/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 23 A turbine blade row is cooled wi... [FREE SOLUTION] | 91Ó°ÊÓ

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A turbine blade row is cooled with a coolant mass fraction of \(3 \%\) and the coolant inlet temperature of \(T_{1 c}=\) \(775 \mathrm{~K}\). The coolant is ejected at the blade trailing edge at a temperature of \(825 \mathrm{~K}\) as it mixes with the hot gas at the temperature of \(1825 \mathrm{~K}\). Assuming gas properties for the coolant and hot gas are \(\gamma_{c}=1.40, c_{p c}=1004 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and \(\gamma_{\mathrm{t}}=1.30\) and \(c_{p t}=1244 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and the coolant flow rate is \(3 \mathrm{~kg} / \mathrm{s}\), calculate (a) the rate of heat transfer to the coolant from the blade in kW (b) the mixed-out specific heat (of the coolant and the hot gas after mixing) (c) the mixed-out temperature (of the coolant and the hot gas after mixing)

Short Answer

Expert verified
(a) The rate of heat transfer to the coolant from the blade is 150.6 kW. (b) The mixed-out specific heat (of the coolant and the hot gas after mixing) is 1213.15 J/(kgK). (c) The mixed-out temperature (of the coolant and the hot gas after mixing) is 1735.77 K.

Step by step solution

01

Calculate the Rate of Heat Transfer to the Coolant from the Blade

The rate of heat transfer from the blade to the coolant (Q_c) can be calculated using the formula: \( Q_c = m_c * c_{pc} * (T_{1c} - T_{2c}) \), where \( m_c \) is the mass flow rate of the coolant, \( c_{pc} \) is the specific heat capacity of the coolant at constant pressure, \( T_{1c} \) is the initial temperature of the coolant and \( T_{2c} \) is the final temperature of the coolant. Insert the given values into the formula: \( Q_c = 3 \, kg/s * 1004 \, J/(kgK) * (825 \, K - 775 \, K) = 150.6 \, kW \).
02

Calculate the Mixed-Out Specific Heat

The mixed-out specific heat (cp_m) can be calculated using the formula: \( cp_m = \frac{m_c * c_{pc} + m_t * c_{pt}}{m_c + m_t} \), where \( m_c \) and \( m_t \) are the mass flow rates of the coolant and the hot gas respectively, and \( c_{pc} \) and \( c_{pt} \) are the specific heat capacities of the coolant and the hot gas respectively. We can find \( m_t \) using the given value of the coolant mass fraction (3%). Consider that the total mass is the sum of the coolant and hot gas masses, hence \( m_t = \frac{1 - 0.03}{0.03} * m_c = 32.33 \, kg/s \). Finally, insert these values into the formula: \( cp_m = \frac{3 \, kg/s * 1004 \, J/(kgK) + 32.33 \, kg/s * 1244 \, J/(kgK)}{3 \, kg/s + 32.33 \, kg/s} = 1213.15 \, J/(kgK) \).
03

Calculate the Mixed-Out Temperature

The mixed-out temperature (T_m) can be calculated using the formula: \( T_m = \frac{m_c * c_{pc} * T_{1c} + m_t * c_{pt} * T_t}{m_c * c_{pc} + m_t * c_{pt}} \), where \( T_t \) is the temperature of the hot gas. Insert the given values into the formula: \( T_m = \frac{3 \, kg/s * 1004 \, J/(kgK) * 825 \, K + 32.33 \, kg/s * 1244 \, J/(kgK) * 1825 \, K}{3 \, kg/s * 1004 \, J/(kgK) + 32.33 \, kg/s * 1244 \, J/(kgK)} = 1735.77 \, K \).

