# Beispielrechnungen zum theoretischen Wärmedurchgangskoeffizienten kth

Tab.1
Berechnungen
KaltwasserkreislaufWarmwasserkreislauf
Gegeben: ρ2=997,12 kg/m3 ρ1=994,39 kg/m3
λ2=0,610 J/(s·m·K) λ1=0,624 J/(s·m·K)
η2=9,46·10-4 kg/(m·s) η1=7,3·10-4 kg/(m·s)
cp,2=4,179 kJ/(kg·K) cp,1=4,176 kJ/(kg·K)
V2=0,8 l/min=0,8·10-3m3/min V1=1,5 l/min=1,5·10-3m3/min
mittlere Strömungsgeschwindigkeit $û 2 = V · 2 A q = 0,8 ⋅ 10 − 3 m 3 / s 60 ⋅ 26,70 ⋅ 10 − 6 m 2$   û2=0,499 m/s $û 1 = V · 1 A q = 1,5 ⋅ 10 − 3 m 3 / s 60 ⋅ 23,76 ⋅ 10 − 6 m 2$   û1=1,052 m/s
Prandtl-Zahl $Pr 2 = c p ,2 ⋅ η 2 λ 2 Pr 2 = ( 4,179 ⋅ 10 3 J / ( k g ⋅ K ) ) ⋅ ( 9,46 ⋅ 10 − 4 k g / ( m ⋅ s ) ) 0,610 J / ( s ⋅ m ⋅ K )$   Pr2=6,5 $Pr 1 = c p ,1 ⋅ η 1 λ 1 Pr 1 = ( 4,176 ⋅ 10 3 J / ( k g ⋅ K ) ) ⋅ ( 7,28 ⋅ 10 − 4 k g / ( m ⋅ s ) ) 0,624 J / ( s ⋅ m ⋅ K )$   Pr1=5,0
Reynolds-Zahl $Re 2 = ρ 2 ⋅ û 2 ⋅ d 2 η 2 Re 2 = ( 997,12 kg / m 3 ) ⋅ ( 0,499 m / s ) ⋅ ( 2 ⋅ 10 − 3 m ) 9,46 ⋅ 10 − 4 kg / ( m ⋅ s ) Re 2 =1052$   Re<2300 (=laminare Strömung) $Re 1 = ρ 1 ⋅ û 1 ⋅ d 1 η 1 Re 1 = ( 994,39 kg / m 3 ) ⋅ ( 1,052 m / s ) ⋅ ( 5,5 ⋅ 10 − 3 m ) 7,3 ⋅ 10 − 4 kg / ( m ⋅ s ) Re 1 =7882$   2300<Re<104 (=turbulente Strömung)
Nusselt-Zahl Laminare Strömung:     Nu2=3,5 Turbulente Strömung:     Nu1=54
Wärmeübergangskoeffizient $α 2 = Nu 2 ⋅ λ 2 d äq,2 = 3,47 ⋅ ( 0,610 J / ( s ⋅ m ⋅ K ) ) 2 ⋅ 10 − 3 m$   α2=1058 J/(s·m2·K) $α 1 = Nu 1 ⋅ λ 1 d äq,1 = 54,47 ⋅ ( 0,624 J / ( s ⋅ m ⋅ K ) ) 5,5 ⋅ 10 − 3 m$   α1=6180 J/(s·m2·K)

Theoretischer Wärmedurchgangskoeffizient

$Δ z = ( d i a − d a i ) 2 = ( 9,5 mm − 7,5 mm ) 2 = 1 mm$

λCu=393 J/(s·m·K)

$k th = 1 1 1058,35 J / ( s ⋅ m 2 ⋅ K ) + 10 − 3 m 393 J / ( s ⋅ m ⋅ K ) + 1 6179,71 J / ( s ⋅ m 2 ⋅ K )$

kth=902 J/(s·m2·K)