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An electrical water \((k=0.61 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) heater uses natural convection to transfer heat from a 1 -cm-diameter by \(0.65\)-m-long, \(110 \mathrm{~V}\) electrical resistance heater to the water. During operation, the surface temperature of this heater is \(120^{\circ} \mathrm{C}\) while the temperature of the water is \(35^{\circ} \mathrm{C}\), and the Nusselt number (based on the diameter) is 5 . Considering only the side surface of the heater (and thus \(A=\pi D L\) ), the current passing through the electrical heating element is (a) \(2.2 \mathrm{~A}\) (b) \(2.7 \mathrm{~A}\) (c) \(3.6 \mathrm{~A}\) (d) \(4.8 \mathrm{~A}\) (e) \(5.6 \mathrm{~A}\)

Short Answer

Expert verified
Question: An electrical water heater operates via natural convection with a surface temperature of 120°C, water temperature of 35°C, thermal conductivity of the water (k) is 0.61 W/mK, heater diameter is 0.01 m, heater length is 0.65 m, a voltage of 110 V, and a Nusselt number of 5. Determine the current passing through the electrical heating element. Answer: (d) \(4.8 \mathrm{~A}\)

Step by step solution

01

Calculate the heat transfer rate (Q)

To find the heat transfer rate, we need to use the following formula: Q = h * A * (T_s - T_w) Where: - Q is the heat transfer rate - h is the convective heat transfer coefficient - A is the surface area of the heater - T_s is the surface temperature of the heater - T_w is the temperature of the water However, we don't have the value of h directly, but we are given the Nusselt number (Nu) and a formula to calculate h. Let's first calculate h using the given parameters.
02

Calculate the convective heat transfer coefficient (h)

We can calculate h using the Nusselt number (Nu) as follows: h = Nu * k / D Where: - h is the convective heat transfer coefficient - Nu is the Nusselt number - k is the thermal conductivity of the water - D is the diameter of the heater Now we can plug in the given values: h = 5 * 0.61 W/mK / 0.01m = 305 W/m²K Now let's calculate the heat transfer rate (Q) using the formula mentioned in Step 1.
03

Calculate the heat transfer rate (Q)

Now that we have the value of h, we can calculate the heat transfer rate (Q) using the formula from Step 1: Q = h * A * (T_s - T_w) Let's first calculate the surface area (A) of the heater: A = π * D * L A = π * 0.01m * 0.65m ≈ 0.02042 m² Now, we can calculate Q: Q = 305 W/m²K * 0.02042 m² * (120°C - 35°C) ≈ 528.73 W
04

Calculate the current (I) through the heating element

Now we can calculate the current (I) using the given voltage and the calculated heat transfer rate (Q) using the following formula: Q = V * I Where: - Q is the heat transfer rate - V is the voltage - I is the current Solving for I: I = Q / V I = 528.73 W / 110 V ≈ 4.81 A
05

Choose the correct option

Based on the calculated current (I) value, we can now choose the correct option from the given choices: (a) \(2.2 \mathrm{~A}\) (b) \(2.7 \mathrm{~A}\) (c) \(3.6 \mathrm{~A}\) (d) \(4.8 \mathrm{~A}\) (e) \(5.6 \mathrm{~A}\) Since our calculated current value is approximately 4.81 A, which is very close to 4.8 A, we can choose option (d) \(4.8 \mathrm{~A}\) as the correct answer.

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Key Concepts

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

Heat Transfer Rate
Understanding the heat transfer rate is critical for grasping the efficiency of heating elements like those used in water heaters. The heat transfer rate, denoted as Q, quantifies the amount of heat that is transferred per unit time from the heater to the surrounding environment—in this case, water. To calculate it, the formula Q = h * A * (T_s - T_w) is used, where h represents the convective heat transfer coefficient, A the area over which heat transfer occurs, T_s the surface temperature of the heater, and T_w the temperature of the water.

In practical applications, maximizing Q is often desirable to achieve efficient heating. As such, increasing the surface area of the heater, improving the thermal conductivity of the material, and maintaining a higher temperature difference between the heater and water can all lead to a more efficient heat transfer rate.
Nusselt Number
When it comes to understanding convection, the Nusselt number (Nu) is an indispensable dimensionless quantity. It is used to measure the convective heat transfer, relating the thermal conductivity of the fluid to the convective heat transfer across the fluid. The formula to determine the convective heat transfer coefficient h, based on the Nusselt number, is h = Nu * k / D, where k is the thermal conductivity and D is the characteristic dimension, such as the diameter of the heater.

