CONCLUSIONES Y RECOMENDACIONES

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 build-ing, and the furnace heating or air conditioner cooling. From this model we would like to answer the following three questions:

(a) How long does it take to change the building temperature substantially?

(b) How does the building temperature vary during spring and fall when there is no furnace heating or air conditioning?

(c) How does the building temperature vary in summer when there is air conditioning or in the winter when there is furnace heating?

A natural approach to modeling the temperature inside a building is to use compartmental analysis. Let represent the temperature inside the building at time tand view the building as a single compartment. Then the rate of change in the temperature is determined by all the factors that generate or dissipate heat.

We will consider three main factors that affect the temperature inside the building. First is the heat produced by people, lights, and machines inside the building. This causes a rate of increase in temperature that we will denote by HAtB. Second is the heating (or cooling) supplied

TAtB

by the furnace (or air conditioner). This rate of increase (or decrease) in temperature will be represented by . In general, the additional heating rate and the furnace (or air condi-tioner) rate are described in terms of energy per unit time (such as British thermal units per hour). However, by multiplying by the heat capacity of the building (in units of degrees temperature change per unit heat energy), we can express the two quantities and in terms of temperature per unit time.

The third factor is the effect of the outside temperature on the temperature inside the building. Experimental evidence has shown that this factor can be modeled using Newton’s law of cooling. This law states that the rate of change in the temperature is proportional to the difference between the outside temperature and the inside temperature . That is, the rate of change in the building temperature due to is

The positive constant Kdepends on the physical properties of the building, such as the number of doors and windows and the type of insulation, but Kdoes not depend on M,T, or t. Hence, when the outside temperature is greater than the inside temperature, 0 and there is an increase in the building temperature due to . On the other hand, when the outside temperature is less than the inside temperature, then 0 and the building temper-ature decreases.

Summarizing, we find (1)

where the additional heating rate is always nonnegative and is positive for furnace heating and negative for air conditioner cooling. A more detailed model of the temperature dynamics of the building could involve more variables to represent different temperatures in different rooms or zones. Such an approach would use compartmental analysis, with the rooms as different compartments (see Problems 35–37, Exercises 5.2).

Because equation (1) is linear, it can be solved using the method discussed in Section 2.3.

Rewriting (1) in the standard form (2)

where (3)

we find that the integrating factor is

To solve (2), multiply each side by and integrate:

e^KtTAtB ^{eKtQAtBdtC .}

e^KtdT

dt^AtBKe^KtTAtBe^KtQAtB , e^Kt m_AtBexpa^{K dtb eKt .}

QAtBJKMAtBHAtBUAtB , PAtBJK ,

dT

dt ^AtBPAtBTAtBQAtB ,

UAtB HAtB

dT

dt K3^MAtBTAtB4 HAtBUAtB ,

TAtB MAtB

MAtB MAtB TAtB K3^MAtBTAtB4 ^.

MAtB TAtB

MAtB TAtB

MAtB

UAtB HAtB

UAtUB AtB HAtB

102 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Solving for gives (4)

Suppose at the end of the day (at time t₀), when people leave the building, the outside tempera-ture stays constant at M₀, the additional heating rate H inside the building is zero, and the furnace/air conditioner rate Uis zero. Determine , given the initial condition . With MM₀,H 0, and U0, equation (4) becomes

Setting tt₀and using the initial value T₀of the temperature, we find that the constant Cis . Hence,

(5) _◆

When M₀T₀, the solution in (5) decreases exponentially from the initial temperature T₀ to the final temperature M₀. To determine a measure of the time it takes for the temperature to change “substantially,” consider the simple linear equation whose solutions

have the form Now, as the function either decays

exponen-tially or grows exponentially . In either case the time it takes for to

change from to is just because

The quantity which is independent of , is called the time constantfor the equation.

For linear equations of the more general form we again refer to as the time constant.

Returning to Example 1, we see that the temperature satisfies the equations

for M₀a constant. In either case, the time constant is just , which represents the time it takes for the temperature difference to change from to .We also call the time constant for the building(without heating or air conditioning). A typical value for the time constant of a building is 2 to 4 hr, but the time constant can be much shorter if windows are open or if there is a fan circulating air. Or it can be much longer if the building is well insulated.

