University of Rome Tor Vergata

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1 University of Rome Tor Vergata Faculty of Engineering Department of Industrial Engineering THERMODYNAMIC AND HEAT TRANSFER HEAT TRANSFER dr. G. Bovesecchi (7249) Accademic Year

2 Ex. 7 A window (.5m x 0.8m), 8mm thin is made of glass (λ=0.78 W/mK), the inner temperature is 20 C and the outer temperature is -0 C. Calculate the heat transfer rate in steady state and the temperature at the interface between air and glass (inside and outside). Assume the inner convective coefficient is 0 W/m 2 K and the outer is 40 W/m 2 K. Solution Assuming that the inner and the outer temperature are constant in time we can assume that the problem is in steady state. The heat transfer rate is given by: Q = UA T 2

3 This equation considers the two mechanisms that occur during heat propagation, convection and conduction. U is the overall heat-transfer coefficient, in this coefficient both of the mechanism are considered. A is the area through the heat is exchanged and T is the temperature difference between inside and outside temperature. The overall heat-transfer coefficient can be evaluated from the equation: U = + s + h i λ g h o U = 0Wm 2 K m 0.78Wm K + 40Wm 2 K = 7.39Wm 2 K 3

4 And so the heat rate is: Q = UA T = 7.39Wm 2 K.2m 2 ( 20 +0) C = W The temperatures (inner and outer) on the interface between air and glass can be determined from: Q = UA T = h i A( T i T ) = h o A( T 2 T ) o T = T i Q h i A = 20 C W = 2.7 C.2m 2 0Wm 2 K T 2 = Q h o A T = W 0 C = 4.46 C o.2m 2 40Wm 2 K 4

5 Ex. 8 Consider the same condition of the previous exercise, but the window (.5 x 0.8 m 2 ), is a multilayer of two glass slabs (λ=0.78 W/mK), 4mm thin, with air between them (s=0mm, λ=0.026 W/mK). Calculate the heat transfer rate in steady state. Solution This problem is similar to the previous one, we can approach it in the same way. Q = UA T U = h i + s g λ g + s a λ a + s g λ g + h o 5

6 U = h i + 2 s g λ g + s a λ a + h o U = 0Wm 2 K m 0.78Wm K m 0.026Wm K + 40Wm 2 K U =.92Wm 2 K Q = UA T =.92Wm 2 K.2m 2 ( 20 +0) C = 69.W The temperature at the interfaces between glass and air (inside) and at the interface between air and glass (outside) can be determined from: 6

7 The temperature at the interfaces between glass and air (inside) and at the interface between air and glass (outside) can be determined from: Q = h i A( T i T ) = s g ( ) = s a A T λ T 2 g so: T = T i Q h i A = 20 C 69.W T 2 = T s g Q λ g A = 4.23 C T 3 = T 2 s a Q λ a A = 3.94 C T 4 = T 3 s g Q λ g A = 8.2 ( ) = s g λ a A T 2 T 3 ( ) λ g A T 3 T 4 = 4.23 C.2m 2 0Wm 2 K m 69.W = 3.94 C.2m Wm K m 69.W = 8.2 C.2m Wm K m 69.W = 8.5 C.2m Wm K 7

8 Ex. 9 The temperature inside an oven (40x40x60 cm 3 ) is 950 C while the outside temperature is 20 C. The oven walls are made of firebrick (λ= W/mK), thickness 24cm. A 3cm thickness-insulating layer (λ=0.05 W/mK) is on the outer side of the wall. The convective coefficient is, inside 30 W/m 2 K and outside 8 W/m 2 K. Calculate the heat transfer rate, the oven wall inner, outer temperature and the temperature at the interface between the firebrick and the insulating layer. 8

9 Solution In this problem we don t consider the edge effect. In order to apply the: Q = UA T We must calculate the oven surface, which is: S = 2 ( )m ( )m 2 =.28m 2 the overall heat-transfer coefficient is: U = h i + 3 i= s i λ i + h o 9

10 U = 30Wm 2 K m Wm K + U = Wm 2 K m 0.05Wm K + 8Wm 2 K = So the heat flux is: Q = UA T = Wm 2 K.28m 2 ( ) C = 90W to determinate the requested temperature value we can take into account that Q = h i A( T i T ) = λ b A( T s T ) 2 = λ i A( T b s 2 T ) 3 = h o A( T 3 T ) o i 0

11 And from this relation we have: Q = h i A T i T T = T i Q h i A = 950 C ( ) Q = λ b A( T s T ) 2 b 90W 30Wm 2 K.28m 2 = 99 C T 2 = T s b Q λ b A = 99 C m 90W Wm K.28m 2 = 696 C T 3 = T 2 s i Q λ i A = 696 C ( ) Q = λ i A T s 2 T 3 i m 90W 0.05Wm K.28m 2 = 38 C

