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Chapter 3: Problem 116

Turbine blades mounted to a rotating dise in a gas turbine engine are cxposed to a gas stream that is at \(T_{a}=1200^{\circ} \mathrm{C}\) and maintains al convection coefficicnt of \(h=250 \mathrm{~W} / \mathrm{m}^{2}\) - K over the blade. The blades, which are fabricated from Inconel, \(k=20 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), have a length of \(L=50 \mathrm{~mm}\). The blade profile has a uniform cross-sectional area of \(A_{c}=6 \times 10^{-4} \mathrm{~m}^{2}\) and a perimeter of \(P=110 \mathrm{~mm}\). A proposed blade- cooling scheme, which involves routing air through the supporting dise, is able to maintain the base of each blade at a temperature of \(T_{b}=300^{\circ} \mathrm{C}\). (a) If the maximum allowable blade temperature is \(1050^{\circ} \mathrm{C}\) and the blade tip may be assumed to be adiabatic, is the proposed cooling scheme satisfactory? (b) For the proposed cooling seheme, what is the rate at which heat is transferred from each blade to the coolant?

Short Answer

Expert verified
Cooling scheme is satisfactory; heat transfer rate is 37.73 W per blade.

Step by step solution

01

Identify Given Data

The problem provides the following values: \(T_a = 1200^{\circ} \mathrm{C}\), \(h = 250 \mathrm{~W/m^2 \, K}\), \(k = 20 \mathrm{~W/m \, K}\), \(L = 0.05 \mathrm{~m}\), \(A_c = 6 \times 10^{-4} \mathrm{~m^2}\), \(P = 0.11 \mathrm{~m}\), and \(T_b = 300^{\circ} \mathrm{C}\). The maximum allowable blade temperature is \(T_{ ext{max}} = 1050^{\circ} \mathrm{C}\).
02

Convert Temperatures to Kelvin

Convert all temperatures from Celsius to Kelvin using the formula: \(T(K) = T(^{\circ}C) + 273.15\). Now, \(T_a = 1473.15 \mathrm{K}\), \(T_b = 573.15 \mathrm{K}\), and \(T_{ ext{max}} = 1323.15 \mathrm{K}\).
03

Calculate the Fin Parameter

The fin parameter \(m\) is determined using: \( m = \sqrt{\frac{hP}{kA_c}} \). Substituting the values: \(m = \sqrt{\frac{250 \times 0.11}{20 \times 6 \times 10^{-4}}} = 115 \mathrm{~m^{-1}}\).
04

Determine Fin Efficiency

The effectiveness \(\eta_f\) of an adiabatic tip fin is given by: \(\eta_f = \frac{\tanh(mL)}{mL}\). Calculate \(mL = 115 \times 0.05 = 5.75\). Then, \(\eta_f = \frac{\tanh(5.75)}{5.75} \approx 0.173\).
05

Calculate Tip Temperature

Using the fin efficiency and the base temperature, find the tip temperature with: \(T_t = T_b + \eta_f (T_a - T_b)\). Substitute: \(T_t = 573.15 + 0.173 \times (1473.15 - 573.15) = 730.29 \mathrm{~K}\). The tip temperature is approximately \(457.14^{\circ} \mathrm{C}\).
06

Evaluate Maximum Temperature Condition

Since \(T_t = 457.14^{\circ} \mathrm{C) < 1050^{\circ} \mathrm{C}\), the cooling scheme maintains the blade below the maximum allowable temperature and is therefore satisfactory.
07

Calculate the Heat Loss Per Blade

Use the equation \(q = hA_c \eta_f (T_a - T_b)\) to compute the heat loss. \(q = 250 \times 6 \times 10^{-4} \times 0.173 \times (1473.15 - 573.15) = 37.73 \mathrm{~W}\).
08

Final Conclusion

The proposed cooling scheme is satisfactory as the maximum blade temperature is kept below the allowable temperature. The heat removed per blade is approximately \(37.73 \mathrm{~W}\).

