91Ó°ÊÓ

Saturated steam at 4 bars absolute pressure with a mean velocity of \(3 \mathrm{~m} / \mathrm{s}\) flows through a horizontal pipe whose inner and outer diameters are 55 and \(65 \mathrm{~mm}\), respectively. The heat transfer coefficient for the steam flow is known to be \(11,000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If the pipe is covered with a \(25-\mathrm{mm}\)-thick layer of \(85 \%\) magnesia insulation and is exposed to atmospheric air at \(25^{\circ} \mathrm{C}\), determine the rate of heat transfer by free convection to the room per unit length of the pipe. If the steam is saturated at the inlet of the pipe, estimate its quality at the outlet of a pipe \(30 \mathrm{~m}\) long. (b) Net radiation to the surroundings also contributes to heat loss from the pipe. If the insulation has a surface emissivity of \(\varepsilon=0.8\) and the surroundings are at \(T_{\text {sur }}=T_{\infty}=25^{\circ} \mathrm{C}\), what is the rate of heat transfer to the room per unit length of pipe? What is the quality of the outlet flow? (c) The heat loss may be reduced by increasing the insulation thickness and/or reducing its emissivity. What is the effect of increasing the insulation thickness to \(50 \mathrm{~mm}\) if \(\varepsilon=0.8\) ? Of decreasing the emissivity to \(0.2\) if the insulation thickness is \(25 \mathrm{~mm}\) ? Of reducing the emissivity to \(0.2\) and increasing the insulation thickness to \(50 \mathrm{~mm}\) ?

Step by step solution

01

Calculate thermal resistance of insulation

Consider the insulation as a cylindrical shell with thickness. Using the formula for thermal resistance of a cylindrical shell, we get: \[ R_\text{insulation} = \frac{\ln \left(\frac{D_\text{outer}}{D_\text{inner}}\right)}{2\pi k} \] Here, \(D_{outer}\) is the outer diameter of the insulation, \(D_{inner}\) is the inner diameter of the insulation and \(k\) is the thermal conductivity of insulation. Thermal conductivity can be calculated as: \[ k = k_0 \cdot \text{purity} \] In this case, \(k_0 = 0.07 \frac{\mathrm{W}}{\mathrm{m\cdot K}}\) and the purity of insulation is \(85\%\).

02

Calculate the rate of heat transfer by free convection to the room per unit length of the pipe

To calculate the heat transfer rate, we use the formula: \[ q = \frac{T_\text{steam} - T_\text{surroundings}}{R_\text{insulation} + R_\text{conv}} \] \(R_{conv}\) is the convective resistance, which can be calculated as \[ R_\text{conv} = \frac{1}{h_\text{conv} A_\text{surf}} \] Here, \(h_\text{conv}\) is the heat transfer coefficient of free convection, and \(A_\text{surf}\) is the surface area per unit length of the pipe. In this case, we are given the heat transfer coefficient \(h_\text{conv} = 11,000 \frac{\mathrm{W}}{\mathrm{m}^2 \cdot \mathrm{K}}\), and the surface area per unit length of the pipe is given by: \[ A_\text{surf} = \pi D_\text{outer} \] We calculate the heat transfer rate per unit length using the given values.

03

Estimate steam quality at the outlet of a 30 m long pipe

We use the formula for heat loss to find the change in enthalpy per unit length: \[ \Delta H = \frac{q}{m_\text{flow}} \] In this case, we have the steam mean velocity (\(3 \frac{\mathrm{m}}{\mathrm{s}}\)) and inner diameter of the pipe (\(55 \mathrm{mm}\)), so we can find the mass flow rate (\(m_\text{flow}\)) \[ m_\text{flow} = G A_{in} = \rho V A_{in} \] Given the absolute pressure of the steam, the density (\(\rho\)) of the saturated steam can be taken from the steam table. Using these values, we estimate the steam quality at the outlet of the 30-meter-long pipe. For part (b):

04

Calculate the rate of heat transfer to the room per unit length of pipe, considering radiation

Now we consider the net radiation heat loss. The formula for net radiation heat transfer rate per unit length is given by: \[ q_\text{rad} = \varepsilon \sigma A_\text{surf} (T_\text{steam}^4 - T_\text{surroundings}^4) \] Here, \(\varepsilon\) is the surface emissivity, \(\sigma\) is the Stefan-Boltzmann constant, and the temperatures should be expressed in Kelvin. We plug in the given values and calculate the total heat transfer rate in both convection and radiation modes. We then find the new outlet flow quality considering the total heat transfer rate. For part (c):

