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A series of ASME SA-193 carbon steel bolts are bolted to the upper surface of a metal plate. The bottom surface of the plate is subjected to a uniform heat flux of \(5 \mathrm{~kW} / \mathrm{m}^{2}\). The upper surface of the plate is exposed to ambient air with a temperature of \(30^{\circ} \mathrm{C}\) and a convection heat transfer coefficient of \(10 \mathrm{~W} / \mathrm{m}^{2}\). K. The ASME Boiler and Pressure Vessel Code (ASME BPVC.IV-2015, HF-300) limits the maximum allowable use temperature to \(260^{\circ} \mathrm{C}\) for the SA-193 bolts. Determine whether the use of these SA-193 bolts complies with the ASME code under these conditions. If the temperature of the bolts exceeds the maximum allowable use temperature of the ASME code, discuss a possible solution to lower the temperature of the bolts.

Short Answer

Expert verified
No, the use of SA-193 carbon steel bolts does not comply with the ASME code under the given conditions. The temperature of the bolts is higher than the maximum allowable use temperature specified by the ASME code. Possible solutions to lower the bolt temperature include increasing the airflow on the upper surface of the plate, using a different material for the plate with higher thermal conductivity, or introducing an insulating layer between the plate and the bolts.

Step by step solution

01

Determine the heat transfer per unit area between the upper surface of the plate and the ambient air

To determine the heat transfer per unit area, we can use the convection heat transfer equation: \(q = h * (T_{plate} - T_{air})\) We're given \(q = 5 \mathrm{~kW} / \mathrm{m}^{2} = 5000 \mathrm{~W} / \mathrm{m}^{2}\) and \(h = 10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{K}\). We don't know the temperature of the upper surface of the plate (\(T_{plate}\)), so we can't directly calculate the heat transfer per unit area. However, we can express the temperature difference as a single variable, which we will do in the next step.
02

Calculate the temperature difference between the upper surface of the plate and the ambient air

With the convection heat transfer equation, we can now express the temperature difference between the upper surface of the plate and the ambient air as: \(T_{plate} - T_{air} = \frac{q}{h}\) Now, we can plug in the given values: \(T_{plate} - 30^{\circ} \mathrm{C} = \frac{5000 \mathrm{~W} / \mathrm{m}^{2}}{10 \mathrm{~W} / \mathrm{m}^{2} \mathrm{K}}\) \(T_{plate} - 30^{\circ} \mathrm{C} = 500 \mathrm{K}\) Now we can solve for \(T_{plate}\): \(T_{plate} = 500 \mathrm{K} + 30^{\circ} \mathrm{C} = 530^{\circ} \mathrm{C}\)
03

Calculate the temperature of the bolts

In this exercise, we can assume that the temperature of the bolts is the same as the temperature of the upper surface of the plate, as they are in direct contact with the plate and will conduct heat from the plate. So, we have: \(T_{bolts} = T_{plate} = 530^{\circ} \mathrm{C}\)
04

Compare the bolt temperature to the maximum allowable use temperature

Now we can compare the bolt temperature to the maximum allowable use temperature specified by the ASME code: \(T_{bolts} = 530^{\circ} \mathrm{C}\) \(T_{max} = 260^{\circ} \mathrm{C}\) As \(T_{bolts} > T_{max}\), the use of these SA-193 bolts does not comply with the ASME code under the given conditions.
05

Discuss a possible solution to lower the bolt temperature

Since the temperature of the bolts exceeds the maximum allowable use temperature according to ASME code, we need to come up with a solution to lower the bolt temperature. One possible solution is to increase the convection heat transfer coefficient, which could be achieved by: 1. Increasing the airflow on the upper surface of the plate, by using fans or vents. 2. Using a different material for the plate with higher thermal conductivity, which can increase the heat transfer resistance and reduce the bolt temperature. 3. Introducing an insulating layer between the plate and the bolts to reduce heat transfer to the bolts. These solutions aim to improve heat dissipation from the plate and reduce the temperature of the bolts, ensuring compliance with the ASME code.

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Most popular questions from this chapter

An engineer who is working on the heat transfer analysis of a house in English units needs the convection heat transfer coefficient on the outer surface of the house. But the only value he can find from his handbooks is $14 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$, which is in SI units. The engineer does not have a direct conversion factor between the two unit systems for the convection heat transfer coefficient. Using the conversion factors between \(\mathrm{W}\) and \(\mathrm{Btu} / \mathrm{h}, \mathrm{m}\) and \(\mathrm{ft}\), and \({ }^{\circ} \mathrm{C}\) and \({ }^{\circ} \mathrm{F}\), express the given convection heat transfer coefficient in Btu/h \(\mathrm{ft}^{2}{ }^{\circ} \mathrm{F}\). Answer: \(2.47\) Btu/h $\cdot \mathrm{ft}^{2}{ }^{\circ} \mathrm{F}$

Consider a sealed 20-cm-high electronic box whose base dimensions are $40 \mathrm{~cm} \times 40 \mathrm{~cm}$ placed in a vacuum chamber. The emissivity of the outer surface of the box is \(0.95\). If the electronic components in the box dissipate a total of \(100 \mathrm{~W}\) of power and the outer surface temperature of the box is not to exceed \(55^{\circ} \mathrm{C}\), determine the temperature at which the surrounding surfaces must be kept if this box is to be cooled by radiation alone. Assume the heat transfer from the bottom surface of the box to the stand to be negligible.

Define emissivity and absorptivity. What is Kirchhoff's law of radiation?

Consider a 20-cm-thick granite wall with a thermal conductivity of $2.79 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The temperature of the left surface is held constant at \(50^{\circ} \mathrm{C}\), whereas the right face is exposed to a flow of \(22^{\circ} \mathrm{C}\) air with a convection heat transfer coefficient of \(15 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). Neglecting heat transfer by radiation, find the right wall surface temperature and the heat flux through the wall.

Heat is lost steadily through a \(0.5-\mathrm{cm}\) thick, $2-\mathrm{m} \times 3-\mathrm{m}\( window glass whose thermal conductivity is \)0.7 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}$. The inner and outer surface temperatures of the glass are measured to be \(12^{\circ} \mathrm{C}\) to \(9^{\circ} \mathrm{C}\). The rate of heat loss by conduction through the glass is (a) \(420 \mathrm{~W}\) (b) \(5040 \mathrm{~W}\) (c) \(17,600 \mathrm{~W}\) (d) \(1256 \mathrm{~W}\) (e) \(2520 \mathrm{~W}\)

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