Question

A 0.6-m-diameter, 1.9-m-long cylindrical tank containing liquefied natural gas (LNG) at \(-160^{\circ} \mathrm{C}\) is placed at the center of a 1.9-m-long \(1.4-\mathrm{m} \times 1.4-\mathrm{m}\) square solid bar made of an insulating material with \(k=0.0002 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). If the outer surface temperature of the bar is \(12^{\circ} \mathrm{C}\), determine the rate of heat transfer to the tank. Also, determine the LNG temperature after one month. Take the density and the specific heat of LNG to be \(425 \mathrm{~kg} / \mathrm{m}^{3}\) and \(3.475 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), respectively.

Short answer

Answer: The rate of heat transfer to the tank is 74.88 W, and the temperature of the LNG after one month is -135.66°C.

Step-by-step solution

Find the rate of heat transfer through the insulating material

The formula for the rate of heat transfer (Q) through a cylinder is given by $$Q = \frac {2\pi kL(T_1-T_2)}{\ln \left(\frac{r_2}{r_1} \right)}$$ where L is the length of the cylinder, k is the thermal conductivity of the material, \(T_1\) and \(T_2\) are the inner and outer surface temperatures, and \(r_1\) and \(r_2\) are the inner and outer radii. Given the problem's parameters: Diameter of the cylinder: \(0.6 \,\text{m}\) Length of the cylinder: \(1.9 \,\text{m}\) Thermal conductivity: \(0.0002 \,\frac{\text{W}}{\text{m}\cdot{^{\circ}\text{C}}}\) Inner surface temperature: \(-160^{\circ} \mathrm{C}\) Outer surface temperature: \(12^{\circ} \mathrm{C}\) First, calculate the inner and outer radii. \(r_1 = \frac{0.6\,\text{m}}{2} = 0.3\,\text{m}\) (radius of LNG tank) The square solid bar is \(1.4\,\text{m}\) wide, so the insulating material thickness is \(\frac{1.4\,\text{m}-0.6\,\text{m}}{2} = 0.4\,\text{m}\). Then, \(r_2 = r_1 + 0.4\,\text{m} = 0.7\,\text{m}\). Now, substitute the given values into the formula and find Q. $$Q = \frac {2\pi (0.0002)(1.9)((12)-(-160))}{\ln \left(\frac{0.7}{0.3} \right)} = 74.88\,\text{W}$$

Calculating the energy gain by the LNG

Find the energy gain by LNG after one month, and determine the temperature rise. The energy gain (E) can be calculated using the formula: $$E = Qt$$ where t is the time in seconds. One month has approximately \(30 \times 24 \times 60 \times 60\) seconds. Calculate the energy gain by LNG after one month: $$E = (74.88\,\text{W}) \times (30 \times 24 \times 60 \times 60\,\text{s}) = 194348160 \, \text{J}$$

Calculating the temperature rise of the LNG

To find the temperature rise, we will use the formula: $$\Delta T = \frac{E}{mc}$$ where m is the mass of the LNG, and c is the specific heat of the LNG. Given the density (\(\rho\)) and specific heat (c) of the LNG are \(425\,\frac{\text{kg}}{\text{m}^3}\) and \(3475\,\frac{\text{J}}{\text{kg}\cdot{^{\circ}\text{C}}}\), respectively. First, find the volume and mass of the LNG: $$V_\text{LNG}= \pi r_\text{LNG}^2L_\text{LNG}=\pi(0.3\,\text{m})^2(1.9\,\text{m})=0.54\,\text{m}^3$$ $$m = \rho V_\text{LNG} = (425\,\frac{\text{kg}}{\text{m}^3})(0.54\,\text{m}^3) = 229.5\,\text{kg}$$ Now, find the temperature rise of the LNG: $$\Delta T = \frac{194348160\,\text{J}}{(229.5\,\text{kg})(3475\,\frac{\text{J}}{\text{kg}\cdot{^{\circ}\text{C}}})} = 24.34^{\circ}\text{C}$$

Calculating the final temperature of the LNG

Now that we have the temperature rise, we can calculate the final temperature of the LNG by adding the initial temperature. $$T_\text{final} = T_\text{initial} + \Delta T$$ $$T_\text{final} = -160^{\circ} \mathrm{C} + 24.34^{\circ}\text{C} = -135.66^{\circ}\text{C}$$ The rate of heat transfer to the tank is \(74.88 \,\text{W}\), and the temperature of the LNG after one month is \(-135.66^{\circ} \mathrm{C}\).

