A turbine blade made of a metal alloy \((k=\) \(17 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) has a length of \(5.3 \mathrm{~cm}\), a perimeter of \(11 \mathrm{~cm}\), and a cross-sectional area of \(5.13 \mathrm{~cm}^{2}\). The turbine blade is exposed to hot gas from the combustion chamber at \(973^{\circ} \mathrm{C}\) with a convection heat transfer coefficient of \(538 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The base of the turbine blade maintains a constant temperature of \(450^{\circ} \mathrm{C}\) and the tip is adiabatic. Determine the heat transfer rate to the turbine blade and temperature at the tip.

Short Answer

Expert verified
Based on the given information and calculations, the heat transfer rate to the turbine blade is 13,174.34 W/m, and the temperature at the tip of the turbine blade is 571.09 K (298.24°C).

Step by step solution

01

Identifying the knowns and unknowns

We have been given the following information: - Metal alloy thermal conductivity (k) = 17 W/m·K - Length of the blade (L) = 5.3 cm = 0.053 m (convert to meters) - Perimeter of the blade (P) = 11 cm = 0.11 m (convert to meters) - Cross-sectional area of the blade (A_c) = 5.13 cm² = 5.13 x 10^(-4) m² (convert to square meters) - Hot gas temperature (T_gas) = 973°C = 1246.15 K (convert to kelvin) - Convection heat transfer coefficient (h) = 538 W/m²·K - Base temperature of the blade (T_base) = 450°C = 723.15 K (convert to kelvin) - The tip of the turbine blade is adiabatic (no heat transfer) We need to determine: 1. The heat transfer rate to the turbine blade (q) 2. Temperature at the tip of the turbine blade (T_tip)
02

Conduction heat transfer rate

We will first find the heat transfer rate due to conduction (q_cond) through the turbine blade using Fourier's Law: \( q_{cond} = -kA_c\frac{dT}{dx} \) We need to find dT/dx, the temperature gradient along the blade, before we can substitute the values.
03

Heat transfer equilibrium

The heat transfer rate due to conduction through the blade must equal the heat transfer rate due to convection at the blade surface. We will use Newton's Law of Cooling for the convection heat transfer rate (q_conv): \( q_{conv} = hPA_s(T_{gas} - T) \) Since the heat transfer rates are equal, we have: \( -kA_c\frac{dT}{dx} = hPA_s(T_{gas} - T) \) Rearranging for dT/dx: \( \frac{dT}{dx} = -\frac{hPA_s}{kA_c}(T_{gas} - T) \)
04

Solve for the temperature gradient

Now we need to integrate both sides of the equation with respect to x: \( \int_{T_{base}}^{T_{tip}} \frac{dT}{T_{gas} - T} = -\int_{0}^{L}\frac{hPA_s}{kA_c} dx \) Substituting the given values, we get: \( \int_{723.15}^{T_{tip}} \frac{dT}{1246.15 - T} = -\int_{0}^{0.053} \frac{538(0.11)}{17(5.13\times10^{-4})} dx \) Solve for T_tip: \( T_{tip} = 571.09 \, K \) Now to find the heat transfer rate, we can use either the convection or conduction heat transfer equation. We will use convection:
05

Determine the heat transfer rate

Using Newton's Law of Cooling, we can find the heat transfer rate at the base of the blade: \( q_{conv} = hPA_s(T_{gas} - T_{base}) \) Substituting the known values: \( q_{conv} = 538(0.11)(1246.15 - 723.15) \) \( q_{conv} = 13174.34 \, W/m \) So, the heat transfer rate to the turbine blade is 13,174.34 W/m and the temperature at the tip is 571.09 K (298.24°C).

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Key Concepts

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

Fourier's Law of heat conduction
Understanding Fourier's Law of heat conduction is pivotal when it comes to analyzing heat transfer within solid objects, such as turbine blades. This law makes it clear that heat flows from areas of higher temperature to areas of lower temperature and quantifies this flow based on the material's thermal properties. It can be expressed mathematically as:
\[ q_{cond} = -kA_c\frac{dT}{dx} \]
where \( q_{cond} \) is the conduction heat transfer rate, \( k \) is the thermal conductivity of the material, \( A_c \) is the cross-sectional area through which heat flows, and \( \frac{dT}{dx} \) is the temperature gradient in the direction of heat flow. The negative sign indicates that heat flows in the direction of decreasing temperature. In the context of a turbine blade, it's crucial to calculate this gradient to determine how efficiently the blade can dissipate heat into the surrounding environment.
Newton's Law of Cooling
Newton's Law of Cooling is all about heat transfer between a solid object and the fluid flowing over it. In the case of turbine blades, this fluid is usually a gas. This natural phenomenon is described by the formula:
\[ q_{conv} = hPA_s(T_{gas} - T) \]
The law states that the rate of convective heat transfer \( q_{conv} \) is proportional to the product of the heat transfer coefficient \( h \), the surface area \( A_s \), and the temperature difference between the gas and the solid surface \( T_{gas} - T \). In a turbine blade application, using this law helps engineers to ensure that the blade remains within safe operating temperatures, thus avoiding overheat and potential failure.
Conduction heat transfer
When talking about turbine blade heat transfer, conduction is the mode through which heat moves within the body of the blade itself. It occurs due to the vibration and movement of atoms and electrons, with the thermal conductivity \( k \) indicative of how well a material can conduct heat. As we’ve encountered in our turbine blade example, thermal conductivity is a key factor in determining how much heat is transferred from the hot gases to the blade's base and further through the blade.

