Calculate Heat Loss

[Arnold Gilgen]

Tocalculate heat loss involves understanding two key types: loss of transmission (heat escaping through surfaces like walls, windows, roofs) and loss of ventilation (heat loss due to air changes per hour).

When evaluating an example house, particularly in the context of a shared party wall and accounting for the surface area, the effective R-value becomes paramount in maintaining a comfortable temperature indoors.

By comprehending the difference in temperature between the inside and outside environments, and considering factors such as windows and doors, engineers can ensure that heat loss is minimised, and energy efficiency is maximised.

As inside air acts akin to water flowing, the intricate dance of R-values along the front wall, surface areas, and points of potential heat transfer like windows and doors can make all the difference in achieving a sustainable and cozy living space.

This calculation helps engineers determine the energy of watts divided by time needed to maintain a comfortable indoor temperature while factoring in the overall energy efficiency of the building, and the appropriate R-value of insulation materials to achieve optimal thermal resistance.

The U-value is calculated by dividing the rate of heat transfer by the internal temperature gradient and external temperature gradient difference between the building, and the surface area through which the heat required is being transferred.

"The software offers precision through detailed result outputs and advanced options for efficient pipe sizing, allowing heating engineers to optimise based on parameters like maximum velocity and pressure drop."

In order to specify the right system, you will need to know how many BTUs (British Thermal Units) are required to replace the heat that is escaping your home through walls and other surfaces. It is determined by performing a heat load calculation which consists of calculating surface heat loss and heat loss due to air infiltration.

It is recommended that a contractor or systems designer be utilized for the final audit, however, you can prepare for an energy audit by sealing up the obvious leaks around windows and doors and addressing areas where insulation is needed.

The first step is to determine the difference between the ideal temperature inside your home and the average temperature that your geographic region never goes below in the winter. The result of this calculation will be called Delta T. If your interior design temperature is around 68 degrees and your average outside winter temperature is 40, then Delta T = 28 degrees which is the difference between the two.

The R-value of the wall will be based on the insulation in a wall. An uninsulated 24 residential wall will have an R value of 4 while the same wall with code approved insulation will have an R value of 14.3. To obtain the U value, divide the R value into 1. The U value in this example would be .07.

The heat loss in the wall is measured in BTUs and the formula is U value x Wall area x Delta T. In our example, this would be: .07 x 164 x 28 = 321.44 BTUH (British Thermal Units per Hour). This is the amount of heat that is escaping through the exterior walls based on the amount of insulation in them. The other interior surface calculation is for the ceiling. Typical ceiling insulation will be an R-19 which has a U value of .53. This results in a loss of 5,565 BTU per hour.

Companies that specialize in energy modeling or energy audits will have experienced technicians that utilize the latest technology to expose points of heat loss as well as air and moisture infiltration. Identifying these areas is often impossible using a visual inspection as they are hidden beneath flooring, behind walls,and above ceilings. That is why it is highly recommended that you contact a professional company to perform your inspection.

Nothing is as comfortable as a warm floor touching cold feet in the winter, and electric radiant floor heating is the ideal system for supplemental heat in a room or for your entire home. Warmup underfloor systems heat up in minutes rather than hours, saving you money and conserving energy. Our systems are UL-approved and provide gentle, even heat to the floor surface to prevent hot and cold spots in a room and leaving the air temperature cooler than other conventional heating methods.

Heat losses occur due to heat being transferred from inside a property through walls & windows etc (fabric heat losses) and being transferred, via drafts of warm air, through gaps in the fabric, which is replaced by cooler outside air (ventilation & infiltration losses).

I guess in newer properties, the specific ventilation requirements are more easily accounted for, since these may be carried out using more accurate means of natural ventilation and the use of mechanical ventilation.

We have a fluid (Air), which has mass (everything has mass) and therefore, has a specific heat capacity (the amount of energy to change a certain mass of the substance by a certain temperature, usually this is joules per kg per each degree of temperature change).

For a basic measure of energy loss via walls/windows/etc you can use the basic formula of: U value x Wall area x Delta T. In your scenario Delta T changes as the outdoor temps drop, but also as your interior temperature drops.

