Quiz – 2014.02.05

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 = 0.4 when fluid is heatedn = 0.3 when fluid is cooled

0.8Nu Re Pr0.023 nN N N

TIME IS UP!!!

Recall

Where:Q = heat flow rateA = heat transfer areah = heat transfer coefficientTw = temperature at solid wallTf = temperature at bulk fluid

𝑄=h 𝐴 (𝑇𝑤−𝑇 𝑓 )Convection Heat Transfer

𝑄=𝑇𝑤−𝑇 𝑓

1h𝐴

Driving force

Thermal Resistance

Combined Heat Transfer

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

Outline

Conduction

Overall Heat Transfer Coefficient

𝑄=−∆𝑇

(∆ 𝑥𝑘𝐴 )𝑄=

−∆𝑇

( 1h𝐴 )

Convection

𝑄=−∆𝑇

1h1𝐴

+∆𝑥𝑘𝐴 +

1h2𝐴

Combined Heat Transfer (flat slab):


𝑄=−∆𝑇

1h1𝐴

+∆𝑥𝑘𝐴 +

1h2𝐴

Combined Heat Transfer (flat slab):

Define: Overall Heat Transfer Coefficient, U

𝑄=−∆𝑇1

𝑈 𝐴

1𝑈𝐴=

1h1𝐴

+∆ 𝑥𝑘𝐴 +

1h2 𝐴



1𝑈𝐴=

1h1𝐴

+∆ 𝑥𝑘𝐴 +

1h2 𝐴

Inside overall heat transfer coefficient, Ui

Outside overall heat transfer coefficient, Uo

1𝑈 𝑖𝐴𝑖

=1h 𝑖𝐴𝑖

+∆ 𝑥

𝑘 𝐴𝐿𝑀+

1h𝑜 𝐴𝑜

1𝑈𝑜𝐴𝑜

=1h𝑖 𝐴𝑖

+∆ 𝑥

𝑘 𝐴𝐿𝑀+

1h𝑜 𝐴𝑜



1𝑈𝐴=

1h1𝐴

+∆ 𝑥𝑘𝐴 +

1h2 𝐴

1𝑈 𝑖𝐴𝑖

=1h 𝑖𝐴𝑖

+∆ 𝑥

𝑘 𝐴𝐿𝑀+

1h𝑜 𝐴𝑜

1𝑈𝑜𝐴𝑜

=1h𝑖 𝐴𝑖

+∆ 𝑥

𝑘 𝐴𝐿𝑀+

1h𝑜 𝐴𝑜

1𝑈 𝑖𝐴𝑖

=1

𝑈𝑜 𝐴𝑜

Relationship between the two:



𝑄=−∆𝑇1

𝑈 𝐴𝑄=𝑈 𝑖 𝐴𝑖 (𝑇 𝑖−𝑇𝑜 )=𝑈𝑜 𝐴𝑜 (𝑇 𝑖−𝑇 𝑜 )


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

𝑄=𝑈 𝑖 𝐴𝑖 (𝑇 𝑖−𝑇𝑜 )=𝑈𝑜 𝐴𝑜 (𝑇 𝑖−𝑇 𝑜 )

Combined Heat Transfer (for Circular Pipe Section)


*The temperature of the fluid and immediate surroundings vary along the length.

TA1 TA2

TB1 TB2

Let:

TA1 = fluid temp. at pt.1


TB1 = surr. temp. at pt.1

TB2 = surr. temp. at pt.21 2

Combined Heat Transfer (for Circular Pipe Section)


TA1 TA2

TB1 TB2

Let:



TB1 = surr. temp. at pt.1

TB2 = surr. temp. at pt.21 2

Combined Heat Transfer (for Circular Pipe)

Making a heat balance across the entire pipe for an area dA:


TA1 TA2

TB1 TB2

1 2


According to the combined heat transfer equation:

𝑑𝑞=𝑈 (𝑇 𝐵−𝑇 𝐴 )𝑑𝐴





According to the combined heat transfer equation:

𝑑𝑞=𝑈 (𝑇 𝐵−𝑇 𝐴 )𝑑𝐴𝑑𝑇 𝐵−𝑑𝑇 𝐴=−𝑑𝑞( 1�̇�𝐵𝑐𝑝𝐵

+1

�̇�𝐴𝑐𝑝𝐴 )



Equating the dq from the 2 equations below:

𝑑𝑞=𝑈 (𝑇 𝐵−𝑇 𝐴 )𝑑𝐴𝑑𝑇 𝐵−𝑑𝑇 𝐴=−𝑑𝑞( 1�̇�𝐵𝑐𝑝𝐵

+1

�̇�𝐴𝑐𝑝𝐴 )

