1D Conduction with no generation
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1D Conduction with no generation The heat diffusion equation, as derived, is given by equation 1. ∂ ∂x k ∂T ∂x + ∂ ∂y k ∂T ∂y + ∂ ∂z k ∂T ∂z +˙ q = ρc p ∂T ∂t (1) If we assume steady 1-D conduction, with no generation, equation 1 reduces to d dx k dT dx =0 (2) The only way that equation 2 can be satisfied if either k = 0, or of dT /dx = 0. A conductivity of zero would imply an ideal thermal insulator, and would of course result in a zero heat flux. This is clearly not of interest. We are therefor interested in the second condition which means that kdT/dx = C, or dT dx = c 1 (3) and, recalling Fourier’s law, that the heat flux is a constant with x. We can easily integrate equation 3, to obtain, T (x)= c 1 x + c 2 (4) Now, we impose the desired boundary conditions, the most simple of which is a Dirichlet condition, or constant T; T (0) = T s,1 and T (L)= T s,2 . Substituting into equation 4 gives T (x)=(T s,2 - T s,1 ) x L + T s,1 (5) Next, we can easily determine the heat flux using Fourier’s law. q x = -kA dT dX = kA T s,1 - T s,2 L (6) Now, let’s condsider our plane wall with convection occuring at each of the surfaces – a much more realistic boundary condition. Later we will see how to determine convection coefficients, but for now we will simply assume that we know the value. [Adjust Figure] Recall the thermal resistances that we defined earlier. We defined a conduc- tion resistance as R T,cond = L kA (7) and a convection resistance as R R,conv = 1 hA (8) Let us imagine now that our wall is replaced a thermal resistance, given by equation 7, and that the two convection boundary conditions are replaced by 1
Transcript of 1D Conduction with no generation
1D Conduction with no generation
The heat diffusion equation, as derived, is given by equation 1.
∂
∂x
(k
∂T
∂x
)+
∂
∂y
(k
∂T
∂y
)+
∂
∂z
(k
∂T
∂z
)+ q̇ = ρcp
∂T
∂t(1)
If we assume steady 1-D conduction, with no generation, equation 1 reducesto
d
dx
(k
dT
dx
)= 0 (2)
The only way that equation 2 can be satisfied if either k = 0, or of dT/dx = 0.A conductivity of zero would imply an ideal thermal insulator, and would ofcourse result in a zero heat flux. This is clearly not of interest. We are thereforinterested in the second condition which means that kdT/dx = C, or
dT
dx= c1 (3)
and, recalling Fourier’s law, that the heat flux is a constant with x.We can easily integrate equation 3, to obtain,
T (x) = c1x + c2 (4)
Now, we impose the desired boundary conditions, the most simple of which isa Dirichlet condition, or constant T; T (0) = Ts,1 and T (L) = Ts,2. Substitutinginto equation 4 gives
T (x) = (Ts,2 − Ts,1)x
L+ Ts,1 (5)
Next, we can easily determine the heat flux using Fourier’s law.
qx = −kAdT
dX= kA
(Ts,1 − Ts,2
L
)(6)
Now, let’s condsider our plane wall with convection occuring at each of thesurfaces – a much more realistic boundary condition. Later we will see how todetermine convection coefficients, but for now we will simply assume that weknow the value. [Adjust Figure]
Recall the thermal resistances that we defined earlier. We defined a conduc-tion resistance as
RT,cond =L
kA(7)
and a convection resistance as
RR,conv =1
hA(8)
Let us imagine now that our wall is replaced a thermal resistance, given byequation 7, and that the two convection boundary conditions are replaced by
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two more thermal resistanceces, each given by equation 8. We are making adirect analagy here to an electric circuit made up of three resistors in serial.The heat rate through the system is analogous to the current in the electriccircuit, and the temperature drop is analogous to the voltage drop. Since allthe resistors are in series, the current which passes through the circuit passesthrough each and every resistor. In the real system, the heat energy must passfrom inside the room, through the boundary layer, then through the wall itselfand then through the boundary layer outside. The thermal boundary layerinside the room bridges the room at T∞,1 and the inside wall temperature Ts,1.
qx =T∞,1 − Ts,1
1/h1A(9)
Where h1 is the convection coefficient representing the convection processadjacent to the inside wall surface.
