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An overhead 25-m-long, uninsulated industrial steam pipe of \(100 \mathrm{~mm}\) diameter is routed through a building whose walls and air are at \(25^{\circ} \mathrm{C}\). Pressurized steam maintains a pipe surface temperature of \(150^{7} \mathrm{C}\), and the coeffcient associated with natural convection is \(h=10\) \(W / m^{2} \cdot K\). The surface emissivity is \(\varepsilon=0.8\). (a) What is the rate of heat loss from the steam line? (b) If the steatm is generated in a gas-fired boiler operating at an efficiency of \(\eta f=0.90\) and natural gas is priced at \(C_{\varepsilon}=50.01\) per MJ, what is the annual cost of heat loss from the line?

Short Answer

Expert verified
(a) 10908.1 W, (b) 19,091,113.47 currency units.

Step by step solution

01

Calculate Convection Heat Transfer

The heat loss by convection from the steam pipe can be calculated using Newton's Law of Cooling: \[ q_{ ext{conv}} = hA(T_s - T_{ ext{infinity}}) \]where:- \( h = 10 \ W/m^2 \cdot K \) is the convection coefficient.- \( A = \pi D L \) is the surface area of the pipe with \(D = 0.1\ m\) and \(L = 25 \ m\).- \( T_s = 150^\circ C \) is the surface temperature of the pipe.- \( T_{\infty} = 25^\circ C \) is the surrounding air temperature.Substituting the given values:\[ A = \pi \times 0.1 \times 25 = 7.85 \ m^2 \]\[ q_{\text{conv}} = 10 \times 7.85 \times (150 - 25) = 9773.75 \ W \]
02

Calculate Radiation Heat Transfer

The heat loss by radiation can be calculated using Stefan-Boltzmann Law: \[ q_{ ext{rad}} = \varepsilon \sigma A (T_s^4 - T_{ ext{surrounding}}^4) \]where:- \( \varepsilon = 0.8 \) is the emissivity.- \( \sigma = 5.67 \times 10^{-8} \ W/m^2 \cdot K^4 \) is the Stefan-Boltzmann constant.- \( T_s = 150 + 273.15 = 423.15\ K \) is the absolute surface temperature.- \( T_{ ext{surrounding}} = 25 + 273.15 = 298.15\ K \) is the absolute surrounding temperature.Substituting the given values:\[ q_{\text{rad}} = 0.8 \times 5.67 \times 10^{-8} \times 7.85 \times (423.15^4 - 298.15^4) = 1134.35 \ W \]
03

Find Total Heat Loss

The total heat loss from the steam line is the sum of the convection and radiation heat losses:\[ q_{ ext{total}} = q_{ ext{conv}} + q_{ ext{rad}} = 9773.75 + 1134.35 = 10908.1 \ W \]
04

Calculate Annual Energy Loss

Convert the power lost to energy lost over a year:\[ E_{ ext{annual}} = q_{ ext{total}} \times 365 \times 24 \times 3600 \] This gives the annual energy loss in Joules. Let's convert this to Megajoules:\[ E_{\text{annual}} = 10908.1 \times 365 \times 24 \times 3600 \times 10^{-6} \ MJ = 343993.72 \ MJ \]
05

Calculate Annual Cost

Using the boiler efficiency and cost per MJ, calculate the annual cost of heat loss:\[ C_{ ext{annual}} = \frac{E_{ ext{annual}}}{\eta_f} \times C_{\varepsilon} \]where:- \( \eta_f = 0.90 \) is the efficiency.- \( C_{\varepsilon} = 50.01 \) per MJ is the cost.Substituting the values:\[ C_{\text{annual}} = \frac{343993.72}{0.90} \times 50.01 = 19091113.47 \ currency \]
06

Final Answer

(a) The rate of heat loss is 10908.1 W. (b) The annual cost of heat loss is approximately 19,091,113.47 currency units.

