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Chapter 2: Problem 99

An electrical current is passed through a \(1 \mathrm{~mm}\)-diameter, \(20 \mathrm{~cm}\)-long copper wire located in an air flow at \(290 \mathrm{~K}\). If the ends of the wire are maintained at \(300 \mathrm{~K}\), determine the maximum current that can be passed if the midpoint temperature is not to exceed \(400 \mathrm{~K}\). The convective heat transfer coefficient is estimated to be 20 \(\mathrm{W} / \mathrm{m}^{2} \mathrm{~K}\). For the copper wire, take \(k=386 \mathrm{~W} / \mathrm{m} \mathrm{K}, \varepsilon=0.8\). and an electrical resistance of \(2.2 \times 10^{-2} \Omega / \mathrm{m}\).

Short Answer

Expert verified
The maximum current is approximately 17.7 A.

Step by step solution

01

Understand the Problem

We have a copper wire with a given diameter, length, and thermal properties. We need to find the maximum electric current it can carry without the midpoint temperature exceeding 400 K. Heat is transferred by convection to the surroundings and by conduction along the wire.
02

Calculate Wire Cross-Sectional Area

The diameter of the wire is given as 1 mm, so the radius is 0.5 mm or 0.0005 m. The cross-sectional area, A, is given by \(A = \pi r^2 = \pi (0.0005)^2 = 7.85 \times 10^{-7} \ m^2\).
03

Calculate Heat Generated in the Wire

Heat generated due to the current is given by \(Q = I^2 R L\). The resistance R of the entire wire is \(R = (2.2 \times 10^{-2})(0.2) = 4.4 \times 10^{-3} \ \Omega \).
04

Determine Temperature Distribution

First, assume a linear temperature distribution along the length of the wire from 300 K to 400 K.
05

Convective Heat Transfer Calculation

The rate of heat transfer by convection is given by \(Q_{conv} = hA_{s}(T_{s} - T_{inf})\), where \(A_{s}\) is the surface area \(A_{s} = \pi DL = \pi(0.001)(0.2) = 0.000628 \ m^2\). Using \(h = 20 \ \rm{W/m^2 K}\), and \(T_s - T_{inf} = 400 - 290 = 110 K\), we calculate \(Q_{conv} = 20 \times 0.000628 \times 110 = 1.3816 \ \rm{W}\).
06

Equate Heat Generation to Heat Loss

At steady state, the heat generated \(Q\) equals the heat lost due to convection \(Q_{conv}\). Thus, \(I^2 \times 4.4 \times 10^{-3} = 1.3816\). Solve for I: \(I^2 = \frac{1.3816}{4.4 \times 10^{-3}}\), hence \(I = \sqrt{314} = 17.72 \ A\).
07

Finalize and Double-check Calculations

Re-evaluate each calculation to ensure accuracy and confirm that assumptions, such as linear temperature distribution, are valid within the problem's constraints.

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Key Concepts

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

Electrical Resistance
Resistance is a key factor in controlling how much current a wire can handle. In our context, the electrical resistance of a conductor depends on its length, cross-sectional area, and resistivity. The formula for resistance is given by:
\(R = \rho \frac{L}{A}\)
For our copper wire, the resistivity, \(\rho\), is specified by the problem parameters. The total resistance also depends on the wire's length and cross-sectional area. The provided resistance value is \(2.2 \times 10^{-2} \ \Omega/m\), for this particular wire.
  • The longer the wire, the more resistance it presents to the electric current.
  • The greater the cross-sectional area, the less resistance there is.
  • This resistance generates heat as electric current flows through the wire.
Understanding how resistance works helps us appreciate limitations on current flow, making it a fundamental concept.
Convective Heat Transfer
Convective heat transfer involves the movement of heat between a solid surface and a fluid in motion, which might be a liquid or a gas. In this exercise, heat is dissipated from the wire into moving air, maintaining the wire's thermal balance. The rate of heat transfer by convection is described by Newton鈥檚 Law of Cooling:
\(Q_{conv} = hA_{s}(T_s - T_\infty)\)
Where:
  • \(Q_{conv}\) is the rate of heat transfer by convection.
  • \(h\) is the convective heat transfer coefficient.
  • \(A_{s}\) stands for the surface area of the wire.
  • \(T_s\) is the surface temperature at the wire鈥檚 midpoint (400 K).
  • \(T_\infty\) is the ambient temperature (290 K).
This setup allows the temperature at the midpoint of the wire to stabilize, ensuring the wire does not overheat.
Temperature Distribution
Temperature distribution refers to how temperatures vary along the length of the wire due to conduction and convective heat loss. Assuming a linear temperature distribution makes calculations manageable. In this problem, both ends of the copper wire are maintained at 300 K, while the temperature at the midpoint is allowed to rise to 400 K.
The assumption of linear temperature variation signifies that the rise in temperature is uniform across the wire鈥檚 length, simplifying the evaluation.
  • Each segment of the wire adds equally to temperature change.
  • This approximation helps match generated and dissipated heat under specified conditions.
  • Understanding distribution aids in predicting how a material behaves under specific thermal constraints.
This modeling forms a simple scenario to study more complex systems.
Conduction
Conduction is one of the primary methods by which heat is transferred through materials. In our problem, it refers to the flow of heat along the copper wire from the warmer center towards the cooler ends. The heat conduction equation depends on the thermal conductivity ( \(k\)) of the material and is given by Fourier鈥檚 Law:
\(Q_{cond} = -kA \frac{dT}{dx}\)
This equation can be simplified in contexts where the temperature change across a small segment is steady over time.
  • \(k\) signifies how well the material conducts heat.
  • In our problem, a higher \(k\) value implies copper conducts heat very efficiently.
  • The heat conduction across the wire balances the heat generated by electrical resistance.
Using conduction analysis helps ensure equilibrium is established quickly, enhancing design safety.

