The temperature of air flowing through a 25-cm-diameter duct whose inner walls are at 320°C is to be measured with a thermocouple soldered in a cylindrical steel wall of 1.2-cm-OD with an oxidized exterior, as shown in the accompanying sketch. The air flows normal to the cylinder at a mass velocity of 17,600 kg/(h m2). If the temperature indicated by the thermocouple is 200°C, estimate the actual temperature of the air.

Question

GIVEN

• Cylindrical thermocouple wall in an air duct

• Duct diameter (Dd) = 25 cm = 0.25 m

• Duct wall temperature (Tds) = 320°C =m 593 K

• Wall outside diameter (Dw) = 1.2 cm = 0.012 m

• Exterior of wall is oxidized

• Air mass velocity /m A

= 17,600 kg/(h m2)

• Thermocouple indicated temperature (Ttc) = 200°C = 473 K

FIND

• Air temperature (T?) ASSUMPTIONS

• Steady state

• Thermal resistance between the thermocouple and the wall exterior surface is negligible

• Inside of duct behaves as a black body enclosure

• Conduction to the thermocouple wall from the duct wall can be neglected

PROPERTIES AND CONSTANTS

the emissivity of oxidized steel (?) = 0.94. From Appendix 1, Table 5, the Stephan-Boltzmann constant (?) = 5.67 × 10–8 W/(m2 K4)

Answer 1

An iterative solution must be used since the rate of heat transfer will depend on the air properties

which are a function of the unknown air temperature. Heat is transferred by radiation from the duct

wall to the thermocouple wall and from the thermocouple wall to the air. Therefore, the air

temperature will be lower then the thermocouple reading. The rate of heat transfer from the wall to the

thermocouple must equal that from the thermocouple to the air



Solving for the air temperature



For the first iteration, let Ta = 150°C. From Appendix 2, Table 28, for air at 150°C

Density (?) = 0.820 kg/m3 Thermal conductivity (k) = 0.0339 W/(m K) Kinematic viscosity (?) = 29.6 × 10–6 m2/s Prandtl number (Pr) = 0.71 At the wall temperature of 200°C Prs = 0.71

The air velocity (U?) is



The Reynolds number based on the well diameter is



The Nusselt number is given by



The air temperature is



The original guess for Ta is close to the above value. Another iteration using air properties at 143°C

would not significantly improve the result.