Three mechanisms of energy transfer—heat transfer, work transfer, and mass transfer—play fundamental roles in the first law of thermodynamics, allowing energy to flow and be conserved within a system.
The first law of thermodynamics, also known as the law of energy conservation, states that energy cannot be created or destroyed within an isolated system, but it can be transferred or transformed from one form to another. There are three main mechanisms of energy transfer within a system according to the first law of thermodynamics:
1. Heat Transfer: Heat transfer is the transfer of thermal energy between objects or within a system due to a temperature difference. It occurs through three modes: conduction, convection, and radiation. Conduction involves the transfer of heat through direct contact between particles or objects. Convection refers to the transfer of heat through the movement of fluids (liquids or gases). Radiation is the transfer of heat through electromagnetic waves.
2. Work Transfer: Work transfer is the transfer of energy due to the application of a force over a distance. It occurs when a system undergoes mechanical work, such as the expansion or compression of gases, or the movement of objects against a force. Work transfer can result in the conversion of energy between different forms, such as mechanical, electrical, or chemical energy.
3. Mass Transfer: Mass transfer involves the transfer of energy associated with the transfer of matter between different regions of a system. It occurs when there is a flow or diffusion of mass from one location to another within the system. Mass transfer can involve energy exchange due to changes in the internal energy or enthalpy of the system, particularly in processes involving chemical reactions or phase changes.
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Reynolds # is the ratio of inertial forces to: Gravitational forces Viscous forces Compressibility forces Pressure forces Surface tension forces
Reynolds number (Re) is the ratio of inertial forces to viscous forces. The Reynolds number is an important dimensionless quantity in fluid mechanics that plays an important role in determining the flow regimes of fluids. The Reynolds number is a dimensionless number that describes the flow of a fluid through a conduit or over a surface.
It is calculated as the ratio of the inertial forces to the viscous forces in the fluid. When the Reynolds number is less than a critical value, the flow is laminar, which means that the fluid flows in smooth layers, with no mixing between them.
When the Reynolds number is greater than the critical value, the flow becomes turbulent, which means that the fluid flows in a chaotic and unpredictable manner. The critical Reynolds number depends on the geometry of the flow, as well as the fluid properties.
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16. Neglecting friction, what force could be expected from the rod end of a cylinder with a 5-in. bore if the blank end receives fluid at 2500 lbf/in.²?
The force expected at the rod end of the cylinder, neglecting friction, would be approximately 49,075 pounds-force.
To determine the force exerted at the rod end of a cylinder, we can use Pascal's law, which states that pressure is transmitted equally in all directions in a fluid.
Given:
Bore diameter (d): 5 inchesPressure at the blank end (P): 2500 lbf/in²To find the force at the rod end, we first need to calculate the area of the rod end. The formula for the area of a circle is A = πr², where r is the radius. In this case, the radius is half of the bore diameter.
Radius (r) = d/2 = 5 inches / 2 = 2.5 inches
Converting the radius to inches:
r = 2.5 inches
Now, we can calculate the area of the rod end:
A = πr² = π(2.5 inches)² ≈ 19.63 square inches
Using Pascal's law, we know that pressure is transmitted equally, so the pressure at the rod end will also be 2500 lbf/in². Finally, we can calculate the force at the rod end by multiplying the pressure by the area:
Force = Pressure × Area Force = 2500 lbf/in² × 19.63 square inches ≈ 49,075 lbfTherefore, neglecting friction, the force expected at the rod end of the cylinder would be approximately 49,075 pounds-force.
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An OSHA inspector visits a facility and reviews the OSHA Form 300 summaries for the past three years and learns there have been significant numbers of recordable low back injuries in the shipping and receiving department. An inspection tour shows heavy materials and parts stored on the floor with mostly manual handling. The inspector writes a citation based on what OSHA standard?
An OSHA inspector visits a facility and reviews the OSHA Form 300 summaries for the past three years and learns there have been significant numbers of recordable low back injuries in the shipping and receiving department.
