The Circle Diagram

Fix the axes and the scale first, because every later reading depends on them. Let \(OY\) lie along the applied voltage and \(OX\) lag it by 90°. A current of magnitude \(I\) lagging by \(\phi\) is then plotted at \((I\sin\phi,\;I\cos\phi)\), and the three-phase power it carries is

Also useful later: with \(\omega_s = 2\pi(1500)/60 = 157.08\) rad/s, one ampere of vertical length is \(692.8/157.08 = 4.411\) N·m of torque.

The no-load point. Its power factor comes from the no-load reading:

The vertical coordinate is simply \(P_0/\sqrt3V_L = 1300/692.8 = 1.88\) A, which is a quicker route and a useful check.

The short-circuit point. The blocked-rotor test was run at 100 V; at standstill the machine is a linear impedance, so current scales with voltage and power with its square:

Again the vertical coordinate checks: \(43200/692.8 = 62.35\) A. Scaling the voltage does not change the power-factor angle, because the impedance is unchanged.

Locate the centre. The diameter is horizontal and passes through \(O'\), so the centre has \(y_C = y_{O'} = 1.88\). Its abscissa follows from \(|CO'| = |CA|\):

Check: \(|CA| = \sqrt{(168.86-100.25)^2+(62.35-1.88)^2} = \sqrt{68.61^2+60.47^2} = 91.45\;\checkmark\)

The construction, as a table to plot from. Both axes to the same scale, say 20 A per large division:

Points on the upper arc, from \(y = 1.88 + \sqrt{91.45^2-(x-100.25)^2}\). The motoring region runs anticlockwise from \(O'\) through the top of the circle to \(A\):

Nine points plus \(O'\) and \(B\) are ample to draw the arc by hand; a compass set to 91.45 A on \(C\) is better still.

Verify the diameter against the reactance. The load component of the short-circuit current is \(A - O'\) taken as a phasor:

Exactly the diameter obtained geometrically, as the theory promises. Note that this \(X_{01}\) is a little larger than the 1.204 Ω that \(V_{ph}/I_{sc}\) would give in Set 29, because the exciting current has been taken out of the short-circuit current first.

Drop a perpendicular from \(A\) to the horizontal axis and mark where the no-load level cuts it. Call the foot \(D\) and the intersection with the horizontal through \(O'\) the point \(E\):

\(AD\) is the whole input at standstill and full voltage; \(ED\) is the fixed loss carried over from the no-load test. What is left, \(AE\), is the copper loss in both windings.

Split \(AE\) with the stator resistance, which is the one piece of information the two tests cannot supply:

Place \(F\) on \(AD\) so that \(AF\) is the rotor copper loss and \(FE\) the stator copper loss:

Draw the two lines from \(O'\). The output line joins \(O'\) to \(A\); the torque line joins \(O'\) to \(F\):

Both pass through \(O'(8.80,\,1.88)\), as they must: at no load the output is zero and the torque is very nearly zero.

What the vertical distances mean. For any operating point \(P = (x,y)\) on the circle, measuring straight down:

Multiply any of these by 692.8 to get watts. The slip follows immediately as \(s = P_{cu2}/P_g\), the ratio of two vertical lengths measured at the same abscissa. Note that all four bands stack in that order at every abscissa, so the four powers always add to the input.

Why the torque line has to be a straight line through \(O'\). At any operating point the stator copper loss is \(3I_1^2R_1\) and the total copper loss is \(3I_1^2(R_1+R_2')\), so the two are always in the same fixed proportion — the proportion measured at standstill. A quantity proportional to the total copper loss is a vertical distance below the output line, so a fixed fraction of it is a vertical distance below a second line drawn through the same starting point. The stator resistance sets the slope and nothing else does.

Convert the required output into a length. The operating point is wherever the circle stands 25 kW above the output line:

Graphically: draw a line parallel to the output line and 36.08 A above it, and mark where it cuts the circle.

That parallel line is

Substituting into \((x-100.25)^2+(y-1.88)^2 = 91.45^2\) and collecting terms:

Two intersections, as there must be: the machine can deliver 25 kW at a small slip on the stable side of the curve, or at a large slip on the unstable side. Only the first is an operating point.

The operating point:

Read the three vertical distances at \(x = 17.76\):

Slip and speed follow from the definition of slip as the fraction of the air-gap power lost in the rotor:

Torque, from the air-gap power and the synchronous speed:

Equivalently \(P_m/\omega_m = 25000/(2\pi\times1428/60) = 167.2\) N·m. The two agree because the diagram treats friction and windage as part of the fixed loss removed on the stator side, so no separate torque is subtracted at the shaft.

Efficiency and the loss balance:

The four bands add exactly: \(1300+1088+1257+25000 = 28645\) W. One caution about the stator copper figure. The construction apportions the standstill copper loss linearly along the horizontal, which is exact for the rotor but only approximate for the stator, because the exciting current is included in \(I_1\) and not in the load component. Computing it directly, \(3I_1^2R_1 = 3(45.0)^2(0.20) = 1215\) W against the diagram's 1088 W. The 127 W difference is 0.4% of the input and 0.5% on the efficiency — the accuracy limit of the method, and the reason the circle diagram is a design and teaching tool rather than a test standard.

