Math 308, Sections 301, 302,
Summer 2008
Lecture 5.
06/6/2008
Chapter 3. Mathematical methods and numerical methods
involving first order equations.
Section 3.3 Heating and cooling of buildings.
Our goal is to formulate a mathematical model that describes the
24-hr temperature profile inside a building as a function of the
outside temperature, the heat generated inside the building, and
the furnace heating or air conditioner cooling. From this model
we’d like to answer three following questions:
◮
How long does it take to change the building temperature
substantially?
◮
How does the building temperature vary during spring and fall
when there is no furnace heating or air conditioning?
◮
How does the building temperature vary in summer when
there is air conditioning or in the winter when there is furnace
heating?
Let T (t) represent the temperature inside the building at time t
and view the building as a single compartment.
We will consider three main factors that affect the temperature
inside the building.
◮
heat produced by people, lights, and machines inside the
building. This causes a rate of increase in temperature that
we will denote by H(t).
◮
the heating (or cooling) supplied by the furnace (or air
conditioner). This rate of increase (or decrease) in
temperature will be represented by U(t).
◮
the effect of the outside temperature M(t) on the
temperature inside the building.
Third factor can be modeled using Newtons law of cooling
dT
dt = K (M(t) − T (t)) .
The positive constant K depends on the physical properties of the
building, K does not depend on M, T or t.
dT
Summarizing, we have
= K (M(t) − T (t)) + U(t) + H(t),
dt
where H(t) ≥ 0 and U(t) > 0 for furnace heating and U(t) < 0
for air conditioning cooling.
Equation in the standard form dT
dt + P(t)T (t) = Q(t) , where
P(t) = K , Q(t)
= KM(t) + U(t) + H(t), integrating factor is
R
µ(t) = exp Kdt = eKt ,
Z
−Kt
Kt
T (t) = e
e Q(t)dt + C =
e
−Kt
Z
Kt
e [KM(t) + U(t) + H(t)]dt + C
.
Example 1. Suppose at the end of the day (at time t0 ), when
people leave the building, the outside temperature stays constant
at M0 , the additional heating rate H inside of the building is zero,
and the furnace/air conditioning rate U is also zero. Determine
T (t), given the initial condition T (t0 ) = T0 .
The solution to this problem is
T (t) = M0 + (T0 − M0 )eKt0 e−Kt = M0 + (T0 − M0 )e−K (t−t0 ) .
When M0 < T0 , the solution T (t) = M0 + (T0 − M0 )e−K (t−t0 )
decreases exponentially from the initial temperature T0 to the final
temperature M0 .
The constant 1/K is called time constant of the building
(without heating or air conditioning). A typical value for the time
constant of the building is 2 to 4 hr.
In the context of Example 1, we can use the notion of time
constant to answer our initial question (a): The building
temperature changes exponentially with a time constant of 1/K .
An answer to question (b) is given in the next example.
Example 2. Find the building temperature T (t) if the additional
heating rate H(t) = H0 , where H0 is a constant, there is no
heating or cooling, and the outside temperature varies as a sine
wave over a 24-hr period, with its minimum at t = 0 (midnight)
and its maximum at t = 12 (noon). That is,
M(t) = M0 − B cos ωt,
where B is a positive constant, M0 is a average outside
temperature, and ω = 2π/24 = π/12 radians/hr.
The solution to this problem is
T (t) = B0 +C e−Kt −B
where F (t) =
cos ωt + (ω/K ) sin ωt
= B0 +C e−Kt −BF (t),
1 + (ω/K )2
cos ωt+(ω/K ) sin ωt
.
1+(ω/K )2
The constant C is chosen so that at midnight (t = 0), the value of
the temperature T is equal to some initial temperature T0 . Thus,
C = T0 − B0 + BF (0) = T0 − B0 +
B
.
1 + (ω/K )2
Notice, that the term C e−Kt tents to zero exponentially. We may
assume that exponential term C e−Kt has died out.
Typical value for the dimensionless ratio ω/K lie between 1/2 and
1. For this range, the lag between inside and outside temperature
is approximately 1.8 to 3 hr and the magnitude of the inside
variation is between 89% and 71% of the variation outside.
Example 3. Suppose, in the building in Example 2, a simple
thermostat is installed that is used to compare the actual
temperature inside the building with a desired temperature TD . If
the actual temperature is below the desired temperature, the
furnace supplies heating; otherwise it is turned off. If the actual
temperature is above the desired temperature, the furnace supplies
cooling; otherwise it is off. Assuming that the amount of heating
or cooling supplies is proportional to the difference in
temperature–that is,
U(t) = KU [TD − T (t)],
where KU is a positive proportionally constant. Find T (t).
The solution to this problem is
Z
−K1 t
K1 t
T (t) = e
e Q(t)dt + C = B2 − B1 F1 (t) + C e−K1 t ,
F1 (t) =
cos ωt + (ω/K1 ) sin ωt
.
1 + (ω/K1 )2
The constant C is chosen so that at time t = 0, the value of the
temperature T is equal to T0 .
