Irregular areas, volumes and mean values of waveforms 209
50
Graph of power/time
40
30
Power (kW)
20
10
0
1
2
3
4
5
6
Time (hours)
7.0 21.5 42.0 49.5 37.0 10.0
Figure 19.10
Area under curve = (width of interval)
× (sum of mid-ordinates)
= (1)[7.0 + 21.5 + 42.0
+ 49.5 + 37.0 + 10.0]
= 167 kWh (i.e. a measure
of electrical energy)
(b) Average value of waveform
=
area under curve
length of base
=
167 kWh
6 h
= 27.83 kW
Alternatively, average value
=
sum of mid-ordinates
number of mid-ordinates
Problem 8. Fig. 19.11 shows a sinusoidal output
voltage of a full-wave rectifier. Determine, using
the mid-ordinate rule with 6 intervals, the mean
output voltage.
10
0 308608908
1808
3608
2
2708
Voltage (V)
3
2
2
Figure 19.11
One cycle of the output voltage is completed in π 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-ordinate
15 ◦
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 V
sum of mid-ordinates =38.636 V
Mean 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. 19.12. 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.
12.0 cm
3.6
4.0
3.5
2.9
2.2
1.7
1.6
Figure 19.12
50
Graph of power/time
40
30
Power (kW)
20
10
0
1
2
3
4
5
6
Time (hours)
7.0 21.5 42.0 49.5 37.0 10.0
Figure 19.10
Area under curve = (width of interval)
× (sum of mid-ordinates)
= (1)[7.0 + 21.5 + 42.0
+ 49.5 + 37.0 + 10.0]
= 167 kWh (i.e. a measure
of electrical energy)
(b) Average value of waveform
=
area under curve
length of base
=
167 kWh
6 h
= 27.83 kW
Alternatively, average value
=
sum of mid-ordinates
number of mid-ordinates
Problem 8. Fig. 19.11 shows a sinusoidal output
voltage of a full-wave rectifier. Determine, using
the mid-ordinate rule with 6 intervals, the mean
output voltage.
10
0 308608908
1808
3608
2
2708
Voltage (V)
3
2
2
Figure 19.11
One cycle of the output voltage is completed in π 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-ordinate
15 ◦
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 V
sum of mid-ordinates =38.636 V
Mean 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. 19.12. 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.
12.0 cm
3.6
4.0
3.5
2.9
2.2
1.7
1.6
Figure 19.12