Where the maxima are. Torque is the vertical distance to the torque line, so it is greatest where the circle is furthest from that line — the point at which the tangent is parallel to it. The same argument applies to output and the output line. For a line of slope \(m\) the point is

and the maximum vertical distance is

Maximum torque, with the torque line \(m = 0.1753\), \(c = 0.333\) and \(\sqrt{1+m^2} = 1.0153\):

Its slip, speed, current and power factor. The output at that point is the distance to the output line:

The breakdown slip is very nearly 20% and the breakdown torque is \(338.8/167.2 = 2.03\) times full load — a typical general-purpose machine.

Maximum output, with the output line \(m = 0.3779\), \(c = -1.450\) and \(\sqrt{1+m^2} = 1.0690\):

And its slip and efficiency:

Maximum output occurs at a smaller slip than maximum torque, because past that point the extra torque no longer compensates for the falling speed.

The three operating points side by side:

Both maxima lie far outside anything the machine could sustain: 110 A and 125 A against a rated 45 A, at efficiencies in the sixties and seventies. They are transient capabilities, not ratings.

Starting conditions are the point \(A\) itself, because \(s = 1\) there. Nothing new needs constructing:

The starting torque is the segment \(AF\) found in Problem 2 — the vertical distance from \(A\) down to the torque line, which at \(x = 168.86\) is exactly \(F\):

At standstill the air-gap power is entirely rotor copper loss — the output is zero and \(A\) lies on the output line. This is why the starting torque is read to the torque line and never to the output line.

Note what the machine is offering. Four times rated current for 86% of rated torque is a poor bargain, and it is the fundamental objection to direct-on-line starting: the current is set by the standstill impedance, which is small, while the torque is set by \(I^2R_2'\), which at \(s = 1\) is only a third of the input.

Reduced-voltage starting scales the diagram, it does not redraw it. Applying a fraction \(x\) of rated voltage divides every current on the circle by \(1/x\) and every power by \(1/x^2\), because the standstill impedance is unchanged. With an autotransformer the supply current is smaller again by \(x\), since the starter transforms it:

At the 60% and 80% tappings:

The autotransformer is the only reduced-voltage method for which the supply current falls as \(x^2\) rather than \(x\) — the same \(x^2\) as the torque. That is its whole advantage over a series resistance or reactance.

The tapping for full-load torque:

Greater than one, which means it cannot be done. This machine does not produce full-load torque at standstill even on full voltage, so no reduced-voltage starter can be asked for it. Any load requiring more than 0.855 pu of torque to break away needs a slip-ring machine with rotor resistance — Set 31.

The tapping for half full-load torque, which is what a fan or a lightly loaded pump asks for:

The last row is included because a star–delta starter is exactly equivalent to a 57.7% autotransformer tapping, on a machine wound for delta running. Set 31 derives that equivalence.

Repeat the intersection of Problem 3 at each output. For an output \(P_m\) the parallel line is \(y = 0.3779x - 1.450 + P_m/692.8\), and the smaller of the two roots is the operating point. The results:

Read the shape of the curves off the table. Three features are worth naming, and all three are properties of every induction motor:

• Power factor climbs steeply from light load, because the magnetising current is fixed while the working current grows, then falls slowly once the leakage reactance begins to dominate.
• Slip is very nearly proportional to torque — 40.2, 81.3, 123.5, 167.2, 212.8 N·m against slips of 0.0103, 0.0214, 0.0337, 0.0479, 0.0652. The small departure from proportionality is the curvature of the torque–slip characteristic setting in.
• Efficiency rises quickly, flattens, and turns over well before full load.

The efficiency maximum. Efficiency is the ratio of two vertical distances, \((y - \text{output line})/y\), and searching the arc gives

Between 12.5 kW and 25 kW the efficiency never leaves the band 87.3–88.2%, which is why induction motors tolerate part loading so well and why the exact position of the peak is of little practical interest.

Maximum power factor. The power-factor angle is the angle between \(OP\) and the vertical, so it is least when \(OP\) is tangent to the circle. With \(|OC| = \sqrt{100.25^2+1.88^2} = 100.27\):

Locate the point of tangency and read what the machine is doing there. The tangent length is \(\sqrt{|OC|^2-r^2} = \sqrt{100.27^2-91.45^2} = 41.12\) A, so

The two points are 8% of rated output apart and their power factors differ in the fourth decimal place. That is not luck: a machine is designed so that its best power factor falls at or just below its rated load, because power factor is what the supply is billed for.

What the whole survey cost. Two test readings, one resistance measurement, one circle and two straight lines — and from them five complete operating points, two maxima, a starting condition and a power-factor optimum. Solving the equivalent circuit of Set 28 for the same nine points would mean nine complex divisions.