222 GRAPHS
Alternatively, average value=sum of mid-ordinates
number of mid-ordinatesProblem 8. Fig. 20.11 shows a sinusoidal out-
put voltage of a full-wave rectifier. Determine,
using the mid-ordinate rule with 6 intervals, the
mean output voltage.Figure 20.11One cycle of the output voltage is completed inπ
radians or 180◦. The base is divided into 6 intervals,
each of width 30◦. The mid-ordinate of each interval
will lie at 15◦,45◦,75◦, etc.
At 15◦ the height of the mid-ordinate is
10 sin 15◦= 2 .588 V.
At 45◦ the height of the mid-ordinate is
10 sin 45◦= 7 .071 V, and so on.
The results are tabulated below:
Mid-ordinate Height of mid-ordinate15 ◦ 10 sin 15◦= 2 .588 V
45 ◦ 10 sin 45◦= 7 .071 V
75 ◦ 10 sin 75◦= 9 .659 V
105 ◦ 10 sin 105◦= 9 .659 V
135 ◦ 10 sin 135◦= 7 .071 V
165 ◦ 10 sin 165◦= 2 .588 Vsum of mid-ordinates=38.636 VMean or average value of output voltage
=sum of mid-ordinates
number of mid-ordinates=38. 636
6
=6.439 V(With a larger number of intervals a more accurate
answer may be obtained.) For a sine wave the actual
mean value is 0.637×maximum value, which in this
problem gives 6.37 V.Problem 9. An indicator diagram for a steam
engine is shown in Fig. 20.12. The base line has
been divided into 6 equally spaced intervals and
the lengths of the 7 ordinates measured with the
results shown in centimetres. Determine (a) the
area of the indicator diagram using Simpson’s
rule, and (b) the mean pressure in the cylinder
given that 1 cm represents 100 kPa.Figure 20.12(a) The width of each interval is12. 0
6cm. Using
Simpson’s rule,area=^13 (2.0)[(3. 6 + 1 .6)+4(4. 0+ 2. 9 + 1 .7)+2(3. 5 + 2 .2)]=^23 [5. 2 + 34. 4 + 11 .4]=34 cm^2
(b) Mean height of ordinates=area of diagram
length of base=34
12= 2 .83 cmSince 1 cm represents 100 kPa, the mean pressure
in the cylinder
= 2 .83 cm×100 kPa/cm=283 kPa.Now try the following exercise.Exercise 92 Further problems on mean or
average values of waveforms- Determine the mean value of the periodic
waveforms shown in Fig. 20.13 over a half
cycle. [(a) 2 A (b) 50 V (c) 2.5 A]