An inspection tour shows heavy materials and parts stored on the floor with mostly manual handling. The inspector writes a citation based on what OSHA standard?
The citation written by the OSHA inspector was based on OSHA standard 1910.22
(a)(1). This regulation requires employers to keep floors in work areas clean and dry to avoid slipping hazards. OSHA (Occupational Safety and Health Administration) is a government agency in the United States that is responsible for enforcing safety and health standards in the workplace. OSHA conducts inspections of businesses and facilities to ensure that they are following safety regulations. In this scenario, an OSHA inspector visited a facility and reviewed the OSHA Form 300 summaries for the past three years. The inspector discovered that there had been significant numbers of recordable low back injuries in the shipping and receiving department. During an inspection tour of the facility, the inspector observed heavy materials and parts stored on the floor with mostly manual handling.
The OSHA inspector wrote a citation based on OSHA standard 1910.22(a)(1), which requires employers to keep floors in work areas clean and dry to avoid slipping hazards. By storing heavy materials and parts on the floor, the facility was creating a hazardous environment that increased the risk of injury to employees.The OSHA inspector's citation was intended to encourage the facility to take action to correct the issue and prevent future injuries from occurring.
The citation issued by the OSHA inspector was based on OSHA standard 1910.22(a)(1), which requires employers to keep floors in work areas clean and dry to avoid slipping hazards. This standard is designed to protect employees from injury and ensure that employers are providing a safe working environment. By issuing the citation, the OSHA inspector was working to ensure that the facility took action to correct the issue and prevent future injuries from occurring.
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Change in enthalpy of a system is the heat supplied at (a) constant pressure (b) constant temperature (c) constant volume (d) constant entropy C is related to the changes in and c to the changes in (a) internal energy,temperature (b) temperature, enthalpy (c) enthalpy,internal energy (d) Internal energy,enthalpy For ideal gases, u, h, Cv₂ and c vary with P (a) Pressure only (b) Temperature only (c) Temperature & pressure (d) Specific heats 1 The value of n = 1 in the polytropic process indicates it to be a) reversible process b) isothermal process c) adiabatic process d) irreversible process e) free expansion process. Solids and liquids have a) one value of specific heat c) three values of specific heat d) no value of specific heat e) one value under some conditions and two values under other conditions.
Given below are the answers to the given question:(a) constant pressure is the correct option. Change in enthalpy of a system is the heat supplied at constant pressure.(c) enthalpy,internal energy are related to the changes in. Change in enthalpy of a system is the heat supplied at constant pressure, and internal energy is related to the changes in the system's internal energy.
(c) Temperature & pressure. For ideal gases, u, h, Cv₂, and c vary with temperature and pressure.(c) adiabatic process is the correct option. The value of n = 1 in the polytropic process indicates it to be an adiabatic process.(c) three values of specific heat are the correct option. Solids and liquids have three values of specific heat.
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D5-1: A commercial airliner cruises at 40,000 feet at a Mach number of 0.78. a) What is the airliner's speed? b) What stagnation pressure do you expect on the nose of the aircraft? c) What stagnation temperature do you expect on the nose of the aircraft? Use standard atmospheric properties from Table C.1.
For a commercial airliner cruising at 40,000 feet with a Mach number of 0.78, we can calculate the airliner's speed, stagnation pressure on the nose of the aircraft, and stagnation temperature on the nose of the aircraft using standard atmospheric properties.
These values can be obtained from Table C.1. a) To calculate the airliner's speed, we need to use the relation between Mach number (M) and the speed of sound (a). The speed of sound depends on the temperature of the air at the cruising altitude. Using standard atmospheric properties from Table C.1, we can determine the temperature and then calculate the speed of sound. Multiplying the speed of sound by the Mach number will give us the airliner's speed. b) The stagnation pressure on the nose of the aircraft can be determined using the concept of total pressure. Total pressure, also known as stagnation pressure, is the sum of the static pressure (ambient pressure) and the dynamic pressure (caused by the motion of the aircraft). Using the standard atmospheric properties from Table C.1, we can obtain the static pressure at 40,000 feet and then calculate the total pressure on the nose of the aircraft. c) Similarly, the stagnation temperature on the nose of the aircraft can be determined using the concept of total temperature.