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 .An answer to question (b) about the temperature inside the building during spring and fall is given in the next example.

1

/

^K

1

/

^K

AT₀M₀B

/

^e

T₀M₀ TM₀

1

/

^K

dT

dt ^AtB KTAtBKM₀ , dATM₀B

dt ^AtB K3^TAtBM₀4 ^, TAtB

1

/

^0a0

dA

/

^{dt aAgAtB,}

AA0B 1

/

^0a0,

Aa_a¹b AA0Be^aA1/^aB AA0B e .

1

/

^a

AA0B

/

^e

A

0.368 AA0B

B

AA0B Aa 6 0B AAtB

Aa 7 0BAAtBAA0Be^at. tSq, AAtB

dA

/

^{dt aA,}

TAtBM₀ ^AT₀M₀Be^KAtt0B . AT₀M₀BexpAKt₀B

M₀Ce^Kt .

TAtBe^Kt c^{eKtKM0 dtCd eKt3M0eKtC4}

TAt₀BT₀ TAtB

e^Kte

^{eKt3KMAtBHAtBUAtB4dtCf .}

TAtBe^Kt

^{eKtQAtBdtCeKt}

TAtB

Section 3.3 Heating and Cooling of Buildings 103

Example 1

Solution

Find the building temperature if the additional heating rate is equal to the constant H₀, there is no heating or cooling and the outside temperature Mvaries as a sine wave over a 24-hr period, with its minimum at t0 (midnight) and its maximum at t12 (noon). That is,

where B is a positive constant,M₀ is the average outside temperature, and

radians/hr. (This could be the situation during the spring or fall when there is neither furnace heating nor air conditioning.)

The function in (3) is now

Setting we can rewrite Qas

(6)

where KB₀represents the daily average value of ; that is,

When the forcing function in (6) is substituted into the expression for the temperature in equation (4), the result (after using integration by parts) is

(7)

where

The constant Cis chosen so that at midnight , the value of the temperature Tis equal to some initial temperature T₀. Thus,

◆

Notice that the third term in solution (7) involving the constant Ctends to zero exponentially.

The constant term B₀in (7) is equal to and represents the daily average temperature inside the building (neglecting the exponential term). When there is no additional heating rate inside the building , this average temperature is equal to the average outside temperature M₀. The term in (7) represents the sinusoidal variation of temperature inside the building responding to the outside temperature variation. Since can be written in the form

(8)

where tan (see Problem 16), the sinusoidal variation inside the building lags behind the outside variation by hours. Further, the magnitude of the variation inside the building is slightly less, by a factor of ^f

/

^v3^{1 A}v

/

^KB24¹

/

²_,than the outside variation. The angular frequency

fv

/

^K

FAtB 3^{1 A}v

/

^KB24^{1/2 cosAvt}f_B , FAtB BFAtABH₀0B

M₀H₀

/

^K

CT₀B₀BFA0BT₀B₀ B 1 ^Av

/

^{KB2 .}

At0B FAtBJ cos vt ^Av

/

^{KBsin vt}

1 ^Av

/

^{KB2 .}

B₀BFAtBCe^Kt ,

TAtBe^Kt c^{eKtAKB0KB cos vtBdtCd}

QAtB KB₀ 1

240²⁴

QAtBdt .

QAtB QAtBKAB₀B cos vtB ,

B₀JM₀H₀

/

^K,

QAtBKAM₀B cos vtBH₀ . QAtB

p

/

¹²

v2p

/

²⁴ MAtBM₀B cos vt ,

A

^UTAAttBB0

B

^{, HAtB}

104 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

Example 2

Solution

of variation radians/hr (which is about ). Typical values for the dimensionless ratio lie between 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. Figure 3.6 shows the 24-hr sinusoidal variation of the outside tempera-ture for a typical moderate day as well as the temperatempera-ture variations inside the building for a dimensionless ratio of unity, which corresponds to a time constant of approximately 4 hr. In sketching the latter curve, we have assumed that the exponential term has died out.

