basic_engineering_mathematics0

114 Basic Engineering Mathematics
From above, ln R R 0
= αθ hence θ = 1 α ln R R 0
When R = 5.4 × 10 3 , α = 1.215477 ...× 10 −4 and R 0 = 5 × 10 3
θ =
=
(
1
5.4 × 10
3
)
1.215477 ...× 10 ln −4 5 × 10 3
10 4
1.215477 ... (7.696104 ...× 10−2 )
= 633 ◦ C correct to the nearest degree.
Problem 18. In an experiment involving Newton’s
law of cooling, the temperature θ( ◦ C) is given by
θ = θ 0 e −kt . Find the value of constant k when θ 0 = 56.6 ◦ C,
θ = 16.5 ◦ C and t = 83.0 seconds.
Transposing θ = θ 0 e −kt gives θ θ 0
= e −kt from which
θ 0
θ = 1 = e kt
e −kt
Taking Napierian logarithms of both sides gives:
ln θ 0
θ = kt
from which,
k = 1 t ln θ 0
θ = 1 ( ) 56.6
83.0 ln 16.5
= 1 (1.2326486 ...)
83.0
Hence k = 1.485 × 10 −2
Problem 19. The current i amperes flowing in a capacitor
at time t seconds is given by i = 8.0(1 − e −t/CR ), where
the circuit resistance R is 25 × 10 3 ohms and capacitance
C is 16 × 10 −6 farads. Determine (a) the current i after
0.5 seconds and (b) the time, to the nearest millisecond,
for the current to reach 6.0A. Sketch the graph of current
against time.
(a) Current i = 8.0(1 − e −t/CR )
= 8.0[1 − e 0.5/(16×10−6 )(25×10 3) ]
= 8.0(1 − e −1.25 )
= 8.0(1 − 0.2865047 ...)
= 8.0(0.7134952 ...)
= 5.71 amperes
(b) Transposing i = 8.0(1 − e −t/CR i
)gives
8.0 = 1 − e−t/CR
from which, e −t/CR = 1 −
i
8.0 = 8.0 − i
8.0
Taking the reciprocal of both sides gives:
e t/CR = 8.0
8.0 − i
Taking Napierian logarithms of both sides gives:
( )
t 8.0
CR = ln 8.0 − i
( ) 8.0
Hence t = CR ln
8.0 − i
( ) 8.0
= (16 × 10 −6 )(25 × 10 3 )ln
8.0 − 6.0
when i = 6.0 amperes,
i.e. t = 400 ( ) 8.0
10 ln = 0.4ln4.0
3 2.0
= 0.4(1.3862943 ...) = 0.5545 s
= 555 ms, to the nearest millisecond
A graph of current against time is shown in Fig. 15.6.
5.71
8
i(A)
Fig. 15.6
6
4
2
i 8.0 (1e t/CR )
0 0.5 1.0 1.5 t(s)
0.555
Problem 20. The temperature θ 2 of a winding which
is being heated electrically at time t is given by:
θ 2 = θ 1 (1 − e −t/τ ) where θ 1 is the temperature (in degrees
Celsius) at time t = 0 and τ is a constant. Calculate.
(a) θ 1 , correct to the nearest degree, when θ 2 is 50 ◦ C, t is
30 s and τ is 60 s
(b) the time t, correct to 1 decimal place, for θ 2 to be half
the value of θ 1
(a) Transposing the formula to make θ 1 the subject gives:
i.e.
θ 2
θ 1 =
(1 − e −t/τ ) = 50
1 − e −30/60
50
=
1 − e = 50
−0.5 0.393469 ...
θ 1 = 127 ◦ C, correct to the nearest degree.