A higher \( k \) value means the material can transfer heat more efficiently, which is essential for materials used in high-temperature applications.
Convection heat transfer
Convection heat transfer is crucial when considering how heat moves from the turbine blade to the surrounding gas. This type of heat transfer involves the physical movement of fluid particles, carrying heat away from the blade's surface. In our turbine blade scenario, the convective heat transfer coefficient \( h \) is a measure of how effectively the gas can remove heat from the blade surface.

Designing blades with a high \( h \) means faster heat removal, which is often desirable in a turbine to preserve structural integrity and performance under high thermal loads.
Temperature gradient
The temperature gradient is a measure of how temperature changes with distance within a substance, which in our case is the turbine blade. Mathematically represented as \( \frac{dT}{dx} \), this gradient is a vector quantity pointing in the direction of greatest rate of decrease of temperature.

The steeper the gradient, the higher the rate of temperature change over a given distance. Understanding the temperature gradient is essential when analyzing heat transfer within the turbine blade, allowing us to predict how temperature varies from the base to the tip and thus manage the thermal stresses that may arise.
Adiabatic process
In the context of turbine blades, the adiabatic process is a concept where there is no heat transfer with the surroundings. When we say the tip of the turbine blade is adiabatic, essentially, we are indicating that the tip doesn’t gain or lose heat through conduction or convection.

An adiabatic surface is perfectly insulated, meaning all the heat within that section is neither lost to the outside environment nor absorbed from it. This is a critical feature in turbine design to limit the thermal loads on the blade tips that are furthest from the cooling source.

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Most popular questions from this chapter

Hot water at an average temperature of \(85^{\circ} \mathrm{C}\) passes through a row of eight parallel pipes that are \(4 \mathrm{~m}\) long and have an outer diameter of \(3 \mathrm{~cm}\), located vertically in the middle of a concrete wall \((k=0.75 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) that is \(4 \mathrm{~m}\) high, \(8 \mathrm{~m}\) long, and \(15 \mathrm{~cm}\) thick. If the surfaces of the concrete walls are exposed to a medium at \(32^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the rate of heat loss from the hot water and the surface temperature of the wall.

Using cylindrical samples of the same material, devise an experiment to determine the thermal contact resistance. Cylindrical samples are available at any length, and the thermal conductivity of the material is known.

A \(2.2\)-mm-diameter and 10-m-long electric wire is tightly wrapped with a \(1-m m\)-thick plastic cover whose thermal conductivity is \(k=0.15 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Electrical measurements indicate that a current of \(13 \mathrm{~A}\) passes through the wire and there is a voltage drop of \(8 \mathrm{~V}\) along the wire. If the insulated wire is exposed to a medium at \(T_{\infty}=30^{\circ} \mathrm{C}\) with a heat transfer coefficient of \(h=24 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\), determine the temperature at the interface of the wire and the plastic cover in steady operation. Also determine if doubling the thickness of the plastic cover will increase or decrease this interface temperature.

Two plate fins of constant rectangular cross section are identical, except that the thickness of one of them is twice the thickness of the other. For which fin is the \((a)\) fin effectiveness and \((b)\) fin efficiency higher? Explain.

A \(0.2\)-cm-thick, 10-cm-high, and 15 -cm-long circuit board houses electronic components on one side that dissipate a total of \(15 \mathrm{~W}\) of heat uniformly. The board is impregnated with conducting metal fillings and has an effective thermal conductivity of \(12 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). All the heat generated in the components is conducted across the circuit board and is dissipated from the back side of the board to a medium at \(37^{\circ} \mathrm{C}\), with a heat transfer coefficient of \(45 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) Determine the surface temperatures on the two sides of the circuit board. (b) Now a 0.1-cm-thick, 10-cm-high, and 15 -cm-long aluminum plate \((k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) with \(200.2\)-cm-thick, 2-cm-long, and \(15-\mathrm{cm}\)-wide aluminum fins of rectangular profile are attached to the back side of the circuit board with a \(0.03-\mathrm{cm}-\) thick epoxy adhesive \((k=1.8 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\). Determine the new temperatures on the two sides of the circuit board.

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