Question

Hot coffee is contained in a cylindrical thermos bottle that is of length L-0.3m and is lying on its side (horizontally). The coffee container consists of a glass flask of diameter DI-7 cm sperated from an aluminum housing of diameter D2-0.08 m by air at atmospheric pressure. If the inner and outer temperatures are 75oC and 35 oC respectively, what is the heat loss from the coffee due to convection?

Answer 1

Note: The diagram attached below is the completion of the given question.

Also calculate the heat loss from the coffee due to radiation and the total heat loss

Answer:

`Q_(conv) = 16.04 W\\Q_(rad) = 3.20 W\\H_(loss) = 19.24 W`

Step-by-step explanation:

Glass diameter of the coffee container,
`D_1 = 7 cm = 0.07 m`

Glass radius of the coffee container, r₁ = 0.07/2 = 0.035 m

Diameter of the aluminium housing,
`D_2 = 0.08 m`

Radius of the aluminium housing, r₂ = 0.08/2 = 0.04 m

Inner temperature, T₁ = 75°C = 348 K

Outer temperature, T₂ = 35°C = 308 K

`\epsilon_1 = 0.25\\\epsilon_2 = 0.25`

`T_f = (T_1 + T_2)/(2) \\T_f = (348+308)/(2) \\T_f = 328 K`

At
`T_f = 328 K`, P = 1 atm

k = 0.0284 w/m-k, v = 23.74 * 10⁻⁶,
`\alpha = 26.6 * 10^(-6)`, pr = 0.703,
`\beta = 3.05 * 10^(-3) K^(-1)`

`H_(loss) = Q_(rad) + Q_(conv)`

`Q_(conv) = (2 \pi k L (T_1 - T_2))/(ln((r_2)/(r_1) )) \\\\Q_(conv) = (2 \pi* 0.0284 * 0.3 (75 - 35))/(ln((0.04)/(0.035) )) \\Q_(conv) = 16.04 W`

`Q_(rad) = (\sigma (\pi D_1 L) (T_1^4 - T_2^4))/((1)/(\epsilon_1 ) + (1 - \epsilon_2)/(\epsilon_2) ((\gamma_1)/(\gamma_2)) )`

`Q_(rad) = (5.67*10^(-8)(\pi * 0.07*0.3) (348^4 - 308^4))/((1)/(0.25 ) + (1 - 0.25)/(0.25) ((0.035)/(0.04)) ) \\\\Q_(rad) = 3.20 W`

`H_(loss) = 3.20 + 16.04\\H_(loss) = 19.24 W`