17
Dinescu Cristian 531-A Treapta Curtis kJ 10 3 J := p 0 40 bar ⋅ := t 0 450 := h 0 3330 kJ kg ⋅ := Calculul termic al treptei de reglare: h 1 2830 kJ kg ⋅ := h 0 h 1 − 500 kJ kg ⋅ = Caderea de entalpie pe treapta de reglare(Curtis): H i1 500 kJ kg ⋅ := H TR H i1 120 kJ kg ⋅ − 380 kJ kg = := Pierderile de presiune pe VIR=ventil cu inchidere rapida, si VR=ventile de reglare: k 1 0.02 := 0. 01 0. 03 .. pt ventile cu inchidere rapida k 2 0.03 := 0. 02 0. 05 .. pt v entie de reglare ∆p k 1 k 2 + ( ) p 0 ⋅ 2 bar ⋅ = := Se alege: ∆p 2 bar ⋅ := Se recalc uleaz a parametrii aburului viu: p' 0 p 0 ∆p − 38 bar ⋅ = := h' 0 h 0 3330 kJ kg ⋅ = := t=temperature(steam, p=38, h=h`0) t' 0 448.6 := v=volume(steam, p=38, t=t`0) v' 0.08 423 m 3 kg ⋅ := 1/17