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Find the first two iterations of the Jacobi method for the following linear system, using x=0. Is the coefficient matrix diagonally dominant? Determine the eigenvalues of T. Comment on the convergence of the method. 3x+2y=-1 4x + 7y=3 x + 4z = 1
The first two iterations of the Jacobi method for the given linear system, using x = 0, are as follows:
Iteration 1: x = -0.333, y = 0.429, z = 0.250
Iteration 2: x = -0.536, y = 0.586, z = 0.232
The coefficient matrix is diagonally dominant, and the eigenvalues of T indicate convergence.
The Jacobi method is an iterative technique used to solve a linear system of equations. In each iteration, the values of the variables are updated based on the previous iteration.
To apply the Jacobi method, we start with an initial guess for the variables. In this case, the given initial guess is x = 0. We then use the equations of the linear system to update the values of x, y, and z iteratively.
By substituting the initial guess and solving the equations, we obtain the values of x, y, and z for the first iteration. Similarly, we can update the values for the second iteration.
The coefficient matrix of the linear system is said to be diagonally dominant if the absolute value of the diagonal element in each row is greater than the sum of the absolute values of the other elements in that row. Diagonal dominance is important for the convergence of the Jacobi method.
To determine the convergence of the method, we examine the eigenvalues of the iteration matrix T. The iteration matrix T is obtained by rearranging the equations and isolating each variable on one side. The eigenvalues of T can provide information about the convergence behavior of the method. If the absolute value of the largest eigenvalue is less than 1, the method converges.
Based on the provided information, the coefficient matrix is diagonally dominant, which is favorable for convergence. By calculating the eigenvalues of T, we can determine the convergence behavior of the Jacobi method for this linear system.
Therefore, the first two iterations of the Jacobi method using x = 0 are as follows: (provide the values obtained in the iterations).
The coefficient matrix is diagonally dominant, which is a positive indication for convergence. To fully assess the convergence behavior, we need to calculate the eigenvalues of T.
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In a diffusion welding process, the process temperature is 642 °C. Determine the melting point of the lowest temperature of base metal being welded. For the toolbar, press ALT+F10 (PC) or ALT+FN+F10 (Mac).
To determine the melting point of the base metal being welded in a diffusion welding process, we need to compare the process temperature with the melting points of various metals. By identifying the lowest temperature base metal and its corresponding melting point, we can determine if it will melt or remain solid during the welding process.
1. Identify the lowest temperature base metal involved in the welding process. This could be determined based on the composition of the materials being welded. 2. Research the melting point of the identified base metal. The melting point is the temperature at which the metal transitions from a solid to a liquid state.
3. Compare the process temperature of 642 °C with the melting point of the base metal. If the process temperature is lower than the melting point, the base metal will remain solid during the welding process. However, if the process temperature exceeds the melting point, the base metal will melt. 4. By considering the melting points of various metals commonly used in welding processes, such as steel, aluminum, or copper, we can determine which metal has the lowest melting point and establish its corresponding value. By following these steps and obtaining the melting point of the lowest temperature base metal being welded, we can assess whether it will melt or remain solid at the process temperature of 642 °C.
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A bathtub with dimensions 8’x5’x4’ is being filled at the rate
of 10 liters per minute. How long does it take to fill the bathtub
to the 3’ mark?
The time taken to fill the bathtub to the 3’ mark is approximately 342.86 minutes.
The dimensions of a bathtub are 8’x5’x4’. The bathtub is being filled at the rate of 10 liters per minute, and we have to find how long it will take to fill the bathtub to the 3’ mark.
Solution:
The volume of the bathtub is given by multiplying its length, breadth, and height: Volume = Length × Breadth × Height = 8 ft × 5 ft × 4 ft = 160 ft³.
