Heating time constant $\tau = 180$ min, time $t = 60$ min
• Total full-load losses = 1800 W, with $\dfrac{\text{copper loss}}{\text{constant loss}} = 1.25$

Step 1: split the losses

$$\begin{aligned} P_i + 1.25P_i &= 1800 \\ P_i\ (\text{constant}) &= \frac{1800}{2.25} = 800\ \text{W} \\ P_{cu}\ (\text{full-load copper}) &= 1800 - 800 = 1000\ \text{W} \end{aligned}$$

Step 2: final temperature rise allowed for the one-hour load

Let $\theta_m'$ be the final steady rise the motor would reach at the one-hour load. For the rise to be just 40 °C after 60 min from cold:

$$\begin{aligned} 40 &= \theta_m'\left(1 - e^{-60/180}\right) \\ \theta_m' &= \frac{40}{1 - e^{-1/3}} = \frac{40}{1 - 0.7165} = \frac{40}{0.2835} = 141.11\ ^\circ\text{C} \end{aligned}$$

Step 3: losses at the one-hour load

The final temperature rise is proportional to the losses ($\theta_m = P/S\lambda$, with the same cooling):

$$\begin{aligned} P' &= 1800 \times \frac{141.11}{40} = 1800 \times 3.5277 = 6349.9\ \text{W} \end{aligned}$$

Step 4: load corresponding to these losses

Let the one-hour output be $x$ times the rated output. Constant loss stays at 800 W, and copper loss varies as $x^2$:

$$\begin{aligned} 800 + 1000\,x^2 &= 6349.9 \\ x^2 &= 5.5499 \\ x &= 2.356 \end{aligned}$$ $$\text{One-hour rating} = 2.356 \times 20 = 47.12\ \text{hp}\ \ (\approx 35.1\ \text{kW})$$

Answer: one-hour rating ≈ 47.1 hp (about 2.36 times the continuous rating), for a 40 °C temperature rise.

(The motor would also need enough pull-out torque to carry this load. Thermally, this is the limit.)

Given

• Temperature rise from cold: $\theta_1 = 25$ °C at $t = 1$ h, $\theta_2 = 37.5$ °C at $t = 2$ h
• On disconnection, the temperature falls from its final steady value to 40 °C in 2.5 h; ambient = 30 °C.

Heating: final rise and heating time constant

From cold, $\theta = \theta_m(1 - e^{-t/\tau_h})$. Let $a = e^{-1/\tau_h}$:

$$\begin{aligned} 25 &= \theta_m(1 - a) \\ 37.5 &= \theta_m(1 - a^2) = \theta_m(1-a)(1+a) \end{aligned}$$

Dividing:

$$\begin{aligned} 1 + a &= \frac{37.5}{25} = 1.5 \;\Rightarrow\; a = 0.5 \\ \theta_m &= \frac{25}{1 - 0.5} = 50\ ^\circ\text{C} \\ \tau_h &= \frac{1}{\ln(1/a)} = \frac{1}{\ln 2} = 1.443\ \text{h} \approx 86.6\ \text{min} \end{aligned}$$

Cooling time constant

Final steady temperature = 30 + 50 = 80 °C (rise 50 °C). After 2.5 h the temperature is 40 °C, i.e. a rise of 40 − 30 = 10 °C above ambient.

$$\begin{aligned} \theta &= \theta_i\, e^{-t/\tau_c} \\ 10 &= 50\, e^{-2.5/\tau_c} \\ \tau_c &= \frac{2.5}{\ln(50/10)} = \frac{2.5}{\ln 5} = \frac{2.5}{1.6094} = 1.553\ \text{h} \approx 93.2\ \text{min} \end{aligned}$$

Answer: final steady temperature rise = 50 °C; heating time constant = 1.443 h (86.6 min); cooling time constant = 1.553 h (93.2 min).

The cooling time constant is larger than the heating one, as expected, because heat dissipation is poorer when the transformer is not loaded (weaker oil circulation).

The strip must have the resistance that gives 1 kW at 250 V, and enough surface area to radiate 1 kW at the allowed temperature.

Assumption: ambient temperature 25 °C, so the strip temperature is 25 + 300 = 325 °C. Heat is lost by radiation only (Stefan's law), from both faces of the strip; the edges are neglected.

