Heat transfer-exercise-book

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Heat transfer-exercise-book

  1. 1. Chris Long & Naser SaymaHeat Transfer: Exercises Download free ebooks at bookboon.com 2
  2. 2. Heat Transfer: Exercises© 2010 Chris Long, Naser Sayma & Ventus Publishing ApSISBN 978-87-7681-433-5 Download free ebooks at bookboon.com 3
  3. 3. Heat Transfer: Exercises Contents Contents Preface 5 1. Introduction 6 2. Conduction 11 3. Convection 35 4. Radiation 60 5. Heat Exchangers 79 360° thinking . 360° thinking . 360° . Please click the advert thinking Discover the truth at www.deloitte.ca/careers D © Deloitte & Touche LLP and affiliated entities. Discover the truth at www.deloitte.ca/careers © Deloitte & Touche LLP and affiliated entities. Download free ebooks at bookboon.com Discover the truth4at www.deloitte.ca/careers© Deloitte & Touche LLP and affiliated entities. © Deloitte & Touche LLP and affiliated entities.
  4. 4. Heat Transfer: Exercises Preface Preface Worked examples are a necessary element to any textbook in the sciences, because they reinforce the theory (i.e. the principles, concepts and methods). Once the theory has been understood, well chosen examples can be used, with modification, as a template to solve more complex, or similar problems. This work book contains examples and full solutions to go with the text of our e-book (Heat Transfer, by Long and Sayma). The subject matter corresponds to the five chapters of our book: Introduction to Heat Transfer, Conduction, Convection, Heat Exchangers and Radiation. They have been carefully chosen with the above statement in mind. Whilst compiling these examples we were very much aware of the need to make them relevant to mechanical engineering students. Consequently many of the problems are taken from questions that have or may arise in a typical design process. The level of difficulty ranges from the very simple to challenging. Where appropriate, comments have been added which will hopefully allow the reader to occasionally learn something extra. We hope you benefit from following the solutions and would welcome your comments. Christopher Long Naser Sayma Brighton, UK, February 2010 Download free ebooks at bookboon.com 5
  5. 5. Heat Transfer: Exercises Introduction 1. Introduction Example 1.1 The wall of a house, 7 m wide and 6 m high is made from 0.3 m thick brick with k  0.6 W / m K . The surface temperature on the inside of the wall is 16oC and that on the outside is 6oC. Find the heat flux through the wall and the total heat loss through it. Solution: For one-dimensional steady state conduction: dT k q  k   Ti  To  dx L 0 .6 q 16  6  20 W / m 2 0 .3 Q  qA  20  6  7   840 W The minus sign indicates heat flux from inside to outside. Download free ebooks at bookboon.com 6
  6. 6. Heat Transfer: Exercises Introduction Example 1.2 A 20 mm diameter copper pipe is used to carry heated water, the external surface of the pipe is subjected to a convective heat transfer coefficient of h  6 W / m 2 K , find the heat loss by convection per metre length of the pipe when the external surface temperature is 80oC and the surroundings are at 20oC. Assuming black body radiation what is the heat loss by radiation? Solution qconv  h Ts  T f   680  20  360 W / m 2 For 1 metre length of the pipe: Qconv  q conv A  qconv  2 r  360  2    0.01  22.6 W / m For radiation, assuming black body behaviour:  q rad   Ts4  T f4   q rad  5.67  10 8 353 4  293 4  q rad  462 W / m 2 For 1 metre length of the pipe Qrad  q rad A  462  2    0.01  29.1 W / m 2 A value of h = 6 W/m2 K is representative of free convection from a tube of this diameter. The heat loss by (black-body) radiation is seen to be comparable to that by convection. Download free ebooks at bookboon.com 7
  7. 7. Heat Transfer: Exercises Introduction Example 1.3 A plate 0.3 m long and 0.1 m wide, with a thickness of 12 mm is made from stainless steel ( k  16 W / m K ), the top surface is exposed to an airstream of temperature 20oC. In an experiment, the plate is heated by an electrical heater (also 0.3 m by 0.1 m) positioned on the underside of the plate and the temperature of the plate adjacent to the heater is maintained at 100oC. A voltmeter and ammeter are connected to the heater and these read 200 V and 0.25 A, respectively. Assuming that the plate is perfectly insulated on all sides except the top surface, what is the convective heat transfer coefficient? Solution Heat flux equals power supplied to electric heater divided by the exposed surface area: V I V I 200  0.25 q    1666.7 W / m 2 A W L 0 .1  0 .3 This will equal the conducted heat through the plate: k q T2  T1  t qt T1  T2   100  1666.7  0.012   98.75C (371.75 K) k 16 The conducted heat will be transferred by convection and radiation at the surface:  q  hT1  T f    T14  T f4  h  q   T14  T f4   1666.7  5.67  10 371.75 8 4  293 4   12.7 W / m 2 K T 1  Tf  371.75  293 Download free ebooks at bookboon.com 8
  8. 8. Heat Transfer: Exercises Introduction Example 1.4 An electronic component dissipates 0.38 Watts through a heat sink by convection and radiation (black body) into surrounds at 20oC. What is the surface temperature of the heat sink if the convective heat transfer coefficient is 6 W/m2 K, and the heat sink has an effective area of 0.001 m2 ? Solution Q q A   hTs  T    Ts4  T4  0.38 0.001  6Ts  293  5.67  10 3 Ts4  293 4   5.67  10 8 Ts4  6Ts  2555.9  0 This equation needs to be solved numerically. Newton-Raphson’s method will be used here: f  5.67  10 8 Ts4  6Ts  2555.9 df  22.68  10 8 Ts3  6 dTs n 1 n f 5.67  10 8 Ts4  6Ts  2555.9 n T s T s T  s  df  22.68Ts3  6      dTs  Start iterations with Ts0  300 K 5.67  10 8  300 4  6  300  2555.9 Ts1  300   324.46 K 22.68  300 3  6 2 5.67  10 8  324.46 4  6  324.46  2555.9 T  324.46  s  323 K 22.68  324.46 3  6 Download free ebooks at bookboon.com 9
