# CSU MECH 344 - HW5 - Solutions (9 pages)

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## HW5 - Solutions

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- Pages:
- 9
- School:
- Colorado State University- Fort Collins
- Course:
- Mech 344 - Heat and Mass Transfer

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PROBLEM 5 12 KNOWN Diameter density specific heat and thermal conductivity of aluminum spheres used in packed bed thermal energy storage system Convection coefficient and inlet gas temperature FIND Time required for sphere to acquire 90 of maximum possible thermal energy and the corresponding center temperature Potential advantage of using copper in lieu of aluminum SCHEMATIC Aluminum sphere D 75 mm Ti 25oC 2700 kg m3 c 950 J kg K k 240 W m K Gas Tg i 300oC h 75 W m2 K ASSUMPTIONS 1 Negligible heat transfer to or from a sphere by radiation or conduction due to contact with other spheres 2 Constant properties ANALYSIS To determine whether a lumped capacitance analysis can be used first compute Bi 2 h ro 3 k 75 W m K 0 025m 150 W m K 0 013 0 1 Hence the lumped capacitance approximation may be made and a uniform temperature may be assumed to exist in the sphere at any time From Eq 5 8a achievement of 90 of the maximum possible thermal energy storage corresponds to Q 0 90 1 exp t t cV i where t Vc hA s Dc 6h 2700 kg m 3 0 075m 950 J kg K 6 75 W m 2 K 427s Hence t t ln 0 1 427s 2 30 984s From Eq 5 6 the corresponding temperature at any location in the sphere is T 984s Tg i Ti Tg i exp 6ht Dc T 984s 300 C 275 C exp 6 75 W m K 984s 2700 kg m 0 075m 950 J kg K T 984 s 272 5 C 2 3 3 Obtaining the density and specific heat of copper from Table A 1 we see that c Cu 8900 kg m 6 3 6 3 400 J kg K 3 56 10 J m K c Al 2 57 10 J m K Hence for an equivalent sphere diameter the copper can store approximately 38 more thermal energy than the aluminum COMMENTS Before the packed bed becomes fully charged the temperature of the gas decreases as it passes through the bed Hence the time required for a sphere to reach a prescribed state of thermal energy storage increases with increasing distance from the bed inlet PROBLEM 5 24 KNOWN Initial and final temperatures of a niobium sphere Diameter and properties of the sphere Temperature of surroundings and or gas flow and convection coefficient associated with the flow FIND a Time required to cool the sphere exclusively by radiation b Time required to cool the sphere exclusively by convection c Combined effects of radiation and convection SCHEMATIC Ti 900oC Tf 300oC Tsur 25oC Niobium 8600 kg m 3 c 290 J kg K k 63 W m K 0 1 or 0 6 Inert gas Too 25oC h 200 W m 2 K D 10 mm ASSUMPTIONS 1 Uniform temperature at any time 2 Negligible effect of holding mechanism on heat transfer 3 Constant properties 4 Radiation exchange is between a small surface and large surroundings ANALYSIS a If cooling is exclusively by radiation the required time is determined from Eq 3 2 5 18 With V D 6 As r D and 0 1 t 8600 kg m3 290 J kg K 0 01m 3 24 0 1 5 67 10 8 W m 2 K 4 298K ln 298 573 298 1173 ln 298 573 298 1173 573 1 1173 2 tan 1 tan 298 298 t 6926s 1 153 0 519 2 1 091 1 322 1190s 0 1 0 6 If 0 6 cooling is six times faster in which case t 199s b If cooling is exclusively by convection Eq 5 5 yields t 3 cD Ti T 8600 kg m 290 J kg K 0 010m 875 ln ln 6h 275 1200 W m 2 K Tf T t 24 1s c With both radiation and convection the temperature history may be obtained from Eq 5 15 D3 6 c dT 4 D 2 h T T T 4 Tsur dt Integrating numerically from Ti 1173 K at t 0 to T 573K we obtain t 21 0s Continued PROBLEM 5 24 Cont Cooling times corresponding to representative changes in and h are tabulated as follows 2 h W m K t s 200 0 6 21 0 200 1 0 19 4 20 0 6 102 8 500 0 6 9 1 For values of h representative of forced convection the influence of radiation is secondary even for a maximum possible emissivity of 1 0 Hence to accelerate cooling it is necessary to increase h However if cooling is by natural convection radiation is significant For a representative natural 2 convection coefficient of h 20 W m K the radiation flux exceeds the convection flux at the surface of the sphere during early to intermediate stages of the transient Heat fluxes W m 2 K 70000 60000 50000 40000 30000 20000 10000 0 0 20 40 60 80 100 Cooling time s Convection flux h 20 W m 2 K Radiation flux eps 0 6 2 2 COMMENTS 1 Even for h as large as 500 W m K Bi h D 6 k 500 W m K 0 01m 6 63 W m K 0 013 0 1 and the lumped capacitance model is appropriate 2 The largest value of hr 8 2 4 corresponds to Ti 1173 K and for 0 6 Eq 1 9 yields hr 0 6 5 67 10 W m K 1173 2 2 2 2 298 K 1173 298 K 73 3 W m K PROBLEM 5 86 KNOWN Thermophysical properties and initial temperature of thick steel plate Temperature of water jets used for convection cooling at one surface FIND Time required to cool prescribed interior location to a prescribed temperature SCHEMATIC Water jets Too 25 oC Ts Too Steel T i 300 oC 7800 kg m 3 c 480 J kg K k 50 W m K x 0 025 m T 0 025 m t 50oC ASSUMPTIONS 1 One dimensional conduction in slab 2 Validity of semi infinite medium approximation 3 Negligible thermal resistance between water jets and slab surface T s T 4 Constant properties ANALYSIS The desired cooling time may be obtained from Eq 5 60 With T 0 025m t 50 C T x t Ts Ti Ts 50 25 C 0 0909 erf x 300 25 C 2 t x 0 0807 2 t t x2 0 0807 4 2 0 025m 2 0 0261 1 34 10 5 m 2 s 3 1793s 5 2 where k c 50 W m K 7800 kg m 480 J kg K 1 34 10 m s 4 2 COMMENTS 1 Large values of the convection coefficient h 10 W m K are associated with water jet impingement and it is reasonable to assume that the surface is immediately quenched to the temperature of the water 2 The surface heat flux may be determined from Eq 5 61 In principle 1 2 the flux is infinite at t 0 and decays as t PROBLEM 5 92 …

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