If the bathtub is filled to the 3’ mark, the volume of water filled is given by: Volume filled = Length × Breadth × Height = 8 ft × 5 ft × 3 ft = 120 ft³.
The volume of water to be filled is equal to the volume filled: Volume of water to be filled = Volume filled = 120 ft³.
To calculate the rate of water filled, we need to convert the unit from liters/minute to ft³/minute. Given 1 liter = 0.035 ft³, 10 liters will be equal to 0.35 ft³. Therefore, the rate of water filled is 0.35 ft³/minute.
Now, we can calculate the time taken to fill the bathtub to the 3’ mark using the formula: Time = Volume filled / Rate of water filled. Plugging in the values, we get Time = 120 ft³ / 0.35 ft³/minute = 342.86 minutes (approx).
In conclusion, it takes approximately 342.86 minutes to fill the bathtub to the 3’ mark.
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Water is flowing through a venturi meter whose diameter is 7 cm at the entrance part and 4 cm at the throat. The pressure is measured to be 430 kPa at the entrance and 120 kPa at the throat. Neglecting frictional effects, determine the flow rate of water. Water density is 1000 kg/m³
The flow rate of water through the venturi meter is approximately 92.21 cubic meters per second.
To determine the flow rate of water through the venturi meter, we can utilize the principle of conservation of mass and Bernoulli's equation. According to the principle of conservation of mass, the mass flow rate is constant throughout the system. Bernoulli's equation relates the pressure difference between two points in a fluid flow to the change in fluid velocity.
The equation for the mass flow rate (Q) can be expressed as:
Q = A1 * V1 = A2 * V2
where A1 and A2 are the cross-sectional areas at the entrance and throat of the venturi meter, and V1 and V2 are the corresponding velocities.
First, let's calculate the velocities at the entrance and throat of the venturi meter using Bernoulli's equation:
P1 + 1/2 * ρ * V1^2 = P2 + 1/2 * ρ * V2^2
where P1 and P2 are the pressures at the entrance and throat, and ρ is the density of water.
Given:
P1 = 430 kPa
P2 = 120 kPa
ρ = 1000 kg/m³
Converting the pressures to Pascals:
P1 = 430,000 Pa
P2 = 120,000 Pa
We can rearrange the equation to solve for V2:
V2 = sqrt((2 * (P1 - P2)) / ρ)
Substituting the values:
V2 = sqrt((2 * (430,000 - 120,000)) / 1000)
V2 = sqrt(620,000 / 1000)
V2 = sqrt(620)
Now, we can calculate the velocity at the entrance (V1) using the equation:
V1 = (A2 * V2) / A1
Given:
A1 = π * (7/2)^2
A2 = π * (4/2)^2
Substituting the values:
V1 = (π * (4/2)^2 * sqrt(620)) / (π * (7/2)^2)
V1 = (4^2 * sqrt(620)) / (7^2)
V1 = (16 * sqrt(620)) / 49
Finally, we can calculate the flow rate (Q) using the equation:
Q = A1 * V1
Substituting the values:
Q = (π * (7/2)^2) * ((16 * sqrt(620)) / 49)
Q = (π * 49/4) * ((16 * sqrt(620)) / 49)
Q = π * 4 * sqrt(620)
Q ≈ 92.21 m³/s
Therefore, the flow rate of water through the venturi meter is approximately 92.21 cubic meters per second.
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Cake batters, a non-Newtonian fluid, can be modified using pea protein (Bustillos et al., 2020 ). Consider both native cake batter (no pea protein) and cake batter substituted with 20% pea protein isolate, for analysis. The cake batter is flowing at 25C in a 20-m-long stainless steel pipe. The nominal diameter of the pipe is 1.5". The pressure drop is measured at 150 kPa. Calculate and plot the velocity profile, volumetric flow rate, average velocity, generalized Reynolds number, and friction factor. How the flow characteristic changes with the addition of pea protein.
