Math 308, Sections 301, 302, Summer 2008 Lecture 5. 06/6/2008

Transcription

1 Math 308, Sections 301, 302, Summer 2008 Lecture 5. 06/6/2008

2 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?

3 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.

4 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. Summarizing, we have dt = 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 µ(t) = exp Kdt = e Kt, { } T(t) = e Kt e Kt Q(t)dt + C = e Kt { } e Kt [KM(t) + U(t) + H(t)]dt + C.

5 Example 1. Suppose at the end of the day (at time t 0 ), when people leave the building, the outside temperature stays constant at M 0, 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(t 0 ) = T 0. The solution to this problem is T(t) = M 0 + (T 0 M 0 )e Kt 0 e Kt = M 0 + (T 0 M 0 )e K(t t 0). When M 0 < T 0, the solution T(t) = M 0 + (T 0 M 0 )e K(t t 0) decreases exponentially from the initial temperature T 0 to the final temperature M 0. 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.

6 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) = H 0, where H 0 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) = M 0 B cos ωt, where B is a positive constant, M 0 is a average outside temperature, and ω = 2π/24 = π/12 radians/hr. The solution to this problem is T(t) = B 0 +Ce Kt cos ωt + (ω/k)sin ωt B 1 + (ω/k) 2 = B 0 +Ce Kt BF(t), where F(t) = cos ωt+(ω/k) sin ωt 1+(ω/K) 2.

7 The constant C is chosen so that at midnight (t = 0), the value of the temperature T is equal to some initial temperature T 0. Thus, C = T 0 B 0 + BF(0) = T 0 B 0 + B 1 + (ω/k) 2. Notice, that the term Ce Kt tents to zero exponentially. We may assume that exponential term Ce 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.

8 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 T D. 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) = K U [T D T(t)], where K U is a positive proportionally constant. Find T(t). The solution to this problem is { } T(t) = e K 1t e K1t Q(t)dt + C = B 2 B 1 F 1 (t) + Ce K1t, F 1 (t) = cos ωt + (ω/k 1)sinωt 1 + (ω/k 1 ) 2.

9 The constant C is chosen so that at time t = 0, the value of the temperature T is equal to T 0. C = T 0 B 2 + B 1 F(0) = T 0 B 2 + B 1 1+(ω/K 1 ) 2. The constant 1/K 1, where K 1 = K + K U, is a time constant for the building with heating and air conditioning. For a typical heating an cooling system, K U 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 K 1 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 Ce K 1t to die off. If we neglect the exponential term, the average temperature inside the building is B 2. Since K 1 is much larger than K and H 0 is small, B 2 is roughly T D. In other words, after certain period of time, the temperature inside of the building is roughly T D 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.

10 Example 4. A red wine is brought up from the wine cellar, which is a cool 10 0 C, and left to breathe in a room of temperature of 23 0 C. If it takes 10 min for the wine to reach 15 0 C, when will the temperature of wine reach 18 0 C? Example 5. On a mild Sunday morning while people are working inside, the furnace keeps the temperature inside the building at 21 0 C. At noon the furnace is turned off and the people go home. The temperature outside is a constant 12 0 C for the whole afternoon. If the time constant for the building is 3 hr, when will the temperature inside the building reach 16 0 C? If some windows are left open and the time constant drops to 2 hr, when will the temperature inside reach 16 0 C?

11 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.

12 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 dt = F(t,v,x), 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 m dv = ma = F(t,v,x), 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 in v(t). m dv dt = F(t,v)

13 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/sec 2 = 9.81 m/sec 2

14 Example 6. An object of mass m is given an initial downward velocity v 0 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 x(t) = mg b t + m b ( v 0 mg b ) (1 e b m t ). The value mg/b is a horizontal asymptote for v(t) is called the bf limiting, or terminal, velocity.

15 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.

16 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 b 1 = 20 N-sec/m when the chute is closed and b 2 = 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?