Question

A cylindrical metal can, 0.1 m high and 0.05 m in diameter, contains liquid helium at its normal boiling point of -452.074 degree Fahrenheit. At this temperature Helium's heat of vaporization is 20.4 kJ/kg. The walls of the helium container are 1.2 cm thick and have a thermal conductivity of 13.889 W/(m K). The helium container is surrounded by liquid nitrogen at a temperature of -327.64 degrees Fahrenheit.

a) What is the conductive surface area of the metal cylinder?

b) What is the temperature of the liquid helium in Kelvin?

c) What is the temperature of the walls in Kelvin?

d) What is the rate of heat flow into the Helium due to conduction?

e) How much Helium is lost per hour due to conduction?   

Answer 1

given
L = 0.1 m
d = 0.05 m
r = d/2 = 0.05/2 = 0.025 m

a) A = 2*pi*r*L + pi*r^2

= 2*pi*0.025*0.1 + pi*0.025^2

= 0.01767 m^2

b) temperature of liquid helium, T_He = (5/9)*(Tf - 32) + 273.15

= (5/9)*(-452.074 - 32) + 273.15

= 4.22 K

c) Temperature of the walls, T_walls = (5/9)*(Tf - 32) + 273.15

= (5/9)*(-327.64 - 32) + 273.15

= 73.35 K

d) rate of heat flow into the helium, dQ/dt = k*A*(T_wall - T_He)/t

= 13.889*0.01767*(73.35 - 4.22)/0.012

= 1413.82 J/s

e) heat gained in one hour time, Q = (dQ/dt)*time

= 1413.82*60*60

= 5089752 J

amount of helium lost per hour due to conduction, m = Q/Lf

= 5089752/(20.4*10^3)

= 249.5 kg per hour