Step 1: resistance required

$$R = \frac{V^2}{P} = \frac{250^2}{1000} = 62.5\ \Omega$$

Step 2: heat dissipated per m² of surface

Stefan's law with emissivity $e$ and radiating efficiency $\eta$:

$$\begin{aligned} q &= 5.7\times10^{-8}\, e\,\eta\,(T_1^4 - T_2^4) \\ T_1 &= 325 + 273 = 598\ \text{K},\quad T_2 = 25 + 273 = 298\ \text{K} \\ q &= 5.7\times10^{-8}\times0.9\times0.75\times(598^4 - 298^4) \\ &= 3.8475\times10^{-8}\times(1.2788\times10^{11} - 0.0789\times10^{11}) \\ &= 4617\ \text{W/m}^2 \end{aligned}$$

Step 3: surface area needed

For length $l$ and width $w$, the radiating surface is $2lw$:

$$\begin{aligned} 2\,l\,w\,q &= 1000 \\ l\,w &= \frac{1000}{2\times4617} = 0.10830\ \text{m}^2 \quad (1) \end{aligned}$$

Step 4: length–width ratio from resistance

$$\begin{aligned} R &= \frac{\rho\, l}{w\,t} \\ \frac{l}{w} &= \frac{R\,t}{\rho} = \frac{62.5\times0.2\times10^{-3}}{1\times10^{-6}} = 12500 \quad (2) \end{aligned}$$

Step 5: solve (1) and (2)

$$\begin{aligned} l^2 &= 12500\times0.10830 = 1353.75 \\ l &= 36.79\ \text{m} \\ w &= \frac{l}{12500} = 2.943\times10^{-3}\ \text{m} = 2.94\ \text{mm} \end{aligned}$$

Check: $R = \dfrac{10^{-6}\times36.79}{2.943\times10^{-3}\times0.2\times10^{-3}} = 62.5\ \Omega$.

Answer: length ≈ 36.8 m, width ≈ 2.94 mm (for 25 °C ambient). With a 20 °C ambient the same method gives about 37.4 m and 2.99 mm.

When losses are produced inside a body, heat must flow through the material to reach its cooling surfaces. The interior is therefore hotter than the surface. The highest (internal or hot-spot) temperature is found as follows.

Assumptions

• A block of homogeneous material of thickness $t$, length $l$ and width $w$, with $l$ and $w$ much larger than $t$. Heat therefore flows only across the thickness ($x$ direction) and leaves equally from the two large faces.
• Uniform loss $p$ W/m³; thermal resistivity $\rho$ °C·m/W (constant).
• Surface heat dissipation coefficient $\lambda$ W/m²°C; ambient temperature $\theta_a$.

          face 1 (area l x w)
   +--------------------------------+   ^
   |   heat <---  centre  ---> heat |   | t
   +--------------------------------+   v
          face 2 (area l x w)

Step 1: surface temperature

Total heat produced $= p\,l\,w\,t$. It leaves from two faces of total area $2lw$:

$$\begin{aligned} p\,l\,w\,t &= 2\,l\,w\,\lambda\,(\theta_s - \theta_a) \\ \theta_s &= \theta_a + \frac{p\,t}{2\lambda} \end{aligned}$$

Step 2: temperature rise inside

Take $x$ from the mid-plane. Heat produced between the mid-plane and a plane at $x$ crosses that plane. Its flow per unit area is $p\,x$. By Fourier's law:

$$\begin{aligned} \frac{d\theta}{dx} &= -\rho\,p\,x \\ \theta(x) &= \theta_h - \frac{\rho\,p\,x^2}{2} \end{aligned}$$

At $x = t/2$, $\theta = \theta_s$:

$$\theta_h - \theta_s = \frac{p\,\rho\,t^2}{8}$$

Step 3: internal (hot-spot) temperature

$$\theta_h = \theta_a + \frac{p\,t}{2\lambda} + \frac{p\,\rho\,t^2}{8}$$

 temperature
     ^        theta_h (hot spot)
     |         _..--.._
     |      .-'        '-.
     |    .'              '.
     |   /                  \
     |  theta_s        theta_s
     +--|---------+---------|--> x
      -t/2        0        +t/2
        <------ thickness t ------>

Remarks

• The first term is the rise of the surface above ambient, set by the cooling. The second is the internal rise, set by conduction.
• The internal rise varies as $t^2$. Dividing the block with a cooling duct at mid-thickness reduces it to one-quarter, which is why cores are split into packets.
• If heat flows in two directions (thickness $t$ and width $w$, with resistivities $\rho_t$, $\rho_w$), the internal rise becomes

$$\theta_h - \theta_s = \frac{p}{8\left[\dfrac{1}{\rho_t t^2} + \dfrac{1}{\rho_w w^2}\right]}$$

When the slot insulation is comparatively very thick, no heat flows from the conductor into the iron. All copper loss of the slot portion flows along the conductor to the overhang, and the overhang dissipates the whole loss of the conductor to the air. The maximum temperature is at the middle of the slot length.

