ME3122E - Tutorial Solution 3

ME3122 Solutions to Tutorial Set 3

1. Using the energy integral equation determine an expression for the heat transfer coefficient by assuming the following velocity and temperature profiles:

constant ( W) /( W) / t uu_  and T T T_T y



where _t is the thermal boundary layer thickness.

Solution:

Energy Integral Equation:

Given W

,

1

.(

)

W W t t

T

T

y

T

T

T

T

_

T



y



















Therefore, W 1 1 W W t T T T T y T T T T



     _{   }  

Substituting the above and

u



u

_ into energy integral equation, we get

Integrating 2 1 ( ) 2 ( ) 4 t p t t p u d k dx C d k dx C u          or t2

4

;

t

4

p p

kx

kx

C u

C u







_



_





Now:

(

w

)

w

T

k

h T

T

y









_





 

qcoduction = qconvection 1/ 2 _{1/ 2}

1

(

)

(

)

4

4

w w t p t p

h T

T

k T

T

C u k

k

kx

h

k

C u

x









    



 

















_



_























0 0  

_











y

y

T

dy

T

T

u

dx

d

t













t p W t W

C

k

T

T

dy

y

dx

d

T

T

u

t









₁

1

0



































_  



2. Glycerine at 30C flows past a 30cm square flat plate at a velocity of 1.5 m/s. The drag force is measured as 10.98 N (for both sides of the plate). Calculate the heat transfer coefficient for such a flow system. At 30C, properties of glycerine are:

3 2 1 2 5 8 / ; 2 .4 5 5 / ; 0 .0 0 0 5 0 / ; 0 .2 6 8 / ; P r 5 3 8 0 p k g m C k J k g K v m s k W m K       Solution: Reynolds Analogy: 2/3 fx x

C

St Pr =

2

; where x x p h St C u



  For one side of the plate:

2 2 1 2 2 2 10.98 61 / 2 2 0.3 2 61 0.0431 1258 1.52 x w w f F N m A C u    _          Substituting: 5 2 / 3

0.5 0.0431

7.015 10

5380

x

St









 Therefore: 5 5 2 7.015 10 7.015 10 1258 2445 1.5 323.68 / x p h C u W m K             

3. Atmospheric air at 25C flows over a plate at a velocity of 60 m/s. The plate of width 1m and length 0.75m is maintained at a uniform temperature of 230C by independently-controlled, electrical strip heaters, each of which is 50mm long in the direction of the airflow.

a) Determine at which heater does the flow undergo a transition from laminar to turbulent flow.

b) Determine at which heater is the heat input a maximum c) Determine the value of this heat input

d) Determine the heat input for the first heater e) Determine the heat input for the first three heaters f) Determine the heat input for the entire plate

Solution: Given: 5 2 3 25 ; 230 ; 1 ; 0.75 60 / ; Heater length 0.05 ( ) / 2 127.5 400 ( ) 1.0135 / ; 2.286 10 / . ; 3.365 10 / ; 0.8824 / ; Pr 0.688 w f w p T C T C w m L m u m s m T T T C K

Rogers and Mayhew tables air gives

c kJ kgK kg m s k W mK kg m





                         a)

Re

c

 

5 10

5





u x

 c

/



. Substituting gives xc =0.216m

Since each heater is 0.05m, therefore transition to turbulent flow occurs at heater No.5

b) hx at 6th heater (turbulent flow) is highest, because at 5th heater hx is the average of the

laminar and turbulent flow values ---as shown in graph.

hx Laminar b.l _h x ~ x-0.5 Turbulent b.l hx ~ x-0.2 Air u∞ = 60 m/s T∞ = 25C x x = L xc xc _x Heaters 1 2 3

c) To find hx over the 6th heater, find the hx at its midpoint, i.e. at x=0.275m.

d) Over first heater --- laminar flow

1/ 2 1/3

2

.

0.664 Re Pr for 0.05 for laminar b.l. flow

: 134.1 / 134.1 (0.05 1)(205) 1375 x x x x h x average Nu k x L Substituting h W m K Q W         

e) Over first 3 heaters--- laminar flow

1/ 2 1/ 3 2 . 0.664 Re Pr 2 3 0.05 0.15 : 77.5 / 77.5 (0.15 1)(205) 2383 L L x x x L L h x Nu Nu k for x L m h W m K Q W             

f) Over the entire plate flow is laminar-turbulent

0.8 1/ 3 . 0.8 1/ 3 0.8 (0.037 Re 871) Pr (mixed) laminar-turbulent b.l. flow

(0.037 Re 871) Pr ; 0.75 : Re 98139 ; 109.3 . .(230 25) 16.81 L L L L L L L L h L Average Nu k for k h L for x L m h Q h L kW            

Note: for Part c, to be precise, apply

0.8 1/ 3 0.8 1/ 3 2 . 0.0296 Re Pr . 0.0296 0.688 @ 0.275 140.6 / .1(230 25) 1441 x x x x x x h x local Nu k

for turbulent b l flow u x k h x x h W m K Q h l W





       _{ }        

4. A light breeze at 4.47 m/s blows across a metal building. The height of the building is 3.7m and the width is 6.1m. A net energy flux of 347 W/m2 from the sun is absorbed in the wall and subsequently dissipated to the surroundings by convection. Assuming that the air is at 27C and 1 atmosphere, estimate the average temperature that the wall will attain under equilibrium conditions.

