Heat And Temperature Ppt Free Download

It surprises many people to learn that the heat index values in the chart above are for shady locations. If you are exposed to direct sunlight, the heat index value can be increased by up to 15F. As shown in the table below, heat indices meeting or exceeding 103F can lead to dangerous heat disorders with prolonged exposure and/or physical activity in the heat.

Heat and temperature are a closely related topic, and as such, the difference between the two can be a bit confusing. The core difference is that heat deals with thermal energy, whereas temperature is more concerned with molecular kinetic energy.

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Heat describes the transfer of thermal energy between molecules within a system and is measured in Joules.[2] Heat measures how energy moves or flows. An object can gain heat or lose heat, but it cannot have heat. Heat is a measure of change, never a property possessed by an object or system. Therefore, it is classified as a process variable.

The second law of thermodynamics is a complex topic that requires intensive study in the field of thermodynamics to truly understand. However, for the purpose of this article, only one small aspect needs to be understood and that is the fact that heat will always flow spontaneously from hotter substances to colder ones. This simple statement explains why an ice cube doesn't form outside on a hot day or why it melts when dropped in a bowl of warm water.

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.

Thanks -- most everything I've seen from previous calculations would indicate that it's probably in the neighborhood of 60,000 BTUs needed for heating & 40,000 BTUs for cooling, which is not terribly far off from the existing 88,000 BTU furnace (110,000 BTU @ 80% efficiency) & 3 ton AC.

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.

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.

The title of this post draw my attention, because I'm also trying to work out the heatloss of my house based on the temperature decrease inside during the night. I have a Ecowitt weather station and continuously measure the inside and outside temperature, but also the windspeed and direction. So I also know the 'feels like' outside temp.

I heat the whole house during the day/evening and my (oil fired) boiler stops every night the same time and the temperature at that point in time is always the same (around 21 degrees C).

I cannot measure the amount of oil I'm using because I have an oil tank. And the boiler is also used to heat a hot water cylinder.

I have just this autumn and winter to figure out if I can switch from an oil boiler to an Air source heat pump (maybe high temperature). I'm running the boiler on a minimum flow temp (55 C) to simulate an ASHP and that works fine most of the time (except on a couple days with a 'feels like' below zero). Location is Montrose Scotland.

My guess would be that the data I collect should be enough to calculate the thermal performance of the house. Any thoughts on that are more than welcome.

There are devices that clamp on to the wiring and measure the time that something is running. Most oil burners are simply on/off, so by measuring the run time you can derive the output. On a cold day measuring the run time and the average temperature over 24 hours should give a pretty good estimate.

I've had a look a the clamp meters. They seem to measure the current, not the on-time. I could however use a smart plug for the boiler and measure the Kwh usage for a day. Knowing the nominal Kwh value of the boiler I can work out the run time over a day (week/month)

I have to check the boiler specs to find out if it's modulating or always burning at max. I guess the flow temperature does not make a difference. I also need to find out the amount of oil that is burned per time unit. The kerosine we use here has around 10.5 Kwh/liter.

So, I've been able to measure the electricity the boiler is using. The boiler pump uses 50w during 16 hours a day = 0.8 kwh a day. When the boiler is running it uses around 160w on top of the pump. The oil usage is around 1.9 l/h. Which means that every hour it's running I get around 1.9 l x10.3 kwh= 19.6 kwh of heat. On 20 dec for instance I used 0.73 kwh/.160=4.6 hour * 1.9 liter =8.7 liters = 90 Kwh on heating. The indoor temperature was 16.3 at 0.800 hr and 20.5 at 23.00.

I'll pick a day where the outside temp is around zero Celsius, no wind, no sunshine. And keep the indoor temp on 19 celcius for 24 hours. I record the outside temp anyway every five minutes. Would that provide enough data to calculate the heat loss?

Oh and another question: I don't seem to get notifications off new posts, even though I checked the box for it.

In my first post, #7, I was looking for a way to calculate the heat loss based on the pace of cooling down inside. Also measuring outside temp and wind chill factor. Then I got nudged into measuring the oil usage.

The energy efficiency of my boiler is around 80%. I forgot to take that into account in my calculation in post#12

I'm currently recording the electricity usage on a daily basis. Not intraday. It might be interesting to know how much kwh is needed to bring the house up to 19.5 celcius in the morning.

The pace of cooling depends upon the heat capacity of the house, the insulation level of the house and the temperature difference between inside and outside. The fuel consumption at a steady temperature just depends upon the insulation level and the temperature difference. 2351a5e196