### Document

```Quiz – 2014.02.05
An organic liquid enters a 0.834-in. ID horizontal steel tube, 3.5 ft
long, at a rate of 5000 lb/hr. You are given that the specific heat,
thermal conductivity, and viscosity of the liquid is 0.565 Btu/lb-°F,
0.0647 Btu/hr-ft-°F, and 0.59 lb/ft-hr, respectively. All these
properties are assumed constant. If the liquid is being cooled,
determine the inside-tube heat transfer coefficient using the
Dittus-Boelter equation:
N N u  0.023  N R e 
0.8
 N Pr 
n = 0.4 when fluid is heated
n = 0.3 when fluid is cooled
n
TIME IS UP!!!
Recall
Convection Heat Transfer
= ℎ  −
Where:
Q = heat flow rate
Driving force
A = heat transfer area
h = heat transfer coefficient
−
=
Tw = temperature at solid wall
1
Tf = temperature at bulk fluid
ℎ
Thermal
Resistance
Combined Heat
Transfer
Outline
3. Conduction Heat Transfer
4. Convection Heat Transfer
5. Combined Heat Transfer
5.1. Overall Heat Transfer Coefficient
5.2. Log-Mean Temperature Difference
6. Overall Shell Heat Balances
Overall Heat Transfer Coefficient
Conduction
Convection
−∆
=
∆

−∆
=
1
ℎ
Combined Heat Transfer
(flat slab):
−∆
=
1
∆
1
+
+
ℎ1   ℎ2
Overall Heat Transfer Coefficient
Define: Overall Heat Transfer Coefficient, U
1
1
∆
1
=
+
+
ℎ1   ℎ2
−∆
=
1

Combined Heat Transfer
(flat slab):
−∆
=
1
∆
1
+
+
ℎ1   ℎ2
Overall Heat Transfer Coefficient
Define: Overall Heat Transfer Coefficient, U
1
1
∆
1
=
+
+
ℎ1   ℎ2
Inside overall heat transfer
coefficient, Ui
1
1
∆
1
=
+
+
ℎ   ℎ
Outside overall heat transfer
coefficient, Uo
1
1
∆
1
=
+
+
ℎ   ℎ
Overall Heat Transfer Coefficient
Define: Overall Heat Transfer Coefficient, U
1
1
∆
1
=
+
+
ℎ1   ℎ2
1
1
∆
1
=
+
+
ℎ   ℎ
1
1
∆
1
=
+
+
ℎ   ℎ
Relationship between the two:
1
1
=

Overall Heat Transfer Coefficient
Define: Overall Heat Transfer Coefficient, U
−∆
=
1

=    −  =    −
Overall Heat Transfer Coefficient
Exercise!
Saturated steam at 267°F is flowing inside a steel pipe with
an ID of 0.824 in. and an OD of 1.05 in. The pipe is insulated
with 1.5 in. of insulation on the outside. The convective heat
transfer coefficient inside and outside the pipe is hi = 1000
Btu/hr/ft2/°F and ho = 2 Btu/hr/ft2/°F, respectively. The mean
thermal conductivity of the metal is 45 W/m/K or 26
Btu/hr/ft/°F, while that of the insulation material is 0.064
W/m/K or 0.037 Btu/hr/ft/°F. Calculate the heat loss for 1 ft
of pipe using resistances if the surrounding air is at 80°F.
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe Section)
=    −  =    −
Log-mean Temperature Difference
Combined Heat Transfer
(for Circular Pipe Section)
*The temperature of the fluid
and immediate surroundings
vary along the length.
Let:
TA1 = fluid temp. at pt.1
TB1
TB2
TA1
TA2 = fluid temp. at pt.2
TA2
1
2
TB1 = surr. temp. at pt.1
TB2 = surr. temp. at pt.2
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Making a heat balance across the entire pipe
for an area dA:
= −
=
Let:
TB1
TA2 = fluid temp. at pt.2
TA1 = fluid temp. at pt.1
TB2
TA1
TA2
1
2
TB1 = surr. temp. at pt.1
TB2 = surr. temp. at pt.2
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Making a heat balance across the entire pipe
for an area dA:
= −
=
TB1
According to the
combined heat
transfer equation:
TB2
TA1
TA2
1
2
=   −
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Making a heat balance across the entire pipe
for an area dA:
=
= −
According to the
combined heat
transfer equation:
1
1
−  = −
+

=   −
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Equating the dq from the 2 equations below:
−
1
1
= −
+

−

1
1
−  = −
+

=   −
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Equating the dq from the 2 equations below:
−
1
1
= −
+

−

Making a heat balance in the inlet and outlet:
=   1 − 2 =   2 − 1
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Equating the dq from the 2 equations below:
−
1
1
= −
+

−

Making a heat balance in the inlet and outlet:
1
1
2 − 1 + (1 − 2 )
+
=

Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Equating the dq from the 2 equations below:
−
1
1
= −
+

−

Substituting:
( − )
2 − 1 + (1 − 2 )
= −

−

Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
2 − 2
2 − 1 + (1 − 2 )
ln
= −
1 − 1

Integrating:
( − )
2 − 1 + (1 − 2 )
= −

−

Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
2 − 2
2 − 1 + (1 − 2 )
ln
= −
1 − 1

Rearranging:
2 − 2 − 1 − 1
=
ln 2 − 2 1 − 1
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Define: Logarithmic Mean Temperature Difference
∆
∆2 − ∆1
=
∆2
ln
∆1
TB1
TB2
TA1
TA2
1
2 − 2 − 1 − 1
=
ln 2 − 2 1 − 1
2
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Define: Logarithmic Mean Temperature Difference
∆
∆2 − ∆1
=
∆2
ln
∆1
TB1
TB2
TA1
TA2
1
2 − 2 − 1 − 1
=
ln 2 − 2 1 − 1
2
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Final Form:
= ∆
TB1
TB2
TA1
TA2
1
2 − 2 − 1 − 1
=
ln 2 − 2 1 − 1
2
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Final Form:
= ∆
But still:
TB1
TB2
TA1
TA2
1
=   ∆ =   ∆
2
Log-mean Temperature Difference
Exercise!
250 kg/hr of fluid A (cp = 5.407 J/gK) is to be cooled
from 150°C using a cooling fluid B which enters a
countercurrent double-pipe heat exchanger at 50°C
and leaves at 85°C. The total heat transfer area
available is 5 m2 and the overall heat transfer
coefficient is 230 W/m2K. Determine the outlet
temperature of fluid A assuming no phase change.
Log-mean Temperature Difference
Solution!
The heat used to increase the
temperature of fluid A is the same
heat transferred across the pipe.
=  ∆ = ∆
∆
150 − 85 − ( − 50)
=
150 − 85

− 50
Log-mean Temperature Difference
Solution!
=  ∆ = ∆
∆
150 − 85 − ( − 50)
=
150 − 85

− 50
Substituting the values:
1000  250
5.407
150 −
1
ℎ

3600
2
= 230
5
∆
2
∙ ∙
1 ℎ
Shift solve for T!
```