The Nusselt number itself encapsulates the effects of both thermal conductivity and convection characteristics of the fluid, providing engineers with a way to predict the convective heat transfer when designing systems such as heaters and radiators. Therefore, by knowing the Nusselt number and thermal properties of the fluid, one can predict how effectively heat will be transferred from a surface to the fluid.
Thermal Conductivity
Thermal conductivity, notated as k, is a fundamental property of materials which measures their ability to conduct heat. In the context of convective heat transfer, thermal conductivity is an essential factor that determines how quickly heat can be transferred from a hot surface, like our electrical heater, to the surrounding fluid, such as water in the problem scenario.

Material with higher thermal conductivity, like metals, are excellent at transferring heat, which is why they are often used in heat exchange applications. In contrast, materials with low thermal conductivity, such as plastics or air, act as insulators. The thermal conductivity of water, which is moderately high, is critical in calculating the convective heat transfer coefficient h, an integral step in resolving heat transfer problems and is directly correlated with the efficiency of heat exchange in various applications.
Electrical Resistance Heating
Electrical resistance heating operates on the principle that when electrical current flows through a resistor, it gets converted into heat—a phenomenon also known as Joule heating. In devices such as electric water heaters, an electrical resistance element serves as the resistor, converting electric energy into heat energy to raise the temperature of water.

The amount of heat generated by electrical resistance heating can be described by the formula Q = V * I, where Q represents the heat transfer rate, V is the voltage across the heating element, and I denotes the current flowing through it. By manipulating electric current or voltage, one can control the amount of heat produced, allowing for precise management of temperature in a variety of heating applications.

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Most popular questions from this chapter

Friction coefficient of air flowing over a flat plate is given as \(C_{f}=0.664(V x / \nu)^{-0.5}\), where \(x\) is the location along the plate. Using EES (or other) software, determine the effect of the air velocity \((V)\) on the wall shear stress \(\left(\tau_{w}\right)\) at the plate locations of \(x=0.5 \mathrm{~m}\) and \(1 \mathrm{~m}\). By varying the air velocity from \(0.5\) to \(6 \mathrm{~m} / \mathrm{s}\) with increments of \(0.5 \mathrm{~m} / \mathrm{s}\), plot the wall shear stress as a function of air velocity at \(x=0.5 \mathrm{~m}\) and \(1 \mathrm{~m}\). Evaluate the air properties at \(20^{\circ} \mathrm{C}\) and \(1 \mathrm{~atm}\).

Liquid water at \(15^{\circ} \mathrm{C}\) is flowing over a \(0.3\)-m-wide plate at \(65^{\circ} \mathrm{C}\) a velocity of \(3.0 \mathrm{~m} / \mathrm{s}\). Using EES, Excel, or other comparable software, plot (a) the hydrodynamic boundary layer and \((b)\) the thermal boundary layer as a function of \(x\) on the same graph for the range of \(x=0.0 \mathrm{~m}\) to \(x=x_{\text {cr. }}\) Use a critical Reynolds number of 500,000 .

A rectangular bar with a characteristic length of \(0.5 \mathrm{~m}\) is placed in a free stream flow where the convection heat transfer coefficients were found to be \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) and \(50 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\) when the free stream velocities were \(25 \mathrm{~m} / \mathrm{s}\) and \(5 \mathrm{~m} / \mathrm{s}\), respectively. If the Nusselt number can be expressed as \(\mathrm{Nu}=C \operatorname{Re}^{m} \operatorname{Pr}^{n}\), where \(C, m\), and \(n\) are constants, determine the convection heat transfer coefficients for similar bars with (a) \(L=1 \mathrm{~m}\) and \(V=5 \mathrm{~m} / \mathrm{s}\), and \((b) L=2 \mathrm{~m}\) and \(V=50 \mathrm{~m} / \mathrm{s}\).

During air cooling of steel balls, the convection heat transfer coefficient is determined experimentally as a function of air velocity to be \(h=17.9 V^{0.54}\) for \(0.5

How does turbulent flow differ from laminar flow? For which flow is the heat transfer coefficient higher?

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