A liquid helium container, shown in Fig. 2.38, contains 1000 cm^3 of liquid helium and has a to

step 1 All right guys, let's take a good question. The question was asking us, how much heating is lost

A liquid helium container, shown in Fig. 2.38, contains 1000...

A liquid helium container, shown in Fig. 2.38, contains $1000 \mathrm{~cm}^3$ of liquid helium and has a total wall area of $600 \mathrm{~cm}^2$. It is insulated from the surrounding liquid nitrogen reservoir by a vacuum jacket of 0.5 cm thickness. Liquid helium is at 4.2 K , while liquid nitrogen is at 77 K . If the vacuum jacket is now filled with helium at a pressure of $10 \mu \mathrm{m} \mathrm{Hg}$, estimate how long it will take for all the $1000 \mathrm{~cm}^3$ liquid helium in the container to disappear. (As a crude approximation, we can assume a constant temperature gradient across the vacuum jacket and evaluate the thermal conduction of the He-filled jacket at its mean temperature. FIGURE CAN'T COPY.

Key Concepts

Fourier's Law of Heat Conduction

This fundamental law states that the rate of heat transfer through a material is proportional to the negative of the temperature gradient and the area through which the heat flows. It is crucial for understanding how heat moves from a warmer region to a cooler one within any medium.

Thermal Conductivity

Thermal conductivity is the intrinsic property of a material that measures its ability to conduct heat. In the context of insulation and cryogenics, understanding the conductivity of different materials (or gases at various pressures) helps in predicting and managing unwanted heat transfer.

Temperature Gradient

A temperature gradient is the rate at which temperature changes with distance within a material. It is the driving force behind heat conduction; a steeper gradient results in a larger heat flux, which is particularly significant in systems where maintaining low temperatures is essential.

Insulation and Vacuum Effects

Insulation minimizes the unwanted transfer of heat by reducing conduction, convection, or radiation. A vacuum provides excellent insulation since it lacks molecules to transfer energy via conduction, making it ideal for cryogenic applications, though filling it with even low-pressure gas can significantly change its insulating properties.

Effect of Gas Pressure on Conduction

The presence of gas, even at low pressures, can enable heat transfer through conduction by providing a medium for molecule-to-molecule energy exchange. Understanding how gas pressure influences thermal conduction is key in estimating heat leaks in otherwise well-insulated systems, particularly in cryogenic environments.

Transcript

00:01 All right guys, let's take a good question.

00:02 The question was asking us, how much heating is lost per hour? well, in order to determine how much heating loss per hour, we need to figure out how much heat was burned.

00:16 And how much heat was burned is related to the heat currents equation here, which is a net heat radiation equation.

00:23 It's h equal to a times e times sigma, times t1 to the power 4 minus t2 to the power 4.

00:30 Surface area of the cylindrical metal can, e is the invisivity.

00:35 Sigma is the stephan constant.

00:37 T1 is the temperature of liquid helium and the t2 is the temperature of liquid hydrogen.

00:44 It seems that we got the value of e, the value of sigma and the value of t1 and t2, but we don't have the value of surface area.

00:53 So in order to determine the value of the net heat radiation, we need to find out the value of surface area? well, there's the equation for the cylindrical surface area, which is a equal to 2xr times r plus h.

01:12 We know the height is h is 0 .25 meter.

01:15 What's the r here? what's the radius here? well, the question the diameter was given.

01:20 So the radius was just simply equal to the diameter divided by 2, which is 0 .09 meter divided by 2.

01:26 And this will give us 0 .045 so the surface area of the cylindrical metal is equal to 2 pi times 0 .045 meter times 0 .045 meter plus 0 .25 meter and this will give us the surface area of the cylindrical metal is 0 .083 meter square.

01:47 So now we have the surface area, now we can determine the value for the net heat radiation, which is equal to such things, h equal to e times sigma, times 8, t1 to the power 4 minus t2 to the power 4.

02:04 If we plug in value, we should get negative 0 .03360 5 watt.

02:12 Because you see at the end it was 4 .22 kelvin to the power 4, minus 77 .3 kelvin to the power 4.

02:18 So if a small value minus a large value, it will give us a negative value.

02:23 But it doesn't matter because eventually you can sell out and i will show you why...

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