Basic Engineering Mathematics, Fifth Edition

(Amelia) #1

260 Basic Engineering Mathematics


16400m^3 = 16400 × 106 cm^3.
Since 1 litre=1000cm^3 ,

capacity of reservoir=

16400 × 106
1000

litres

= 16400000 = 16. 4 × 106 litres

Now try the following Practice Exercise

PracticeExercise 111 Volumes of irregular
solids (answers on page 352)


  1. The areas of equidistantly spaced sections of
    the underwater form of a small boat are as fol-
    lows:
    1.76, 2.78, 3.10, 3.12, 2.61, 1.24 and 0.85m^2.
    Determine the underwater volume if the
    sections are 3m apart.

  2. To estimate the amount of earth to be
    removed when constructing a cutting, the
    cross-sectional area at intervals of 8m were
    estimated as follows:
    0, 2.8, 3.7, 4.5, 4.1, 2.6 and 0m^3.
    Estimate the volume of earth to be excavated.

  3. Thecircumferenceofa 12mlonglogoftimber
    of varying circular cross-section is measured
    at intervals of 2m along its length and the
    results are as follows. Estimate the volume of
    the timber in cubic metres.


Distance from
one end(m) 0 2 4 6
Circumference(m) 2.80 3.253.94 4.32

Distance from
one end(m) 8 10 12
Circumference(m) 5.16 5.82 6.36

28.3 Mean or average values of waveforms


The mean or average value,y,of the waveform shown
in Figure 28.6 is given by

y=

area under curve
length of base,b

If the mid-ordinaterule is used to find the area under the
curve, then

y=

sum of mid-ordinates
number of mid-ordinates
(
=

y 1 +y 2 +y 3 +y 4 +y 5 +y 6 +y 7
7

for Figure 28.6

)

y 1 y 2 y 3 y 4 y 5 y 6 y 7

dddd
b

ddd

y

Figure 28.6

For asine wave, the mean or average value
(a) over one complete cycle is zero (see
Figure 28.7(a)),
(b) overhalfacycleis 0. 637 ×maximum valueor
2
π

×maximum value,

(c) of a full-wave rectified waveform (see
Figure 28.7(b)) is 0. 637 ×maximum value,
(d) of a half-wave rectified waveform (see
Figure 28.7(c)) is 0. 318 ×maximum value
or

1
π

×maximum value.

V

(^0) t
Vm
V
0
(a) (b)
t
Vm
(c)
V
(^0) t
Vm
Figure 28.7

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