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Key Concepts

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

Heat Transfer Calculation
Understanding the heat transfer calculation is crucial when analyzing the cooling of turbine blades or other engineering applications. In simple terms, heat transfer is the process of thermal energy moving from a warmer object to a cooler one. This is expressed mathematically by the equation:
\( Q = m \cdot c_p \cdot (T_2 - T_1) \)
where \( Q \) is the heat transfer rate, \( m \) is the mass flow rate, \( c_p \) is the specific heat capacity, and \( T_1 \) and \( T_2 \) are the initial and final temperatures. In the case of turbine blade cooling, we calculated the rate of heat transfer to the coolant by plugging in the given values for the mass flow rate of the coolant, its specific heat capacity, and the temperature change as the coolant absorbs heat from the blade. It gives us insight into how efficiently the coolant absorbs heat, which is vital for preventing blade damages due to high temperatures.
The step-by-step solution provided a clear method to calculate this heat transfer rate. However, when striving for a profound understanding, note that factors such as airflow patterns, blade geometry, and material properties can also influence heat transfer efficiency but are beyond the scope of this particular calculation.
Specific Heat Capacity
The specific heat capacity, a property of materials, indicates the amount of heat needed to raise the temperature of a unit mass of a substance by one degree Celsius (or one Kelvin). We denote specific heat capacity with the symbol \( c_p \) when at constant pressure which is often the case in gas-phase reactions or processes like the cooling of turbine blades. In the provided exercise, the specific heat capacities of the coolant and the hot gas are important inputs for calculating the heat transfer and the mixed-out temperature after mixing.

In practical terms, a high specific heat capacity means that a substance can absorb more heat without experiencing a significant rise in temperature. This makes such materials good candidates for coolants. The specific heat capacity is a fundamental concept in thermodynamics and plays an integral role in designing thermally efficient systems. It's particularly important for materials that experience temperature fluctuations, as the stability of a system can depend on its ability to maintain a consistent temperature under heat flux. To further improve understanding, one might explore how specific heat capacity varies with temperature and how this variation can affect heat transfer calculations.
Coolant Mass Fraction
The coolant mass fraction is a way to describe the ratio of the mass of the coolant to the total mass of both coolant and hot gas in a system. It is expressed as a percentage and in our case, it's given as 3%. To put it plainly, for every 100 kg of the mixture flowing past the turbine blades, 3 kg is coolant and the rest is hot gas.

This proportion is important in determining the effectiveness of the cooling process and is used in calculations for heat transfer and the resultant temperature after mixing. A small change in the coolant mass fraction can notably impact the system's thermal management. For instance, if the coolant mass fraction is too low, the system may not effectively regulate temperature; on the flip side, too much coolant could mean inefficiencies or increased costs. When the mass fraction and flow rates are given, as in the preceding exercise, we can calculate the actual mass flow rates of both the coolant and the hot gas, which are crucial for finding other parameters like the mixed-out temperature and specific heat after mixing.
Mixed-Out Temperature
The mixed-out temperature is a term used to describe the uniform temperature that results when two streams at different temperatures mix together. It's an essential concept in thermodynamics, especially within the context of a cooling process, as it indicates the final equilibrium temperature of the combined streams. In turbine blade cooling, this equilibrium temperature must stay within certain limits to ensure the structural integrity of the blades.

The calculation of the mixed-out temperature relies on the principle of energy balance, which follows the law of conservation of energy. By accounting for the masses and specific heat capacities of the coolant and hot gas, as well as their respective temperatures before mixing, we are able to determine the temperature after mixing. The mixed-out temperature is representative of the thermal state of the gas after it has absorbed the heat of the coolant and is a direct indicator of the cooling system's performance. When tutoring students or crafting educational content, it's beneficial to include examples that vary the input variables to show the impact on the mixed-out temperature, further enhancing comprehension.

