# Heat Exchangers

Topics: Heat transfer, Trigraph, Gh Pages: 8 (469 words) Published: March 28, 2013
Problelm 3.6
Given data: Thi := 250 °C Tci := 30 °C mh := 0.15 kg s mc := 0.6 kg s

cph := 2000

J kg⋅ K

cpc := 1000

J kg⋅ K

U := 1000

W m K
2

Finding Heat Exchanger Area(A0):
2

Cc := mc⋅ cpc = 600

m ⋅ kg K⋅ s
3

Ch := mh⋅ cph = 300

m ⋅ kg K⋅ s
3

2

Tco :=

mh⋅ cph⋅ Thi + mc⋅ cpc⋅ Tci mh⋅ cph + mc⋅ cpc
4

= 103.333 °C

Tho := Tco

Q := Ch⋅ ( Thi − Tho) = 4.4 × 10 W Thi − Tco − ( Tho − Tci) Thi − Tco  ln  Tho − Tci    Cc Ch Ch Cc ε0 := if ( Cc < Ch) otherwise = 0.5

∆Tlm :=

= 105.798 ∆°C

C0 :=

Taking C0=Cmin/Cmax

Cmin :=

Cc if Cc < Ch Ch otherwise

Q Cmin⋅ ( Thi − Tci)

= 0.667

HEX effectiveness

ln  NTU :=

  1 − ε0  = 1.386
1 − C0

1 − C0⋅ ε0 

Hence A0 := NTU⋅

Cmin U

= 0.416 m

2

Problelm 3.7
Given data: Do := 25.4mm Tci := −7 °C mh := 0.6 kg s mc := 2.4 kg s Tco := 0 °C

cph := 4180

J kg⋅ K

cpc := 5200

J kg⋅ K

U := 120

W m K
2

Thi := 85 °C

a) Finding Heat Exchanger Area(A0) for a Cross Flow:
2 4 m ⋅ kg 2 3 m ⋅ kg

Cc := mc⋅ cpc = 1.248 × 10

K⋅ s

3

Ch := mh⋅ cph = 2.508 × 10

K⋅ s

3

Q := Cc⋅ ( Tco − Tci) = 8.736 × 10 W

4

Cmin :=

Cc if Cc < Ch Ch otherwise

Qmax := Cmin⋅ ( Thi − Tci) = 2.307 × 10 W

5

C0 :=

Cc Ch Ch Cc

if ( Cc < Ch) otherwise

= 0.201

Taking C0=Cmin/Cmax

ε0 :=

Q Qmax

= 0.379

HEX effectiveness

ln  NTU :=

  1 − ε0  = 0.496
1 − C0

1 − C0⋅ ε0 

Hence A0 := NTU⋅

Cmin U

= 10.375 m

2

Problelm 3.7
Given data: Do := 25.4mm Tci := −7 °C mh := 0.6 kg s mc := 2.4 kg s Tco := 0 °C

cph := 4180

J kg⋅ K

cpc := 5200

J kg⋅ K

U := 120

W m K
2

Thi := 85 °C

b) Finding Heat Exchanger Area(A0) for a 1-2 TEMA E:
2 4 m ⋅ kg 2 3 m ⋅ kg

Cc := mc⋅ cpc = 1.248 × 10

K⋅ s

3

Ch := mh⋅ cph = 2.508 × 10

K⋅ s

3

Q := Cc⋅ ( Tco − Tci) = 8.736 × 10 W

4

Cmin :=

Cc if Cc < Ch Ch otherwise

Qmax := Cmin⋅ ( Thi − Tci) = 2.307 × 10 W

5

C0 :=

Cc Ch Ch Cc

if ( Cc < Ch) otherwise

= 0.201

Taking C0=Cmin/Cmax

ε0 :=

Q Qmax

= 0.379

HEX effectiveness

Guess: given

NTU' := 0.1

( 1 + C0) + 1 + C0

(

2

)

0.5

0.5   1 + e− NTU'⋅ ( 1+C02)  2   = ⋅ 0.5    ε0 1 − e− NTU'⋅ ( 1+C02)   

NTU := find( NTU')

NTU = 0.501 Cmin U
2

Hence A0 := NTU⋅

= 10.463 m

,

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