  9. 9. Heat Transfer: Exercises Introduction The difference between the last two iterations is small, so: Ts0  323 K  50C The value of 300 K as a temperature to begin the iteration has no particular significance other than being above the ambient temperature. Increase your impact with MSM Executive EducationPlease click the advert For almost 60 years Maastricht School of Management has been enhancing the management capacity of professionals and organizations around the world through state-of-the-art management education. Our broad range of Open Enrollment Executive Programs offers you a unique interactive, stimulating and multicultural learning experience. Be prepared for tomorrow’s management challenges and apply today. For more information, visit www.msm.nl or contact us at +31 43 38 70 808 or via admissions@msm.nl the admissions@msm.nl www.msm.nl or contact us at +31 43 38 70 808 globally networked management school For more information, visit or via Executive Education-170x115-B2.indd 1 Download free ebooks at bookboon.com 18-08-11 15:13 10
  10. 10. Heat Transfer: Exercises Conduction 2. Conduction Example 2.1 Using an appropriate control volume show that the time dependent conduction equation in cylindrical coordinates for a material with constant thermal conductivity, density and specific heat is given by:  2T 1 T  2T 1 T    r 2 r r z 2  t k Were   is the thermal diffusivity. c Solution Consider a heat balance on an annular control volume as shown the figure above. The heat balance in the control volume is given by: Heat in + Heat out = rate of change of internal energy u Q r  Q z  Q r  r  Q z  z  (2.1) t Q Q r  r  Q r  r r Q Q z  z  Q z  z z Download free ebooks at bookboon.com 11
  11. 11. Heat Transfer: Exercises Conduction u  mcT Substituting in equation 2.1: Q Q  (mcT )  r z (2.2) r z t Fourier’s law in the normal direction of the outward normal n: Q T  k A n T T Qr   kA   k  2 r z ( A  2 r z ) r r T T Q z   kA   k  2 r r ( A  2 r r ) z z Equation 2.1 becomes   T    T  T   k  2 r z  r   k  2 r r  z  mc (2.3) r  r  z  z  t Noting that the mass of the control volume is given by: m   2 r r z Equation 2.3 becomes   T    T  T k r  r  k r  z  cr r  r  z  z  t Dividing by r, noting that r can be taken outside the brackets in the second term because it is not a function of z. Also dividing by k since the thermal conductivity is constant: 1   T   2T c T r   r r  r  z 2 k t k Using the definition of the thermal diffusivity:   and expanding the first term using the product rule: c 1    2T T r   2T 1 T r    which gives the required outcome: r r  r 2 r r  z 2  t Download free ebooks at bookboon.com 12
  12. 12. Heat Transfer: Exercises Conduction  2T 1 T  2T 1 T    r 2 r r z 2  t Example 2.2 An industrial freezer is designed to operate with an internal air temperature of -20oC when the external air temperature is 25oC and the internal and external heat transfer coefficients are 12 W/m2 K and 8 W/m2 K, respectively. The walls of the freezer are composite construction, comprising of an inner layer of plastic (k = 1 W/m K, and thickness of 3 mm), and an outer layer of stainless steel (k = 16 W/m K, and thickness of 1 mm). Sandwiched between these two layers is a layer of insulation material with k = 0.07 W/m K. Find the width of the insulation that is required to reduce the convective heat loss to 15 W/m2. See the light! The sooner you realize we are right, the sooner your life will get better!Please click the advert A bit over the top? Yes we know! We are just that sure that we can make your media activities more effective. Get “Bookboon’s Free Media Advice” Email kbm@bookboon.com Download free ebooks at bookboon.com 13
  13. 13. Heat Transfer: Exercises Conduction Solution q  UT where U is the overall heat transfer coefficient given by: q 15 U   0.333W / m 2 K T 25  (20) 1  1 L p Li Ls 1  U        0.333  hi k p k i k s ho     1 L p Li Ls 1  1        hi k p k i k s ho  0.333    1   1 L p Ls 1     1  1 0.003 0.001 1   Li  k i         0.07        0.333  hi k p k s ho       0.333 12 1 16 8  Li  0.195m (195 mm) Example 2.3 Water flows through a cast steel pipe (k = 50 W/m K) with an outer diameter of 104 mm and 2 mm wall thickness. Download free ebooks at bookboon.com 14
  14. 14. Heat Transfer: Exercises Conduction i. Calculate the heat loss by convection and conduction per metre length of uninsulated pipe when the water temperature is 15oC, the outside air temperature is -10oC, the water side heat transfer coefficient is 30 kW/m2 K and the outside heat transfer coefficient is 20 W/m2 K. ii. Calculate the corresponding heat loss when the pipe is lagged with insulation having an outer diameter of 300 mm, and thermal conductivity of k = 0.05 W/m K. Solution Plain pipe Q Q  2 r1 Lhi Ti  T1   Ti  T1  2 r1 Lhi 2Lk T1  T2  Q Q  T2  T1  lnr2 / r1  2 Lk / ln(r2 / r1 ) Q Q  2 r2 Lho T2  To   T2  To  2 r2 Lho Adding the three equations on the right column which eliminates the wall temperatures gives: 2LTi  To  Q 1 ln r2 / r1  1   hi r1 k ho r2 Q 2 15  (10)    163.3W / m L 1 ln0.052 / 0.05 1   30000  0.05 50 20  0.052 Download free ebooks at bookboon.com 15
