Lecture9 Pre Thermal Resistance

Rframe

T1,¥

Rtot,3 pane

T2,¥

Rframe 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 reality two dimensional, and not one dimensional as we have assumed in this entire derivation. This may be ok, or is at least a useful approximation when the parallel resistances are not too dissimilar, but as the difference between them increases, the approximation will be worse and worse. We will explore this in the next tutorial.

R Value of Insulation In practice, when you are shopping for insulation, it is usually rated using a R-value. This number is very closely related to the concept of a conduction resistance except that it is normalized by area, or based on a heat flux instead of a heat rate. q  =

T1 − T2 RT

(45)

L (46) RT = k   2 . This is essentially the R-value, but where RT , in SI, was units of Km W o 2  f t hr R-values are usually stated in British units of FBtu . Note that the Btu, or British thermal unit, is a unit of energy, not power. This is a time unit, hours, appears in the units of R-value. Special care should be taken when using R-values, as they are sometimes even  on a per unit length basis, which resulting in the wonderful units of  o given F f t2 hr Btuin . This is a seeminly bizarre thing to do, since what this is is essentially the inverse of the conductivity of the material (see equation ??) expressed in really strange units. All in all, great care must be taken when using numbers such as R-values, in that the units should be carefully checked to ensure that you are getting what you think you are getting. See http://www.dow.com/styrofoam/na/dow home/choose.htm for example insulation specifications.

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The overall Heat Transfer Coefficient We have seen a few examples of resitance networks now, and have seen then even relatively simple systems can have quite a number of individual resitances. Another way of expressing the total heat transfer through the system is to use an overall heat transfer coefficient, which is actually the inverse of the R-value, except that we will express it in much more familiar units. If we consider, say a triple pane window system and we measure the heat rate through the system, we can use something very much like Newton’s law of cooling to express the heat rate. qx = U A(T∞,1 − T∞,2 )

(47) 2

where U is the overall heat transfer coefficient, and has units of [W/m K] as does a convection coefficient which is based on the same form of equation. The overall heat transfer coefficient is specifically not given the symbol h so so as not to confuse the mechanisms of heat transfer. It is worth noting though that both the overall heat transer coefficient and convection coefficients are simply convenient ways of relating temperature differences to heat fluxes. They are convenient tools that we can use, but they are not intrinsic to the problem, and we could analyze or describe these systems in other ways. This is in stark contrast to the thermal conductivity appearing in Fourier’s law, which is a material property and is clearly intrinsic to a conduction problem. In any event, the overall heat transfer coefficient clearly related to the total thermal resistance and could easily be calculated from the total resistance. U=

1 Rtot A

(48)

Thermal Contact Resistance

  qx = qcontact + qgap  Rt,c =

TA − TB Rt,c =  qx A 12

(49) (50)

If we want to calculate the the heat flux through a composite material as shown above and include the effect of contact resistance, we can simply add an extra resistance to the resistance network, paying attention to the units.

Rtot = RA +

 Rt,c R LB LA + RB = + tc + A kA A A kB A

(51)

Using Rt ot we can calculate the total heat rate through our material and then we could calculate the temperature on either side of the interface, TA or TB , considering RA or RB exactly as we did to detemermine the window surface temperatures. 0.01[m] L m2 K  W  = 0.025 = k W 0.04 mK

(52)

0.01[m] m2 K L  W  = 2.6 × 10−5 = k W 386 mK

(53)

R”t,c = R”t,c =

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Radial Systems So far we have looked at conduction problems in planar materials like walls and windows. This is not adequate to describe conduction heat transfer through pipe walls as in heat exchangers, or through the insulation on piping systems. In order to descrbibe these cylindrical systems, we have to determine a conduction resistance in the radial direction of a cylindrical system. Consider the steady heat conduction equation in cylidrical coordinates, with no generation


 1 ∂ ∂T 1 ∂ ∂T ∂ ∂T kr + 2 k + k =0 (54) r ∂r ∂r r ∂φ ∂φ ∂z ∂z
 1 d dT kr =0 (55) r dr dr dT = c1 dr

(56)

T = c1 ln r + c2

(57)

r

T (r) =

Ts,1 − Ts,2 ln ln(r1 /r2 )

qr =




r r2

 + Ts,2

2πLk(Ts,1 − Ts,2 ) ln(r2 /r1 )

Rtcond =

ln r2 /r1 2πLk

Critical Thickness of Insulation

14

(58) (59) (60)