12 Ex. 20 Hot water flows in a pipe at 80 C and.5m/s. The convection coefficient is 6540 W/m 2 K. The pipe is made of stainless steel (λ=50 W/mK), the inside radius is 78mm, and the thickness is 6mm. The pipe is surrounded by air at 27 C and h= 23 W/m 2 K. Calculate the thermal heat rate per meter. Solution the heat exchange direction is radial, through the lateral surface of the pipe (cylinder): Q = U A T = U ( π d o L) T where d0 is the outer diameter. 2

13 The overall heat-transfer coefficient is: U = r o + r o ln r o + r i h i λ ss r i h o = m m 6540Wm 2 K + ln m m m 50Wm K + 23Wm 2 K = U = 22.84Wm 2 K And the heat flux per unit length is: Q L = U ( π d o ) T Q L = 22.84Wm 2 K π m ( 80 27) C = W m 3

14 Ex. 2 To a cylindrical metallic conductor (0mm diameter and long m), made of stainless steel (ρ= Ωm, λ=5 W/m K), is applied 00A current and is cooled by an airflow rate (u=0m/ s). The airflow rate is normal to the cylinder axis. The air is at 27 C. calculate the convective coefficient, the temperature on the cylinder axis and on the surface. Solution Nu d = 0.93 Re d 0.68 Pr /3 the convective coefficient comes from the Nusselt empirical equation using the thermophisical properties of air ad 27 C. 4

15 Nu d = h D λ air = 0.93 Re 0.68 d Pr /3 Nu d = h D = 0.93 λ air u D 0.93 υ h = D from the table for air at 27 C: 0.68 u D υ 0.68 Pr /3 λ air Pr /3 λ air = 0.026Wm K υ air = m 2 s Pr =

16 0.68 0m s 0 2 m m 2 s / W m K h = 0 2 m h = 00.4Wm 2 K The temperature propagation in the cylinder is radial. From the center to the border. We must apply the Fourier Law to the cylinder using the cylindric coordinate. = r d dr r dt dr + q g λ = 0 where q g is the heat flof in the cylinder per unit volume. Q q g = Q V = π r 2 H 6

17 r d dr r dt dr + q g λ = 0 integrating the previous equation: T ( r) = q g 4 λ r 2 + C lnr + C 2 Now we have to apply the boundary condition: r = 0 r = R dt ( r) dr T r = 0 ( ) = T ( R) 7

18 from the first of the 2 condition C=0 from the second: T ( R) = q g 4 λ R + C 2 C 2 = T ( R) + q g 4 λ R T(R) is the condition on the cylinder surface. Q = h S T = h S ( T ( R) T ) T ( R) Q = h S + T 8

19 Now we can write T as a function of radius r T ( r) = q g 4 λ r 2 + q g 4 λ R2 + Q h S + T T ( r) = q g ( 4 λ R2 r 2 ) Q + h S + T taking into account the definition of becomes: T ( r) = q g ( 4 λ R2 r 2 ) + q V g h S + T = q g the above equation ( ) + q ( π r 2 H ) g = q g 4 λ R2 r 2 ( ) + T = h 2 π r H 9

20 T ( r) = q g ( 4 λ R2 r 2 ) + q R g 2 h + T and q g is: q g = Q V = Q π R 2 H = ρ I 2 H π R 2 π R 2 H = ρ I 2 = Ωm 00 2 A 2 = W m 3 ( π ( m) 2 )2 ( π R 2 ) = 2 20

21 Ex. 22 A spherical tank (d=.5m th=3cm λ=5 W/m K) has ice at 0 C inside. The ambient temperature is 24 C. The outer convective coefficient is 8 W/m 2 K and the ice latent heat of melting is 334 kj/kg. Calculate the heat flux, and the melting ice mass per second. Solution The heat flux given from the ambient to the tank is: where A is the surface of the sphere. The conduction in through the thick of the tank is given by the Fourier equation using sphere coordinate. Q = h A T = h A ( T amb T ) e () Q = λ A dt dr = λ ( 2 4 π r ) dt dr 2

22 that can be solved dividing the variable and integrating between ri and re and between Ti and Te. Q dr r 2 = 4 λ π dt Q r e dr T = 4 λ π e dt r i r 2 T i Q = 4 λ π ( T T ) e i r i r e (2) the boundary conditions is given by equation () and the difference Tamb Te can be determined. 22

23 ( T amb T ) Q e = h A = Q ( ) h 4 π r e 2 (3) if in equation (2) we put into evidence the T. T e T i = Q 4 λ π r i r e (4) summing equation 3 to 4 T amb T i = Q ( ) Q h 4 π r e 2 4 λ π r i r e Q = ( ) 4 π T amb T i h r 2 e λ r i r e 23

24 assuming the inner temperature of the tank equal to the ice temperature we can calculate the heat flux. Q = 4 π ( 24 0) C 8W m 2 K 0.78m 5W m K 0.75 = 493kW 0.78 m ( ) 2 the melting ice mass per second is: Q = m h mel m = Q h mel = 493kW = 4.47 kg s 334kJ kg 24

