Q.59Consider the flow of a gas with density 1kg /m3, viscoity 1.5×10−5kg /(ms), speific heat C p=846 J / (kg K ) and thermal conductivity k=0.01665W / (mK ), in a pipe of diameter D=0.01m and lengthL=1m, and assume the viscosity does not change with temperature. The Nusselt number of a pipe with (L/D) ratio greater than 10 and reynolds number greater than 20000 is given by Nu=0.026ℜ0.8Pr1 /3 while the Nusselt number for a laminar flow for Reynolds

number less than 2100 and (ℜPrDL )<10 is Nu=1.86 [ℜPr (DL )]

1/3

. If the gas flows through the

pipe with an average velocity of 0.1m /s, the heat transfer coefficient is

(A) 0.68 W/(m2K)(B) 1.14 W/(m2K)(C) 2.47 W/(m2K)(D) 24.7 W/(m2K)

SOLUTION:

Nu={ 0.026ℜ0.8 Pr13 L

D>10 ;ℜ>20000

1.86 [ ℜPr (D /L ) ]13 ℜ<2100 ;ℜ Pr (D /L )<10

LD

= 1m0.01m

=100

ℜ= ρuDμ

=1×0.1×0.011.5×10−5

=66.67

<First condition for Nu is not valid>

Pr=Cp μ

k=846×1.5×10

−5

0.01665=0.7622

ℜPr (D /L )=66.67×0.7622×100−1=0.508

<Second condition for Nu is valid>

∴Nu=1.86 [ ℜPr (D /L ) ]13=1.86 [0.508 ]1/3=1.4843

Nu=h Dk

Heat transfer coefficient , h= NukD

=1.4843×0.01665[ WmK ]

0.01 [m ]=2.47 W

m2K

ANSWER: (C) 2.47 W/(m2K)

Q.61A countercurrent flow double pipe heat exchanger is used to heat water flowing at 1 kg/s from 40°C to 80°C. Oil is used for heating and its temperature changes from 100°C to 70°C. The overall heat transfer coefficient is 300 W/(m2 °C).

If it is replaced by a 1-2 shell and tube heat exchanger with countercurrent flow configuration with water flowing in shell and oil flowing in the tube, what is the excess area required with respect to the double pipe heat exchanger ?

The correction factor, Ft for LMTD (log mean temperature difference) based on the above double pipe heat exchanger is 0.5. The heat transfer coefficient remains unchanged, and the same inlet and outlet conditions are maintained.

Cp,water = 4180 J/kg °C, Cp,oil = 2000 J/kg °C

(A) 0 m2

(B) -20.15 m2

(C) 22.6 m2

(D) 9.69 m2

SOLUTION:

Case 1:

q=m c p∆T

q=1[ kgs ]×4180[ Jkg℃ ]× (80−40 ) [℃ ]=167.2kW

q=UA (LMTD )

LMTD=(100−80 )− (70−40 )

ln(100−80 )(70−40 )

=24.66℃

Counter current Flow

Cold fluid

Hot fluid

40°C

80°C

100°C

70°C

A1=167200 [W ]

300 [ Wm2℃ ]×24.66 [℃ ]

=22.6m2

Case2 :

corrected LMTD=F t×LMTD=0.5×24.66=12.33℃

A2=167200 [W ]

300 [ Wm2℃ ]×12.33 [℃ ]

=45.2m2

Excess arearequired=45.2−22.6=22.6m2

ANSWER: (C) 22.6 m2