*Notes to a video lecture on http://www.unizor.com*

__Heat Transfer - Problems__

*Problem 1*

Determine the power of the heat source inside the room required to

maintain a certain difference between inside and outside air

temperature, given the following:

(a) the difference between inside and outside temperature

**δ=T**_{out}−T_{home}(b) the room has only one wall facing the outside air, and the area of this wall is

**A**(c) the thickness of the wall is

**L**(d) the wall is made of solid material with a coefficient of

*thermal conductivity*

*.*

**k***Solution*

Let

*be the distance from a point inside the wall to its surface facing the room.*

**x**The

*is a temperature at this point as a function from this distance.*

**T(x)**To be equal to

*at the surface facing the room and to be*

**T**_{home}*at the surface facing the outside and to be linearly changing from one value to another inside the wall of the thickness*

**T**_{out}*, the temperature inside the wall at distance*

**L***from the surface facing the room should be*

**x**

**T(x)=T**_{home}+(T_{out}−T_{home})·x/LFrom this we can determine the

*heat flux*through the wall at a distance

*, using the Fourier's law of thermal conduction:*

**x**

**Q(x,A) = −k·A·**d**T(x)/**d**x =**

= −k·A·(T

= −k·A·δ/L= −k·A·(T

_{out}−T_{home})/L == −k·A·δ/L

That is exactly how much heat we need to maintain the difference between temperature in the room and outside.

*Problem 2*

Calculate the heating requirement of the room with only one concrete

wall facing outside with no windows, assuming the following:

(a) the temperature in the room must constantly be

*T*

_{room}=20°C(b) the temperature outside is also constant

*T*

_{out}=5°C(c) the thickness of the concrete wall facing outside is

*L=0.2m*

(d) the area of the wall facing outside is

*A=12m²*

(e)

*heat conductivity*of concrete is

*k=0.6W/(m·°K)*

*Solution*

Using the above, we can determine the heat flux through the wall at a distance

*:*

**x**

**Q(x,A) = −k·A·δ/L**(for any distance

*from the surface of the wall that faces the room)*

**x**Outside surface of the wall is at 5°C, inside is at 20°C.

So,

*.*

**δ=15°**Therefore, the

*heat flux*through the wall (at any distance from the inside surface) will be

**Q = 0.6·12·15/0.2 = 540W**That is exactly how much heat we need to maintain the temperature in the room.

*Problem 3*

Our task is to determine the law of cooling of a relatively small hot

object immersed in the cool infinitely large reservoir with liquid or

gaseous substance.

We assume the volume of substance this object is immersed in to be

"infinite" to ignore its own change of temperature related to heat

emitted by our hot object.

This law of cooling should be expressed in terms of object's temperature

*as a function of time*

**T***, that is, we have to find the function*

**t***.*

**T(t)**Assumptions:

(a) the initial temperature of the object at time

*is assumed to be*

**t=0**

**T**_{0}(b) the object has a shape of a thin flat square (so, its temperature is changing simultaneously in all its volume) of size

*x*

**L***and mass*

**L**

**m**(c) the

*specific heat capacity*of the object's material is

**C**(d) the temperature of the substance surrounding our object is constant and equals

**T**_{s}(e) the

*convective heat transfer coefficient*of the substance around our object is

*.*

**h***Solution*

Consider a small time interval from

*to*

**t***Δ*

**t+***.*

**t**The temperature of an object is

*, while the temperature of the substance around it is constant*

**T(t)***.*

**T**_{s}Then the amount of heat transferred from our object to the substance

around it per unit of time per unit of its surface area is proportional

to the difference in temperatures with the

*convective heat transfer coefficient*as a factor:

*[*

**q = −h·***]*

**T(t)−T**_{s}Since total surface area of our thin flat square, ignoring its thickness, is

*and the time interval we consider is Δ*

**2L²***, the total amount of heat transferred by our object through its surface during this time period is*

**t**Δ

*[*

**Q(t) = −2L²·h·***]*

**T(t)−T**_{s}*Δ*

**·**

**t**This is amount of heat taken away by the substance from our object during a time interval Δ

*.*

**t**The same amount of heat is lost by the object, taking its temperature from

*to*

**T(t)***Δ*

**T(t+***.*

**t)**As we know, changes of heat and temperature are proportional and related to the object mass and specific heat capacity:

Δ

*Δ*

**Q(t) = C·m·**

**T**where Δ

*Δ*

**T = T(t+**

**t)−T(t)**.Equating the amount of heat lost by our object to the amount of heat

carried away by convection of the substance around it, we have come to

an equation

*[*

**−2L²·h·***]*

**T(t)−T**_{s}*Δ*

**·***[*

**t =**

= C·m·= C·m·

*Δ*

**T(t+***]*

**t)−T(t)**Dividing both parts by Δ

*, diminishing this time interval to zero and using a derivative by time*

**t***to express the limit, we get the following differential equation*

**t***[*

**−2L²·h·***]*

**T(t)−T**_{s}

**=**

= C·m·d= C·m·

**T(t)/**d**t**To solve it, let's make two simple substitutions:

(a)

**A = 2h·L²/(C·m)**(b)

**X(t) = T(t) − T**_{s}Then, since

*is constant,*

**T**_{s}*d*

**T(t)/**d**t =**d**X(t)/**d**t**and our differential equation looks like

*d*

**X(t)/**d**t = −A·X(t)**To solve this, we convert it as follows:

*d*

**X(t)/X(t) = −A·**d**t***d*[

*ln*]

**(X(t))**

**= −A·**d**t**Integrating:

*ln*]

**(X(t))***,*

**= −A·t + B**where

*is any constant, defined by initial condition*

**B***.*

**X(0)=T(0)−T**_{s}=T_{0}−T_{s}From the last equation:

**X(t) = e**^{B}·e^{−A·t}Using initial condition mentioned above,

**X(t) = (T**_{0}−T_{s})·e^{−A·t}Since

**X(t)=T(t)−T**_{s}

**T(t) − T**_{s}= (T_{0}−T_{s})·e^{−A·t}or

**T(t) = T**_{s}+ (T_{0}−T_{s})·e^{−A·t}where

**A = 2h·L²/(C·m)**So, the difference in temperature between a hot object and infinitely

large surrounding substance is exponentially decreasing with time.

## No comments:

Post a Comment