262 Basic Engineering Mathematics
Problem 7. The power used in a manufacturing
process during a 6 hour period is recorded at
intervals of 1 hour as shown below.
Time (h)
0 1 2 3 4 5 6
Power (kW) 0 14 29 51 45 23 0
Plot a graph of power against time and, by using the
mid-ordinate rule, determine (a) the area under the
curve and (b) the average value of the power
The graph of power/time is shown in Figure 28.10.
Graph of power/time
Power (kW)
50
40
30
20
10
0
1
2
3
Time (hours)
4
5
6
21.5
7.0
42.0 49.5 37.0 10.0
Figure 28.10
(a) The time base is divided into 6 equal intervals, each
of width 1 hour. Mid-ordinates are erected (shown
by broken lines in Figure 28.10) and measured.
The values are shown in Figure 28.10.
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. Figure 28.11 shows a sinusoidal
output voltage of a full-wave rectifier. Determine,
using the mid-ordinate rule with 6 intervals, the
mean output voltage
0 308608908
1808
2708
3608
2
10
Voltage (V)
3
2
2
Figure 28.11
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 Figure 28.12. The base line has
Problem 7. The power used in a manufacturing
process during a 6 hour period is recorded at
intervals of 1 hour as shown below.
Time (h)
0 1 2 3 4 5 6
Power (kW) 0 14 29 51 45 23 0
Plot a graph of power against time and, by using the
mid-ordinate rule, determine (a) the area under the
curve and (b) the average value of the power
The graph of power/time is shown in Figure 28.10.
Graph of power/time
Power (kW)
50
40
30
20
10
0
1
2
3
Time (hours)
4
5
6
21.5
7.0
42.0 49.5 37.0 10.0
Figure 28.10
(a) The time base is divided into 6 equal intervals, each
of width 1 hour. Mid-ordinates are erected (shown
by broken lines in Figure 28.10) and measured.
The values are shown in Figure 28.10.
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. Figure 28.11 shows a sinusoidal
output voltage of a full-wave rectifier. Determine,
using the mid-ordinate rule with 6 intervals, the
mean output voltage
0 308608908
1808
2708
3608
2
10
Voltage (V)
3
2
2
Figure 28.11
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 Figure 28.12. The base line has