  15. 15. Heat Transfer: Exercises Conduction Insulated pipe Q 2 Ti  To   L 1 lnr2 / r1  ln(r3 / r2 ) 1    hi r1 k k ins ho r3Please click the advert GOT-THE-ENERGY-TO-LEAD.COM We believe that energy suppliers should be renewable, too. We are therefore looking for enthusiastic new colleagues with plenty of ideas who want to join RWE in changing the world. Visit us online to find out what we are offering and how we are working together to ensure the energy of the future. Download free ebooks at bookboon.com 16
  16. 16. Heat Transfer: Exercises Conduction Q 2 15  (10)    7.3W / m L 1 ln0.052 / 0.05 ln(0.15 / 0.052) 1   30000  0.05 50 0.05 20  0.15 For the plain pipe, the heat loss is governed by the convective heat transfer coefficient on the outside, which provides the highest thermal resistance. For the insulated pipe, the insulation provides the higher thermal resistance and this layer governs the overall heat loss. Example 2.4 Water at 80oC is pumped through 100 m of stainless steel pipe, k = 16 W/m K of inner and outer radii 47 mm and 50 mm respectively. The heat transfer coefficient due to water is 2000 W/m2 K. The outer surface of the pipe loses heat by convection to air at 20oC and the heat transfer coefficient is 200 W/m2 K. Calculate the heat flow through the pipe. Also calculate the heat flow through the pipe when a layer of insulation, k = 0.1 W/m K and 50 mm radial thickness is wrapped around the pipe. Solution The equation for heat flow through a pipe per unit length was developed in Example 2.3: 2LTi  To  Q 1 ln r2 / r1  1   hi r1 k ho r2 Hence substituting into this equation: 2  10080  20 Q  0.329  10 6 W 1 ln50 / 47  1   0.047  2000 16 0.05  200 For the case with insulation, we also use the equation from Example 2.3 2LTi  To  Q 1 lnr2 / r1  ln(r3 / r2 ) 1    hi r1 k k ins ho r3 2  10080  20  Q  5.39  10 3 W 1 ln50 / 47  ln(100 / 50) 1    0.047  2000 16 0.1 0.1  200 Notice that with insulation, the thermal resistance of the insulator dominates the heat flow, so in the equation above, if we retain the thermal resistance for the insulation and ignore all the other terms, we obtain: Download free ebooks at bookboon.com 17
  17. 17. Heat Transfer: Exercises Conduction 2LTi  To  2  10080  20 Q   5.44  10 3 W ln(r3 / r2 ) ln(100 / 50) k ins 0.1 This has less than 1% error compared with the full thermal resistance. Example 2.5 A diagram of a heat sink to be used in an electronic application is shown below. There are a total of 9 aluminium fins (k = 175 W/m K, C = 900 J/kg K,   2700kg / m 3 ) of rectangular cross-section, each 60 mm long, 40 mm wide and 1 mm thick. The spacing between adjacent fins, s, is 3 mm. The temperature of the base of the heat sink has a maximum design value of Tb  60C , when the external air temperature T f is 20oC. Under these conditions, the external heat transfer coefficient h is 12 W/m2 K. The fin may be assumed to be sufficiently thin so that the heat transfer from the tip can be neglected. The surface temperature T, at a distance, x, from the base of the fin is given by: Tb  T f cosh m( L  x) hP T  Tf  where m 2  and Ac is the cross sectional area. sinh mL kAc Determine the total convective heat transfer from the heat sink, the fin effectiveness and the fin efficiency. Download free ebooks at bookboon.com 18
  18. 18. Heat Transfer: Exercises Conduction Solution Total heat fluxed is that from the un-finned surface plus the heat flux from the fins. Q  Qu  Q f Qu  Au h (Tb  T f )  w  s  N  1) h Tb  T f  Qu  0.04  0.0039  1)   12 60  20  0.461 W For a single fin:  dT  Q f  kAc    dx  x 0 Where Ac is the cross sectional area of each fin Since Tb  T f cosh m( L  x) T  Tf  sinh mL Who is your target group? And how can we reach them? At Bookboon, you can segment the exact rightPlease click the advert audience for your advertising campaign. Our eBooks offer in-book advertising spot to reach the right candidate. Contact us to hear more kbm@bookboon.com Download free ebooks at bookboon.com 19
  19. 19. Heat Transfer: Exercises Conduction Then dT  sinh m( L  x)  m Tb  T f  dx cosh mL Thus  dT    sinh mL  Q f  kAc    kAc   m Tb  T f   dx  x 0  cosh mL  Q f  kAc m Tb  T f  tanh(mL)  hpkAc  T  T f  tanh mL  1/ 2 b Since 1  hP 2 m  kA    c  P  2( w  t )  2(0.04  0.001)  0.082 m Ac  w  t  0.04  0.0001  40  10 6 m 2 1  12  0.082  2 m 6   11.856 m 1  175  40  10  mL  11.856  0.06  0.7113 tanhmL   tanh0.7113  0.6115  Q f  12  0.082  175  40  10 6  1/ 2  60  20  0.6115  2.03 W / fin So total heat flow: Q  Qu  Q f  0.461  9  2.03  18.7 W Finn effectiveness Fin heat transfer rate Qf  fin   Heat transfer rate that would occur in the absence of the fin hAc Tb  T f  2.03  fin   106 12  40  10 6 60  20  Download free ebooks at bookboon.com 20
  20. 20. Heat Transfer: Exercises Conduction Fin efficiency: Actual heat transfer through the fin  fin  Heat that would be transferred if all the fin area were at the base temperature Qf  fin  hAs Tb  T f  As  wL  wL  Lt  Lt  2 L( w  t ) As  2  0.06(0.04  0.001)  4.92  10 3 m 2 2.03  fin   0.86 12  4.92  10 3 60  20  Example 2.6 For the fin of example 4.5, a fan was used to improve the thermal performance, and as a result, the heat transfer coefficient is increased to 40 W/m2 K. Justify the use of the lumped mass approximation to predict the rate of change of temperature with time. Using the lumped mass approximation given below, calculate the time taken,  , for the heat sink to cool from 60oC to 30oC.  hA   T  T   T f i  T f exp  s   mC  Solution Consider a single fin (the length scale L for the Biot number is half the thickness t/2) hL h  t / 2 40  0.0005 Bi     10  4 k k 175 Since Bi  1 , we can use he “lumped mass” model approximation. T  T  f  hA   s  exp    T  T  i f  mC  mC  T  T f    ln  hAs  Ti  T f    Download free ebooks at bookboon.com 21