The wall conducts heat from the inner surface at Ts,1 to the outer surfaceat Ts,2.
qx =Ts,1 − Ts,2
L/kA(10)
and finally, the convection process on the outer wall surface convects theheat energy from the outside wall surface temperature, Ts,1 to the outside airtemperature, T∞,2.
qx =Ts,2 − T∞,2
1/h2A(11)
Since qx is constant through each process, we can combine the above equa-tions in order to arrive at an expression containing only temperatures that weknow. In this case it is reasonable to assume that we would know the inside airT, T∞,1, and the outside air T, T∞,2.
qx =T∞,1 − T∞,2
Rtot(12)
where Rtot is simply the sum of the series resistances.
Rtot =1
h1A+
L
kA+
1h2A
(13)
We will now use these concepts to explore heat transfer through variousfenestration systems.
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A Single Pane of Glass
Consider first a window composed of a single pane of glass, as shown in figure1. The window is 0.8 m high, 1.5 m wide and 8 mm thick and is made of glasswith a thermal conductivity k = 0.78W/mK. The inside temperature is heldat T∞,1 = 20oC, while the outside temperature is T∞,2 = −10oC. Assumingh1 = 10W/m2K and h2 = 40W/m2K let’s calculate the heat loss through thewindow, and the inner glass surface temperature.
T1,�
Ts,1
Ts,2
T2,�kglass
T1,�
Ts,1Ts,2
T2,�
Rconv,1 RglassRconv,2
Figure 1: Schematic of a single pane window, with thermal resistance netwoekshown.
Assumptions
• Steady state heat transfer
• 1D heat transfer
• No internal generation
• constant termal conductivity
• Negligibe radiation heat transfer
Owing to the above assumptions, we can employ a thermal resistance net-work to solve the problem. The window can be replaced by the circuit shown infigure 1, made up of a convection resistance at the inner surface, a conductionresistance through the glass, and a convection resistance at the outer surface.
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Rconv,1 =1
h1A=
110
[W
m2K
]0.8[m]1.5[m]
= 0.0833[K/W ] (14)
Rglass =L
kglassA=
0.8 × 10−3[m]0.78
[W
mK
]0.8[m]1.5[m]
= 0.00855[K/W ] (15)
Rconv,2 =1
h2A=
140
[W
m2K
]0.8[m]1.5[m]
= 0.0208[K/W ] (16)
Since the resistances are all in series, the total resistance is simply given asthe sum of the individual resistances.
Rtot = Rconv,1 + Rglass + Rconv,2 (17)= 0.0833 + 0.00855 + 0.0208 (18)= 0.113[K/W ] (19)
The total heat rate through the window is then given by
q =T∞,1 − T∞,2
Rtot=
20[oC] − (−10)[oC]0.113[K/W ]
= 266W (20)
Note that the units in equation 42 look very odd in that we have [oC] inthe numerator, and [K/W ] in the denominator. This is perhaps a little sloppy,but remember that a temperature change of 1 K is identical to a temperaturechange of 1 oC. As long as we are considering only temperature differences, wecan use either units.
Now that we know the heat rate passing through our circuit, we can de-termine the temperature at any point in the curcuit using equations 9, 10 and11. Here we are interested in the inner surface temperature, Ts,1 and thereforeequation 9 will suffice.
q =T∞,1 − Ts,1
1/h1A(21)
Ts,1 = T∞,1 − q1
h1A= 20[oC] − 266[W ]0.0833[K/W ] = −2.2[oC] (22)
We see that even though the room is well above freezing, the glass surfacetemperature is below freezing, and you can expect to see frost forming on thiswindow.
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Double Pane Window
Let’s now split this pane of glass into to 4 mm thick panes, and separate themby a 10 mm air gap (kair = 0.026W/mK), as shown in figure 2. Assuming thesame conditions as above, we will determine the heat rate through this window,and the inner surface temparature. The same assumptions are necessary here inorder to replace the window system with the thermal resistance network shownin the figure. We will additionally assume that the air in the enclosure is per-fectly stationary, such that heat transfer through the air layer is by conductiononly.
T1,�
Ts,1
Ts,3
Ts,4
Ts,2
kglass
kglass
T1,�
Ts,1Ts,2 Ts,3 Ts,4
T2,�
Rconv,1 Rglass,1 RairRglass,2 Rconv,2
T2,�
Figure 2: Schematic of a double pane window, with thermal resistance networkshown.
There are now five resistances to define. The first is the convection resistanceat the inner surface,
Rconv,1 =1
h1A=
110
[W
m2K
]0.8[m]1.5[m]
= 0.0833[K/W ] (23)
which is exactly as above. Next is a 4 mm section of glass
Rglass,1 =L
kglassA=
0.4 × 10−3[m]0.78
[W
mK
]0.8[m]1.5[m]
= 0.00427[K/W ] (24)
followed by a conduction resistance through the air layer.