Notation

• $J$ = current density, $\rho_e$ = electrical resistivity, so loss per unit volume $p = J^2\rho_e$
• $A$ = conductor cross-section, $S_p$ = perimeter of the overhang surface (W dissipated per unit length per °C is $S_p\lambda$)
• $L$ = slot (embedded) length, $l_o$ = overhang length at each end, $\rho_t$ = thermal resistivity of copper
• $\lambda$ = specific heat dissipation of the overhang surface (W/m²°C), $\theta_a$ = ambient temperature

  l_o            L (slot)            l_o
 ~~~~~ |=====================| ~~~~~
 theta_o      theta_max        theta_o

Step 1: overhang temperature

The total loss of one conductor, $p A (L + 2l_o)$, is dissipated from the two overhang surfaces, of area $2 l_o S_p$ (overhang temperature taken as uniform):

$$\begin{aligned} p A (L + 2 l_o) &= 2 l_o S_p \lambda (\theta_o - \theta_a) \\ \theta_o &= \theta_a + \frac{p A (L + 2 l_o)}{2 l_o S_p \lambda} \end{aligned}$$

Step 2: rise of slot portion above the overhang

At distance $x$ from the slot centre, the heat crossing the conductor section is $p A x$. So

$$\frac{d\theta}{dx} = -\rho_t\,p\,x \;\Rightarrow\; \theta = \theta_{max} - \frac{\rho_t p x^2}{2}$$

At $x = L/2$, $\theta = \theta_o$:

$$\theta_{max} - \theta_o = \frac{p\,\rho_t\,L^2}{8}$$

Step 3: maximum conductor temperature

$$\theta_{max} = \theta_a + \frac{J^2\rho_e A (L + 2 l_o)}{2 l_o S_p \lambda} + \frac{J^2\rho_e\,\rho_t\,L^2}{8}$$

The maximum temperature is the ambient, plus the rise of the overhang (set by its cooling), plus the parabolic rise along the slot. A long core or a high current density increases it sharply. This is why large machines use ventilating ducts, thin high-conductivity insulation or direct conductor cooling.

A slot conductor produces loss $p = J^2\rho_e$ per unit volume ($J$ = current density, $\rho_e$ = electrical resistivity). The heat leaves along the conductor to the overhang, or across the slot insulation to the iron. Part (ii) of the question is cut off in the source; it is read here as the case where the overhang is not considerable (short, with little cooling), so that the heat must leave through the slot insulation.

(i) Slot insulation thick

No heat passes through the insulation, so heat flows axially to the overhang. With slot length $L$, copper thermal resistivity $\rho_t$ and the centre at $x = 0$:

$$\begin{aligned} \frac{d\theta}{dx} &= -\rho_t\,p\,x \\ \theta_{max} - \theta_o &= \frac{p\,\rho_t\,L^2}{8} \end{aligned}$$

The gradient is zero at the slot centre and largest at the slot ends. The temperature varies parabolically along the conductor, with its maximum at the middle.

(ii) Overhang not considerable (heat through slot insulation)

Little heat can flow to the short overhang, so the heat of the slot portion, $Q = p A L$, crosses the insulation (thickness $t_i$, thermal resistivity $\rho_i$, perimeter $S_p$, area $S_p L$):

$$\begin{aligned} \theta_c - \theta_{iron} &= Q\,\frac{\rho_i\,t_i}{S_p L} = \frac{p\,A\,\rho_i\,t_i}{S_p} \\ \frac{d\theta}{dy} &= \frac{p\,A\,\rho_i}{S_p}\ \ (\text{constant}) \end{aligned}$$

The conductor is at nearly uniform temperature along its length. The gradient is linear across the insulation thickness.

Iron losses (hysteresis and eddy) are produced throughout the volume of the core. This heat must flow through the iron to the cooling surfaces and ducts, so the centre of each core packet is hotter than its surface. The resulting variation of temperature inside the core is the temperature gradient.

Anisotropy of a laminated core

• Along the laminations, heat flows through continuous iron, so the thermal resistivity $\rho_x$ is low.
