Question

1. (10 points) A spherical thermocouple junction 1 mm in diameter is inserted in a combustion chamber to measure the temperature of the products of combustion. The hot gases are at 1000 K and have a velocity of 5 m/s. The thermocouple junction has an emissivity of 50% and the cooled walls of the combustor are at 500 K. The properties of the thermocouple junction are: p = 9000 kg/m”, Cp = 500 J/kg:K, k = 100 W/mK. The properties of the combustion gases are: V = 50-10% m/s, k = 0.05 W/m-K, and Pr = 1. The thermocouple was at room temperature when it was inserted in the chamber. If radiation and conduction through the lead wires can be neglected, determine: a. The Reynolds number of the hot gases moving over the thermocouple junction. b. The average Nusselt number. c. The average convection heat transfer coefficient, in W/m2.K. d. The time required for the AT between the hot gases and the thermocouple junction to reach 5% of the initial AT (explicitly: 0/6 = [T«-T]/[T«-Ti] = 5%), in s. e. The steady-state temperature of the thermocouple junction (solve: conv = Grad), in K. Combustion gases, V, T. Combustor wall, Thermocouple junction, D. T

Answer 1

q con grad Tce book V = 5 m/s T=1000K D=0.oom OD assump tions are Wise -> lumped heat Capolitance is valid properties constont conduction loss leod is negligible Radiation exchange in blu Small surface and large enclosure Given for Theomo couple data K= 100w/mk I = 9000 kg/m² + c = 500 J/kg k

number of hot gas Given 99ses data px = 1 k 0.05 Wimk 50 x10 bmls a) Re: VD 5 m/s (0.001m) 50x106 m²l Re = 100 Average nusselt number b) correlation whitaker use ot Nu 2+(0.4 Rela +0.06 Ré 2 + (0.4(100) + 0.06(033) (19* = Nu 7-29

Coefficent trans for heat average 7.29 Nuo. h 7.29 x 0.05w/mk ooolm h = 364 w/mk d) time required Bi- alo 364 x 0.001 100 3:64x10 3 Biot numbero <0.1 hence lumped analysis is valid T-Too = 0.05 given 5% Ti-Too

t= fvc en T.-To ñ As T-TO 11 for sphere 6 ta byc en I 6h 0.05 0.00lm ( 2000 kg/m) 500 og lagk ln (20) 6x364 W/m²K 6.17 S temperature e) steady stote given q conv= qrad 5 ASCToo-T) = eAs 6 (157) (1000-T) 0.5% 5.67010 [14-500] 36+ T = 943 k Using Trial and error

here i have taken Whitaker relation /many other correlations exist you can take if it seems appropriate you will get the same answer

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