Rair =L
kairsA=
10 × 10−3[m]0.026
[W
m2K
]0.8[m]1.5[m]
= 0.321[K/W ] (25)
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The second pane of glass has a resistance identical to the first,
Rglass,2 = Rglass,1 = 0.00427[K/W ] (26)
which means that the total resistance of the two panes of glass will the sameas the 8 mm single pane above. Finally, there is a convective resistance at theexteriour surface of the window, which is the same as in the single pane example.
Rconv,2 =1
h2A=
140
[W
m2K
]0.8[m]1.5[m]
= 0.0208[K/W ] (27)
The total thermal resistance is
Rtot = Rconv,1 + Rglass,1 + Rair + Rglass,2 + Rconv,2 (28)= 0.0833 + 0.00427 + 0.321 + 0.00427 + 0.0208 (29)= 0.434[K/W ] (30)
Notice that the conduction resistance in the static air layer is by far the largestresistance.
The heat rate through the window is then
q =T∞,1 − T∞,2
Rtot=
20[oC] − (−10)[oC]0.433[K/W ]
= 69.2W (31)
Finally, the temperature at the inside surface, Ts,1 can be calculated consideringthe convection resistance between the room and the window, as above.
Ts,1 = T∞,1 − q1
h1A= 20[oC] − 69.2[W ]0.0833[K/W ] = 14.2[oC] (32)
The inner surface is considerably warmer, even though all we have added is alayer of air. It much less likely in this case that frost will form on the window.Notice that the conductivity of air is an order of magnitude lower that of theglass, and it is therefore not at all surprising that the heat transfer has decreasedsignificantly. This helps to explain how thermal insulation works. The ideabehind many insulations is to trap a significant quantity of air indside a porousstructure of relatively low conductivity. As long as the air is held stationary, sothat convection does not occur, then the major resistance to heat transfer willbe that of the air, and the material will be a good insulator.
What is the limit to how low heat rate through the window can be for ourwindow which now occupies 18 mm. Imagine that the window is made up purelyof an 18 mm air gap, Perhaps held in place with a thin film of plastic wrap ateach end. If this is the case, then we can imagine that the convection resistanceswill be the same as in the above two calculations, and we can easily add theconduction resistance due to our 18 mm air gap to find the total resistance,
Rtot = Rconv,1 + Rair,18mm + Rconv,2 (33)= 0.0833 + 0.577 + 0.0208 (34)= 0.681[K/W ] (35)
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resulting in a heat rate through the window of 44.1 W and a correspondinginside surface temperature of 16.3 oC. Clearly the double pane window is agood solution.
Imagine now a house that has a single pane window, and is experiencingserious condensation and frost. The owner has not taken a heat transfer courseand decides that the windows must be too thin. They see the double panewindows and figure that they can do much better by filling the whole spacewith glass – they are not cheap you see and figure glass should be better thanair. So, they replace the 8 mm glass windows with 18 mm glass windows atconsiderable expense. The new thermal resisitance will be
Rtot = Rconv,1 + Rglass,18mm + Rconv,2 (36)= 0.0833 + 0.0192 + 0.0208 (37)= 0.123[K/W ] (38)
for a heat rate of 244 W and an inside surface temperature of -0.3 oC. Theywill have spent a lot of money for absolutely nothing! They see their neighbor’sdouble pane windows without frost, and decide that they are going to outdothem by replacing the windows again, this time using triple pane windows. Ifdouble pane windows are better than single pane windows, then clearly triplepane windows must be even better, right?
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Triple pane window
We want the window to fit in the same package, of 18 mm width, and the glassis already pretty thin at 4 mm per pane. This means the triple pane windowsystem will be as shown in figure 3, with two air gaps of 3 mm each.
T1,�
Ts,1
kglass
kglasskglass
T2,�
T1,� T2,�
Rconv,1 Rair RairRglass,2Rglass,2Rglass,2 Rconv,2
Figure 3: Schematic of a triple pane window, with thermal resistance networkshown.