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Most popular questions from this chapter

An explosive is to be stored in large slabs of thickness \(2 L\) clad on both sides with a protective sheath. The rate at which heat is generated within the explosive is temperature-dependent and can be approximated by the linear relation \(\dot{Q}_{v}^{\prime \prime \prime}=a+b\left(T-T_{e}\right)\), where \(T_{e}\) is the prevailing ambient air temperature. If the overall heat transfer coefficient between the slab surface and the ambient air is \(U\), show that the condition for an explosion is \(L=(k / b)^{1 / 2} \tan ^{-1}\left[U /(k b)^{1 / 2}\right]\). Determine the slab thickness if \(k=0.9 \mathrm{~W} / \mathrm{m} \mathrm{K}, U=0.20 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}, a=60 \mathrm{~W} / \mathrm{m}^{3}\), \(b=6.0 \mathrm{~W} / \mathrm{m}^{3} \mathrm{~K}\).

A test rig for the measurement of interfacial conductance is used to determine the effect of surface anodization treatment on the interfacial conductance for aluminum-aluminum contact. The specimens themselves form the heat flux meters because they are each fitted with a pair of thermocouples as shown. The heat flux is determined from the measured temperature gradient and a known thermal conductivity of the aluminum of \(185 \mathrm{~W} / \mathrm{m} \mathrm{K}\) as $$ q_{1}=\frac{k\left(T_{1}-T_{2}\right)}{L_{1}} ; \quad q_{2}=\frac{k\left(T_{3}-T_{4}\right)}{L_{2}} ; \quad q=\frac{1}{2}\left(q_{1}+q_{2}\right) $$ and the interfacial temperatures obtained by linear extrapolation. $$ \begin{gathered} T_{5}=\frac{1}{2}\left(T_{1}+T_{2}\right)-\left[\frac{L_{3}+\frac{L_{1}}{2}}{L_{1}}\right]\left(T_{1}-T_{2}\right) ; \\ T_{6}=\frac{1}{2}\left(T_{3}+T_{4}\right)+\left[\frac{L_{4}+\frac{L_{2}}{2}}{L_{2}}\right]\left(T_{3}-T_{4}\right) \end{gathered} $$ The interfacial conductance is then obtained as $$ h_{i}=\frac{q}{\left(T_{5}-T_{6}\right)} $$ The main possible sources of error in the values of \(h_{i}\) so determined are due to uncertainty in temperature measurement and thermocouple locations. Since only temperature differences are involved, the absolute uncertainty in the individual temperature measurements is not of concern; rather it is the relative uncertainties. Previous calibrations of similar type and grade thermocouples indicate that the relative uncertainties are \(\pm 0.2^{\circ} \mathrm{C}\). Also, the techinician who drilled the thermocouple holes and installed the thermocouples estimates an uncertainty of \(\pm 0.5 \mathrm{~mm}\) in the thermocouple junction locations. At a particular pressure, the temperatures recorded are \(T_{1}=338.7 \mathrm{~K}, T_{2}=328.7 \mathrm{~K}, T_{3}=305.3 \mathrm{~K}\), \(T_{4}=295.0 \mathrm{~K}\). (i) Assuming that the uncertainties in temperature measurement and thermocouple location can be treated as random errors, estimate the uncertainty in the interfacial conductance. (ii) In reality, the uncertainties in temperature and location are bias errors (for example, the location of a thermocouple does not vary from test to test). Thus it is more appropriate to determine bounds on the possible error in \(h_{i}\), by considering best and worst cases. Determine these bounds.