Suppose, in the building of 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 tempera-ture is below the desired temperatempera-ture, the furnace supplies heating; otherwise, it is turned off.

If the actual temperature is above the desired temperature, the air conditioner supplies cooling;

otherwise, it is off. (In practice, there is some dead zone around the desired temperature in which the temperature difference is not sufficient to activate the thermostat, but that is to be ignored here.) Assuming that the amount of heating or cooling supplied is proportional to the difference in temperature—that is,

where KUis the (positive) proportionality constant—find .

If the proportional control is substituted directly into the differential equation (1) for the building temperature, we get

(9)

A comparison of equation (9) with the first-order linear differential equation (2) shows that for this example the quantity Pis equal to KKUand the quantity representing the forcing function includes the desired temperature TD. That is,

QAtBKMAtBHAtBKUTD . PKKU ,

QAtB dTAtB

dt K3^MAtBTAtB4 HAtBKU3^TDTAtB4 ^. UAtB

TAtB UAtBKU3^TDTAtB4 ^,

1

/

^K

v

/

^K

1

/

²

v

/

^K

1

/

⁴

v is 2p

/

²⁴

Section 3.3 Heating and Cooling of Buildings 105

Example 3

Solution

90°

80°

70°

60°

50° 0 Midnight

6 12 Noon

18 24 Midnight Outside

Inside

T, °F

Figure 3.6 Temperature variation inside and outside an unheated building

When the additional heating rate is a constant H₀ and the outside temperature M varies as a sine wave over a 24-hr period in the same way as it did in Example 2, the forcing function is

The function has a constant term and a cosine term just as in equation (6). This equiva-lence becomes more apparent after substituting

(10) where

(11)

The expressions for the constant Pand the forcing function of equation (10) are the same as the expressions in Example 2, except that the constants K, B₀, and B are replaced, respectively, by the constants K₁,B₂, and B₁. Hence, the solution to the differential equation (9) will be the same as the temperature solution in Example 2, except that the three constant terms are changed. Thus,

(12) where

The constant Cis chosen so that at time t0 the value of the temperature Tequals T₀. Thus,

◆

In the above example, the time constant for equation (9) is , where K₁ KKU. Here is referred to as the time constant for the building with heating and air conditioning. For a typical heating and cooling system,KUis somewhat less than 2; for a typical building, the constant Kis between and . Hence, the sum gives a value for K₁of about 2, and the time constant for the building with heating and air conditioning is about hr.

When the heating or cooling is turned on, it takes about 30 min for the exponential term in (12) to die off. If we neglect this exponential term, the average temperature inside the building is B₂. Since K₁is much larger than Kand H₀is small, it follows from (11) that B₂is roughly TD, the desired temperature. In other words, after a certain period of time, the temperature inside the building is roughly TD, with a small sinusoidal variation. (The outside average M₀ and inside heating rate H₀have only a small effect.) 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. These observa-tions provide an answer to question (c), regarding the temperature inside the building during summer and winter, that was posed at the beginning of this section.

1

/

²

1

/

⁴

1

/

²

1

/

^K1

1

/

^P1

/

^K1

CT₀B₂B₁F₁A0B . F₁AtBJ

cos vt ^Av

/

^{K1Bsin vt}

1 ^Av

/

^K1B² . TAtBB₂B₁F₁AtBC expAK₁tB ,

QAtB B₂J KUTDKM₀H₀

K₁ , B₁J BK

K₁ . vJ 2p

24 p

12 , K₁JKKU , QAtBK₁AB₂B₁ cos vtB ,

QAtB

QAtBKAM₀B cos vtBH₀KUTD .

106 Chapter 3 Mathematical Models and Numerical Methods Involving First-Order Equations

The assumption made in Example 3 that the amount of heating or cooling is may not always be suitable. We have used it here and in the exercises to illustrate the use of the time constant. More adventuresome readers may want to experiment with other models for , especially if they have available the numerical techniques discussed in Sections 3.6 and 3.7. Group Project F, page 148, addresses temperature regulation with fixed-rate controllers.