05

Analyzing the effect of changing insulation thickness and emissivity

We repeat the calculations of heat transfer rates from Steps 2 and 4 using the modified values of insulation and emissivity for each case: 1. Increasing the insulation thickness to \(50 \mathrm{mm}\) and keeping the emissivity at \(0.8\) 2. Decreasing the emissivity to \(0.2\) and keeping the insulation thickness at \(25 \mathrm{mm}\) 3. Reducing the emissivity to \(0.2\) and increasing the insulation thickness to \(50 \mathrm{mm}\) After calculating the respective heat transfer rates, we compare them and infer the effect of changing insulation thickness and emissivity on heat loss and steam quality.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thermal Resistance

Thermal resistance is a key concept when exploring heat transfer through materials, such as the insulation covering a pipe. It quantifies a material's ability to resist the flow of heat. Think of it as the thermal equivalent of electrical resistance. In the case of a cylindrical shell like the pipe insulation in our exercise, thermal resistance is calculated using the formula: \[ R_\text{insulation} = \frac{\ln \left(\frac{D_\text{outer}}{D_\text{inner}}\right)}{2\pi k} \] where \( D_\text{outer} \) and \( D_\text{inner} \) are the outer and inner diameters of the insulation, and \( k \) is the thermal conductivity of the insulation material. Thermal conductivity \( k \) derives from the material purity, typically given by \( k = k_0 \cdot \text{purity} \). A material with high thermal resistance is a good insulator, reducing the heat flow per unit surface area and, consequently, the overall heat loss. In this way, increasing the thickness of the insulation generally increases the thermal resistance, thereby reducing heat transfer.

Convective Heat Transfer

Convective heat transfer describes the process where heat moves with the flow of fluids, such as steam or air, across a surface. In heat transfer problems, the rate at which heat is transferred through convection is characterized by the convective heat transfer coefficient, \( h \). In this exercise, steam flows inside a cylindrical pipe, and convection primarily occurs along the pipe’s interior surface. The formula for convective heat transfer per unit length is given by: \[ R_\text{conv} = \frac{1}{h_\text{conv} A_\text{surf}} \] where \( A_\text{surf} \) is the surface area available for heat exchange and \( h_\text{conv} \) is the coefficient for atmospheric air in our setup. This method allows us to calculate how effectively heat is being exchanged between the steam and the pipe walls, influencing the steam's temperature and phase changes as it travels down the pipe.

Radiation Heat Transfer

Radiation heat transfer happens through electromagnetic waves and does not require a medium, unlike convection. Energy is transferred by thermal radiation between two surfaces at different temperatures. This emission of energy has a significant effect on the overall heat loss from a pipe. This transfer depends not only on the surface's temperature but also on its emissivity, \( \varepsilon \), a measure of its ability to emit thermal radiation. The exercise considers the heat lost by radiation with a formula: \[ q_\text{rad} = \varepsilon \sigma A_\text{surf} (T_\text{steam}^4 - T_\text{surroundings}^4) \] Here, \( \sigma \) is the Stefan-Boltzmann constant, a universal constant, which allows the computation of radiant heat energy transferred per unit area. Adjusting the emissivity value and the surface characteristics can significantly alter the net heat loss by radiation, which is crucial in designing energy-efficient heating systems.

Steam Quality

Steam quality refers to the proportion of steam in a liquid-steam mixture, expressed as a percentage. It plays a vital role in thermodynamic calculations, affecting the energy transfer characteristics of the steam. If steam quality is 100%, it's all vapor, with no liquid content; at 0%, it's fully saturated with liquid. In this exercise, as steam travels through the pipe and loses heat, its quality decreases. By calculating the steam's heat loss, you can determine how much of the steam condenses as it flows, which is key to predicting the outlet steam quality. If \[ \Delta H = \frac{q}{m_\text{flow}} \] represents the change in enthalpy, understanding this loss helps in evaluating the efficiency and output condition of the steam system – whether the steam exits fully vaporized or with a portion condensed.

Saturated Steam

Saturated steam is steam at a temperature where any heat loss will result in condensation to water at that same temperature. It occurs at specific pressures, where the water is in equilibrium with its vapor. In industrial applications, maintaining steam in a saturated state ensures maximum efficiency since any heat addition or extraction will lead to phase changes. Knowing the saturated steam conditions is crucial for performing precise calculations in heat transfer operations. Using steam tables, we can find the properties of steam at different pressures, such as density, enthalpy, and specific volume. This data is crucial for comprehensively understanding the behavior of steam, assessing the impact of thermal resistance and other losses, and making informed decisions on system designs and operational parameters.