A test technique for measuring the thermal conductivity of copper-nickel alloys is based on the measurement of the tip temperature of pin fins made from the alloys. The standard fin dimensions are a diameter of \(5 \mathrm{~mm}\) and a length of \(20 \mathrm{~cm}\). The test fin and a reference brass fin \((k=111 \mathrm{~W} / \mathrm{m} \mathrm{K})\) are mounted on a copper base plate in a wind tunnel. The test data include \(T_{B}=100^{\circ} \mathrm{C}, T_{e}=20^{\circ} \mathrm{C}\), and tip temperatures of \(64.2^{\circ} \mathrm{C}\) and \(49.7^{\circ} \mathrm{C}\) for the brass and test alloy fins, respectively. (i) Determine the conductivity of the test alloy. (ii) If the conductivity should be known to \(\pm 1.0 \mathrm{~W} / \mathrm{m} \mathrm{K}\), how accurately should the tip temperatures be measured?

The convective heat transfer coefficient around a cylinder held perpendicular to a flow varies in a complicated manner. A test cylinder to investigate this behavior consists of a \(0.001\) in-thick, \(12.7 \mathrm{~mm}\)-wide stainless steel heater ribbon (cut from shim stock) wound around a \(2 \mathrm{~cm}\)-O.D., \(2 \mathrm{~mm}\)-wall-thickness Teflon tube. A single thermocouple is located just underneath the ribbon and measures the local ribbon temperature \(T_{s}(\theta)\). The cylinder is installed in a wind tunnel, and a second thermocouple is used to measure the ambient air temperature \(T_{e}\). The power input to the heater is metered, from which the electrical heat generation per unit area \(\dot{Q} / A\) can be calculated ( \(A\) is the surface area of one side of the ribbon). As a first approximation, the local heat transfer coefficient \(h_{c}(\theta)\) can be obtained from $$ h_{c}(\theta)=\frac{\dot{Q} / A}{T_{s}(\theta)-T_{e}} $$ Hence, by rotating the cylinder with the power held constant, the variation of \(h_{c}\) can be obtained from the variation of \(T_{s}\) : where \(T_{s}(\theta)\) is low, \(h_{c}(\theta)\) is high, and vice versa. A typical variation of \(T_{s}(\theta)\) is shown in the graph. A problem with this technique is that conduction around the circumference of the tube causes the local heat flux \(q_{s}(\theta)\) to not exactly equal \(\dot{Q} / A\). (i) Derive a formula for \(h_{c}(\theta)\) that approximately accounts for circumferential conduction. (ii) The following table gives values of \(T_{s}(\theta)\) in a sector where circumferential conduction effects are expected to be large. Use these values together with \(\dot{Q} / A=5900 \mathrm{~W} / \mathrm{m}^{2}\) and \(T_{e}=25^{\circ} \mathrm{C}\) to estimate the conduction effect at \(\theta=110^{\circ}\). \begin{tabular}{cc} Angle (degrees) & \(T_{s}\left({ }^{\circ} \mathrm{C}\right)\) \\ \hline 100 & \(65.9\) \\ 110 & \(65.7\) \\ 120 & \(64.4\) \end{tabular} (iii) Comment on the design of the cylinder. Would a \(3 \mathrm{~mm}\)-thick brass tube, directly heated by an electric current, be a suitable alternative? Use \(k=15 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for the stainless steel and \(0.38 \mathrm{~W} / \mathrm{m} \mathrm{K}\) for Teflon.

An experimental boiling water reactor is spherical in shape and operates with a water temperature of \(420 \mathrm{~K}\). The shell is made from nickel alloy steel \((k=21\) \(\mathrm{W} / \mathrm{m} \mathrm{K})\) and has an inside radius of \(0.7 \mathrm{~m}\) with a wall thickness of \(7 \mathrm{~cm}\). The reactor is surrounded by a layer of concrete \(20 \mathrm{~cm}\) thick. If the outside heat transfer coefficient is \(8 \mathrm{~W} / \mathrm{m}^{2} \mathrm{~K}\) and the ambient air is at \(300 \mathrm{~K}\), what are the temperatures of the internal and external surfaces of the concrete? Also, if the reactor operates at a power level of \(30 \mathrm{~kW}\), what fraction of the power generated is lost by heat transfer through the shell? The resistance to heat flow from the water to the shell can be taken to be negligible.
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