Problem 2 A highly conducting thin wall L- 1 m long separates the hot and cold a

Question

Problem 2 A highly conducting thin wall L- 1 m long separates the hot and cold airstreams flowing on both sides parallel to the plate surface. The hot stream is at atmospheric pressure and has temperature Thot 200 C and velocity Uhor 50 m/s. The cold airstream (1 atm) has temperature Tcold-0C and velocity uc-15 m/s. Determine the following: (1) The average heat transfer coefficients for both airstreams (2) The average wall temperature, Ts (3) The total heat transfer rate between the streams per meter width of the separating plate. (Assume the wall is the arithmetic mean of Thor and Toold for calculation of physical properties)

Explanation / Answer

from properties table of air at atmospheric pressure

Temp (0C)

density (kg/m3)

thermal

conductivity(W/m.k)

thermal

diffusivity(m2s)

dynamic

viscosity(kg/m.s)

kinematic

viscosity(m2/s)

prandtl

number

first we will check whether the flow is laminar or turbulent or is in transition zone

Reynolds number at hot stream side = velocity*length/kinematic viscosity

substituting values from the table

Re = 50×1/3.45×10-5 = 14.49×105

similarly Reynolds number at cold stream side = 11.27×105

As on both sides Reynolds number is greater then critical reynolds number. So flow will be laminar up to some length of the plate and than will become turbulent i.e. flow is mixed type.

for mixed type flow reynolds number is given by, NuL =( 0.036ReL0.8-835)Pr0.33

Nusselt number on hot stream side =(0.036×(14.49×105)0.8-835)×(0.697)0.33

                                                        = 1969.47

Average heat transfer coefficient on hot stream side = Nu.k/L(Nu is nusselt number and k is thermal conductivity)

                                                                                    = 1969.47×0.037

                                                                                    = 72.87W/m2k

Nusselt number on cold stream side = (0.036×(11.27×105)0.8-835)×(0.736)0.33

                                                         = 1502.92

Average heat transfer coefficient on cold stream side = 1502.92×0.023

                                                                                     = 34.56W/m2k

Let the width of the conducting wall is 1m (you can use any width as it will be cancelled out in calculations)

Let the average wall temperature is T

Heat transferred from hot stream to wall = heat transferred from cold stream to wall

so the average temperature of the wall is 135.66C

Total heat transferred across the plate assuming unit width of plate is

= hh×A×(Th-Ts)

= 72.87×1×(200-135.666)

= 4688.42J

properties of air at 1 atm pressure

Temp (0C)

density (kg/m3)

specific heat (J.kg/k)

thermal

conductivity(W/m.k)

thermal

diffusivity(m2s)

dynamic

viscosity(kg/m.s)

kinematic

viscosity(m2/s)

prandtl

number

0 1.292 1006 0.023 1.81*10-5 1.72*10-5 1.33*10-5 0.736 200 0.7459 1023 0.037 4.9*10-5 2,57*10-5 3.45*10-5 0.697

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