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Key Concepts

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

Convection Coefficient
Convection is the process of heat transfer that occurs when a fluid moves over a surface. Imagine how air cools you when it blows past on a hot day. In turbines, this fluid is typically the gas stream that surrounds the turbine blades. The convection coefficient, denoted as \( h \), helps quantify this cooling effect.
- **Definition**: The convection coefficient is measured in \( ext{W/m}^2 ext{K} \) and represents the heat absorbed or released by a unit area of a surface per unit time per degree of temperature difference between the surface and the fluid. - **Importance**: A higher \( h \) implies more effective heat transfer, helping keep the blades cooler.
In our exercise, the convection coefficient is given as \( h = 250 ext{ W/m}^2 ext{K} \). It reflects the ability of the gas stream to transfer heat away from or to the turbine blade.
Thermal Conductivity
Thermal conductivity, symbolized as \( k \), is a property of the material that describes how well it can conduct heat.
- **Nature of Thermal Conductivity**: Materials with high thermal conductivity, like copper, transfer heat quickly. Materials with low thermal conductivity, like wood, act as insulators.- **Relevance in Turbines**: For turbine blades made of Inconel, with \( k = 20 ext{ W/m K} \), this informs us how the material spreads the heat itself from the hotter areas exposed to the gas to cooler areas possibly reaching the coolant.
This parameter helps determine how temperature gradients develop within the blade during operation. In the scenario given, the thermal conductivity contributes to determining the fin parameter and eventually influences the rate of cooling.
Fin Efficiency
Fin efficiency, or \( \eta_f \, \), measures the effectiveness of a fin in transferring heat relative to an ideal situation where the whole fin is at the base temperature.
- **Concept**: A fin's role is similar to that of a radiator, which increases the surface area for heat transfer. But efficiency isn't always 100% because the end (or tip) of the fin is less effective in transferring heat than the base.- **Calculating Efficiency**: For an adiabatic tip, we use the formula \( \eta_f = \frac{\tanh(mL)}{mL} \, \), indicating that the efficiency depends on the fin parameter \( m \) and the length \( L \) of the fin.
In our problem, the computed fin efficiency \( \eta_f \) was about 0.173, showing that only 17.3% of the potential heat transfer occurs along the fin's length. This reduced efficiency must be accounted for in determining the cooling effectiveness of the turbine blade.
Adiabatic Process
An adiabatic process is one where no heat is transferred into or out of the system.
- **Meaning**: In simplest terms, it's like a perfectly insulated thermos bottle holding liquid without losing or gaining heat from the surroundings. - **Application in Turbines**: The tip of the blade is considered under adiabatic conditions. This means while evaluating the heat transfer, the tip doesn't gain or lose heat to the environment, simplifying the temperature calculations.
In this exercise, assuming an adiabatic tip means we can focus on how heat transfers from the base towards the thinner, less effective parts without needing to consider external heat exchanges at the end. This is useful when setting up equations for computing temperature distributions along the blade.

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Most popular questions from this chapter

One modality for destroying malignant tissue involves imbedding a small spherical heat source of radius \(r_{e}\) within the tissue and maintaining local temperatures above a critical value \(T_{z}\) for an extended period. Tissue that is well removed from the source may be assumed to remain at nomal body temperature \(\left(T_{b}=37^{\circ} \mathrm{C}\right.\). Obtain a general expression for the radial temperature distribution in the tissue under steady- state conditions for which heat is dissipated at a rate \(q\). If \(r_{e}=0.5 \mathrm{~mm}\), what heat rate must be supplied to maintain a tissue temperature of \(T \geq T_{e}=42^{\circ} \mathrm{C}\) in the domain \(0.5 \leq r \leq\) \(5 \mathrm{~mm}\) ? The tissue thermal cceductivity is approximately \(0.5 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\). Assume negligible perfusion.

As a means of enhancing heat transfer from highperformance logic chips, it is common to attach a heot sink to the chip surface in order to increase the surface area available for convection heat transfer. Because of the ease with which it may be manufactured (by taking orthogonal sawcuts in a block of material), an attmctive option is to use a heat sink consisting of an atmy of square fins of width w on s side. The spacing between adjoining fins would be determined by the width of the sawblade, with the sum of this spacing and the fan width designated as the fin pitch \(S\). The method by which the hent sink is joined to the chip would detesmine the interfacial contact resistance, \(R_{\mathcal{U}^{*}}^{*}\) Consider a square chip of width \(W_{c}=16 \mathrm{~mm}\) and conditions for which cooling is provided by a diclectric liquid with \(T_{w}=25^{\circ} \mathrm{C}\) and \(h=1500 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The heat sink is fahricated from copper \((k=400\) Whe \(\mathrm{K})\), and its characteristic dimensions are \(w=0.25\) \(\operatorname{mmn}_{,} S=0.50 \mathrm{~mm}, L_{y}=6 \mathrm{~mm}\), and \(L_{1}=3 \mathrm{~mm}\). The preccribed values of \(w\) and \(S\) represent minima imposed by manufacturing constraints and the need to maintain adequate flow in the passages between fins. (a) If a metallurgical joint provides a contact resistance of \(R_{\mathrm{W}}^{*}=5 \times 10^{-6} \mathrm{~m}^{2}+\mathrm{K} / \mathrm{W}\) and the maximum allowable chip temperature is \(85^{\circ} \mathrm{C}\), what is the maximum allowable chip power dissipation \(q, ?\) Assume all of the heat to be transferred through the heat sink. (b) It may be possible to increase the heat dissipation by increasing \(w\), subject to the constraint that \((S-w) \geq 0.25 \mathrm{~mm}\), and/or increasing \(L_{f}\) (suhject to manufacturing constraints that \(L_{y} \leq 10 \mathrm{~mm}\). Assess the effect of such changes.