  21. 21. Heat Transfer: Exercises Conduction m  As t / 2 Ct  T  Tf    ln    2700  900  0.001 ln 30  20   42 seconds   2h  Ti  T f    2  40  60  20  Example 2.7 The figure below shows part of a set of radial aluminium fins (k = 180 W/m K) that are to be fitted to a small air compressor. The device dissipates 1 kW by convecting to the surrounding air which is at 20oC. Each fin is 100 mm long, 30 mm high and 5 mm thick. The tip of each fin may be assumed to be adiabatic and a heat transfer coefficient of h = 15 W/m2 K acts over the remaining surfaces. Estimate the number of fins required to ensure the base temperature does not exceed 120oCPlease click the advert THE BEST MASTER IN THE NETHERLANDS Download free ebooks at bookboon.com 22
  22. 22. Heat Transfer: Exercises Conduction Solution Consider a single fin: P  2( w  t )  2(0..005  0.03)  0.07 m Ac  w  t  0.005  0.03  150  10 6 m 2 1 1  hP  2  15  0.07 2 m  kA    6   6.2361 m 1  c   180  150  10  mL  6.2361 0.1  0.62361 tanhmL   0.5536 Q f  hPkAc  1/ 2 T b  T f  tanh(mL) (From example 2.5)  Q f  15  0.07  180  150  10 6  1/ 2  120  20  0.5536  9.32 W So for 1 kW, the total number of fins required: 1000 N  108 9.32 Download free ebooks at bookboon.com 23
  23. 23. Heat Transfer: Exercises Conduction Example 2.8 An air temperature probe may be analysed as a fin. Calculate the temperature recorded by a probe of length L = 20 mm, k = 19 W/m K, D = 3 mm, when there is an external heat transfer coefficient of h = 50 W/m2K, an actual air temperature of 50oC and the surface temperature at the base of the probe is 60oC. Solution The error should be zero when Ttip  T . The temperature distribution along the length of the probe (from the full fin equation) is given by: htip cosh m( L  x)  sinh m( L  x) Tx  T mk  Tb  T htip cosh mL  sinh mL mk 1/ 2  hP  m  A   D 2 / 4, P  D  kA  At the tip, x  L , the temperature is given by ( cosh( 0)  1 , sinh( 0)  0 ): Ttip  T 1   Tb  T htip cosh mL  sinh mL mk Where  is the dimensionless error:,   0, Ttip  T (no error) Download free ebooks at bookboon.com 24
  24. 24. Heat Transfer: Exercises Conduction   1, TL  Tb (large error) For L  20mm, k  19W / m K , D  3mm, h  htip  50 W / m 2 K T  50C , Tb  60C A   D 2 / 4, P  D 1/ 2 1/ 2 1/ 2 1/ 2  hP   h D  4   4h   4  50  m    2        59.235 m 1  kA   k D   kD   19  0.003  mL  59.235  0.02  1.185 h 50   0.0444 mk 59.235  19 Tx  T 1   0.539 Tb  T cosh 1.185  0.0444  sinh 1.185 With us you can shape the future.Please click the advert Every single day. For more information go to: www.eon-career.com Your energy shapes the future. Download free ebooks at bookboon.com 25
  25. 25. Heat Transfer: Exercises Conduction Ttip  0.539Tb  T   T Ttip  0.53960  50  50  55.39C Hence error  5.39C Example 2.9 A design of an apartment block at a ski resort requires a balcony projecting from each of the 350 separate apartments. The walls of the building are 0.3 m wide and made from a material with k = 1 W/m K. Use the fin approximation to examine the implications on the heat transfer for two separate suggestions for this design. In each case, the balcony projects 2 m from the building and has a length (parallel to the wall) of 4 m. Assume an inside temperature of 20oC and an outside temperature of - 5oC; the overall (convective + radiative) heat transfer coefficient on the inside of the building is 8 W/m2 K and on that on the outside of the building is 20 W/m2 K a) A balcony constructed from solid concrete and built into the wall, 0.2 m thick, k = 2 W/m K. b) A balcony suspended from 3 steel beams, k = 40 W/m K, built into the wall and projecting out by 2 m each of effective cross sectional area Ae  0.01 m 2 , perimetre P  0.6 m (The actual floor of the balcony in this case may be considered to be insulated from the wall c) No balcony. Solution a) For the concrete balcony ho t / 2 Treat the solid balcony as a fin Bi  kb Download free ebooks at bookboon.com 26
  26. 26. Heat Transfer: Exercises Conduction 20  0.1 Bi  1 2 Not that Bi is not << 1, thus 2D analysis would be more accurate. However, treating it as a fin will give an acceptable result for the purpose of a quick calculation. P  2 ( H  t )  2 ( 4  0 .2)  8 .4 m Ac  H  t  4  0.2  0.8 m 2 To decide if the fin is infinite, we need to evaluate mL (which is in fact in the notation used here is mW) 1/ 2 1/ 2  hP   20  8.4  mW    W    2  20.5  kA   2  0 .8  This is large enough to justify the use of the fin infinite equation. Qb  ho Pk b Ac  T2  To  1/ 2 1/ 2   qb  1 ho Pk b Ac 1 / 2 T2  To    ho Pk b  A   T2  To  (1) Ac  c  Also assuming 1-D conduction through the wall: qb  hi (Ti  T1 ) (2) kb qb  (T1  T2 ) (3) L Adding equations 1, 2 and 3 and rearranging: (To  Ti ) qb  1/ 2 (4) 1 L  Ac     hi k b  ho Pk b    This assumes 1D heat flow through the wall, the concrete balcony having a larger k than the wall may introduce some 2-D effects. Download free ebooks at bookboon.com 27