• Across the laminations, heat must cross the insulation (varnish or oxide) and the air films between sheets, so the thermal resistivity $\rho_y$ is many times higher (roughly 20–50 times).

       cooling surface / duct
   +----------------------------+  ^
   | ========================== |  |
   | ======== hot spot ======== |  | l_y (across)
   | ========================== |  |
   +----------------------------+  v
   <----------- l_x ------------>  (along)

One-direction flow

For loss $p$ per unit volume flowing across a length $l$ between two cooled faces, consider a plane at $x$ from the centre. Heat crossing it per unit area is $p x$, so $\dfrac{d\theta}{dx} = -\rho p x$ and

$$\theta_h - \theta_s = \frac{p\,\rho\,l^2}{8}$$

The temperature is parabolic, with the hot spot at the centre.

Two-direction flow (actual core)

Let the loss divide into $p_x$ (along) and $p_y$ (across). Both paths share the same hot spot and surface temperature:

$$\theta_h - \theta_s = \frac{p_x\rho_x l_x^2}{8} = \frac{p_y\rho_y l_y^2}{8}, \qquad p_x + p_y = p$$ $$\theta_h - \theta_s = \frac{p}{8\left[\dfrac{1}{\rho_x l_x^2} + \dfrac{1}{\rho_y l_y^2}\right]}$$

Conclusions

• Most heat flows along the laminations because $\rho_x$ is small.
• The hot-spot rise varies as the square of the path length. Ventilating ducts that divide the core into packets of 40–80 mm therefore reduce the gradient sharply.
• Iron loss density $p$ should be kept moderate (a suitable $B_m$) in large cores.

A machine heats up because its losses produce heat faster than it can be dissipated at first. The temperature rises exponentially towards a final steady value, at which all the heat produced is dissipated.

Assumptions

• The machine is a homogeneous body with uniform temperature.
• Heat produced $P$ (W) is constant; heat dissipation is proportional to the temperature rise (Newton's law of cooling).

Notation

$P$ = losses (W), $G$ = mass (kg), $h$ = specific heat (J/kg°C), $S$ = cooling surface (m²), $\lambda$ = specific heat dissipation (W/m²°C), $\theta$ = temperature rise above ambient at time $t$.

Derivation

In a small time $dt$ the temperature rises by $d\theta$. Energy balance:

$$\underbrace{P\,dt}_{\text{produced}} = \underbrace{G h\,d\theta}_{\text{stored}} + \underbrace{S\lambda\theta\,dt}_{\text{dissipated}}$$

Rearranging:

$$\begin{aligned} \frac{dt}{Gh} &= \frac{d\theta}{P - S\lambda\theta} \\ \frac{S\lambda}{Gh}\,dt &= \frac{d\theta}{\dfrac{P}{S\lambda} - \theta} \end{aligned}$$

Let $\theta_m = \dfrac{P}{S\lambda}$ and $\tau = \dfrac{Gh}{S\lambda}$. Integrating, with $\theta = \theta_i$ at $t = 0$:

$$\begin{aligned} \frac{t}{\tau} &= \ln\frac{\theta_m - \theta_i}{\theta_m - \theta} \\ \theta &= \theta_m\left(1 - e^{-t/\tau}\right) + \theta_i\,e^{-t/\tau} \end{aligned}$$

Starting cold ($\theta_i = 0$):

$$\theta = \theta_m\left(1 - e^{-t/\tau}\right)$$

As $t \to \infty$, $\theta \to \theta_m = P/(S\lambda)$, the final steady temperature rise: all heat produced is then dissipated.

 rise
  ^          theta_m ---------------------
  |               _..--~~~~~~
  |          _.-'
  |       .-'  <- 0.632 theta_m at t = tau
  |     /
  |   /  initial slope = theta_m / tau
  | /
  +---|------|------|------|------> t
     tau    2tau   3tau   4tau

Heating time constant

$$\tau_h = \frac{G h}{S\lambda}$$

It is defined as:

1. the time to reach the final rise $\theta_m$ if the initial rate of rise continued unchanged (no heat dissipated); or
2. the time to reach 63.2 % of the final rise, since $1 - e^{-1} = 0.632$.