The convection resistances will be the same as in the previous cases; Rconv,1 =0.0833[K/W ] and Rconv,2 = 0.0208[K/W ]. Each pane of glass will also be thesame as in the case of the double pane window, Rglass = 0.00427[K/W ] andthe air layers will have resistances of Rair = 3 × 10−3/(0.026 · 1.5 · 0.8) =0.0962[K/W ]. The total resistance is then
Rtot = Rconv,1 + Rglass,1Rair,1 + Rglass,2 + Rair,2 + Rglass,3Rconv,2 (39)= 0.0833 + 3(0.00427) + 2(0.0962) + 0.0208 (40)= 0.309[K/W ] (41)
The heat rate through the window is then
q =T∞,1 − T∞,2
Rtot=
20[oC] − (−10)[oC]0.309[K/W ]
= 97.0W (42)
Finally, the temperature at the inside surface, Ts,1 is .
Ts,1 = T∞,1 − q1
h1A= 20[oC] − 97[W ]0.0833[K/W ] = 11.9[oC] (43)
The frost problem will probably be alleviated, but there are noticeablygreater heat losses through this triple pane window. When we consider nat-ural convection in enclosures, we may revisit this problem, and calculate what
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the heat transfer rate through the window is when considering natural con-vection. The cases here are limiting cases where the heat transfer occurs byconduction only throught the window system. Natural covection will result inthe movement of air in the gap, and hence increased heat transfer (or a decreasedresistance). It may actually make sense to have a triple pane window when weconsider these effects. As the thickness of the air gap decreases, viscous effectsbecome increasingly important, and can supress the onset of natural convection.
Summary of Window Calculations
CASE q [W ] Ts,1[oC] % change8 mm Glass 266 -2.2 -18 mm Glass 244 -0.2 -8.3Double Pane 69.3 14.2 -74Triple Pane 97.0 11.9 -63.5
18 mm Air Gap 44.0 16.3 -83.5
Table 1: Summary of window calculations. Percent change is the change in heatrate referenced to the 8 mm single pane window.
A double pane window is clearly the best choice of what we have lookedat so far. It reduces the heat rate compared to the single pane window by 74%. What does this mean for energ costs? There is a saving of 196.7 W insteady operation with a 30oC temperature difference accross the window. Foreach such day, this corresponds to 196.7 W * 24 h = 4.73 kWh. If this heatenergy is generated using resistance heaters, and electricity costs $0.10 / kWhthen $0.47 a day are saved. If there are only 100 such days in a year, then thiscorresponds to $47.00 /year in savings. In addition, the window will save energin the summer, when you are cooling the house and the temperature gradientis reversed. The payback time for replacing the window is then at least severalyears, but the window will last 20-100+ years. It is clearly beneficial both inthe sort term, and in the long term to replace the windows.
When you are ready to do this though it will be important to consider alsothe effect of coatings on the windows which relate to radiation heat transferthrough the system. We will consider this towards the end of the course.
Parallel Resistances
What if we need to consider the heat transfer through the frame of the windowas well. A schematic of such a case is shown in figre4. In this case there are twoparallel paths that heat energy can take to escape through the window.
We can extend our analysis to these more complicated analyses once werecall how to combine parallel resistances. Using the total resistance of the three
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T1,�
T2,�
Ts,1
kglass
kglasskglass
T1,� T2,�
Rconv,1 Rair RairRglass,2Rglass,2
Rframe
Rframe
Rglass,2 Rconv,2
Figure 4: Schematic of a triple pane window with a frame. A thermal resistancenetwork is also shown.
pane window, determined above, Rtot,3pane we can represent the new resistancenetwork as three resistances in parallel, as shown in figure 5.
These can now easily be combined into a total resistance,
1Rtot
=1
Rframe+
1Rtot,3pane
+1
Rframe(44)
taking into account the cross sectional area of each section. Note that theabove shown network is not unique, because we are now pushing our assump-tions. The network of figure 5 assumes that the frame and the window sectionshare the same temperatures at their respective faces – an isothermal assump-tion. Another possibility is to assume that surfaces parallel to the x axis areadiabatic, and this will result in a different resistance network. The idea is thatthe actual heat transfer through the system will be bracketed by these two ap-proximations. Why do we have this ambiguity? In actuality, the heat transferis now two dimensional – the temperature gradient throught the frame will bedifferent thatn that through the window system, and therefor there must besome heat transfer between the frame and the window. This is heat transfer in
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T2,�
Rtot,3 pane
Rframe
Rframe
T1,�
Figure 5: A parallel thermal resistance network representing the above window.
a direction normal to our axis, and means that the actual problem is in realitytwo dimensional, and not one dimensional as we have assumed in this entirederivation. This may be ok, or is at least a useful approximation when theparallel resistances are not too dissimilar, but as the difference between themincreases, the approximation will be worse and worse. We will explore this inthe next tutorial.
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