Calculate the annual heat loss (in U.S. units) through the following wall assemblies using U -fa

Calculate the annual heat loss (in U.S. units) through the...

Key Concepts

Wall Assembly Characteristics

Different wall constructions (such as wood-framed walls, concrete, brick, or insulating panels) have varying thermal properties. Understanding the specific assembly type is essential in evaluating its U-factor or R-value and thus its overall heat loss characteristics.

Heat Loss Calculation

Calculating heat loss involves using the U-factor and the temperature difference (often represented by degree-days) along with the area of the building component. This process is fundamental in energy modeling and assessing building performance.

Heating Degree-Days

Heating degree-days aggregate the difference between the outdoor temperature and a baseline temperature over a heating season. They are a key metric in evaluating the energy demand for heating a building and are used to estimate the annual heat loss.

R-value

R-value quantifies the thermal resistance of a material or building element. The higher the R-value, the greater the material's insulating effectiveness. It plays a critical role in the design of building components to minimize heat loss.

U-factor

The U-factor is a measure of the rate of heat transfer through a building component, expressed in BTU/hr-ft²-°F. It is essentially the inverse of the thermal resistance (R-value) of an assembly, and is used to calculate how much heat is lost per degree of temperature difference between the inside and outside of a building.

Thermal Resistance of Assemblies

This concept considers the cumulative insulating properties of all layers in a building element like a wall. By summing the R-values of individual layers, one can estimate the overall resistance to heat flow, which directly influences energy efficiency.

Transcript

00:01 In this question, the setup is that we are looking at the insulation of a house.

00:09 And the main concept that we'll be dealing with here is the heat transfer, right? the rates of heat transfer based on the thermal conductivity of different materials at thermal equilibrium.

00:24 So thermal equilibrium is a very important condition and we'll see why later on.

00:28 So we're going to start off with the setup of the house.

00:32 We are given that the house contains three layers in its walls.

00:37 So the three layers starting from the dry wall that is at the innermost layer of the house in which the temperature is actually given as 22 degrees celsius.

00:50 This is the inner temperature of the house.

00:54 Then it has a second layer of fiberglass and the final layer is the insulated.

01:00 Siding that is exposed to the environment right the outside at a temperature of negative 2 degree celsius we want to find what is the overall rate of heat transfer right through the drywall all the way to the insulated siding so the information that we're given is the r factors for the first layer the dry wall over here the r factor 0 .56 whereas for the insulated siding this r factor is also given as 2 .6 all this in sire units now for the fiberglass we also need to find why is the r factor right so just a recap the r factor is equals to the ratio of the thickness of the material divided by the thermal conductivity of this material.

02:12 So this is the r factor in units of meter square degree celsius per watts.

02:23 Right, so as are units.

02:27 The first thing that we will want to do is to find what is our r value of this fiberglass layer.

02:34 Alright, so we're given that the thickness of this fiberglass layer is 3 .5 inches.

02:43 Let me just call this r value r are r2.

02:46 All right.

02:49 R2 we have the thickness which is 3 .5 inches and the thermal conductivity of fiberglass is about 0 .042 meter square degrees celsius per watts.

03:19 Sorry this unit should be slightly different.

03:23 This is in terms of watts per meter degree celsius sorry for the wrong units alright, so what we have to do is to first convert our inches into meters and we can do that by multiplying by 2 .54 times 10 power minus 2 meters per inches so this will give us about 8 .89 times 10 power minus 2 meters divided by 0 .042 and we should get about 2 .12 square degrees celsius per watts.

04:12 I'm going to just ignore the units over here.

04:14 This is just 2 .12.

04:16 All right.

04:19 So for convenience.

04:21 Now we are ready to look at the rate of heat transfer.

04:26 So at thermal equilibrium, why is it so important that we're looking at thermal equilibrium? because at thermal equilibrium, the rates of heat transfer throughout all the layers must be the same.

04:39 Right? the rate of heat transfer from the dry wall to the fiberglass let me just call this p1 is equal to the rate of heat transfer from the fiberglass to the insulated sliding which i shall call p2 as well as the weight of heat transfer from the insulated sliding all the way to the environment is p3 right this is at thermal equilibrium because if any of one of this is not equal to each other then the one of the materials will be gaining extra heat which means that it will change its temperature but that's not at thermal equilibrium right at thermal equilibrium their temperatures must be at a constant which means that the rate of heat coming in and rate of heat going out is the same so that's why p1 must be equal to p2 and p2 must be equals to p3 and therefore p1 and p2 and p3 are all equal to each other.

05:57 So this is an important condition that will help us to solve for the temperatures which we can then finally solve for one of the rates of heat transfer.

06:10 If you are to just solve for any one p2, p1 or p3 we can get that will be answer.

06:18 So what is p1? right so p1 quite simple we know the equation it's just k times the area right the area of the dry wall times the temperature difference between the two ends so i'm going to call the temperature at this and over here t1 and the temperature over at this other n t 2 and these two are unknowns so the temperature difference would be 22 degrees minus t 1 we're going to keep it at degrees celsius so we divide this by d all right now we know that k over d is just the r value sorry it's just 1 over the r value so this can be replaced with just r 1 where r1 is given as 0 .56 equate this to p2 right p2 is equal to so i'm going to equate this to p3 right p3 will make it easier to solve for p3 the rate of heat transfer is once again a times d temperature difference which is t2 minus negative 2 degrees so this plus 2 divided by r 3 so we see that there's a common factor of a which we can cancel out and we can just rearrange this a little bit so r1 is 0 .56 we can multiply this this r1 over over here, get me off this, alright, and then we open up the brackets, right, so r1 over r3, 0 .56 divided by 2 .6, all right, and we arranging a bit, can actually get t1.

09:35 T1 would just be taking 22 minus away this right hand side.

10:00 So, now we have our first equation now to get a second equation we equate p1 and p2 so once again p1 is a times 22 degrees minus t1 over r1 equate this to p2 which is a times t1 minus t 2 r2.

11:03 So t2 over here, sorry i'm going to replace t1 in first.

11:09 All right, substitute in t1 expression for t1 which is 21 .57 minus 0 .215 t2.

11:53 All right.

11:54 Remember that r2 is calculated to be 2 .12 over here.

12:00 So 2 .12 divided by r1 which is point 5 -6 and now putting all the terms with t2 to one side which is just this term and this term and then we're gonna solve for t2 and we will get t2 to be about 9 .8 3 degree celsius with this we can just look at any one of the rate of heat transfer either p1, p2 or p3 to get the rates of heat transfer right both of all of them are they same so p3 right because we have t2 so the easiest one will be using p3 should be just a times t2 plus 2 divided by r3 so the area they're looking at for the house is 3 times 10 so that's 30, then t2 plus 2 gives us about 11 .83 degrees celsius and r3 is given as 2 .6.

14:39 So the rate of heat transfer gets is about 136 watts.

14:48 Now for the second part of this question, we are modifying the second layer a little bit.

14:55 So the second layer where we're looking at the fiberglass, we're going to add wooden planks into it, kind of like substituting the fiberglass, some of the fiberglass with wooden planks, which we call studs.

15:14 So instead of just a full slab of fiberglass, some of it has been replaced with this studs of wood, right? which has dimensions of 3 .5 times 1 .5 inches...

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