In practice the machine is considered thermally steady after about $4\tau$–$5\tau$ (98–99 %). The time constant is large for large, heavy machines with poor cooling, and is reduced by better ventilation (larger $\lambda$).

Convection is the transfer of heat from a hot surface by a moving fluid (air, hydrogen, oil). It is the main way heat leaves an electrical machine. Depending on what drives the fluid, it is called natural or artificial (forced) convection.

Natural vs artificial convection

The dissipation coefficient for forced air flow is often written $\lambda_c = \lambda_{c0}(1 + 0.1\,v_a)$, where $v_a$ is the air velocity in m/s. Doubling the cooling therefore lets a much larger loss be removed for the same rise.

Temperature rise–time curve of an electrical machine

Assumptions: the machine is a homogeneous body with uniform temperature; losses $P$ are constant; heat dissipated is proportional to the temperature rise.

Notation: $G$ = mass (kg), $h$ = specific heat (J/kg°C), $S$ = cooling surface (m²), $\lambda$ = specific heat dissipation (W/m²°C), $\theta$ = temperature rise at time $t$.

In time $dt$, heat produced = heat stored + heat dissipated:

$$P\,dt = G h\,d\theta + S\lambda\theta\,dt$$ $$\begin{aligned} \frac{d\theta}{dt} + \frac{S\lambda}{Gh}\theta &= \frac{P}{Gh} \end{aligned}$$

Let $\theta_m = \dfrac{P}{S\lambda}$ (final steady rise) and $\tau = \dfrac{Gh}{S\lambda}$ (heating time constant). The equation becomes

$$\tau\frac{d\theta}{dt} + \theta = \theta_m$$

Separating variables and integrating from $\theta_i$ at $t = 0$:

$$\begin{aligned} \int_{\theta_i}^{\theta}\frac{d\theta}{\theta_m - \theta} &= \int_0^t\frac{dt}{\tau} \\ \ln\frac{\theta_m - \theta_i}{\theta_m - \theta} &= \frac{t}{\tau} \\ \theta &= \theta_m\left(1 - e^{-t/\tau}\right) + \theta_i\,e^{-t/\tau} \end{aligned}$$

From cold: $\theta = \theta_m(1 - e^{-t/\tau})$.

 rise
  ^      theta_m - - - - - - - - - - - - -
  |              ___....-------~~~~
  |         _.-''
  |      .-'   0.632 theta_m at t = tau
  |    /
  |  /
  | /
  +---|------|------|------|------> t
     tau    2tau   3tau   4tau

Cooling (losses removed): $0 = Gh\,d\theta + S\lambda_c\theta\,dt$, giving $\theta = \theta_i e^{-t/\tau_c}$ with $\tau_c = Gh/(S\lambda_c)$.

Link between convection and the curve

• Forced convection raises $\lambda$. This lowers both the final rise $\theta_m = P/(S\lambda)$ and the time constant $\tau = Gh/(S\lambda)$.
• For the same permitted rise, a forced-cooled machine can carry larger losses, i.e. a higher output from the same frame.
• In self-ventilated machines the fan stops at standstill, so $\lambda_c < \lambda$ and the cooling time constant is longer than the heating time constant.

The statement is true. Iron loss is produced throughout the volume of a core, and the heat must flow through the iron to a cooling surface. The internal (hot-spot) temperature rise depends on the square of the heat path length. Ventilating ducts shorten this path and add cooling surface, so they reduce the temperature rise inside the material sharply.

Justification

For a core packet of length $l$ between two cooled faces, with loss $p$ per unit volume and thermal resistivity $\rho$, the hot spot is at the middle:

$$\theta_h - \theta_s = \frac{p\,\rho\,l^2}{8}$$

If the core is divided by $n$ radial ducts into $(n+1)$ packets, each packet has length $\dfrac{l}{n+1}$:

$$\theta_h' - \theta_s = \frac{p\,\rho}{8}\left(\frac{l}{n+1}\right)^2 = \frac{\theta_h - \theta_s}{(n+1)^2}$$

For example, one central duct ($n = 1$) reduces the internal rise to one-quarter. Three ducts reduce it to one-sixteenth.