Solution:

2

27 300 ; 6.1 ( 6.1 3.70) 4.47 / ; Constant 347 /

and ( ) / 2

are unknown and to be determined.

w f w

T C K L m area

u m s q W m

wall temperature T film temperature T T T

             5 2 3 Approach: / . ) 300 Properties of air at 300K: 1.846 10 / . ; 2.264 10 / ; 1.177 / ; Pr 0.707 1.177 4.47 6.1 Re 1. f L

Approximation assumption is reqd

i First assume T T K or any reasonable value

kg m s k W mK kg m u L









                6 5 5 1.74 10 5 10 846 10

laminar - turbulent flow over flat wall

      x Air u∞= 4.47m/s T∞= 27C, Tw=? qw” =347 W/m2

0.8 1/ 3 .

2

(0.037 Re 871) Pr

above expression is valid for isothermal wall but is assumed also valid for constant heat flux wall

10.68 / . 347 ( ) .( 27) 59.5 L L L L L w w h L average Nu k h W m K q A A h A T T C           

) To improve accuracy, we may repeat calculations with ( ) / 2 (27 59.5) / 2 43.25 316.25 f w ii T T T C K        

5. Engine oil at the rate of 0.02 kg/s flows through a 3-mm diameter tube, 30m long. The oil has an inlet temperature of 60C, while the tube wall temperature is maintained at 100C by a stream condensing on its outer surface.

a) Estimate the average heat transfer coefficient for the internal flow of oil. b) Determine the outlet temperature of the oil.

Take the properties of engine oil to be: cp = 2131 J/kgK,  = 852 kg/m3,  = 0.375x10-4

m2/s, k = 0.138 W/mK, Pr = 490.

Solution:

Tube flow, constant Tw case (neglect entrance effects)

Properties of engine oil are given as : cp = 2131 J/kgK,  = 852 kg/m

3_{,  = 0.375x10}-4 _m2_{/s, k} = 0.138 W/mK, Pr = 490. 1 2 Tb1=600C Tb2=? Tw=100C=const d=0.003m L=30m Engine oil =0.02 kg/s Tb1=60C Tb2=? Tw=const

7 w 2 4 Re 4 0.02 10 _{256.6 <2000} 0.374 852 3

the flow is laminar

3.66 (const T case) 3.66 0.138 168.4 / 0.003 d u d m d hd Nu k h W m K                      2 1 2 2 0 Lecture notes ( ) exp . 100 0.003 30 168.4 exp 100 60 0.02 2131 86.9 p b b LMTD i p b b q mC T T hA T T dL h T mC T T C





       _       _{ }  _               

6. Water at 25C enters a tube of diameter 0.02 m at mass flow rate of 0.01 kg/s and is to be heated to a temperature of 65C. The outside of the tube is wrapped with an electric heating element that produces a uniform heat flux of 20 kW/m2 over its entire length. Determine the length of the tube required and the inner surface temperature of the tube at the outlet. Determine the length of the tube required and the inner surface temperature of the tube at the outlet if the flow rate of water is increased to 0.08kg/s. Comment on the use of your equation to determine h.

Tb1 Tb2 q"=20 kW/m2 Water T∞ = 25 C Tb1=25 C Tb2=65 C Tw T x q"=const

Solution: 1 2 6 6 0.02 , 0.01 / 25 ; 65 (1 / 2)(25 65) 45

45 : from the tables for water 4181 / ; 594 10 / 0.638 / , Pr 3.89 4 4 0.01 Re 594 10 0.02 1071.6 2000 the fl b b b p d d m m kg s T C T C T C At C c J kg C kg ms k W mK u d m d                                      w 2 ow is laminar 4.36 for constant q 4.36 0.638 / 0.02 139 / d hd Nu k h W m K       2 2 " 1 ) ( ) 20000 0.02 0.01 4181 40 1.331 ) ( ) ;

( ) constant for constant q" tube flow " 20000 ( ) 143.8 139.08 143.8 65 208.8 p b b w b w b w b w i q dL mC T T L L m ii Q hA T T T T Q q T T hA h T C                            0.8 0.4 0.8 0.4 2 ) 0.08 / 8 0.01 / Re 8 1071.6 8572.8 2000 flow is turbulent, Nu number eqn is

0.023Re Pr

--Dittus & Boelter eqn for turbulent tube flow 0.638 (8572.8) (3.89) 0.023 0.02 1770 / ' 8 10 d d d iii m kg s kg s hd Nu k h h W m K L L                  ' ' .64

for the same entry and exit bulk temperatures ( w b)

m