  27. 27. Heat Transfer: Exercises Conduction From (4) 20  (5) qb  1/ 2  77.2 W / m 2 1 0 .3  0 .8     8 2  20  8.4  2  Compared with no balcony: (To  Ti ) 20  (5) qb    52.6 W / m 2 1 L 1 1 0.3 1     hi k w ho 8 1 20 The difference for one balcony is Ac (77.2  52.6)  0.8  24.6  19.7 W For 350 apartments, the difference is 6891 W. For the steel supported balcony where Ac  0.01 m 2 and P  0.6 m Do your employees receive the right training? Bookboon offers an eLibrairy with a wide range ofPlease click the advert Soft Skill training & Microsoft Office books to keep your staff up to date at all times. Contact us to hear more kbm@bookboon.com Download free ebooks at bookboon.com 28
  28. 28. Heat Transfer: Exercises Conduction As before, however, in this case Bi << 1 because k s  k b 1/ 2 1/ 2  hP   20  6  mW    w   2  11  kA   40  0.1  mW  2 , so we can use the infinite fin approximation as before (To  Ti ) 20  ( 5) qb  1/ 2  1/ 2  182 W / m 2 1 L  Ac  1 0.3  0.01        hi k s  ho Pk s    8 40  20  6  40  Qb  Ac qb  0.01  182  1.82 W / beam For 350 apartments, Qb  1915 W Example 2.10 In free convection, the heat transfer coefficient varies with the surface to fluid temperature difference T s  T f  . Using the low Biot number approximation and assuming this variation to be of the form h  G Ts  T f  Where G and n are constants, show that the variation of the dimensionless n temperature ratio with time will be given by   n  1  nhinit   t Where T s  Tf  Area  ,  Tinit  T f Mass  Specific Heat Capacity and hinit = the heat transfer coefficient at t = 0. Use this expression to determine the time taken for an aluminium motorcycle fin (   2750 kg / m 3 , C  870 J / kgK ) of effective area 0.04 m2 and thickness 2mm to cool from 120oC to 40oC in surrounding air at 20oC when the initial external heat transfer coefficient due to laminar free convection is 16 W/m2 K. Compare this with the time estimated from the equation (   e  ht ) which assumes a constant value of heat transfer coefficient. Solution Low Biot number approximation for free convection for Bi  1 Heat transfer by convection = rate of change of internal energy Download free ebooks at bookboon.com 29
  29. 29. Heat Transfer: Exercises Conduction d (Ts  T f ) hA(Ts  T f )  mC (1) dt n We know that h  G (Ts  T f ) Where G is a constant. (Note that this relation arises from the usual Nusselt/Grashof relationship in free convection; for example: Nu  0.1Gr Pr  in turbulent flow or Nu  0.54Gr Pr  1/ 3 1/ 4 for laminar flow) Equation 1 then becomes: mC d (Ts  T f ) G Ts  T f  (T n s  Tf )   A dt t  GA t d (Ts  T f ) 0 mC t dt   t  0 (Ts  T f ) n 1 GnAt  Ts  T f   Ts  T f  n n (2) mC t 0 At t  0, T s  T f   Ts ,i  T f  If we divide equation 2 by Ts ,i  T f   n T s   Tf And use the definition  T s ,i T  f We obtain GnAt GnAt   n  1  Ts ,i  T f  n mC Ts ,i  T f  mC n   Since G Ts ,i  T f  h i , the heat transfer coefficient at time t = 0, then hi At  n  1 mC Download free ebooks at bookboon.com 30
  30. 30. Heat Transfer: Exercises Conduction Or   n  nhi  t  1 For aluminium   2750 kg / m 3 , C  870 J / kg K For laminar free convection, n = ¼ m   A X  2750  0.04  0.002  0.22 kg A 0.04    2.1 10  4 m 2 K / J mC 0.22  870  n  nhi  t  1 which givesPlease click the advert www.job.oticon.dk Download free ebooks at bookboon.com 31
  31. 31. Heat Transfer: Exercises Conduction t  n  1 nhi  40  20 When T  40C   0 .2 120  20 Then t 0.2  1 1 / 4  590 s 1 / 4  16  2.1  10 4 For the equation   e  h t which assumes that the heat transfer coefficient is independent of surface-to-fluid temperature difference. ln  ln 0.2 t   479 s  h  16  2.1  10  4 590  479 Percentage error =  100  19% 590 Example 2.11 A 1 mm diameter spherical thermocouple bead (C = 400 J/kg K, � � ���������� ) is required to respond to 99.5% change of the surrounding air � � ��������� � � � ��� � ���� ����������� , � � ������� � ⁄��� and Pr = 0.77) temperature in 10 ms. What is the minimum air speed at which this will occur? Download free ebooks at bookboon.com 32
  32. 32. Heat Transfer: Exercises Conduction Solution Spherical bead: ���� � ��� � ������� � � �� � ⁄6 Assume this behaves as a lumped mass, then �� � �� � ����� �� � �� (given) For lumped mass on cooling from temperature Ti �� � �� � ��������� � ����� �� � �� �� �� ������������ � ������� �� ��� � ����� �� � ��� Which gives the required value of heat transfer coefficient �� � ��� ��� So Download free ebooks at bookboon.com 33
  33. 33. Heat Transfer: Exercises Conduction �� � �� 0.� � � � � � 0.� � 6 �� � 6 0.� � 10�� � 400 � 7800 �� � 260 � ⁄�� � 6 �� 260 � 10�� ��� � � � �.� � 0.0262 For a sphere ��� � 2 � �0.4��� � 0.06��� � �� �.� ��� ��� From which with Pr = 0.707 � � 0.4��� � 0.06��� � �.4 � 0 ��� ��� � � � 0.2��� ���� � 0.04��� ���� Using Newton iteration ���� � ����� � � � � � ���� Starting with ReD = 300 �0.4√300 � 0.06�300���� � �.4� 0.222 ��� � 300 � � 300 � ��� 0.2 0.04 0.01782 � � � √300 300��� Which is close enough to 300 From which ��� �� � � 4.� ��� �� Download free ebooks at bookboon.com 34
  34. 34. Heat Transfer: Exercises Convection 3. Convection Example 3.1 Calculate the Prandtl number (Pr = Cp/k) for the following a) Water at 20C:  = 1.002 x 103 kg/m s, Cp = 4.183 kJ/kg K and k = 0.603 W/m K b) Water at 90C:  = 965 kg/m3,  = 3.22 x 107 m2/s, Cp = 4208 J/kg K and k = 0.676 W/m K c) Air at 20C and 1 bar: R = 287 J/kg K,  = 1.563 x 105 m2/s, Cp = 1005 J/kg K and k = 0.02624 W/m K 1.46  10 6 T 3 2 d) Air at 100C:  kg/m s 110  T  C p  0.917  2.58  10 4 T  3.98  10 8 T 2 kJ / kg K (Where T is the absolute temperature in K) and k = 0.03186 W/m K. e) Mercury at 20C:  = 1520 x 106 kg/m s, Cp = 0.139 kJ/kg K and k = 0.0081 kW/m K f) Liquid Sodium at 400 K:  = 420 x 106 kg/m s, Cp = 1369 J/kg K and k = 86 W/m K g) Engine Oil at 60C:  = 8.36 x 102 kg/m s, Cp = 2035 J/kg K and k = 0.141 W/m K Is your recruitment website still missing a piece? Bookboon can optimize your current traffic. ByPlease click the advert offering our free eBooks in your look and feel, we build a qualitative database of potential candidates. Contact us to hear more kbm@bookboon.com Download free ebooks at bookboon.com 35