 without duct            with one duct
 +-----------+          +----+  +----+
 |    hot    |          |    |  |    |
 +-----------+          +----+  +----+
   rise ~ l^2           rise ~ (l/2)^2 = 1/4
                               ^ air

Why the ducts cut across the laminations

• Heat flows easily along the laminations (through iron), but with difficulty across them, through the insulation between sheets, where the thermal resistivity is much higher.
• A duct "across the lamination" (between packets, at right angles to the core length) gives every sheet its own short path along its plane to cooling air, so heat does not have to cross many insulation layers.

Further benefits

1. Extra cooling surface: the duct faces dissipate heat, so $\theta_s$ (surface temperature) also falls.
2. Air is blown through the ducts by the rotor (radial ventilation), raising the heat transfer coefficient.
3. Lower core temperature protects the winding insulation and permits higher flux density or current density, giving more output from the same frame.

Typical practice: packets of 40–80 mm with ducts about 10 mm wide in the stator and rotor cores of large machines, and oil ducts in transformer cores.

Cooling removes the heat produced by losses so that the temperature of the insulation stays within its class limit. Better cooling allows a higher output from the same machine size. Cooling methods are classified as follows.

1. By the way the coolant moves

• Natural cooling: no fan; heat leaves by natural convection and radiation from the frame. Used for small and totally enclosed non-ventilated (TENV) machines.
• Self ventilation: a fan mounted on the machine's own shaft circulates air. Most common for induction motors. Cooling depends on speed.
• Separate (forced) ventilation: an independently driven fan or blower. Used for variable-speed and large machines.

2. By the path of the cooling air

• Radial ventilation: air enters at the ends, flows radially outward through radial ducts between core packets. Used in medium and large machines (above about 20 kW).
• Axial ventilation: air flows along axial holes in the rotor/stator cores. Used in small, high-speed machines.
• Mixed (radial-axial) ventilation: a combination, used in large machines.

 radial ventilation (section)
  frame |^^^^^^^^^^^^^^^^^^|  air out
 stator |==| |==| |==| |==|
        |  duct  duct  duct|
  rotor |==| |==| |==| |==|
  air in-> --- shaft --- <-air in

3. By the cooling circuit

• Open circuit: fresh air is drawn from the surroundings and thrown out (drip-proof, screen-protected machines).
• Closed circuit: the same air recirculates through a heat exchanger (air-to-air or air-to-water). Used where the air is dirty or for large machines. TEFC (totally enclosed fan-cooled) motors have an external fan blowing over a finned frame.

4. Other coolants

• Hydrogen cooling of large turbo-alternators. Compared with air, hydrogen has about 7–14 times the heat transfer ability and 1/14 the density, giving lower windage loss, no oxidation and less noise.
• Direct (inner) conductor cooling: hydrogen or water flows through hollow conductors. Used in very large alternators (above about 300 MW).

5. Transformer cooling methods

The choice depends on rating, enclosure, environment and cost. Larger ratings need more intense cooling because losses grow faster (∝ linear dimension³) than surface area (∝ linear dimension²).

Given

• From cold: $\theta_1 = 25$ °C after 1 h, $\theta_2 = 40$ °C after 2 h
