Question

A 6-m-diameter spherical tank is filled with liquid oxygen \(\left(\rho=1141 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=1.71 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\) at \(-184^{\circ} \mathrm{C}\). It is observed that the temperature of oxygen increases to \(-183^{\circ} \mathrm{C}\) in a 144-hour period. The average rate of heat transfer to the tank is (a) \(249 \mathrm{~W}\) (b) \(426 \mathrm{~W}\) (c) \(570 \mathrm{~W}\) (d) \(1640 \mathrm{~W}\) (e) \(2207 \mathrm{~W}\)

Short answer

Answer: The average rate of heat transfer to the tank is approximately 426 Watts.

Step-by-step solution

Calculate the volume of the spherical tank

The diameter of the tank is given as 6 meters, so the radius (r) is half of that, which is 3 meters. The volume (V) of a sphere can be calculated using the formula: \(V = \dfrac{4}{3}\pi r^3\). Plugging in the radius, we get: \(V = \dfrac{4}{3}\pi (3)^3 = 36\pi\) cubic meters.

Find the mass of liquid oxygen

The density (\(\rho\)) of the liquid oxygen is given as \(1141 \mathrm{~kg/m^3}\). To find the mass (m) of the liquid oxygen in the tank, use the formula: \(m = \rho V\). Plugging in the calculated volume and given density, we get: \(m = 1141 \times 36\pi = 129956.12 \mathrm{~kg}\).

Calculate the change in temperature

The initial temperature of the liquid oxygen is \(-184^{\circ}\mathrm{C}\) and the final temperature is \(-183^{\circ}\mathrm{C}\). The change in temperature (\(\Delta T\)) can be found using the formula: \(\Delta T = T_{final} - T_{initial} = -183 - (-184) = 1^{\circ}\mathrm{C}\).

Calculate the heat transferred

Now, we can use the formula for heat transfer \(Q = mc_p\Delta T\) to find the total heat transferred to the tank during the given time period. Plugging in the values for mass, specific heat, and change in temperature, we get: \(Q = 129956.12 \times 1.71 \times 1 = 222192.5452 \mathrm{~kJ}\).

Find the average heat transfer rate

To calculate the average heat transfer rate (P), divide the total heat transfer by the total time in seconds. The total time given is 144 hours. Convert hours to seconds: 144 hours * 3600 seconds/hour = 518400 seconds. Therefore: \(P = \dfrac{Q}{time} = \dfrac{222192.5452 \times 10^3}{518400} = 428.57 \mathrm{W}\). The answer is closest to the option (b) \(426 \mathrm{W}\), which means the average rate of heat transfer to the tank is around 426 Watts.

Key concepts

Spherical Tank Heat Transfer

Understanding heat transfer involving a spherical tank is critical not only in industrial applications but also in studying thermodynamics. Due to their shape, spherical tanks minimize the surface area for a given volume, which affects how heat is transferred into or out of the liquid contained within.

The thermal conductivity of the tank material, the temperature difference between the tank's contents and the surroundings, and the thickness of the tank walls are key factors determining the rate at which heat is transferred. In the case of a liquid oxygen tank, as in the exercise, insulation likely plays a key role in maintaining the cryogenic temperatures by reducing heat influx.

For a basic estimate of the heat transfer rate without intricate details of tank material or insulation properties, we can assume that the heat transferred primarily serves to change the temperature of the liquid oxygen. From this point, we can apply the specific heat capacity of the substance to determine the energy needed for a certain temperature change, which leads us to the next important concept: thermal energy in heat transfer.

Thermal Energy in Heat Transfer

Thermal energy refers to the internal energy present in a system due to the movement of its particles. In the context of heat transfer, we're interested in how this energy is exchanged between a system and its surroundings.

The amount of heat transferred (\(Q\text{ in kJ}\) in the exercise) is a product of the mass (\(m\text{ in kg}\)), specific heat capacity (\(c_p\text{ in kJ/kg}\text{°C}\)), and the change in temperature (\(ΔT\text{ in °C}\)). This relationship is described by the equation: \[Q = mc_pΔT\]

The specific heat capacity is a material's property describing how much energy is needed to raise the temperature of one kilogram of that material by one degree Celsius. In the case of liquid oxygen, its high specific heat means it can absorb a lot of energy before changing temperature significantly, contributing to its use in applications that require temperature stability.

Change in Temperature Calculation

The change in temperature is a straightforward yet crucial concept in thermodynamics and heat transfer calculations. In our exercise, it's the difference between the final and initial temperatures of the liquid oxygen within the tank.

To calculate this change in temperature (\(ΔT\text{ in °C}\)), subtract the initial temperature (\(T_{initial}\text{ in °C}\)) from the final temperature (\(T_{final}\text{ in °C}\)):\[ΔT = T_{final} - T_{initial}\]
In our example, the temperature increased by one degree Celsius. This seemingly small change requires a significant amount of energy due to the mass and specific heat of the liquid oxygen. Calculating the change in temperature is essential for determining the thermal energy transferred in processes such as heating, cooling, or maintaining the substance at a constant temperature in thermal management systems.