  35. 35. Heat Transfer: Exercises Solution Convection a) Solution Solution a)  Cp 1.002  10 3  4183 a) Pr    6.95 k 0.603  C p 1.002  10 3 4183 3 Pr   C p  1.002  10  4183 6.95 b) Pr  k   6.95 k 0.603 0.603 b)  Cp  C p 965  3.22  10 7  4208 b) Pr     1.93 k k 0.676  C p  C p 965  3.22  10 7 4208 7 Pr   C p   C p  965  3.22  10  4208  1.93 c) Pr  k  k   1.93 k k 0.676 0.676 c)  C p c) Pr  k  C p Pr   C p Pr  Pk 100000  k  1.19 kg / m 3 RT 287  293 P 100000   P  100000 51.19 kg //m 33   RT19  1.563  10  1.19 kg m 1.  287  293   1005 Pr  RT 287  293  0.712 0.02624 5 1.19  1.563  10 5 1005 Pr  1.19  1.563  10  1005  0.712 d) Pr   0.712 0.02624 0.02624 d) d) 1.46  10 6 T 3 2 1.46  10 6  3733 / 2    2.18  10 5 kg / m s 110  T  110  373 1.46  10 6T 3322 1.46  10 6 37333/ /22 6 6 1.46  10 T  1.46  10  373  2.18  10 5 kg / m s  5   0110  T 58  10 4 T 110.98  10 8 T 2  2..18  102.58 /10 s  373  3.98  10 8  3732 Cp .110  2. 917  T   3  373 110  373  0 917  kg m 4  1007.7 J / kg K C p  0.917  2.58  10 4T  3.98  10 8T 22  0.917  2.58  10 4 373  3.98  10 8 37322 4 8 4 C p  0.917  2.58  10 T  3.98  10 T  0.917  2.58  10  373  3.98  10  373 8  1007. 10/ 5kg 1007.7  2.18 7 J  K  Pr  1007.7 J / kg K  0.689 0.03186 5 2.18  10 5 1007.7 Pr  2.18  10  1007.7  0.689 e) Pr   0.689 0.03186 0.03186 e)  Cp 1520  10 6  139 e) Pr    0.0261 k 0.0081  10 3  C p 1520  10 6 139 6 Pr   C p  1520  10 3139  0.0261 Pr  k  0.0081  10 3  0.0261 k 0.0081  10 Download free ebooks at bookboon.com 36
  36. 36. Heat Transfer: Exercises Convection f)  Cp 420  10 6  1369 f) Pr    0.0067 k 86  Cp 420  10 6  1369 g) Pr    0.0067 k 86  Cp 8.36  10 2  2035 g) Pr    1207 k 0.141  Cp 8.36  10 2  2035 Pr  Comments:   1207 k 0.141  Large temperature dependence for water as in a) and b); Comments:  small temperature dependence for air as in c) and d);  use of Sutherland’s law for viscosity as in part d);  Large temperature dependence for water as in a) and b);  difference between liquid metal and oil as in e), f) and g);  small temperature dependence for air as in c) and d);  units of kW/m K for thermal conductivity;  use of Sutherland’s law for viscosity as in part d);  use of temperature dependence of cp as in part a).  difference between liquid metal and oil as in e), f) and g);  units of kW/m K for thermal conductivity; Example 3.2  use of temperature dependence of cp as in part a). Calculate the appropriate Reynolds numbers and state if the flow is laminar or turbulent for Example 3.2 the following: Calculate the appropriate Reynolds numbers and state if the flow is laminar or turbulent for a) A 10 m (water line length) long yacht sailing at 13 km/h in seawater  = 1000 kg/m3 and the following: 3  = 1.3 x 10 kg/m s, b) A compressor disc of radius 0.3 m rotating at 15000 rev/min in air at 5 bar and 400C and a) A 10 m (water line length) long yacht sailing at 13 km/h in seawater  = 1000 kg/m3 and 6 3 2 1.3 x 1010 T  = 1.46  3 kg/m s, kg/m s  b) A compressor T  of radius 0.3 m rotating at 15000 rev/min in air at 5 bar and 400C and 110 disc c) 0.05 1.46  10 6 T 3 2dioxide gas at 400 K flowing in a 20 mm diameter pipe. For the viscosity kg/s of carbon  1.56 T  6 kg/m s 32 take  110   10 T kg/m s c) 0.05 kg/s of carbon T  233 dioxide gas at 400 K flowing in a 20 mm diameter pipe. For the viscosity 3 5 1. a  10 6 3 long, travelling at 100 km/hr in air ( = 1.2 kg/m and  = 1.8 x 10 d) The roof of56coach6 Tm 2 take   kg/m s) kg/m s 233  T  e) The flow of exhaust gas (p = 1.1 bar, T = 500ºC, R = 287 J/kg K and  = 3 3.56 x 105 kg/m s) over d) The roof of a coach 6 m long, travelling at 100 km/hr in air ( = 1.2 kg/m and  = 1.8 x 105 a valve guide of diameter 10 mm in a 1.6 litre, four cylinder four stroke engine running at kg/m s) 3000 rev/min (assume 100% volumetric efficiency an inlet density of 1.2 kg/m3 and an exhaust e) The flow of exhaust gas (p = 1.1 bar, T = 500ºC, R = 287 J/kg K and  = 3.56 x 105 kg/m s) over port diameter of 25 mm) a valve guide of diameter 10 mm in a 1.6 litre, four cylinder four stroke engine running at 3000 rev/min (assume 100% volumetric efficiency an inlet density of 1.2 kg/m3 and an exhaust port diameter of 25 mm) Download free ebooks at bookboon.com 37
  37. 37. Heat Transfer: Exercises Convection Solution 13  10 3 10 3   10 uL 3600 a) Re    2.78  10 7 (turbulent)  1.3  10 3 b) T  400  273  673 K 1.46  10 6  6733 2   3.26  10 5 kg / m s 110  673 15000   2  1571 rad / s 60 u   r  1571  0.3  471 .3 m / s P 100000    2.59 kg / m 3 RT 287  673 Turning a challenge into a learning curve. Just another day at the office for a high performer. Accenture Boot Camp – your toughest test yet Choose Accenture for a career where the variety of opportunities and challenges allows you to make aPlease click the advert difference every day. A place where you can develop your potential and grow professionally, working alongside talented colleagues. The only place where you can learn from our unrivalled experience, while helping our global clients achieve high performance. If this is your idea of a typical working day, then Accenture is the place to be. It all starts at Boot Camp. It’s 48 hours packed with intellectual challenges and intense learning experience. that will stimulate your mind and and activities designed to let you It could be your toughest test yet, enhance your career prospects. You’ll discover what it really means to be a which is exactly what will make it spend time with other students, top high performer in business. We can’t your biggest opportunity. Accenture Consultants and special tell you everything about Boot Camp, guests. An inspirational two days but expect a fast-paced, exhilarating Find out more and apply online. Visit accenture.com/bootcamp Download free ebooks at bookboon.com 38