• On disconnection, temperature falls from its final steady value to 45 °C in 90 min (1.5 h); ambient = 30 °C

Heating: final rise and heating time constant

$\theta = \theta_m(1 - e^{-t/\tau_h})$. Let $a = e^{-1/\tau_h}$:

$$\begin{aligned} 25 &= \theta_m(1 - a), \qquad 40 = \theta_m(1 - a^2) \\ 1 + a &= \frac{40}{25} = 1.6 \;\Rightarrow\; a = 0.6 \\ \theta_m &= \frac{25}{1 - 0.6} = 62.5\ ^\circ\text{C} \\ \tau_h &= \frac{1}{\ln(1/0.6)} = \frac{1}{0.5108} = 1.958\ \text{h} \approx 117.5\ \text{min} \end{aligned}$$

Cooling time constant

Final temperature = 30 + 62.5 = 92.5 °C (rise 62.5 °C). After 1.5 h the temperature is 45 °C, a rise of 45 − 30 = 15 °C.

$$\begin{aligned} 15 &= 62.5\,e^{-1.5/\tau_c} \\ \tau_c &= \frac{1.5}{\ln(62.5/15)} = \frac{1.5}{\ln 4.1667} = \frac{1.5}{1.4271} = 1.051\ \text{h} \approx 63.1\ \text{min} \end{aligned}$$

Answer: final steady temperature rise = 62.5 °C; heating time constant = 1.958 h (117.5 min); cooling time constant = 1.051 h (63.1 min).

Given

• Rating $S$ = 350 kVA; maximum efficiency at 85 % of full load
• Heat run at full load: $\theta_1 = 24$ °C after 1 h, $\theta_2 = 34$ °C after 2 h
• Fan cooling raises heat dissipation per unit area per °C ($\lambda$) by 15 %

Step 1: time constant and final temperature rise

With $a = e^{-1/\tau}$, $\theta_1 = \theta_m(1-a)$ and $\theta_2 = \theta_m(1-a^2)$:

$$\begin{aligned} 1 + a &= \frac{34}{24} = 1.4167 \;\Rightarrow\; a = 0.4167 \\ \tau &= \frac{1}{\ln(1/0.4167)} = \frac{1}{\ln 2.4} = 1.142\ \text{h} \approx 68.5\ \text{min} \\ \theta_m &= \frac{24}{1 - 0.4167} = 41.14\ ^\circ\text{C} \end{aligned}$$

Step 2: losses in terms of full-load copper loss $P_c$

At maximum efficiency, copper loss = iron loss:

$$\begin{aligned} P_i &= (0.85)^2 P_c = 0.7225\,P_c \\ \text{Full-load losses} &= P_i + P_c = 1.7225\,P_c \end{aligned}$$

The final rise is $\theta_m = \dfrac{\text{losses}}{S\lambda}$, so the losses allowed $\propto \lambda\,\theta_m$. At a new load of $x$ times full load, the losses are $P_i + x^2 P_c$ (iron loss is constant).

(i) Same final temperature rise ($41.14$ °C)

Losses allowed rise by 15 %:

$$\begin{aligned} 0.7225 + x^2 &= 1.15 \times 1.7225 = 1.9809 \\ x^2 &= 1.2584 \;\Rightarrow\; x = 1.1218 \\ \text{New rating} &= 1.1218 \times 350 = 392.6\ \text{kVA} \end{aligned}$$

(ii) Allowable temperature rise 50 °C

Losses allowed are now in the ratio $1.15 \times \dfrac{50}{41.14}$:

$$\begin{aligned} 0.7225 + x^2 &= 1.15 \times \frac{50}{41.14} \times 1.7225 = 2.4073 \\ x^2 &= 1.6848 \;\Rightarrow\; x = 1.2980 \\ \text{New rating} &= 1.2980 \times 350 = 454.3\ \text{kVA} \end{aligned}$$

Answer: time constant ≈ 1.14 h (68.5 min); final rise ≈ 41.1 °C; new rating (i) ≈ 393 kVA, (ii) ≈ 454 kVA.

Given

• Rating $S$ = 400 kVA; maximum efficiency at 80 % of full load
• Heat run at full load: $\theta_1 = 24$ °C after 1 h, $\theta_2 = 34$ °C after 2 h
• Fan cooling raises heat dissipation per unit area per °C ($\lambda$) by 15 %

Step 1: time constant and final temperature rise

With $a = e^{-1/\tau}$, $\theta_1 = \theta_m(1-a)$ and $\theta_2 = \theta_m(1-a^2)$:

$$\begin{aligned} 1 + a &= \frac{34}{24} = 1.4167 \;\Rightarrow\; a = 0.4167 \\ \tau &= \frac{1}{\ln(1/0.4167)} = \frac{1}{\ln 2.4} = 1.142\ \text{h} \approx 68.5\ \text{min} \\ \theta_m &= \frac{24}{1 - 0.4167} = 41.14\ ^\circ\text{C} \end{aligned}$$

Step 2: losses in terms of full-load copper loss $P_c$

At maximum efficiency, copper loss = iron loss:

$$\begin{aligned} P_i &= (0.80)^2 P_c = 0.6400\,P_c \\ \text{Full-load losses} &= P_i + P_c = 1.6400\,P_c \end{aligned}$$