A plane wall of thickness \(2 L\) and thermal conductivity \(k\) experiences a uniform volumetric generation rate 4. As shown in the sketch for Case I. the surface at \(x=-L\) is perfectly insulated, while the other surface is maintained at a uniform, constant temperature \(T_{a}\). For Case 2, a very thin dielectric strip is inserted at the midpoint of the wall \((x=0)\) in order to electrically isolate the two sections, \(A\) and B. The thermal resistance of the strip is \(R_{7}^{*}=0.0005 \mathrm{~m}^{2} \cdot \mathrm{K} / \mathrm{W}\). The parameters associated with the wall are \(k=50 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}, L=20 \mathrm{~mm}\), \(q=5 \times 10^{6} \mathrm{~W} / \mathrm{m}^{3}\), and \(T_{e}=50^{\circ} \mathrm{C}\). (a) Sketch the temperature distribution for case I on \(T-x\) coordinates. Deseribe the key features of this distribution. Identify the location of the maxirmum temperature in the wall and calculate this temperature. (b) Sketch the temperature distribution for Cuse 2 on the same \(T-x\) coordinates. Describe the key features of this distribution. (c) What is the temperature difference between the two walls at \(x=0\) for Case 2? (d) What is the location of the maximum temperature in the composite wall of Case 2? Calculate this temperature.

A high-temperature, gas-cocled nuclear reactor consists of a composite cylindrical wall for which a thorium fuel elerment \((k=57 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K})\) is encused in graphite \((k=3\) \(W / m \cdot K)\) and gaseous helium flows through an annular coolant channel. Consider conditions for which the helium temperature is \(T_{w}=600 \mathrm{~K}\) and the convection coefficien at the outer surface of the graphite is \(h=2000 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). (a) If thermal energy is uniformly gencrated in the fuel element at a rate \(q=10^{\text {n }} \mathrm{W} / \mathrm{m}^{3}\), what are the temperatures \(T_{1}\) and \(T_{2}\) at the inner and outer surfaces, respectively, of the fuel element? (b) Compute and plot the temperature distribution in the composite wall for selected values of \(\dot{q}\). What is the maximum allowable value of \(q\) ?

A firefighter's protective clothing, referred to as a humout cout, is typically constructed as an ensemble of three layexs separated by air gaps, as shown schematically. Representative dimensions and thermal conductivities for the layers are as follows. \begin{tabular}{lcc} \hline Layer & Thickness (mu) & \(k(\mathrm{~W} / \mathrm{m}-\mathrm{K})\) \\ \hline Shell (s) & \(0.8\) & \(0.047\) \\ Moisture barricr \((\mathrm{mb})\) & \(0.55\) & \(0.012\) \\ Thermal liner (t) & \(3.5\) & \(0.038\) \\ \hline \end{tabular} The air gaps between the layers are \(1 \mathrm{~mm}\) thick, and heat is transferred by conduction and radiation ex. change through the stagnant air. The linearized radiation coefficient for a gap may be approximated as, \(h_{\text {rat }}=\sigma\left(T_{1}+T_{2}\right)\left(T_{1}^{2}+T_{2}^{2}\right)=4 o T_{m z}^{3}\), where \(T_{\text {ax }}\) represents the average temperature of the surfaces comprising the gap, and the radiation flux across the gap may be expressed as \(q_{n d}^{*}=h_{\text {ad }}\left(T_{1}-T_{2}\right)\). (a) Represent the turnout coat by a thermal circuit, labeling all the thermal resistances. Calculate and tabulate the thermal resistances per unit area \(\left(\mathrm{m}^{2}\right.\). \(\mathrm{K} / \mathrm{W})\) for each of the layers, as well as for the conduction and radiation processes in the gaps. Assume that a value of \(T_{\text {es }}=470 \mathrm{~K}\) may be used to approximate the radiation resistance of both gaps. Comment on the relative magnitudes of the resistances. (b) For a pre-flash-over fire environment in which firefighters often work, the typical radiant heat flux on the fire-side of the tumout coat is \(0.25 \mathrm{~W} / \mathrm{cm}^{2}\). What is the outer surface temperature of the tumout coat if the inner surface temperature is \(66^{\circ} \mathrm{C}\), a condition that would result in bum injury?
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