Ana gabatar da Gabatarwa zuwa Modey mai sanyaya
Fasahar hana kai tsaye shine dabarar sarrafawa mai mahimmanci a kan tasirin Petelies. Jaccan Peltier ya gano a shekara ta 1834, wannan abin mamaki ya ƙunshi dumama ko sanyaya-kayan kayan masarufi biyu (bismuth da sanarwa) ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa ta hanyar wucewa. A yayin aiki, kai tsaye yana gudana ta hanyar TEC module yana haifar da zafi da za a canza daga wannan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefe zuwa wancan gefen. Kirkirar da sanyi da zafi. Idan shugabanci na yanzu an juyawa yanzu, sanyi da zafi bangarorin. Hakanan ana iya daidaita ikonsa ta hanyar canza aikinta na yanzu. Tsarin mai sanyaya guda ɗaya (adadi. 1) ya ƙunshi farantin yanki biyu tare da p da n-rubuta) tsakanin faranti na yumɓu. Abubuwan da ake amfani da kayan semiconondiktorcory suna da alaƙa da lantarki a cikin jerin da na therery a layi daya.
Za'a iya ɗaukar Module mai sanyaya, Petelier Na'ura, za a iya ɗauka a matsayin nau'in famfo mai ƙarfi da ƙarfi, kuma saboda yawan nauyinsa, ya dace da amfani a matsayin ɓangare na sanyaya kayan sanyaya tsarin (saboda iyakance sarari). Tare da fa'idodi kamar yin shuru, hujjoji na zamani, na fi dacewa da rayuwa da sauki avaspace, jiyya, magani, magani, annoba Yin rigakafi, samfuran masu amfani (samfuran masu amfani (sanyaya ruwa, mai sanyaya mota, sanyaya mota, mai sanyaya mai sanyaya, Cool & zafi barci pad, da sauransu).
A yau, saboda karancin nauyi, karamin girman ko karfin kaya da ƙarancin ruwa, Aerctpace, kayan aiki, Aertospace, Samfuran Kasuwanci (kamar kayan ruwa na ruwa (kamar kayan aikin ruwa, masu sanyawa. carcooler da sauransu
| Sigogi | |
| I | Yin aiki na yanzu zuwa TEC Module (a cikin Amps) |
| Imax | Yin aiki na yanzu wanda ya sa matsakaicin bambancin zafin jiki △ tmax(a cikin Amututtukan) |
| Qc | Yawan zafi da za a iya tunawa da fuskar bakin ciki na TEC (a Watts) |
| Qmax | Matsakaicin adadin zafin da za a iya tunawa a gefen sanyi. Wannan yana faruwa a i = imaxkuma lokacin da Delta T = 0. (a Watts) |
| Tm | Zazzabi na zafi gefen fuska lokacin da TEC module ke aiki (a cikin ° C) |
| Tsanyi | Zazzabi na yanayin sanyi lokacin da sec module aiki (a cikin ° C) |
| △T | Bambanci a cikin zafin jiki tsakanin gefen zafi (th) da gefen sanyi (tc). Delta T = th-Tc(a cikin ° C) |
| △Tmax | Matsakaicin bambanci a cikin zafin jiki A tec module zai iya cimma tsakanin gefen da (th) da gefen sanyi (tc). Wannan yana faruwa (mafi yawan ƙarfin sanyaya) a i = imaxda qc= 0. (A cikin ° C) |
| Umax | Wadatar warkewa a i = imax(a Volts) |
| ε | TEC module sanyaya (%) |
| α | Duba madaidaicin madaidaitan abin da ya kamata (V / ° C) |
| σ | Lantarki mafi inganci na kayan masarufi (1 / CM olhm) |
| κ | The Thermo of the Thermevery Thrmeclecom abu (w / cm · c) |
| N | Yawan Sermeelecle |
| Iεmax | A halin yanzu da gefen zafi da tsofaffin gefen zafin jiki na TEC module shine ƙayyadadden darajar kuma yana buƙatar samun iyakar ƙarfin (a amps) |
Gabatar da Tsarin aikace-aikacen zuwa TEC module
Qc= 2n [α (tc+273) - -li²/ 2σs-κs / lx (tha \ h- tc)
△ t = [iα (tc+273) -li /²2σs] / (κs / l + i α]
U = 2 n [I il / σs + α (tha \ h- tc)]
ε = Qc/ UI
Qha \ h= Qc + IU
T tmax= Tha \ h+ 273 κ / σα² x [1-√2σα² / κx (th+273) + 1]
Imax =Bs / lαx [√2σα² / κx (th+273) + 1-1]
Iεmax =ασs (tha \ h- tc) / L (√ 14 + 0.5σ 0.5σα² (546+ tha \ h- tc)/ κ-1)
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