𝑑𝑇 𝐵−𝑑𝑇 𝐴

𝑇 𝐵−𝑇 𝐴=−𝑈 ( 1

�̇�𝐵𝑐𝑝𝐵+ 1�̇�𝐴𝑐𝑝𝐴 )𝑑𝐴



Equating the dq from the 2 equations below:𝑑𝑇 𝐵−𝑑𝑇 𝐴

𝑇 𝐵−𝑇 𝐴=−𝑈 ( 1


Making a heat balance in the inlet and outlet:

𝑞=�̇�𝐵𝑐𝑝𝐵 (𝑇 𝐵1−𝑇 𝐵 2)=�̇�𝐴𝑐𝑝𝐴 (𝑇 𝐴2−𝑇 𝐴1 )




𝑇 𝐵−𝑇 𝐴=−𝑈 ( 1


Making a heat balance in the inlet and outlet:Adding the 2 equations:

1�̇�𝐵𝑐𝑝𝐵

+ 1�̇�𝐴𝑐𝑝𝐴

=(𝑇 𝐴2−𝑇 𝐴1 )+(𝑇 𝐵1−𝑇 𝐵2)

𝑞




𝑇 𝐵−𝑇 𝐴=−𝑈 ( 1


Substituting:

𝑑 (𝑇 ¿¿𝐵−𝑇 𝐴)𝑇 𝐵−𝑇 𝐴

=−𝑈 [ (𝑇 𝐴2−𝑇 𝐴1 )+(𝑇 𝐵1−𝑇 𝐵2)𝑞 ]𝑑𝐴 ¿



ln (𝑇 𝐵2−𝑇 𝐴2

𝑇 𝐵1−𝑇 𝐴1 )=−𝑈𝐴[ (𝑇 𝐴2−𝑇 𝐴1 )+(𝑇 𝐵1−𝑇 𝐵2)𝑞 ]

Integrating:

𝑑 (𝑇 ¿¿𝐵−𝑇 𝐴)𝑇 𝐵−𝑇 𝐴

=−𝑈 [ (𝑇 𝐴2−𝑇 𝐴1 )+(𝑇 𝐵1−𝑇 𝐵2)𝑞 ]𝑑𝐴 ¿



ln (𝑇 𝐵2−𝑇 𝐴2

𝑇 𝐵1−𝑇 𝐴1 )=−𝑈𝐴[ (𝑇 𝐴2−𝑇 𝐴1 )+(𝑇 𝐵1−𝑇 𝐵2)𝑞 ]

Rearranging:

𝑞=𝑈𝐴[ (𝑇 𝐵2−𝑇 𝐴2 )− (𝑇 𝐵1−𝑇 𝐴1 )ln [ (𝑇 𝐵2−𝑇 𝐴2 )/ (𝑇 𝐵1−𝑇 𝐴1) ] ]



Define: Logarithmic Mean Temperature Difference


∆ 𝑇 𝐿𝑀=∆𝑇 2−∆𝑇 1

ln( ∆𝑇2

∆𝑇1 ) TA1 TA2

TB1 TB2

1 2



Define: Logarithmic Mean Temperature Difference


∆ 𝑇 𝐿𝑀=∆𝑇 2−∆𝑇 1

ln( ∆𝑇2

∆𝑇1 ) TA1 TA2

TB1 TB2

1 2




TA1 TA2

TB1 TB2

1 2

Final Form:

𝑞=𝑈𝐴∆𝑇 𝐿𝑀



TA1 TA2

TB1 TB2

1 2

Final Form:

𝑞=𝑈𝐴∆𝑇 𝐿𝑀

𝑞=𝑈𝑜 𝐴𝑜∆𝑇 𝐿𝑀=𝑈 𝑖 𝐴𝑖∆𝑇 𝐿𝑀

But still:


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.


Solution!

∆ 𝑇 𝑙𝑚=(150−85 )−(𝑇 −50)

𝑙𝑛( 150−85𝑇 −50 )

𝑞=𝑚𝑐𝑝∆𝑇=𝑈𝐴 ∆𝑇 𝑙𝑚

The heat used to increase the temperature of fluid A is the same heat transferred across the pipe.


Solution!∆ 𝑇 𝑙𝑚=

(150−85 )−(𝑇 −50)

𝑙𝑛( 150−85𝑇 −50 )𝑞=𝑚𝑐𝑝∆𝑇=𝑈𝐴 ∆𝑇 𝑙𝑚

5.407 𝐽𝑔𝐾 (1000 𝑔1𝑘𝑔 )( 250𝑘𝑔h𝑟 ) (150−𝑇 )=230 𝐽

𝑠 ∙𝑚2 ∙𝐾(5𝑚2 )( 3600 𝑠1h𝑟 )∆𝑇 𝑙𝑚

Substituting the values:

Shift solve for T!

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Quiz – 2014.02.05