Liquid helium is commonly used in low temperature scientific studies because it

Question

Liquid helium is commonly used in low temperature scientific studies because it is chemically inert and has the lowest operating temperature of any refrigerant (< 4K). The boiling temperature of helium at sea level atmospheric pressure (1 atm pressure) is -268°C and can it be assumed a tank open to the atmosphere will remain constant at -268°C until the liquid helium in the tank is fully evaporated. Therefore any heat transfer to the tank will result in the evaporation of some helium, which has a heat of vaporization of 22 kJ/kg and a density of 146 kg/m3 at 1 atm. Consider a thick-walled cylindrical tank with inner radius of 1 m, outer radius of 1.1 m, height of 1 m, and constant thermal conductivity of 18 W/m°C. The tank is initially filled with liquid helium at 1 atm and -268°C, and is exposed to ambient air at 22°C with a convective heat transfer coefficient of 25 W/m2 °C. The inner surface temperature of the cylindrical tank is observed to be the same as the temperature of the helium inside. Ignoring heat transfer effects at the end caps of the tank and assuming steady one-dimensional radial heat transfer, (a) express the differential equation and the boundary conditions for heat conduction through the tank, (b) obtain a relation for the variation of temperature in the tank material by solving the differential equation, c) determine the rate of evaporation of the liquid helium in the tank as a result of the heat transfer from the ambient air, and d) discuss how this example relates to the 1st and 2nd laws of thermodynamics and how the rate of evaporation from the helium tank could be reduced?

Explanation / Answer

GIVEN:- r1 = 1m, r2 = 1.1m, L = 1m, h = 18 W/mC, T1 = -268C, T2 = 22C, k = 25 W/m2C, Heat of vaporization (qv) = 22 kJ/kg

SOLUTION:-

(a) Differential equation & boundary conditions--->

As per 1st law of thermodynamics, when there is no shaft work and no mass flow then the heat transfer is zero.

i.e. k(A x dT/dX) = 0 i.e. k(X x dT/dX) = 0

In the context of this problem, the above eq can be written as k x d/dr[ A(r) dT/dr] = 0

Since A = 2 x r x L, then (k x 2 x L) x d/dr[r x dT/dr] = 0 i.e. d/dr[r x dT/dr] = 0 ---- eq-1

Integrating eq-1, we get rdT/dr = a

i.e. dT = a x dr/r ----eq-2 (ANSWER - Differential equation)

Under boundary conditions of r1 & r2, we divide RHS numerator & denominator by r1

Then dT = a d(r/r1) / (r/r1) ---- eq-3 (ANSWER - Differential eq under boundary condition)

(b) Relation for variation of temperature---

Integrating eq-3 yields T = a ln(r/r1) + b --- eq-4

Under boundary conditions T = T1 @ r1 yields T1 = b and T = T2 @ r2 yields T2 = a ln(r2/r1) + T1

or a = (T2 - T1) / ln(r2/r1)

Substituting values of a & b in eq-4 gives--> T = [(T2 - T1) / ln (r2/r1) x ln (r/r1)] + T1 (ANSWER)

(c) The heat transfer rate per unit length Q = -k x A x dT/dr

i.e. Q = - 2 x r1 x k x (T2 - T1) / ln(r2/r1) r1 = 2k (T1 - T2) / ln(r2/r1)

thus Q = 2k (T1-T2) / ln (r2/r1) ----- eq-5

The thermal resistance 'R' is given by R = ln(r2/r1) / 2k i.e. Q = (T1-T2) / R --- eq-6

Substituting values in eq-5, we get- - Q = 2(25) (-268-22) / ln(1.1/1) = 478058.76 W/m (ANSWER)

(d) From eq-6, we observe that the rate of evaporation 'Q' is inversely proportional to thermal resistance 'R'.

And hence Q can be reduced by increasing the R. This can be acheived by coating the outer surface of cylinder by an insulator of high thermal resistance. (ANSWER)

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