A testing lab is contracted to measure the thermal conductivity of various liquids as a function of the liquid temperature. Typically, the lab would measure the thermal conductivity and its temperature dependence by performing many time-consuming experiments at various operating temperatures. A new experimental design is proposed whereby the temperature dependence may be determined in one experiment. The proposed apparatus consists of multiple layers, with each layer consisting of a t_1cm-thick square sheet of low thermal conductivity material sandwiched between two t_ss = 1-mm-thick stainless steel plates with k_ss = 15 W/m ∙ K. The resulting stainless steel-low thermal conductivity-stainless steel sandwiches then separate N = 5, t₁, = 2-mm-thick layers of the liquid. The entire structure is heated from above to eliminate natural convection within the liquid, and cooled from below with a flowing liquid. The temperature of each stainless steel sheet is measured with a thermocouple, and the device is encased in insulation. The temperature range over which the thermal conductivity of a particular liquid is to be measured is 300 K < T < 400 K. To resolve the temperature dependence of the liquid's thermal conductivity, the temperature difference across each liquid layer is to be held to within L1 T = 2°C. The nominal thermal conductivity of the liquid is k, = 0.8 W/m ∙ K.

(a) Consider the low thermal conductivity material to be Bakelite. Determine the overall height, H, of the experimental apparatus.

(b) Consider replacing the Bakelite with an aero-gel characterized by ka = 0.0065 W/m ∙ K. What is the overall height of the apparatus?

(c) To minimize heat losses through the sides of the device, the area of the heater (A_h) is made 10 times larger than the area of the sides (A_s) of the device. Compare the required heater area and required electrical power for devices constructed using Bakelite and aero-gel low thermal conductivity materials.

Transcribed Image Text:

Stainless steel Liquid Stainless steel Heater leads i = 1 fiem Low-k 4. material -N=5 T10 Coolant in Coolant out L. Insulation