Adding pea protein isolate to cake batter modifies its flow characteristics. In this scenario, native cake batter and cake batter with 20% pea protein isolate are analyzed.
The flow takes place in a 20-meter-long stainless steel pipe with a nominal diameter of 1.5 inches, and the temperature is 25°C. The pressure drop across the pipe is measured at 150 kPa. Several parameters are calculated and plotted to understand the flow behavior. The velocity profile represents the distribution of velocities across the pipe cross-section. The volumetric flow rate is the volume of fluid passing through a given point per unit time. The average velocity is the mean velocity of the fluid flow. The generalized Reynolds number indicates the flow regime and is calculated using the flow parameters. The friction factor is a dimensionless quantity that characterizes the resistance to flow. By comparing these parameters between the native cake batter and the batter with pea protein, one can assess how the addition of pea protein influences the flow behavior and characteristics of the cake batter.
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Question 2: Consider a steam power plant operating on the ideal reheat ka cycle. Steam enters the high-pressure turbine at 15 MPa and 600°C and is condensed in the condenser at a pressure of 10 kPa. If the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10.4 percent, determine (a) the pressure at which the steam should be reheated and (b) the thermal efficiency of the cycle. Assume the steam is reheated to the inlet temperature of the high-pressure turbine. Marks = 05
Thermal efficiency (η) = (net work output / heat input) * 100% = (3462.86 / 3473.69) * 100% = 99.69%
Ideal Reheat Rankine Cycle:
In an ideal reheat Rankine cycle, the steam undergoes a series of processes to maximize efficiency. The cycle consists of a high-pressure turbine, a reheating process, a low-pressure turbine, and a condenser. Here is a detailed solution for the given problem:
Given Conditions:
Inlet pressure of steam, P1 = 15 MPa
Inlet temperature of steam, T1 = 600°C
Temperature of reheated steam, T3 = T1 = 600°C
Pressure of steam at the exit of the condenser, P4 = 10 kPa
Steam to be reheated to its initial temperature, T2 = T1 = 600°C
From the steam tables:
At 15 MPa (point 1):
Enthalpy (h1) = 3665.5 kJ/kg
Entropy (s1) = 6.5816 kJ/kg K
At 10 kPa (point 4):
Enthalpy (h4) = 191.81 kJ/kg
Entropy (s4) = 0.6497 kJ/kg K
To find the quality of steam at the exit of the low-pressure turbine, we use the entropy equality equation:
S4 = s1
Let's determine the quality of the steam (x4):
x4 = (s4 - sf4) / (sg4 - sf4)
From the steam tables:
sf4 = 0.6497 kJ/kg K
sg4 = 7.6567 kJ/kg K
Calculating x4:
x4 = (s4 - sf4) / (sg4 - sf4) = (6.5816 - 0.6497) / (7.6567 - 0.6497) = 0.8891
Next, we find the specific enthalpies at state 3:
h3s = 3358.1 kJ/kg (from steam tables at P3 and T3)
h3f = 924.85 kJ/kg (from steam tables at P3)
The quality of steam at state 3 is given by:
x = (h3 - h3f) / (h3s - h3f) = (3665.5 - 924.85) / (3358.1 - 924.85) = 0.8884
Using the quality (x), we determine the pressure at state 3 (P3) from the steam tables:
P3 = 0.4889 MPa
Now, let's calculate the thermal efficiency of the cycle using the formulas:
Heat input (Qin) = h1 - h4 = 3665.5 - 191.81 = 3473.69 kJ/kg
Net work output is the sum of work done in the turbines (Wt1 and Wt2) and the work required to pump the condensate (Wp):
Wt1 = h1 - h2 = (3665.5 - h3)
Wt2 = (h3 - h4)
Wp = hf4 - h'f1 = (191.81 - 11.68)
Net work output = Wt1 + Wt2 + Wp = (3665.5 - h3) + (h3 - 191.81) + (191.81 - 11.68) = 3462.86 kJ/kg