  38. 38. Heat Transfer: Exercises Convection Characteristic length is r not D uD 2.59  471.3  3 Re    1.12  10 7 (turbulent)  3.26  10 5 D 2 c) m   uA  u   4 4m  u D 2 uD   4mD 4m   Re     D  D 2 1.56  10 6  400 3 2   1.97  10 5 kg / m s 233  400 4  0.05 Re   1.6  10 5 (turbulent)   0.02  1.97  10 5 100  10 3 d) u  27.8 m / s 3600 uL 1.2  27.8  6 Re    11.1  10 7 (turbulent)  1.8  10 5 e) Let m be the mass flow through the exhaust port  m = inlet density X volume of air used in each cylinder per  second 1.6  10 3 3600 1 m  1.2      0.012 kg / s 4 60 2 4m  u  D2 ud Re d   Download free ebooks at bookboon.com 39
  39. 39. Heat Transfer: Exercises Convection 4  0.01  0.012 Re   6869 (laminar)   3.56  10 5  0.025 Comments:  Note the use of D to obtain the mass flow rate from continuity, but the use of d for the characteristic length  Note the different criteria for transition from laminar flow (e.g. for a pipe Re  2300 plate Re  3  10 5 ) Example 3.3 Calculate the appropriate Grashof numbers and state if the flow is laminar or turbulent for the following: a) A central heating radiator, 0.6 m high with a surface temperature of 75C in a room at 18C ( = 1.2 kg/m3 , Pr = 0.72 and  = 1.8 x 105 kg/m s)] b) A horizontal oil sump, with a surface temperature of 40C, 0.4 m long and 0.2 m wide containing oil at 75C ( = 854 kg/m3 , Pr = 546,  = 0.7 x 103 K1 and  = 3.56 x 102 kg/m s) c) The external surface of a heating coil, 30 mm diameter, having a surface temperature of 80C in water at 20C ( = 1000 kg/m3, Pr = 6.95,  = 0.227 x 103K1 and  = 1.00 x 10-3kg/m s) d) Air at 20ºC ( = 1.2 kg/m3 , Pr = 0.72 and  = 1.8 x 105 kg/m s) adjacent to a 60 mm diameter vertical, light bulb with a surface temperature of 90C Solution  2 g  T L3 a) Gr  2 T  75  18  57 K 1 1 1    K 1 T 18  273 291 1.2 2  9.81  57  0.6 3 Gr   1.84  10 9 291  1.8  10  3 2 Gr Pr  1.84  10 9  0.72  1.3  10 9 (mostly laminar) Area 0.4  0.2 b) L   0.0667 m Perimeter 2  0.4  0.2  Download free ebooks at bookboon.com 40
  40. 40. Heat Transfer: Exercises Convection T  75  40  35 K  2 g  T L3 854 2  9.81  0.7  10 3  35  0.0667 3 Gr    4.1  10 4  2 3.56  10 2 2 Gr Pr  4.1  10 4  546  2.24  10 7 Heated surface facing downward results in stable laminar flow for all Gr Pr c) The Wake the only emission we want to leave behindPlease click the advert .QYURGGF PIKPGU /GFKWOURGGF PIKPGU 6WTDQEJCTIGTU 2TQRGNNGTU 2TQRWNUKQP 2CEMCIGU 2TKOG5GTX 6JG FGUKIP QH GEQHTKGPFN[ OCTKPG RQYGT CPF RTQRWNUKQP UQNWVKQPU KU ETWEKCN HQT /#0 KGUGN 6WTDQ 2QYGT EQORGVGPEKGU CTG QHHGTGF YKVJ VJG YQTNFoU NCTIGUV GPIKPG RTQITCOOG s JCXKPI QWVRWVU URCPPKPI HTQO VQ M9 RGT GPIKPG )GV WR HTQPV (KPF QWV OQTG CV YYYOCPFKGUGNVWTDQEQO Download free ebooks at bookboon.com 41
  41. 41. Heat Transfer: Exercises Convection T  80  20  60 K  2 g  T L3 1000 2  9.81  0.227  10 3  60  0.033 Gr    3.6  10 6  2 1 10  3 2 Gr Pr  3.6  10 6  6.95  25  10 6 (laminar) Area D 2 D d) L   Perimeter 4D 4 T  90  20  70 K 1 1 1    K 1 T 20  273 293  2 g  T L3 1.2 2  9.8  70  0.0153 Gr    3.5  10 4  293  1.8  10  2 5 2 Gr Pr  3.5  10 4  0.72  2.5  10 4 (laminar) Comments:  Note evaluation of  for a gas is given by   1 / T  For a horizontal surface L  A / p Example 3.4 Calculate the Nusselt numbers for the following: a) A flow of gas (Pr = 0.71,  = 4.63 x 105 kg/m s and Cp = 1175 J/kg K) over a turbine blade of chord length 20 mm, where the average heat transfer coefficient is 1000 W/m2 K. b) A horizontal electronics component with a surface temperature of 35C, 5 mm wide and 10 mm long, dissipating 0.1 W by free convection from one side into air where the temperature is 20C and k = 0.026 W/m K. c) A 1 kW central heating radiator 1.5 m long and 0.6 m high with a surface temperature of 80ºC dissipating heat by radiation and convection into a room at 20C (k = 0.026 W/m K assume black body radiation and  = 56.7 x 109 W/m K4) d) Air at 4C (k = 0.024 W/m K) adjacent to a wall 3 m high and 0.15 m thick made of brick with k = 0.3 W/m K, the inside temperature of the wall is 18C, the outside wall temperature 12C Download free ebooks at bookboon.com 42
  42. 42. Heat Transfer: Exercises Convection Solution  Cp a) Pr  k Solution  Cp a)  C p 4.63  10 5  1175 k  Pr  k  0.0766 W / m K Pr 0.71  C p 4.63  10 5  1175 khL  1000  0.02   0.0766 W / m K Nu   Pr 0.71 261  k 0.0766 h L 1000  0.02 Nu L h q L0.0766  261  b) Nu  k k T k hL q L b) Nu   Q k 0 .1 T k  q   2000 W / m 2 A 0.01  0.005 Q 0 .1 q   2000 W / m 2 A 0.01  0.005 T  35  20  15 C T  35  20  15 C Area 50 5 L   mm  0.001667 m Perimeter 30 50 3 5 Area L   mm  0.001667 m Perimeter 30 3 h L 2000  0.001667 Nu    8 .5 Nuk h L  15  0.026 2000  0.001667  8 .5 k 15  0.026 qc L c) Nu  q L c) Nu T k c  T k In this case,case, q mustthe the convective heat flux––radiative heat flux In this q must be be convective heat flux radiative heat flux Ts  Ts  80  273  353 K 80  273  353 K T T20 20  273  293 K   273  293 K Q R  ATs4 TsT4     .56 1010   .1.5  06 3534  293 4416 WW Q R  A  4  T4 56 7 .7  9 9 1 5  0. .6 353 4  2934 416 T  80  20  60 K T  80  20  60 K Qc  Q  QR  1000  416  584 W Qc  Q  QR  1000  416  584 W Download free ebooks at bookboon.com 43
