A large thermos flask consists of two thin-walled coaxial cylinders with a narrow gap such that the radius ratio R1/R2 = 0.95. The surfaces that face each other are silvered and have an emissivity of 0.05 each. The annulus is evacuated. The length of the flask is L = 0.3 m and the inner radius is R

Question

A large thermos flask consists of two thin-walled coaxial cylinders with a narrow gap such that the radius ratio [latex]\frac{R_{1} }{R_{2}} = 0.95[/latex]. The surfaces that face each other are silvered and have an emissivity of 0.05 each. The annulus is evacuated. The length of the flask is L = 0.3 m and the inner radius is [latex]R_{1} = 0.05 m[/latex]. Calculate the heat loss when the thermos flask is filled with hot water such that [latex]T_{1} = 373K [/latex]while the outer cylinder is at a temperature of [latex]T_{2} = 300 K[/latex]. Also calculate the instantaneous rate of change of the temperature of hot water, assuming it to be a lumped system.

Accepted Answer

Step 1 The radius of the inner wall is specified as [latex]R_{1} = 0.05 m[/latex]. The radius ratio is 0.95 and hence the radius of the outer wall is [latex]R_{2} =\frac{R_{1} }{0.95} =\frac{0.05 }{0.95} = 0.0526 m[/latex]. The annular gap is thus obtained as [latex] s = R_{2} −R_{1} = 0.0526 − 0.05 = 0.0026 m[/latex]. Since the length of the flask is some 6 times the radius and 100 times the gap, it is admissible to ignore end effects in the calculation. Thus, the two surfaces may be assumed to form a two-dimensional duct-like enclosure. The area ratio may hence be assumed as[latex] \frac{A_{1} }{A_{2}} =\frac{R_{1} }{R_{2}} = 0.95[/latex]. The given data is written down as [latex] ε_{1} = ε_{2} = 0.05; T_{1} = 373K; T_{2}= 300K[/latex].

Step 2 Expression 10.56 may now be used to calculate the heat loss.

[latex]q_{1}=\frac{\sigma (T^{4}_{1}-T^{4}_{2} )}{\frac{1}{\varepsilon_{1} }+\frac{A_{1}}{A_{2}}\left(\frac{1}{\varepsilon_{2} }-1 ight) }[/latex] (10.56)

[latex]q_{1}=\frac{5.67 imes 10^{-8}(373^{4}-300^{4} ) }{\frac{1}{0.05} imes 0.95 imes \left(\frac{1}{0.05}-1 ight) } =16.77 W/m^{2}[/latex]

The total heat loss is a product of the heat flux [latex]q_{1}[/latex] and the area [latex]A_{1}[/latex]. The area is given by

[latex]A_{1} = 2π R_{1}L = 2 × π × 0.05 × 0.3 = 0.0942 m^{2}[/latex]

Hence, the heat loss from surface 1 (and thus the heat loss from hot water within the flask) is

[latex]Q_{1} = q_{1}A_{1} = 16.77 × 0.0942 = 1.58 W[/latex]

Step 3 The initial rate of cooling of hot water (temperature =373 K or 100° C) contained inside the flask may be calculated now. The properties of water at 373 K are taken as

Density of water: [latex]ρ = 958.4 kg/m^{3}[/latex], Specific heat of water [latex]C_{p} = 4211 J/kg K[/latex]. Volume of water in the tank is calculated as

[latex]V = π R^{2}_{1} L = π × 0.05^{2} × 0.3 = 0.002356 m^{3}[/latex]

Hence, the mass of water in the flask is

M = ρV = 958.4 × 0.002356 = 2.258 kg

Using the lumped system approximation, we then have the initial cooling rate given by

[latex]\frac{dT }{dt}\mid _{t=0}=-\frac{Q_{1} }{MC_{p}} =-\frac{1.58}{2.258 imes 4211} =-1.662 imes 10^{-4} °C/s or = 0.6 °C/h[/latex]

Question 10.8: A large thermos flask consists of two thin-walled coaxial cy...

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