The final rise is $\theta_m = \dfrac{\text{losses}}{S\lambda}$, so the losses allowed $\propto \lambda\,\theta_m$. At a new load of $x$ times full load, the losses are $P_i + x^2 P_c$ (iron loss is constant).

(i) Same final temperature rise ($41.14$ °C)

Losses allowed rise by 15 %:

$$\begin{aligned} 0.6400 + x^2 &= 1.15 \times 1.6400 = 1.8860 \\ x^2 &= 1.2460 \;\Rightarrow\; x = 1.1162 \\ \text{New rating} &= 1.1162 \times 400 = 446.5\ \text{kVA} \end{aligned}$$

(ii) Allowable temperature rise 50 °C

Losses allowed are now in the ratio $1.15 \times \dfrac{50}{41.14}$:

$$\begin{aligned} 0.6400 + x^2 &= 1.15 \times \frac{50}{41.14} \times 1.6400 = 2.2920 \\ x^2 &= 1.6520 \;\Rightarrow\; x = 1.2853 \\ \text{New rating} &= 1.2853 \times 400 = 514.1\ \text{kVA} \end{aligned}$$

Answer: time constant ≈ 1.14 h (68.5 min); final rise ≈ 41.1 °C; new rating (i) ≈ 446 kVA, (ii) ≈ 514 kVA.

Given

• 150 kVA; temperature rise from cold: 25 °C after 1 h, 37.5 °C after 2 h
• Fan improves heat dissipation by 20 %; maximum efficiency at 80 % of full load

Step 1: time constant and final rise

With $a = e^{-1/\tau}$:

$$\begin{aligned} 1 + a &= \frac{37.5}{25} = 1.5 \;\Rightarrow\; a = 0.5 \\ \tau &= \frac{1}{\ln 2} = 1.443\ \text{h} \approx 86.6\ \text{min} \\ \theta_m &= \frac{25}{1 - 0.5} = 50\ ^\circ\text{C} \end{aligned}$$

Step 2: losses

At maximum efficiency (80 % load), iron loss = copper loss at that load:

$$\begin{aligned} P_i &= 0.8^2 P_c = 0.64\,P_c \\ \text{Full-load losses} &= 0.64P_c + P_c = 1.64\,P_c \end{aligned}$$

Step 3: new rating for the same final rise

Since $\theta_m = \dfrac{\text{losses}}{S\lambda}$, a 20 % higher $\lambda$ allows 20 % more losses for the same $\theta_m$. At $x$ times the old full load:

$$\begin{aligned} 0.64 + x^2 &= 1.2 \times 1.64 = 1.968 \\ x^2 &= 1.328 \;\Rightarrow\; x = 1.1524 \\ \text{New rating} &= 1.1524 \times 150 = 172.9\ \text{kVA} \end{aligned}$$

Answer: heating time constant = 1.443 h (86.6 min); final steady rise = 50 °C; new rating with fan ≈ 172.9 kVA.

Given

• Heat run at quarter load: rise 30 °C after 2 h, 40 °C after 3 h (from cold)
• Maximum efficiency at 70 % of full load
• Find the final steady rise at full load, and the time on 30 % overload (1.3 × full load) to reach that rise

Step 1: time constant and final rise at quarter load

With $a = e^{-1/\tau}$:

$$\begin{aligned} 30 &= \theta_q(1 - a^2), \qquad 40 = \theta_q(1 - a^3) \\ \frac{1 - a^3}{1 - a^2} &= \frac{1 + a + a^2}{1 + a} = \frac{40}{30} \\ 3a^2 - a - 1 &= 0 \;\Rightarrow\; a = \frac{1 + \sqrt{13}}{6} = 0.7676 \\ \tau &= \frac{1}{\ln(1/0.7676)} = 3.781\ \text{h} \\ \theta_q &= \frac{30}{1 - 0.7676^2} = 73.03\ ^\circ\text{C} \end{aligned}$$

Step 2: losses

Maximum efficiency at 70 % load gives $P_i = 0.7^2 P_c = 0.49\,P_c$ ($P_c$ = full-load copper loss).

Step 3: final rise at full load

The final rise is proportional to losses (same cooling):

$$\theta_{FL} = 73.03 \times \frac{1.49}{0.5525} = 196.9\ ^\circ\text{C}$$

Step 4: time on 30 % overload

Final rise on overload: $\theta_{OL} = 73.03 \times \dfrac{2.18}{0.5525} = 288.1$ °C. Time (from cold, same $\tau$) to reach 196.9 °C:

$$\begin{aligned} 196.9 &= 288.1\left(1 - e^{-t/3.781}\right) \\ e^{-t/3.781} &= 1 - 0.6835 = 0.3165 \\ t &= 3.781 \times \ln\frac{1}{0.3165} = 4.35\ \text{h} \approx 261\ \text{min} \end{aligned}$$

Answer: heating time constant = 3.78 h; final steady rise at full load ≈ 197 °C; on 30 % overload the same rise is reached in ≈ 4.35 h (about 4 h 21 min).

(The data give an unusually high full-load rise. This follows directly from the figures given; the method is what matters.)