Fisica Tecnica - Lezione Teorica 23

Scambiatori di calore

Q̇ > 0

C = ṁC_p

Q̇ = C_c(T_Cin - T_Cout) = C_H(T_Hin - T_Hout)

UA Conduttanza totale

Q̇ = UA · ΔT_H

ΔT_12H = ΔT_I - ΔT_O

ln ^ΔT_I/_ΔTO

Applico il bilancio scritto in precedenza

δQ̇ = C_C(T_C(x) - T_C(x + dx))

δQ̇ = C_H(T_H(x) - T_H(x + dx))

δQ = c_CdT_C

δQ = -c_HdT_H

δQ = USA (T_H(x) - T_C(x))

ΔT(x) = T_H(x) − T_C(x)

dΔT(x) = dT_H(x) − dT_C(x)

dT_C = - δQ̇ / c_C

dT_H = - δQ̇ / c_H

δQ̇ ( 1 / c_C − 1 / c_H )

dΔT(x)( 1 / c_C − 1 / c_H ) = ΔT(x) USA

dΔT(x) / ΔT(x) ( 1 / c_C − 1 / c_H ) USA

∫_x=0^L dΔT(x) / ΔT = ∫_x=0^L ( 1 / c_C − 1 / c_H ) USA

ln( ΔT_L / ΔT_0 ) = ( 1 / c_C − 1 / c_H ) UA

Q̇ = c_C (T_CM − T_Ci)

Q = c_H (T_Hi − T_HM)

1 / c_C = (T_CM − T_Ci ) / Q̇

1 / c_H = (T_Hi − T_HM ) / Q

1 / c_C − 1 / c_H

= 1 / Q̇

= ΔT_L − ΔT_0

Q̇ = UA( (ΔT_L − ΔT_0) / ln( ΔT_L / ΔT_0 ) ) = ΔT_LM

CICLO J-B

im T₃oT₂ ne verso un po' x riscaldato