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Chapter 3: Problem 203

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

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

Step by step solution

01

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.
02

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: \(m = 1141 \times 36\pi = 129956.12 \mathrm{~kg}\).
03

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: \(\Delta T = T_{final} - T_{initial} = -183 - (-184) = 1^{\circ}\mathrm{C}\).
04

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: \(Q = 129956.12 \times 1.71 \times 1 = 222192.5452 \mathrm{~kJ}\).
05

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 (\(c_p\text{ in kJ/kg}\text{掳C}\)), and the change in temperature (\(螖T\text{ in 掳C}\)). This relationship is described by the equation: \[Q = mc_p螖T\]

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 (\(螖T\text{ in 掳C}\)), subtract the initial temperature (\(T_{initial}\text{ in 掳C}\)) from the final temperature (\(T_{final}\text{ in 掳C}\)):\[螖T = T_{final} - T_{initial}\]
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.

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Most popular questions from this chapter

Two 3-m-long and \(0.4-\mathrm{cm}\)-thick cast iron \((k=\) \(52 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) steam pipes of outer diameter \(10 \mathrm{~cm}\) are connected to each other through two 1 -cm-thick flanges of outer diameter \(20 \mathrm{~cm}\). The steam flows inside the pipe at an average temperature of \(200^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(180 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The outer surface of the pipe is exposed to an ambient at \(12^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Disregarding the flanges, determine the average outer surface temperature of the pipe. (b) Using this temperature for the base of the flange and treating the flanges as the fins, determine the fin efficiency and the rate of heat transfer from the flanges. (c) What length of pipe is the flange section equivalent to for heat transfer purposes?

A mixture of chemicals is flowing in a pipe \(\left(k=14 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, D_{i}=2.5 \mathrm{~cm}, D_{o}=3 \mathrm{~cm}\right.\), and \(L=10 \mathrm{~m}\) ). During the transport, the mixture undergoes an exothermic reaction having an average temperature of \(135^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(150 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). To prevent any incident of thermal burn, the pipe needs to be insulated. However, due to the vicinity of the pipe, there is only enough room to fit a \(2.5\)-cm-thick layer of insulation over the pipe. The pipe is situated in a plant, where the average ambient air temperature is \(20^{\circ} \mathrm{C}\) and the convection heat transfer coefficient is \(25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Determine the insulation for the pipe such that the thermal conductivity of the insulation is sufficient to maintain the outside surface temperature at \(45^{\circ} \mathrm{C}\) or lower.

A spherical vessel, \(3.0 \mathrm{~m}\) in diameter (and negligible wall thickness), is used for storing a fluid at a temperature of \(0^{\circ} \mathrm{C}\). The vessel is covered with a \(5.0\)-cm-thick layer of an insulation \((k=0.20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). The surrounding air is at \(22^{\circ} \mathrm{C}\). The inside and outside heat transfer coefficients are 40 and \(10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), respectively. Calculate \((a)\) all thermal resistances, in \(\mathrm{K} / \mathrm{W},(b)\) the steady rate of heat transfer, and \((c)\) the temperature difference across the insulation layer.

What is the difference between the fin effectiveness and the fin efficiency?

Consider a 25-m-long thick-walled concrete duct \((k=\) \(0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) of square cross section. The outer dimensions of the duct are \(20 \mathrm{~cm} \times 20 \mathrm{~cm}\), and the thickness of the duct wall is \(2 \mathrm{~cm}\). If the inner and outer surfaces of the duct are at \(100^{\circ} \mathrm{C}\) and \(30^{\circ} \mathrm{C}\), respectively, determine the rate of heat transfer through the walls of the duct. Answer: \(47.1 \mathrm{~kW}\)
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