Sciąga pozary, r = P/Rz ×T \\\ T = 273,15 + ts \\\ Rz = Mr/Mz \\\ Mr = 8314,7 J/kmol×K \\\ Mz = 0,2×(2×16)+0,78×2×14+0,015×(1×12+2×16)+0,005


ρ = P/Rz ⋅T \\\ T = 273,15 + ts \\\ Rz = Mr/Mz \\\ Mr = 8314,7 J/kmol⋅K \\\ Mz = 0,2თ(2თ16)+0,78თ2თ14+0,015თ(1თ12+2თ16)+0,005(12+4თ1) \\\ ၲ = P/RaთTv\\\ Tv= (1+0,6 x)T \\\\ Ra = 287,04 J/kgთk \\\\ X= 0,622 Pp/P-Pp \\\\ Pp = 610,5 exp (17,27თtw/237,3+tw)-0,000644თP(ts-tw)\\\\ Wrz = Wm თ (A - 0,4)/A\\\\ A>8 m2 k=1 A<8 m2 k=1,14\\\ Wrz=Wzm*k\\\\ V=A*Wm.\\\\ Vn*ၲn=V*ၲ \\\\\ m.=V*ၲ\\\\ Wm.=ၓAiWi / ၓAi\\\\ Vi = Ai*Wi\\\\\\ gdy w < 1 m/s,Ksr=F/ၴ=(0,008+0,485ზw)(36,5-ts) gdy w > 1 m/s,Ksr=F/ၴ=(0,108+0,385ზw)(36,5-ts)\\\\ c= ზၣRT\\\\ R=288,3 J/(kgK),\\\\\ V =c*(t2-t1/t2+t1) \\\\\ Pc = 300,5თ9,80655 = 2916,8986 PaPs = 280თ9,80655 = 2745,8986 Pa\\\\\ pc = ps+pd \\\\\ pd = ၲთw2max)/2\\\\\ 1Atm = 101324,72 Pa\\\\\ 1 bar = 105Pa\\\\\ 1 at = 9,80655თ104Pa\\\\\ 1 mm H2O = 9,80665 Pa\\\\\ h= L თ1/5\\\\\\ Rozw: ၄t=t2-t1 ; ၄t= 75ႰC\\\\\75ႰC- 5,4 mV (przed pomiarem)\\\\\ x - 3,12 mV (pomiar)\\\\\\x= (3,12თ75)/ 5,4 = 43,3 ႰC\\\\\\ 1 Tr = 133,3241 Pa \\\\\ ၪ=(Pp/Ppn);\\\\\ Pp = 800 თ133,322 = 106657,6 Pa\\\\\ Ppn = 610,5 exp (17,27თts/237,3+ts)\\\\\ ၪ=(Pp/Ppn);\\\\\ =0,622 Pp/P-Pp \\\\ Ppn = 610,5 exp (17,27თts/237,3+ts)\\\\ Ppn = 610,5 exp (17,27თts/237,3+ts)\\\\\\ ၪ=(Pp/Ppn);\\\\ 1 mm H2O = 9,80655Pa\\\\ ) Rfn = ၄p/Vn2\\\\\ ) Rf = (ၲn/ၲ) თ Rfn \\\\\\\ R = Rf/ၲn\\\\\\ ၡf = RfთA3/BთL\\\\\ ၡf = ၬfთ(ၲn/8) \\\\\ ၬf = 8ၡf / ၲn\\\\\ h = cpa თ ts + x თ (cpw თ ts + ၧp)



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