UAtB

K_U3^TDTAtB4 ^UAtB

Section 3.3 Heating and Cooling of Buildings 107

1. A cup of hot coffee initially at 95ºC cools to 80ºC in 5 min while sitting in a room of temperature 21ºC.

Using just Newton’s law of cooling, determine when the temperature of the coffee will be a nice 50ºC.

2. A cold beer initially at 35ºF warms up to 40ºF in 3 min while sitting in a room of temperature 70ºF.

How warm will the beer be if left out for 20 min?

3. A white wine at room temperature 70ºF is chilled in ice (32ºF). If it takes 15 min for the wine to chill to 60ºF, how long will it take for the wine to reach 56ºF?

4. A red wine is brought up from the wine cellar, which is a cool 10ºC, and left to breathe in a room of tem-perature 23ºC. If it takes 10 min for the wine to reach 15ºC, when will the temperature of the wine reach 18ºC?

5. It was noon on a cold December day in Tampa: 16ºC.

Detective Taylor arrived at the crime scene to find the sergeant leaning over the body. The sergeant said there were several suspects. If they knew the exact time of death, then they could narrow the list. Detec-tive Taylor took out a thermometer and measured the temperature of the body: 34.5ºC. He then left for lunch. Upon returning at 1:00 P.M., he found the body temperature to be 33.7ºC. When did the murder occur? [Hint:Normal body temperature is 37ºC.]

6. On a mild Saturday morning while people are work-ing inside, the furnace keeps the temperature inside the building at 21ºC. At noon the furnace is turned off, and the people go home. The temperature outside is a constant 12ºC for the rest of the afternoon. If the time constant for the building is 3 hr, when will the temperature inside the building reach 16ºC? If some windows are left open and the time constant drops to 2 hr, when will the temperature inside reach 16ºC?

7. On a hot Saturday morning while people are work-ing inside, the air conditioner keeps the temperature inside the building at 24ºC. At noon the air condi-tioner is turned off, and the people go home. The temperature outside is a constant 35ºC for the rest of

the afternoon. If the time constant for the building is 4 hr, what will be the temperature inside the building at 2:00 P.M.? At 6:00 P.M.? When will the tempera-ture inside the building reach 27ºC?

8. A garage with no heating or cooling has a time con-stant of 2 hr. If the outside temperature varies as a sine wave with a minimum of 50ºF at 2:00 A.M. and a maximum of 80ºF at 2:00 P.M., determine the times at which the building reaches its lowest temperature and its highest temperature, assuming the exponen-tial term has died off.

9. A warehouse is being built that will have neither heating nor cooling. Depending on the amount of insulation, the time constant for the building may range from 1 to 5 hr. To illustrate the effect insulation will have on the temperature inside the warehouse, assume the outside temperature varies as a sine wave, with a minimum of 16ºC at 2:00 A.M. and a maxi-mum of 32ºC at 2:00 P.M. Assuming the exponential term (which involves the initial temperature T₀) has died off, what is the lowest temperature inside the building if the time constant is 1 hr? If it is 5 hr?

What is the highest temperature inside the building if the time constant is 1 hr? If it is 5 hr?

10. Early Monday morning, the temperature in the lec-ture hall has fallen to 40ºF, the same as the tempera-ture outside. At 7:00 A.M., the janitor turns on the furnace with the thermostat set at 70ºF. The time constant for the building is 2 hr and that for the building along with its heating system is hr. Assuming that the outside temperature remains constant, what will be the temperature inside the lecture hall at 8:00 A.M.? When will the temperature inside the hall reach 65ºF?

11. During the summer the temperature inside a van reaches 55ºC, while that outside is a constant 35ºC.

When the driver gets into the van, she turns on the air conditioner with the thermostat set at 16ºC. If the time constant for the van is 1

/

^K2 hr and that for 1

/

²

1

/

^K1

1

/

^K

In document Estudio sobre las familias migrantes y la incidencia en las relaciones escolares y familiares de los hijos, realizado en el sexto y séptimo año de educación general básica, de la escuela dos de agosto de la parroquia de ciano cantón Puyango provincia de L (página 150-154)