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Fins and Lumped Analysis most Important Problems in GATE
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Here we have just started the fins and unsteady State heat transfer concepts through gate problems

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1. HEAT TRANSFER By, SAKTHI PRASAD

2. CONTENT FINS and LUMPED ANALYSIS By, SAKTHI PRASAD

3. The heat transfer process between a body and its ambient is governed by a Internal Conductive resistance(ICR) and an external convective resistancetECR). The body can be considered to be a lumped heat capacity system if, a)ICR > ECR c)ICR ECHR b)ICR is marginally smaller than ECR d)ICR is negligible

4. ICR is negligible as, internal conductive resistance of solid external convective resistance . Bi Bi < 0.1 (lumped analysis) For problem to be lumped Bi < 0.1, so that surface to core has same temperature. Option (d)

5. Biot number signifies, a)Ratio of heat conducted to heat convected b)Ratio of heat convected to heat conducted c)Ratio of external convective resistance to internal conductive resistance d)Ratio of internal conductive resistance to external convective resistance

6. We know that, internal conductive resistance of solid external convective resistance Bi Option (d)

7. Lumped heat transfer analysis of a solid object suddenly exposed to a fluid medium at a different temperature is valid when, a)Biot No. < 0.1 c)Fourier No. < 0.1d)Fourier No. > 0.1 b) Biot No. > 0.1

8. We know that, internal conductive resistance of solid external convective resistance Bi- So, Option (a)

9. The value of biot number is very small (less than 0.1), when a) Convective resistance of fluid is negligible b) Conductive resistance of fluid is negligible c) Conductive resistance of the solid is negligible d) None of the above

10. We know that if biot no. is less than 0.1 it is lumped analysis. For lumped analysis, conductive resistance of solid is negligible Option (c)

11. Which one the following configuration has the highest fin effectiveness? a) Thin, Closely spaced fins b) Thin, Widely spaced fins c) Thick, Widely spaced fins d)Thick, Closely spaced fins

12. We know that effectiveness of fin is, PK For infinitely long fin - hA PK tanh(ml) VhA For insulated tip-t From this ,perimeter to area ratio should be high, to have more effectiveness. So, Option(a)

13. We know that, internal conductive resistance of solid external convective resistance Bi So from this expression we conclude Option(b)

14. A spherical thermocouple junction of diameter 0.706mm is to be used for the measurement of temperature of a gas steam. The convective heat transfer co-efficient of the bead surface is 400W/m2K. The thermo-physical properties of the thermocouple material are, k-200W/mK C-400/Kgk and p 8500kg/m3. If the thermocouple initially at 30 C is placed in a hot steam of 300 C, time taken by the bead to reach 298 C is[GATE - 2004] a)2.35s b)4.9s c)14.7s d)29.4s

15. 298 300hA )T 30 300 pvCp hA -270 taking (ln)both side -2 400 6 In( -270) 0.706 10-3 8500 400 4.906 (seconds) option (b)

16. A small copper ball of 5mm diameter at 500K is dropped into an oil bath whose temperature is 300K. The thermal conductivity of copper is 400W/mK, its density 9000kg/m3 and specific heat 385J/kgK. If heat transfer co-efficient is 250W/m2K and lumped analysis is assumed to be valid, the rate of fall of the temperature of the ball at the beginning of cooling will be in, K/s a)8.7 b)13.9 c)17.3 d)27.7 [GATE 2005]

17. Given data: Diameter of sphere (d) 0.005 m Convective heat transfer coefficient (h) 250 W/m2k Thermal conductivity (k) 400 W/mk Specific heatcp 385 J/kg k Density p 9000 kg/m3 Initial temperatureTo 500k Final temperatureTo 300k 6 To find: Rate of fall of temperature of ball at beginning of cooling OT

19. A fin has 5mm diameter and 100mm long. The thermal conductivity of the material is 400W/mK. One end of the fin is maintained at 130 C and its remaining surface is exposed to ambient air at 30 C. If the convective heat transfer co-efficient is 40W/m2K, the heat loss(in W) from the fin is--[GATE-2010] a)o.o8 b)5.0 c7.0 d)7.8

20. 4 -100 + 1.25 = = 101.25 hp KA 40 d 4 40 4 V400x5x10-3 = 8.94427

21. A spherical steel ball of 12mm diameter is initially at 100K. It is slowly cooled in a surrounding of 300K. The heat transfer co-efficient between the steel ball and the surrounding is 5W/m2K. The thermal conductivity of steel is 20W/mK. The temperature difference between the center and the surface of the steel ball is _[GATE 2011] a) Large because conduction resistance is far high than the convective resistance b) Large because conduction resistance is far less than the convective resistance c) Small because conduction resistance is far high than the convective resistance d) Small because conduction resistance is far less than the convective resistance

22. Given data: Sphere diameter 12 mm Initial temperature T = 100 k Surrounding temperature To 300k Convective heat transfer (h)- 5 w/m2k Thermal conductivity (k) = 20 w/mk To find: Temperature difference between center and surface of ball? Solution: hLc 5 d 6x20 = 0.0005

23. Since Bi <0.1, So conduction resistance is small, and there is no temperature variation between centre and outside surface Option (d)

24. A steel ball of diameter 60mm is initially at thermal equilibrium at 1030 C in a furnace. It is suddenly removed from the furnace and cooled in ambient air at 30 C, with convective heat transfer coefficient h-20W/m2K. The thermo-physical properties of steel are: density p 7800Kg/m3, conductivity k 40W/mK and specific heat c 600J/KgK. The time required in seconds to cool the steel ball in air from 1030 C to 430 C is_[GATE 2013] a)519 b)931 c)1195 d)2144

25. Solution: hLc Bi-_- 4 3 23 6 hA To-Too 430 30A T 1030 30 pvc hA 400 1000

26. taking (ln) on both side -20 6 100060x10-3 x7800x60 T 2144(seconds) Option(d)

27. Given data: Diameter 0.01 x 103m Initial temperature 750 C Surrounding temperature 100 C Intermediate temperature 30 C Convective heat transfer coefficient (h) 250 W/m2k Thermal conductivity (k) 43 W/mk Specific heat cp 473J/kg k Density -7801 kg/m2 To find T (time in seconds) -?

28. Solution: hLc LD2 4 4 hA 300 100 750 100 hA pvc -0-(pn )T

29. hA 200 650 taking (ln)both side -250 4 0.01 10-3 7801 473 200 650' = 43.49(seconds)