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Key Concepts

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

Convection
Convection is a vital heat transfer process that occurs through fluid motion, such as air or water. It involves the transfer of heat from the surface of a solid object, like a steam pipe, to the surrounding fluid. This happens due to temperature differences and results in heat being carried away by the fluid. In cases like our steam pipe scenario, we're looking at natural convection, where the fluid motion is induced by buoyancy forces. These forces arise because warm air expands and rises, creating a flow that transports heat away from the hotter surface. To calculate the heat loss due to convection, we use Newton's Law of Cooling. The formula is given by:\[ q_{\text{conv}} = hA(T_s - T_{\text{infinity}}) \]Here:
  • \( h \) is the convection heat transfer coefficient, indicating the efficiency of heat transfer in the fluid.
  • \( A \) is the surface area of the steam pipe in direct contact with the air.
  • \( T_s \) is the temperature of the pipe's surface, and \( T_{\infty} \) is the environmental air temperature.
Using these parameters allows us to quantify the amount of heat the pipe loses via convection, an essential part of total heat loss in thermal analysis.
Radiation Heat Loss
Radiation is another key mechanism of heat transfer, enabling energy emission in the form of electromagnetic waves. Unlike conduction and convection, radiation does not require a medium; thus, it can occur in a vacuum. In our steam pipe situation, heat is radiated from the pipe's hot surface to the cooler surroundings. Radiation heat loss is calculated using the Stefan-Boltzmann Law, which considers the surface temperature of the emitting body and its surrounding:\[ q_{\text{rad}} = \varepsilon \sigma A (T_s^4 - T_{\text{surrounding}}^4) \]Where:
  • \( \varepsilon \) is the emissivity of the surface, a measure of a material's ability to emit energy as radiation.
  • \( \sigma \) is the Stefan-Boltzmann constant, approximately \(5.67 \times 10^{-8} \, W/m^2 \cdot K^4\).
  • \( A \) is the surface area.
  • \( T_s \) and \( T_{\text{surrounding}} \) are absolute temperatures of the surface and surroundings, respectively.
This formula helps us determine the amount of heat lost from the steam pipe through radiation, important for energy efficiency calculations.
Emissivity
Emissivity is a material-specific property influencing how effectively a surface can emit thermal radiation. It's expressed as a dimensionless number between 0 and 1. A surface with an emissivity value of 1 is a perfect blackbody, meaning it absorbs and emits all the radiation possible. Conversely, a value of 0 indicates no emission.In our example, the steam pipe has an emissivity of \( \varepsilon = 0.8 \). This suggests it is a good emitter, but not perfect. Different materials and surface finishes have varying emissivity levels, affecting their rate of radiation heat loss. Knowing the emissivity helps in the precise calculation of radiation heat loss using the Stefan-Boltzmann Law. Different applications require specific emissivity values for optimal thermal management, making it a crucial factor in engineering and design.
Stefan-Boltzmann Law
The Stefan-Boltzmann Law is fundamental in understanding thermal radiation. It provides the total energy radiated per unit surface area of a blackbody as a function of its temperature. The law is expressed as:\[ q = \varepsilon \sigma T^4 \]Here:
  • \( q \) is the radiated energy flux.
  • \( \varepsilon \) is the emissivity of the surface.
  • \( \sigma \) is the Stefan-Boltzmann constant.
  • \( T \) is the absolute temperature in Kelvin.
In practical terms, for non-blackbody surfaces, the emissivity adjusts the blackbody radiation rate to match real-world materials. The law helps us quantify how much heat, via radiation, a surface releases based on its temperature, aiding in the assessment of heat loss in thermal systems like the steam pipe example. This understanding is crucial for energy efficiency in industrial and environmental contexts, informing the design choices regarding heat management.

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Most popular questions from this chapter

Under conditions for which the same room temperature is maintained by a heating or cooling system, it is not uncommon for a person to feel chilled in the winter but comfortable in the summer. Provide a plausible explanation for this situation (with supporting calculations) by considering a room whose air temperature is maintained at \(20^{\circ} \mathrm{C}\) throughout the year, while the walls of the room are nominally at \(27^{\circ} \mathrm{C}\) and \(14^{\circ} \mathrm{C}\) in the summer and winter, respectively. The exposed surface of a person in the room may be assumed to be at a temperature of \(32^{\circ} \mathrm{C}\) throughout the year and to have an emissivity of \(0.90\). The coefficient associated with heat transfer by natural convection between the person and the room air is approximately \(2 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A small sphere of reference-grade iron with a specific heat of \(447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and a mass of \(0.515 \mathrm{~kg}\) is suddenly immersed in a water-ice mixfure. Fine thermoceuple wires suspend the sphere, and the temperature is observed to change from 15 to \(14^{\circ} \mathrm{C}\) in \(6.35 \mathrm{~s}\). The experiment is repeated with a metallic sphere of the same diameter, but of unknown compoxition with a mass of \(1.263 \mathrm{~kg}\). If the same observed temperature change occurs in \(4.59 \mathrm{~s}\), what is the specific heat of the unknown material?

The roof of a car in a parking lot absorbs a solar radiant flux of \(800 \mathrm{~W} / \mathrm{m}^{2}\), while the underside is perfectly insulated. The convection coefficient between the roof and the ambient air is \(12 \mathrm{~W} / \mathrm{m}^{2}+\mathrm{K}\). (a) Neglecting radiation exchange with the surruundings, calculate the temperiture of the roof under seady. state conditions if the ambient air temperature is \(20^{\circ} \mathrm{C}\). (b) For the sume ambient air temperature, calculate the temperature of the roof if its surface emissivity is \(0.8\). (c) The convection coefficient depends on airfiow conditions over the roof, increasing with increasing air specd. Compute and plot the roof temperature as a function of \(h\) for \(2 \leq h \leq 200 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\).

A freezer compartment consists of a cubical cavity that is \(2 \mathrm{~m}\) on a side. Assume the bottom to be perfectly insulated. What is the minimum thickness of styrofoam insulation \((k=0.030 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that muse be applied to the top and side walls to ensure a heat loed of less than \(500 \mathrm{~W}\). when the inner and outer surfaces are \(-10\) and \(35^{\circ} \mathrm{C}\) ?

An instrumentation package has a spherical outer surface of diameter \(D=100 \mathrm{~nm}\) and emissivity \(c=0.25\). The package is placed in a large space simulation chamber whose walls are maintained at \(77 \mathrm{~K}\). If operation of the electronic components is restricted to the lemperature range \(40 \leq 7 \leq 85^{\circ} \mathrm{C}\), what is the range of acceptable power dissipation for the package? Display your results graphically, showing also the effect of variations in the emissivity by censidering values of \(0.20\) and \(0.30\).

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