B1
C = T0 − B2 + B1 F (0) = T0 − B2 + 1+(ω/K
2.
1)
The constant 1/K1 , where K1 = K + KU , is a time constant for
the building with heating and air conditioning. For a typical
heating an cooling system, KU is somehow less than 2; for a
typical building, constant K is between 1/2 an 1/4. Hence, the
sum gives a value for K1 of about 2, and the time constant for the
building with heating and air conditioning is about 1/2.
When the heating or cooling is turned on, it takes about 30
minutes for the exponential term C e−K1 t to die off. If we neglect
the exponential term, the average temperature inside the building
is B2 . Since K1 is much larger than K and H0 is small, B2 is
roughly TD . In other words, after certain period of time, the
temperature inside of the building is roughly TD with a small
sinusoidal variation. Thus, to save energy, the heating or cooling
system may be left off during the night. When it is turned on in
the morning, it will take roughly 30 min for the inside of the
building to attain the desired temperature.
Example 4. A red wine is brought up from the wine cellar, which
is a cool 100 C, and left to breathe in a room of temperature of
230 C. If it takes 10 min for the wine to reach 150 C, when will the
temperature of wine reach 180 C?
Example 5. On a mild Sunday morning while people are working
inside, the furnace keeps the temperature inside the building at
210 C. At noon the furnace is turned off and the people go home.
The temperature outside is a constant 120 C for the whole
afternoon. If the time constant for the building is 3 hr, when will
the temperature inside the building reach 160 C? If some windows
are left open and the time constant drops to 2 hr, when will the
temperature inside reach 160 C?
Section 3.4 Newtonian mechanics
Mechanics is the study of the motion of objects and the effect of
forces acting on those objects. Newtonian or classical,
mechanics deals with the motion of ordinary objects – that is,
objects that are large compared to an atom and slow moving
compared with the speed of light. A model for Newtonian
mechanics can be based on Newton’s laws of motion:
1. When a body is subject to no resultant external force, it
moves with a constant velocity.
2. When a body is subject to one or more external forces, the
time rate of change of the body’s momentum is equal to the
vector sum of the external forces acting on it.
3. When one body interacts with a second body, the force of the
first body on the second is equal in magnitude, but opposite
in direction, to the force of the second body on the first.
These laws are extremely useful for studying of ordinary objects in
an inertial reference frame–that is frame in which an
undisturbed body moves with a constant velocity. We can express
Newtons second law as
dp
= F (t, v , x),
dt
where F (t, v , x) is the resultant force on the body at time t,
location x, and velocity v , and p(t) is the momentum of the body
at time t. The momentum is the product of the mass of the body
and its velocity – p(t) = mv (t). We can express second Newton’s
law as
dv
= ma = F (t, v , x),
m
dt
where a = dv
dt is the acceleration of the boy at time t.
We will focus on situations where the force F does not depend on
x. We can regard the first order equation
m
in v (t).
dv
= F (t, v )
dt
Procedure for Newtonian models
1. Determine all relevant forces acting on the object being
studied. It is helpful to draw a simple diagram of the object
that depicts these forces.
2. Choose an appropriate axis or coordinate system in which to
represent the motion of the object and the forces acting on it.
3. Apply Newton’s second law to determine the equations of
motion for the object.
g = 32 ft/sec2 = 9.81 m/sec2
Example 6. An object of mass m is given an initial downward
velocity v0 and allowed to fall under the influence of gravity.
Assuming the gravitational force is constant and the force due to
air resistance is proportional to the velocity of the object,
determine the equation of motion for this body.
Hence, the equation of the motion is
b
mg
mg m
x(t) =
v0 −
(1 − e− m t ).
t+
b
b
b
The value mg /b is a horizontal asymptote for v (t) is called the bf
limiting, or terminal, velocity.
Example 7. A 400-lb object is released from rest 500 ft above the
ground and allowed to fall under the influence of gravity. Assuming
that the force in pounds due to air resistance is −10v , where v is
the velocity of the object in ft/sec, determine the equation of
motion of the object. When will the object hit the ground?
Example 8. An object of mass 8 kg is given an upward initial
velocity of 20 m/sec and then allowed to fall under the influence of
gravity. Assume that the force in newtons due to air resistance is
−16v , where v is the velocity of the object in m/sec. Determine
the equation of motion of the object. If the object is initially 100
m above the ground, determine when the object will strike the
ground.
Example 9. A parachutist whose mass is 100 kg drops from a
helicopter hovering 3000 m above the ground and falls under the
influence of gravity. Assume that the force due to air resistance is
proportional to the velocity of the parachutist, with the
proportionality constant b1 = 20 N-sec/m when the chute is closed
and b2 = 100 N-sec/m when the chute is open. If the chute does
not open until 30 sec after the parachutist leaves the helicopter,
after how many second will he hit the ground? If the chute does
not open until 1 min after the parachutist leaves the helicopter,
after how many second will he hit the ground?