Re: The lost formula  Printable Version + HP Forums (https://archived.hpcalc.org/museumforum) + Forum: HP Museum Forums (https://archived.hpcalc.org/museumforum/forum1.html) + Forum: Old HP Forum Archives (https://archived.hpcalc.org/museumforum/forum2.html) + Thread: Re: The lost formula (/thread177254.html) 
Re: The lost formula  Csaba Tizedes (Hungary)  01202011 In last December I used an equation and works very well. I want to made a housewalltemperaturemeter to calculate heat loss of my house. This was a very simple polystirene cube (bad thermal conductivity for more precise measure) and a little PSION Revo thermistor. This "device" works very simple: Push the thermistor to the wall, wait for 1520 seconds, measure the resistance then calculate temperature. A pic about it and a calibration calculation:
Re: The lost formula  designnut  01202011 That seems precise. What method of regression analysis did you use? Re: The lost formula  John B. Smitherman  01202011 Good work Csaba and thanks for sharing. Where did you get the temperature data used to develop the correlation? By the way, what did you learn about your house's heat loss?
John
Re: The lost formula  John B. Smitherman  01222011 Hi Sam. I used a 35s to work through Csaba's example and of course it only does linear regression. I had to remember to take the ln of resistance and 1/Temp in degrees K before entering them as y and x. Regards,
John
Re: The lost formula  Csaba Tizedes (Hungary)  01262011 Hi Sam,
I usually use Excel to fit functions, because it's simple and compatible with everything.
This regression a simple leastsquares method. The temperature need to calculate from Celsius to Kelvins!
Re: The lost formula  Csaba Tizedes (Hungary)  01262011 I have a good mercuryfilled thermometer (scale divided into 0.5C from 0C100C, and approximate the temperature +0.1C). I used a simple plasticware, filled with ~half liter water then I was cooled slowly with snow. If temperature not changed I was read a measuring point data.
1.) I assume that the wall have four layer (see the table: 2nd is the brick layer, 3rd is the heatisolation layer and 1st and 4th the inner and outer mortar/grout (I dont know how to say it in English)). 2.) I want to know the inner heat transfer coefficient (alpha_i [W/(m2*K)]), so I wrote a heat flux from diningroom wall "inner" side to "outer" (ambient) , then I wrote some equation from "inner" side to inner wall surface. In this points the temperatures measurable (Ti, Tiw, To), and then I solved this equation (1) to alpha_i. 3.) With this coefficient the heat flux calculated: q'A [W/m2]=alpha_i*(TiTiw) 4.) Then I plot this heat flux as function of temperature difference. Unfortunately I have not measuring points yet at dT=22...24C.
The coloumn alpha_i_A calculated at alpha_o=5 W/(m2*K) (no wind, cloudy weather) and alpha_i_B calculated at alpha_o=15 W/(m2*K) (windy, cloudy weather)