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Most popular questions from this chapter

An axial-flow turbine stage, at its pitchline radius, is shown. The rotor exit is swirl free, i.e., \(C_{\theta 3}=0\). The axial velocity is constant throughout the stage and is equal to \(C_{2}=\) \(300 \mathrm{~m} / \mathrm{s}\). The flow to the turbine stage is purely axial, i.e., \(C_{1}=C_{21}\) and the gas total temperature is \(T_{\mathrm{t} 1}=1500 \mathrm{~K}\) with \(\gamma_{1}=1.33\) and \(c_{p t}=1,156 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\). Assume that the turbine is uncooled and the stage degree of reaction is \(20 \%\). Calculate (a) the nozzle exit flow angle, \(\alpha_{2}\), in degrees (b) the stage specific work in \(\mathrm{kJ} / \mathrm{kg}\) (c) total temperature relative to the rotor, \(T_{12, r}\), in \(\mathrm{K}\) (d) absolute Mach number at nozzle exit, \(M_{2}\) (e) speed of sound, \(a_{3}\), in \(\mathrm{m} / \mathrm{s}\) (f) relative rotor exit Mach number, \(M_{3}\)

The Stanton number is in general a function of Prandtl and Reynolds numbers, among other nondimensional parameters such as Mach number, roughness, curvature, freestream turbulence intensity, etc. Eckert-Livingood model for a flat plate with constant wall temperature, excluding all other effects except Prandtl number and the Reynolds number, is $$ \mathrm{St}_{\mathrm{g}}=0.0296 \operatorname{Pr}_{\mathrm{g}}^{-2 / 3} \mathrm{Re}_{x}^{-1 / 5} $$ for a turbulent boundary layer. The Prandtl number for the gas is \(0.704\) and remains constant along the plate. Make a spreadsheet calculation of Stanton number \(\mathrm{St}_{\mathrm{g}}\) with respect to Reynolds number in the range of \(200,000 \leq \operatorname{Re}_{x} \leq 500,000\). Graph the Stanton number versus Reynolds number. Also, calculate the wall-averaged Stanton number.

An axial-flow turbine stage at the pitchline is shown. The flow entering and exiting the turbine stage is axial, i.e., \(\alpha_{1}=\alpha_{3}=0\) The nozzle exit flow is \(\alpha_{2}=65^{\circ}\). The shaft speed is \(\omega=\) \(5500 \mathrm{rpm}\) and the pitchline radius is \(r_{\mathrm{m}}=50 \mathrm{~cm}\). Assuming \(C_{z}=250 \mathrm{~m} / \mathrm{s}=\) constant. Calculate (a) turbine-specific work \(w_{1}(\mathrm{~kJ} / \mathrm{kg})\) (b) \(\beta_{3}\) (degrees) (c) \({ }^{\circ} R_{\mathrm{m}}\)

The free-stream gas temperature is \(T_{\infty}=1600 \mathrm{~K}\) and the free- stream gas speed is \(V_{\infty}=850 \mathrm{~m} / \mathrm{s}\). The gas properties are \(\gamma_{1}=1.30, c_{p e}=1244 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\) and Prandtl number \(\operatorname{Pr}=0.73\). Consider the flow of this gas over a flat plate at a high Reynolds number corresponding to turbulent flow. Calculate (a) the gas total temperature \(T_{\text {tos }}\) in \(\mathrm{K}\) in the freestream (b) the adiabatic wall temperature \(T_{a w}\) in \(\mathrm{K}\) (c) percent error if we assume adiabatic wall temperature is the same as total temperature of the gas

The combustor discharges gas into a turbine with total pressure and temperature of \(1.5 \mathrm{MPa}\) and \(1650 \mathrm{~K}\), respectively. The turbine nozzle turns the purely axial flow at its entrance to \(62^{\circ}\) while maintaining a constant axial velocity of \(C_{z}=360 \mathrm{~m} / \mathrm{s}\). The rotor blade rotational speed at the pitchline is \(U_{\mathrm{m}}=400 \mathrm{~m} / \mathrm{s}\). For an uncooled turbine stage with \(\gamma=1.33\) and \(c_{\mathrm{p}}=1157 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\), calculate (a) Mach number downstream of the nozzle, \(M_{2}\) (b) relative flow angle to the rotor, \(\beta_{2 \mathrm{~m}}\), in degrees (c) total temperature sensed by the rotor, \(T_{12 r}\), in \(\mathrm{K}\) (d) degree of reaction at pitchline (note \(\alpha_{3}=0\) )

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