The information you have is how many degrees per hour your house cools at a given temperature. What you are trying to get is how many BTU's per hour it loses at a given temperature. The quantity you are missing is BTU/degree, which is called the heat capacity. (Sometimes erroneously called "thermal mass," don't get me started.)

You could estimate the heat capacity of a building if you knew the specific heat of the materials, and how much the building weighed. Most common building materials -- drywall, lumber, concrete, tile -- have specific heats of an order of magnitude of 0.5. A single family house weighs on the order of 100,000 lbs. So a ballpark guess would be it takes 50,000 BTU to raise a house one degree, or a house loses 50,000 BTU when it cools by one degree.

You should know the basic heat loss equation is Q = - k A (T2-T1), similar to the one agm gave you.. You can use this simply to figure out a quick heat loss buy just knowing the surface Area and the change in indoor temperature over a specific time. k doesn't vary, constant, since it is the properties of the materials/walls/roof etc.

So A would be the surface area of the house, walls, roof etc. Use the inside temperature for T2 and T1 over a period of time, hours.

Q would normally end up being BTUs/hr so when you multiply by the time you end up with a basic overall heat loss.

eg. Q = -k (Btu/ hr*ft^2*F) (1000ft^2) x (70 - 60F) = k x -10,000 Btu/hr x let's say over 5 hrs time = k * -50,000 Btus....which is close to what you came up with. Your area may be larger and the time less. k shouldn't be that big. So pretty close for a quick, overall estimate. If you figure out k, conductivity for portion of a wall you can plug that in but you may have to use the outside temperature for T2.

Tagging on to DC's comment below using this example to solve for k

So we have Q = k(constant) x -10000 BTU/hr and A (Tout -Tin) = (1000ft^2)(40 - 70F) then when you solve for k = Q/ -A (T2 -T1)

k= (-10000 BTU/hr)/( 30000 ft^2*F) =- 0.3 BTU/hr*ft^2*F

You could do a rough calculation of heat out=heat in. IE, if the heat was out for 24 hours and then had to run for 6 to catch up, overall 30 hours of heating required 6 hours of capacity, so your load at that temperature is 20% of capacity. That's a first approximation.

What that leaves out is that heat loss at 57F is going to be less than at 72F. And the outside temperature may not have been constant. So a better approximation is to create a spreadsheet with a column for each hour. For each hour you have outside temperature, which you get from observations or estimate; beginning inside temperature, which is 72F for the first column and the ending temperature of the preceding hour for every other column. Then you calculate heat flow out, which is beginning inside temperature minus outside temperature, times the thermal constant of the house[1], and heat flow in, which is the full capacity of your heating system in the hours that the heat was running. You substract those to get net heat flow, and divide by the heat capacity of the house[2] to get net temperature change, which you then add to the beginning temperature to get the ending temperature.

The heat constant of the house and the heat capacity of the house are the unknowns that you are trying to solve for. You know three points on the curve -- 72F at the beginning, 57F at the middle and 72F at the end -- so there is a unique solution for the two unknowns. The way I would solve[3] it is to plug in guesses and adjust them until the spreadsheet balances. In the second paragraph of this post I showed how to do a rough estimate of heat constant, use that as your first guess and adjust from there.

[1] The heat constant of the house is in BTU/hr/degree, it's kA in the formula Tom gave in post #5, and it's what a Manual J is ultimately trying to figure out. The Manual J process is basically measuring A and estimating k.

[2] The heat capacity of a house is in degrees/BTU, it's how much heat it takes to raise the temperature of the house.

[3] I have an honors degree in applied mathematics and I solve problems like this by plugging numbers into a spreadsheet in successive approximation. It's easier and less error-prone than trying to solve them analytically.

You make a good point about heat loss rate being different at lower temperatures. My house is currently empty, holding 60F, with a flat 35F exterior temp. One 4 hour cycle consists of 26 minute runtime of the boiler, which outputs 153k, so a 10.8% duty rate or 16.5k per hour. With a 25 delta, roughly 660 btu/degree hour.

By convention the base temperature is 65F, the assumption is that occupant activity provides about five degrees worth of heating. That's a crude assumption. I believe it's used because heating degree-days -- itself a crude measure -- is reported with a 65F baseline.

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