25 University of Rome Tor Vergata Faculty of Engineering Department of Industrial Engineering THERMODYNAMIC AND HEAT TRANSFER LUMPED PARAMETER dr. G. Bovesecchi (7249) Accademic Year

26 Lumped Parameter Ex. 23 A 5mm diameter sphere is made of copper and is at 70 C (ρ=8928 kg/m 3, λ=398 W/m K, cp=376 J/kg K) is cooled by a mass air flow rate at 30 C. Calculate the convective coefficient if the sphere temperature after 2 minute is 50 C. Furthermore derive the analytical expression of T function of time. Note: assume that the problem can be represented by lumped parameter system. Solution First of all we must derive the analatical expression of T as function of time. From the thermal balance: ρcv dt dt = ha ( T T ) 2

27 Lumped Parameter separating the variable: dt ( T T ) = ha ρcv dt ( ) ( T T ) = ha d T T ρcv dt integrating both therm of the equation: ln( T T ) = ha ρcv t + C () to determine the integration constant C we consider that: for t = 0 then T = T 0 3

28 Lumped Parameter And C is: C = ln( T 0 T ) (2) Putting (2) in equation () we can derive the temperature function of time. ln( T T ) = ha ρcv t + ln ( T T ) 0 ln T T T 0 T = ha ρcv t T T = exp ha T 0 T ρcv t 4

29 Lumped Parameter The convective coefficient is: h = ρ c V A t ln T T T 0 T = h = 8928kg m J kg K 4 3 π ( ) m 3 4 π ( ) 2 m 2 20s ln = h = 48.5 W m 2 K This result is working only is the assumption of lumped parameter is respected. 5

30 Lumped Parameter We must check if the Biot number is lower than 0. Bi < 0. Bi = h V λ A < 0. Bi = 48.5W m 2 K 4 3 π ( ) m 3 = W m K 4π ( ) 2 m 2 6

31 Lumped Parameter Ex. 24 A 00mm (diameter) stainless steel sphere (ρ=7900 kg/m 3, λ=43 W/m K, cp=477 J/kg K) is at 80 C and is at 40m from the ground level. The sphere is cooled by a mass air flow from the bottom, the air temperature is 20 C and the convective coefficient is 50 W/m 2 K. Calculate the temperature of the sphere when it reaches the ground. Assume that the air temperature is constant and the sphere speed is the average between the initial velocity and the speed at the ground. Solution This problem looks like a lumped parameter problem, we must check if it is so. Bi < 0. 7

32 Lumped Parameter We must check if the Biot number is lower than 0. Bi < 0. Bi = h V λ A < 0. Bi = 50W m 2 K 4 3 π ( ) m 3 = W m K 4π ( ) 2 m 2 So we can use the temperature vs time equation derived in ex. 23. T T = exp ha T 0 T ρcv t 8

33 Lumped Parameter Time depends on the speed of the sphere. Using the assumption given by the problem we can calculate it from the sphere average speed. v ave = v final v initial 2 Final speed is a function of the height and acceleration of gravity. v final = 2 g H Being the initial speed 0: v ave = 2 g H 2 = 4m s 9

34 Lumped Parameter the falling time is: t = H v ave = 40m = 2.86s 4m s The temperature when the sphere reaches the ground. T = T + ( T 0 T )exp ha ρcv t = 50W m = 20 C + ( 2 K 4π ( ) 2 m ) C exp 7900kg m J kg K 4 3 π ( ) 3 m s = T = 79.6 C 0

35 Lumped Parameter Ex. 25 The heat exchanger of a car works between 88 C and 93 C, if the heat produced is 33kW. Water is the working fluid (4 liter, cp=4.9 kj/kg K). When the water temperature reaches 93 C a fan starts and the convective coefficient is 35 W/m 2 K, when the temperature reaches 88 C the fan stops and the convective coefficient is 5 W/m 2 K. Air is at 20 C Calculate the interval time in which the fan is on and off and the asymptotic temperature in both cases. Solution The T trend is given by the equation: Q T T = exp h A h A ρ c p V τ + T T ( i )exp h A ρ c p V τ

36 Lumped Parameter the interval time can be calculated from the above equation: τ = ρ c V T T p h A ln ( ) Q h A Q ( T i T ) h A τ = 975 kg m J kg K 0.04 m 3 5W m 2 K 20 m 2 ln = 6.9s ( 93 25) ( 88 25) W 5W m 2 K 20 m W 5W m 2 K 20 m 2 = 2

37 Lumped Parameter τ = 975 kg m J kg K 0.04 m 3 5W m 2 K 20 m 2 ln = 5.s ( 88 25) ( 93 25) W 5W m 2 K 20 m W 5W m 2 K 20 m 2 = the asymptotic temperature is for t to: Q T = T i ± h A 3

38 Lumped Parameter the interval time can be calculated from the above equation: T = 88 C + 33kW 5W m 2 K 00 m 2 = 98 C T = 93 C 33kW 35W m 2 K 20 m 2 = 46 C 4