Finanzas

CALDERA

TURBINA

CONDENSADOR

1

2

3

4
CALDERA

TURBINA

CONDENSADOR

1

2

3

4
1.1-2 = copres. Isentrópicas
2-3 = adic. de calor P.const.
3-4 = expansiónisentrópicas
4-1 = sale q a P.const.
P1 = 0.07 Bar = 7 KPa
H1=h5 a 7 KPa
7.5-7168.75-x=7.5-7168.75-x x = h1= 162.55 KJKg
V1 =0.00107 m3Kg
En 2 P2 = 30 bar = 3MPa
Wbba = 0.00107 m3Kg ( 3000KPa – 7 KPa) = 3.203KJKg
H2 = h1 + Wbba = 162.55 KJKg + 3.203KJKg
= 165.75KJKg
En 3 S3 =S4P2 = P3
Considerando que el vapor sobrecalentado esta en el tiempo de saturación tenemos a 350 º C.
H3= 3116.1 KJKgS3 = 6.1856 KJKg
En 4 = P4 =P1 = 7 KPa
S4 =S3 = 6.1856KJKg
Interpolando:
Sf = 0.55628 KJKg
Sfg =7.7225 KJKg
La calidad es de X=72.89 %
hf = 245.2KJKg
hfg = 3862.64 KJKg
h4 = 245.2 KJKg + 0.7279 * 3862.64 KJKg
h4 = 3060.68 KJKg
Q entrante = h3 - h2 = 3116.1 KJKg - 165.75KJKg = 2950.35 KJKgQ salida = h4 - h1 = 3060.68 KJKg - 162.55 KJKg = 2898.13 KJKg
µ = 1 - Q entrante Q salida = 1.77 %
2.
En 1 = p1=5 KPa
Hf =h1 =137.75KJKg
Vf = V1 = 0.001005 m3Kg
En 2 = P2 = 2 MPa

Wbba = 0.001005 m3Kg ( 2000KPa – 5 KPa) = 2.005KJKg
H2= h1 + Wbba h2= 139.755 KJKg

En 3 P3 = 2 MPa a 300ºCH3= 3024.2KJKg
S3 = 6.7684 KJKg
En 4 :
P4=P1 = 5 KPa
S4=S3 = 6.7684KJKg
X= 0.7987

H4 =xHfg + Hf= 2063.31 KJKg
W turbina = H3 – H4
W turbina = 960.89 KJKg