  43. 43. Heat Transfer: Exercises Convection Qc 584 qc    649 W / m 2 A 1.5  0.6 q L 649 0.6 Nu  c    249 T k 60 0.026 d) T  12  4  8 K k b T1  T2  q  60 C W (assuming 1-D conduction) 0.318  12  q  12 W / m 2 0.18 q L 12 3 Nu  c    188 T k 8 0.024 Comments:  Nu is based on convective heat flux; sometimes the contribution of radiation can be significant and must be allowed for.  The value of k is the definition of Nu is the fluid (not solid surface property).  Use of appropriate boundary layer growth that characterises length scale. Brain power By 2020, wind could provide one-tenth of our planet’s electricity needs. Already today, SKF’s innovative know- how is crucial to running a large proportion of the world’s wind turbines. Up to 25 % of the generating costs relate to mainte- nance. These can be reduced dramatically thanks to our systems for on-line condition monitoring and automatic lubrication. We help make it more economical to createPlease click the advert cleaner, cheaper energy out of thin air. By sharing our experience, expertise, and creativity, industries can boost performance beyond expectations. Therefore we need the best employees who can meet this challenge! The Power of Knowledge Engineering Plug into The Power of Knowledge Engineering. Visit us at www.skf.com/knowledge Download free ebooks at bookboon.com 44
  44. 44. Heat Transfer: Exercises Convection Example 3.5 In forced convection for flow over a flat plate, the local Nusselt number can be represented by the general expression Nu x  C1 Re n . In free convection from a vertical surface the local Nusselt number x is represented by Nu x  C 2 Grxm , where C1, C2, n and m are constants a) Show that the local heat transfer coefficient is independent of the surface to air temperature difference in forced convection, whereas in free convection, h, depends upon (Ts  T)m b) In turbulent free convection, it is generally recognised that m = 1/3. Show that the local heat transfer coefficient does not vary with coordinate x. Solution hx a) Nu x  k ux Re x   For forced convection: Nu x  C1 Re n x n k ux Hence h  C1   x      This shows that the heat transfer coefficient for forced does not depend on temperature difference. For free convection Nu x  C 2 Grxm  2 g  T x 3 Grx  2 m k   2 g  T x 3  Hence h  C2    (1) x   2  So for free convection, heat transfer coefficient depends on T m b) From (1), with m = 1/3 for turbulent free convection: Download free ebooks at bookboon.com 45
  45. 45. Heat Transfer: Exercises Convection 1/ 3 k   2 g  T x 3  h  C2    x   2  1/ 3 k   2 g  T  h  C2    x x   2  1/ 3   2 g  T  h  kC 2      2  Hence the convective heat transfer coefficient does not depend on x Example 3.6 An electrically heated thin foil of length L = 25 mm and width W = 8 mm is to be used as a wind speed metre. Wind with a temperature T and velocity U  blows parallel to the longest side. The foil  is internally heated by an electric heater dissipating Q (Watts) from both sides and is to be operated in air with T  20C , C p  1.005 kJ / kg K ,   1.522  10 m / s   1.19 kg / m 3 and Pr  0.72 5 2 . The surface temperature, T of the foil is to be measured at the trailing edge – but can be assumed to  be constant. Estimate the wind speed when T  32C and Q  0.5 W . Solution Firstly, we need to estimate if the flow is laminar or turbulent. Assuming a critical (transition) Reynolds number of Re  3  10 5 the velocity required would be: 3  10 5 3  10 3 3  10 5  1.522  10 5 u turb     304 m / s L L 25  10 3 Wind speed is very unlikely to reach this critical velocity, so the flow can be assumed to be laminar. Nu x  0.331 Re 1x/ 2 Pr 1 / 3 q av L Nu av  0.662 Re1 / 2 Pr 1 / 3  L Ts  T k Download free ebooks at bookboon.com 46
  46. 46. Heat Transfer: Exercises Convection q av L Re1 / 2  L Ts  T k  0.662 Pr 1 / 3 0 .5 / 2 q av   1250 W / m 2 0.025  0.008 1250  0.025 Re1 / 2   173.5 L 32  20  0.0253  0.662  0.721 / 3 Re L  3  10 4 Re L  3  10 4  1.522  10 5 u    18.3 m / s L 25  10 3Please click the advert Download free ebooks at bookboon.com 47
  47. 47. Heat Transfer: Exercises Convection Example 3.7 The side of a building of height H = 7 m and length W = 30 m is made entirely of glass. Estimate the heat loss through this glass (ignore the thermal resistance of the glass) when the temperature of the air inside the building is 20C, the outside air temperature is -15C and a wind of 15 m/s blows parallel to the side of the building. Select the appropriate correlations from those listed below of local Nusselt numbers to estimate the average heat transfer coefficients. For air take: ρ= 1.2 kg / m3, μ = 1.8 x 10-5 kg / m s, Cp = 1 kJ / kg K and Pr = 0.7.  Free convection in air, laminar (Grx 109): Nux = 0.3 Grx1/4  Free convection in air, turbulent (Grx 109): Nux = 0.09 Grx1/3  Forced convection, laminar (Rex 105): Nux = 0.33 Rex0.5 Pr1/3  Forced convection, turbulent (Rex 105): Nux = 0.029 Rex0.8 Pr1/3 Solution  Cp  Cp 1.8  10 5  1000 Pr  gives: k    0.026 W / m K k Pr 0.7 First we need to determine if these flows are laminar or turbulent. For the inside (Free convection): 1 1 1    K 1 T 20  273 293 Download free ebooks at bookboon.com 48

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