36 Higher Engineering Mathematics
0
2
4
6
5.71
8
0.5
0.555
i (A)
1.0 1.5 t(s)
i 5 8.0 (1 2 e^2 t/CR)
Figure 4.6
Problem 20. The temperatureθ 2 of a winding
which is being heated electrically at timetis given
by:θ 2 =θ 1 ( 1 −e
−t
τ)whereθ 1 is the temperature (in
degrees Celsius) at timet=0andτis a constant.
Calculate,
(a) θ 1 , correct to the nearest degree, whenθ 2 is
50 ◦C,tis 30s andτis 60s
(b) the timet, correct to 1 decimal place, forθ 2 to
be half the value ofθ 1.
(a) Transposing the formula to makeθ 1 the subject
gives:
θ 1 =
θ 2
( 1 −e
−t
T)
=
50
1 −e
− 30
60
=
50
1 −e−^0.^5
=
50
0. 393469 ...
i.e. θ 1 = 127 ◦C, correct to the nearest degree.
(b) Transposing to maketthe subject of the formula
gives:
θ 2
θ 1
= 1 −e
−t
τ
from which, e
−t
τ = 1 −
θ 2
θ 1
Hence −
t
τ
=ln
(
1 −
θ 2
θ 1
)
i.e. t=−τln
(
1 −
θ 2
θ 1
)
Since θ 2 =
1
2
θ 1
t=−60ln
(
1 −
1
2
)
=−60ln0. 5 = 41 .59s
Hence the time for the temperatureθ 2 to be
one half of the value ofθ 1 is 41.6s, correct to 1
decimal place.
Now try the following exercise
Exercise 18 Further problems on the laws
of growth and decay
- The temperature,T◦C, of a cooling object
varies with time,tminutes, according to the
equation:T=150e−^0.^04 t. Determine the tem-
perature when (a)t=0, (b)t=10 minutes.
[(a) 150◦C (b) 100. 5 ◦C] - The pressureppascals at heighth metres
above ground level is given by p=p 0 e
−h
C,
where p 0 is the pressure at ground level
andCis a constant. Find pressurepwhen
p 0 = 1. 012 × 105 Pa, heighth=1420m, and
C=71500. [99210]
- The voltage drop,vvolts, across an induc-
tor L henrys at time t seconds is given
by v=200e
−Rt
L ,whereR= 150 and
L= 12. 5 × 10 −^3 H. Determine (a) the voltage
whent= 160 × 10 −^6 s, and (b) thetime for the
voltage to reach 85V.
[(a) 29.32volts (b) 71. 31 × 10 −^6 s]
- The lengthlmetres of a metal bar at tem-
peraturet◦Cisgivenbyl=l 0 eαt,where
l 0 and αare constants. Determine (a) the
value of αwhenl= 1 .993m,l 0 = 1 .894m
andt= 250 ◦C, and (b) the value ofl 0 when
l= 2 .416,t= 310 ◦Candα= 1. 682 × 10 −^4.
[(a) 2. 038 × 10 −^4 (b) 2.293m] - The temperatureθ 2 ◦C of an electrical conduc-
tor at timetseconds is given by:
θ 2 =θ 1 ( 1 −e−t/T),whereθ 1 is the initial
temperature andT seconds is a constant.
Determine:
(a) θ 2 when θ 1 = 159. 9 ◦C,t=30s and
T=80s, and
(b) the timetforθ 2 to fall to half the value
ofθ 1 ifTremains at 80s.
[(a) 50◦C (b) 55.45s ] - A belt is in contact with a pulley for a
sector ofθ= 1 .12radians and the coefficient