## The basics of shaded-pole motors

First, let’s say few words about shaded-pole motors, and then dive into calculation procedure. A shaded pole motor is the simplest form of a single phase motor and is very low in cost.

It develops a rotating field **by delaying the build up of magnetic flux through part of the pole structure**. The shaded portion of the pole is isolated from the rest of the pole by a copper conductor that forms a single turn around it.

The magnetic flux in the unshaded portion increases with the current through its winding. Magnetic flux increases in the shaded portion.

However, it is delayed by the current induced in the copper field.

The magnetic field sweeps across the pole face from the unshaded portion to the shaded portion, developing a torque in the squirrel cage.

To maximize torque, the rotor is made with relatively high resistance.

**where low torque is acceptable (such as fans)**and are usually

**less than 1/4 HP**. Due to their very low efficiency, shaded pole motors should only be used in applications where the motor is either very small or operates for very short periods of time (e.g. shower fan motor).

### Calculation Procedure

**A four-pole shaded-pole motor (Figure 2) has the following data associated with it:**

- Nominal voltage 120 V,
- Full-load delivered power of 2.5 mhp (millihorsepower),
- Frequency 60 Hz,
- 350-mA full-load curent,
- 12-W full-load power input,
- 1525 r/min full-load speed,
- 1760 r/min no-load speed,
- 6.6 W no-load power input,
- 235 mA no-load current, and
- Stator resistance measured with DC of 30.

**Calculate the losses and efficiency at full load.**

#### Step #1 – Calculate the Rotational Losses

From the no-load conditions, consider that the rotational losses of friction and windage are equal to the power input less the stator-copper loss. The stator resistance, measured with dc, was found to be **30 Ω**.

**The increase of resistance to ac as compared with dc may vary from 10 to 30 percent**, the lower values being for small or stranded conductors and the higher values for large, solid conductors.

**Assume a value of 15 percent for this problem. The rotational losses P _{fw} equal:**

**P _{fw} =** P

_{NL}– I

_{NL}

^{2}(R

_{dc}) = 6.6 W – (235 × 10

^{-3}A)

^{2}(30 Ω)(1.15) = 6.6 W – 1.905 W =

**4.69 W**

(**R _{dc}** is dc-to-ac resistance-correction factor)

#### Step #2 – Calculate Stator-Copper Loss at Full Load

At full load the stator-copper loss P_{scu} is:

**P _{scu} =** I

_{FL}

^{2}(R

_{dc})(350 ×10

^{-3}A)2(30 Ω)(1.15) =

**4.23 W**

#### Step #3 – Calculate the Slip

The synchronous speed for a four-pole machine is obtained from the relation **n = 120 f/p**, where **f** frequency in Hz and **p** the number of poles.

Thus, **n = (120)(60)/4 = 1800 r/min**.

**1800 – 1525 = 275 r/min**, and the slip is

**(275 r/min)/(1800 r/min) = 0.153 or 15.3 percent**.

#### Step #4 – Calculate Rotor-Copper Loss at Full Load

In induction machines, the rotor-copper loss is equal to the power transferred across the air gap multiplied by the slip. The power transferred across the air gap equals the input power minus the stator-copper loss. Thus, at full load:

**P _{rcu} =** (12 W – 4.23 W) (0.153) =

**1.2 W**

#### Step #5 – Summarize the Full-Load Losses

Stator-copper loss | 4.23 W |

Rotor-copper loss | 1.2 W |

Friction and windage loss | 4.69 W |

Total losses | 10.12 W |

#### Step #6 – Calculate the Efficiency

The motor delivers **2.5 mhp**. The input is **12 W** or (12 W)(1 hp/746 W) = **16.1 mhp**.

Therefore, the efficiency is:

**η =** (output)(100 percent) / input = (2.5 mhp)(100 percent) / 16.1 mhp = **15.5 percent**

Alternatively, the developed mechanical power is equal to the power transferred across the air gap multiplied by:

(1 – s), or (12 W – 4.23 W)(1 – 0.153) = **6.58 W**.

From this must be subtracted the friction and windage losses:

6.58 – W 4.69 W = 1.89 W, or (1.89 W)(1 hp/746 W) = **2 .53 mhp**.

**The efficiency is:**

**η =**(input – losses)(100 percent) / input = (12 W – 10.12 W)(100 percent)/12 W

**η =**

**15.7 percent**.

**References:**

- Electric Motors – Energy Efficiency Reference Guide by CEA Technologies Inc.
- Handbook Of Electric Power Calculations by H. Wayne Beaty

in induction motor FLUKE 438-II claim to measure efficiency by just hooking meter on i/p side of motor and calculate efficiency without knowing no load losses and resistance of motor winding, how is possible, can anybody know any formula.

If you accurately measure the current of the motor with it’s normal load to find the

impedance. calculate the capacitance and parallel it with the motor. Non polarized 159V AC

caps work very well. My squirrel cage motor was near 250 ohms which was calculated

at 12.5uF. The current dropped from 500 ma to 320 ma. A parallel resonant circuit.

The motor is on it’s 4th year in a wood stove. I used 2 27 uf caps in series/ Do not use

back to back electrolytics, Mylar or polypropylene.

Thanks a lot to the help to understand the reasons of the so little efficiency of these motors.

Thanks a lot for these nice notes

Thanks for sharing this notes