Key concepts

Thermal Conductivity

Thermal conductivity is a measure of a material's ability to conduct heat. In essence, it tells us how fast heat can pass through a material.
In this case, we're looking at an insulating material that surrounds a tank of liquefied natural gas (LNG).
The thermal conductivity of this material is given as a very low value, 0.0002 W/m·K, which means it is excellent at resisting heat transfer.
This value is crucial because it helps us calculate how much heat will seep through the insulating material over time.
Using the formula for the rate of heat transfer through a cylinder given by: \[ Q = \frac{2\pi kL(T_1-T_2)}{\ln \left(\frac{r_2}{r_1} \right)} \] where:

• \(k\) is the thermal conductivity of the material.
• \(L\) is the length of the cylinder which is 1.9 m in this problem.
• \(T_1\) and \(T_2\) are the inner and outer temperatures, respectively.
• \(r_1\) and \(r_2\) are the radii of the inner and outer surfaces.

This formula allows us to predict the rate at which heat is transferred from the warmer exterior surface to the cooler interior space of the LNG tank.

Specific Heat Capacity

Specific heat capacity is a property of a material that indicates how much heat energy the material can store.
When applied to LNG, it means the amount of heat required to raise the temperature of one kilogram of LNG by one degree Celsius.
For LNG, this value is given as \(3.475\, \text{kJ/kg}\cdot{^{\circ} ext{C}}\).
This value plays an important role in determining how the temperature within the LNG tank will change as it gains heat over time.
In our example, the specific heat capacity helps us calculate the final temperature of the LNG after it has absorbed heat for a month.
The larger this capacity, the more energy is required to increase the temperature of the LNG, implying better thermal stability. This is reflected in the formula:\[ \Delta T = \frac{E}{mc} \]Here:

• \(E\) is the total energy gain by the LNG, which we calculate by multiplying the heat transfer rate \((Q)\) by the time.
• \(m\) is the mass calculated based on liquid density and tank volume.
• \(c\) is the specific heat, which determines how much temperature change results from the energy absorbed.

Temperature Change

Temperature change is the measure of how much the temperature of a material increases or decreases.
In the context of our LNG tank, we are interested in how much the LNG warms up when heat transfers into it over one month.
This change is the result of the energy gained which was transferred through the insulating material.
We use the equation \(\Delta T = \frac{E}{mc}\) to calculate this change.
Calculated as \(24.34^{\circ}\text{C}\), this temperature shift indicates how much warmer the LNG will become after one month.
Understanding this concept is important, especially in contexts where precise temperature control is crucial, such as in cryogenic storage or other temperature-sensitive environments.
Even with effective insulation, a surprising amount of heat can gradually transfer into a well-sealed container over time.

Energy Transfer

Energy transfer in this scenario refers to the movement of heat energy from the external environment through the insulating material to the LNG inside the tank.
This transfer occurs because of the temperature difference between the outer surface of the insulating material at \(12^{\circ} \text{C}\) and the LNG inside at \(-160^{\circ} \text{C}\).
Heat moves from warmer areas to cooler ones, so naturally, the heat from the outer environment seeks to flow into the colder LNG environment.
The energy gained by the LNG is crucial for calculating the temperature change.
The formula used to compute this energy transfer is: \[ E = Qt \]Where:

• \(Q\) is the rate of heat transfer.
• \(t\) is the time over which the transfer happens — in this instance, the time span is approximately one month.

After calculating the energy energy gained as \(194348160 \text{J}\), we can see the importance of time in energy transfer calculations—small consistent transfers accumulate into significant energy changes.