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 2951 4523 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

0 123

Time (hours)

456

7.0 21.5 42.0 49.5 37.0 10.0

Figure 28.10

(a) Thetimebaseis 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]

=167kWh(i.e. a measure of electrical energy)

(b) Average value of waveform=

area under curve

length of base

=

167kWh

6h

=27.83kW

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

(^0308608908180827083608)
2 
10
Voltage (V)
3 
2

2


Figure 28.11
Onecycleoftheoutputvoltageiscompletedinπradians
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
10sin 15◦= 2 .588V,
At 45◦the height of the mid-ordinate is
10sin 45◦= 7 .071V, and so on.
The results are tabulated below.
Mid-ordinate Height of mid-ordinate
15 ◦ 10sin15◦= 2 .588V
45 ◦ 10sin45◦= 7 .071V
75 ◦ 10sin75◦= 9 .659V
105 ◦ 10sin105◦= 9 .659V
135 ◦ 10sin135◦= 7 .071V
165 ◦ 10sin165◦= 2 .588V
Sum of mid-ordinates= 38 .636V
Mean or average value of output voltage

sum of mid-ordinates
number of mid-ordinates

38. 636
    6
    =6.439V
    (With a larger number of intervals a more accurate
    answer may be obtained.)
    For a sine wave the actual mean value is


39. 637 ×maximum value, which in this problem
    gives 6.37V.
    Problem 9. An indicator diagram for a steam
    engine is shown in Figure 28.12. The base line has