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
𝑄 = ℎ𝐴 𝑇𝑤 − 𝑇𝑓
Where:
Q = heat flow rate
Driving force
A = heat transfer area
h = heat transfer coefficient
𝑇𝑤 − 𝑇𝑓
𝑄=
Tw = temperature at solid wall
1
Tf = temperature at bulk fluid
ℎ𝐴
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
−∆𝑇
𝑄=
∆𝑥
𝑘𝐴
−∆𝑇
𝑄=
1
ℎ𝐴
Combined Heat Transfer
(flat slab):
−∆𝑇
𝑄=
1
∆𝑥
1
+
+
ℎ1 𝐴 𝑘𝐴 ℎ2 𝐴
Overall Heat Transfer Coefficient
Define: Overall Heat Transfer Coefficient, U
1
1
∆𝑥
1
=
+
+
𝑈𝐴 ℎ1 𝐴 𝑘𝐴 ℎ2 𝐴
−∆𝑇
𝑄=
1
𝑈𝐴
Combined Heat Transfer
(flat slab):
−∆𝑇
𝑄=
1
∆𝑥
1
+
+
ℎ1 𝐴 𝑘𝐴 ℎ2 𝐴
Overall Heat Transfer Coefficient
Define: Overall Heat Transfer Coefficient, U
1
1
∆𝑥
1
=
+
+
𝑈𝐴 ℎ1 𝐴 𝑘𝐴 ℎ2 𝐴
Inside overall heat transfer
coefficient, Ui
1
1
∆𝑥
1
=
+
+
𝑈𝑖 𝐴𝑖 ℎ𝑖 𝐴𝑖 𝑘𝐴𝐿𝑀 ℎ𝑜 𝐴𝑜
Outside overall heat transfer
coefficient, Uo
1
1
∆𝑥
1
=
+
+
𝑈𝑜 𝐴𝑜 ℎ𝑖 𝐴𝑖 𝑘𝐴𝐿𝑀 ℎ𝑜 𝐴𝑜
Overall Heat Transfer Coefficient
Define: Overall Heat Transfer Coefficient, U
1
1
∆𝑥
1
=
+
+
𝑈𝐴 ℎ1 𝐴 𝑘𝐴 ℎ2 𝐴
1
1
∆𝑥
1
=
+
+
𝑈𝑖 𝐴𝑖 ℎ𝑖 𝐴𝑖 𝑘𝐴𝐿𝑀 ℎ𝑜 𝐴𝑜
1
1
∆𝑥
1
=
+
+
𝑈𝑜 𝐴𝑜 ℎ𝑖 𝐴𝑖 𝑘𝐴𝐿𝑀 ℎ𝑜 𝐴𝑜
Relationship between the two:
1
1
=
𝑈𝑖 𝐴𝑖 𝑈𝑜 𝐴𝑜
Overall Heat Transfer Coefficient
Define: Overall Heat Transfer Coefficient, U
−∆𝑇
𝑄=
1
𝑈𝐴
𝑄 = 𝑈𝑖 𝐴𝑖 𝑇𝑖 − 𝑇𝑜 = 𝑈𝑜 𝐴𝑜 𝑇𝑖 − 𝑇𝑜
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)
𝑄 = 𝑈𝑖 𝐴𝑖 𝑇𝑖 − 𝑇𝑜 = 𝑈𝑜 𝐴𝑜 𝑇𝑖 − 𝑇𝑜
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:
𝑑𝑞 = −𝑚𝐴 𝑐𝑝𝐴 𝑑𝑇𝐴
= 𝑚𝐵 𝑐𝑝𝐵 𝑑𝑇𝐵
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:
𝑑𝑞 = −𝑚𝐴 𝑐𝑝𝐴 𝑑𝑇𝐴
= 𝑚𝐵 𝑐𝑝𝐵 𝑑𝑇𝐵
TB1
According to the
combined heat
transfer equation:
TB2
TA1
TA2
1
2
𝑑𝑞 = 𝑈 𝑇𝐵 − 𝑇𝐴 𝑑𝐴
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Making a heat balance across the entire pipe
for an area dA:
𝑑𝑞 = 𝑚𝐴 𝑐𝑝𝐴 𝑑𝑇𝐴
= −𝑚𝐵 𝑐𝑝𝐵 𝑑𝑇𝐵
According to the
combined heat
transfer equation:
1
1
𝑑𝑇𝐵 − 𝑑𝑇𝐴 = −𝑑𝑞
+
𝑚𝐵 𝑐𝑝𝐵 𝑚𝐴 𝑐𝑝𝐴
𝑑𝑞 = 𝑈 𝑇𝐵 − 𝑇𝐴 𝑑𝐴
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Equating the dq from the 2 equations below:
𝑑𝑇𝐵 − 𝑑𝑇𝐴
1
1
= −𝑈
+
𝑑𝐴
𝑇𝐵 − 𝑇𝐴
𝑚𝐵 𝑐𝑝𝐵 𝑚𝐴 𝑐𝑝𝐴
1
1
𝑑𝑇𝐵 − 𝑑𝑇𝐴 = −𝑑𝑞
+
𝑚𝐵 𝑐𝑝𝐵 𝑚𝐴 𝑐𝑝𝐴
𝑑𝑞 = 𝑈 𝑇𝐵 − 𝑇𝐴 𝑑𝐴
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Equating the dq from the 2 equations below:
