210 Higher Engineering Mathematics
(a) The width of each interval is
12.0
6
cm. Using
Simpson’s rule,
area =
1
3 (2.0)[(3.6 + 1.6) + 4(4.0
+ 2.9 + 1.7) + 2(3.5 + 2.2)]
=
2
3 [5.2 + 34.4 + 11.4]
= 34 cm
2
(b) Mean height of ordinates
=
area of diagram
length of base
=
34
12
= 2.83 cm
Since 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 84 Further problems on mean or
average values of waveforms
1. Determine the mean value of the periodic
waveforms shown in Fig. 19.13 over a half
cycle.
[(a) 2 A (b) 50 V (c) 2.5 A]
0
10
20
22
2
Current (A)
t (ms)
0
5
1 0
2100
100
Voltage (V)
t (ms)
(a)
(b)
Figure 19.13
0
1 5
3 0
25
5
Current (A)
t (ms)
(c)
Figure 19.13 (Continued )
2. Find the average value of the periodic waveforms shown in Fig. 19.14 over one complete
cycle.
[(a) 2.5 V (b) 3 A]
(b)
0
5
Current (A)
(a)
0
6
8
10
2
4
10
Voltage (mV)
t (ms)
6
8
10
2
4
t (ms)
Figure 19.14
3. An alternating current has the following values
at equal intervals of 5 ms
Time (ms) 0 5 10 15 20 25 30
Current (A) 0 0.9 2.6 4.9 5.8 3.5 0
Plot a graph of current against time and estimate the area under the curve over the 30 ms
period using the mid-ordinate rule and determine its mean value.
[0.093 As, 3.1 A]
4. Determine, using an approximate method, the
average value of a sine wave of maximum
value 50 V for (a) a half cycle and (b) a
complete cycle.
[(a) 31.83 V (b) 0]