𝑑𝑇𝐵 − 𝑑𝑇𝐴
1
1
= −𝑈
+
𝑑𝐴
𝑇𝐵 − 𝑇𝐴
𝑚𝐵 𝑐𝑝𝐵 𝑚𝐴 𝑐𝑝𝐴
Making a heat balance in the inlet and outlet:
𝑞 = 𝑚𝐵 𝑐𝑝𝐵 𝑇𝐵1 − 𝑇𝐵2 = 𝑚𝐴 𝑐𝑝𝐴 𝑇𝐴2 − 𝑇𝐴1
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Equating the dq from the 2 equations below:
𝑑𝑇𝐵 − 𝑑𝑇𝐴
1
1
= −𝑈
+
𝑑𝐴
𝑇𝐵 − 𝑇𝐴
𝑚𝐵 𝑐𝑝𝐵 𝑚𝐴 𝑐𝑝𝐴
Making a heat balance in the inlet and outlet:
Adding the 2 equations:
1
1
𝑇𝐴2 − 𝑇𝐴1 + (𝑇𝐵1 − 𝑇𝐵2 )
+
=
𝑚𝐵 𝑐𝑝𝐵 𝑚𝐴 𝑐𝑝𝐴
𝑞
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Equating the dq from the 2 equations below:
𝑑𝑇𝐵 − 𝑑𝑇𝐴
1
1
= −𝑈
+
𝑑𝐴
𝑇𝐵 − 𝑇𝐴
𝑚𝐵 𝑐𝑝𝐵 𝑚𝐴 𝑐𝑝𝐴
Substituting:
𝑑(𝑇𝐵 −𝑇𝐴 )
𝑇𝐴2 − 𝑇𝐴1 + (𝑇𝐵1 − 𝑇𝐵2 )
= −𝑈
𝑑𝐴
𝑇𝐵 − 𝑇𝐴
𝑞
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
𝑇𝐵2 − 𝑇𝐴2
𝑇𝐴2 − 𝑇𝐴1 + (𝑇𝐵1 − 𝑇𝐵2 )
ln
= −𝑈𝐴
𝑇𝐵1 − 𝑇𝐴1
𝑞
Integrating:
𝑑(𝑇𝐵 −𝑇𝐴 )
𝑇𝐴2 − 𝑇𝐴1 + (𝑇𝐵1 − 𝑇𝐵2 )
= −𝑈
𝑑𝐴
𝑇𝐵 − 𝑇𝐴
𝑞
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
𝑇𝐵2 − 𝑇𝐴2
𝑇𝐴2 − 𝑇𝐴1 + (𝑇𝐵1 − 𝑇𝐵2 )
ln
= −𝑈𝐴
𝑇𝐵1 − 𝑇𝐴1
𝑞
Rearranging:
𝑇𝐵2 − 𝑇𝐴2 − 𝑇𝐵1 − 𝑇𝐴1
𝑞 = 𝑈𝐴
ln 𝑇𝐵2 − 𝑇𝐴2 𝑇𝐵1 − 𝑇𝐴1
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Define: Logarithmic Mean Temperature Difference
∆𝑇𝐿𝑀
∆𝑇2 − ∆𝑇1
=
∆𝑇2
ln
∆𝑇1
TB1
TB2
TA1
TA2
1
𝑇𝐵2 − 𝑇𝐴2 − 𝑇𝐵1 − 𝑇𝐴1
𝑞 = 𝑈𝐴
ln 𝑇𝐵2 − 𝑇𝐴2 𝑇𝐵1 − 𝑇𝐴1
2
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Define: Logarithmic Mean Temperature Difference
∆𝑇𝐿𝑀
∆𝑇2 − ∆𝑇1
=
∆𝑇2
ln
∆𝑇1
TB1
TB2
TA1
TA2
1
𝑇𝐵2 − 𝑇𝐴2 − 𝑇𝐵1 − 𝑇𝐴1
𝑞 = 𝑈𝐴
ln 𝑇𝐵2 − 𝑇𝐴2 𝑇𝐵1 − 𝑇𝐴1
2
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Final Form:
𝑞 = 𝑈𝐴∆𝑇𝐿𝑀
TB1
TB2
TA1
TA2
1
𝑇𝐵2 − 𝑇𝐴2 − 𝑇𝐵1 − 𝑇𝐴1
𝑞 = 𝑈𝐴
ln 𝑇𝐵2 − 𝑇𝐴2 𝑇𝐵1 − 𝑇𝐴1
2
Log-mean Temperature Difference
Combined Heat Transfer (for Circular Pipe)
Final Form:
𝑞 = 𝑈𝐴∆𝑇𝐿𝑀
But still:
TB1
TB2
TA1
TA2
1
𝑞 = 𝑈𝑜 𝐴𝑜 ∆𝑇𝐿𝑀 = 𝑈𝑖 𝐴𝑖 ∆𝑇𝐿𝑀
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.
𝑞 = 𝑚𝑐𝑝 ∆𝑇 = 𝑈𝐴∆𝑇𝑙𝑚
∆𝑇𝑙𝑚
150 − 85 − (𝑇 − 50)
=
150 − 85
𝑙𝑛
𝑇 − 50
Log-mean Temperature Difference
Solution!
𝑞 = 𝑚𝑐𝑝 ∆𝑇 = 𝑈𝐴∆𝑇𝑙𝑚
∆𝑇𝑙𝑚
150 − 85 − (𝑇 − 50)
=
150 − 85
𝑙𝑛
𝑇 − 50
Substituting the values:
𝐽 1000 𝑔 250 𝑘𝑔
5.407
150 − 𝑇
𝑔𝐾 1 𝑘𝑔
ℎ𝑟
𝐽
3600 𝑠
2
= 230
5𝑚
∆𝑇𝑙𝑚
2
𝑠∙𝑚 ∙𝐾
1 ℎ𝑟
Shift solve for T!