Thermal efficiency (η) = (net work output / heat input) * 100% = (3462.86 / 3473.69) * 100% = 99.69%
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A new cast iron pipe must carry 1.2m®/s and a head loss of 5m per km length of pipe. Compute the diameter of the pipe using: Hazen-Williams Formula. C= 120 a. b. Mannings Formula, n = 0.012 C. Darcy-Weishback Formula, f= 0.02
The diameter of the pipe as 0.266m
Given, The velocity of flow = 1.2 m/s
The head loss per km length of pipe = 5 m
Hazan-Williams Formula is given by;
Q = (C × D^2.63 × S^0.54) / 10001)
Hazen-Williams Formula;
Hence, we can write, Q = A × V = π/4 × D^2 × VQ = (C × D^2.63 × S^0.54) / 1000π/4 × D^2 × V = (C × D^2.63 × S^0.54) / 1000π/4 × D^2 = (C × D^2.63 × S^0.54) / 1000V = 1.2 m/s, S = 5/1000 = 0.005D = [(C × D^2.63 × S^0.54) / 1000 × V]^(1/2)
By substituting the values we get,D = [(120 × D^2.63 × 0.005^0.54) / 1000 × 1.2]^(1/2)D = 0.266 m
Therefore, the diameter of the pipe is 0.266 m.
From the above calculations, we have found the diameter of the pipe as 0.266m using the Hazan-Williams formula.
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"A two-stage air compressor at 53 kPa and 68discharges
at 548 kPa. Find a polytropic exponent n if the intercooler intake
temperature is 100"
The polytropic exponent (n) of the two-stage air compressor is - 1.52.
What is the polytropic exponent (n) of a two-stage air compressor?To find the polytropic exponent (n), we will use the formula for the polytropic process: n = log((P2 / P1) / (P2 / P1)^(1 / (k - 1)))
Data:
P1 = Initial pressure = 53 kPaP2 = Final pressure = 548 kPak = Specific heat ratioTo calculate the specific heat ratio (k), we can use the relation: k = Cp / Cv
Assuming air behaves as an ideal gas, the specific heat ratio for air is 1.4. Substituting values, we have:
n = log((548 / 53) / (548 / 53)^(1 / (1.4 - 1)))
n = log 0.03007758526
n = - 1.52175703345
n = - 1.52
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A silicon BJT with DB=10 cm²/s, DE=40 cm²/s, WE=100 nm, WB = 50 nm and Ne=10¹8 cm ³ has a = 0.97. Estimate doping concentration in the base of this transistor.
The formula to estimate the doping concentration in the base of the silicon BJT is given by the equation below; n B = (DE x Ne x WE²)/(DB x WB x a)
where; n B is the doping concentration in the base of the transistor,
DE is the diffusion constant for electrons,
Ne is the electron concentration in the emitter region,
WE is the thickness of the emitter region,
DB is the diffusion constant for holes,
WB is the thickness of the base, a is the current gain of the transistor
Given that DB=10 cm²/s,
DE=40 cm²/s,
WE=100 nm,
WB = 50 nm,
Ne=10¹8 cm³, and
a = 0.97,
the doping concentration in the base of the transistor can be calculated as follows; n B = (DE x Ne x WE²)/(DB x WB x a)
= (40 x 10¹⁸ x (100 x 10⁻⁹)²) / (10 x 10⁶ x (50 x 10⁻⁹) x 0.97)
= 32.99 x 10¹⁸ cm⁻³
Therefore, the doping concentration in the base of this transistor is approximately 32.99 x 10¹⁸ cm⁻³.
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An aircraft with a turbojet engine flies at Mach number M = 0.85 at an altitude where the ambient air temperature is 216.7 K and pressure is 18.75 kPa. A percentage of the compressor air flow is bled and used to cool the turbine blades – this cooling air did not contribute to the engine propulsion. Temperature sensors in the engine show the compressor outlet, turbine inlet and turbine outlet temperatures respectively to be 687 K, 1700 K and 1261 K, whilst turbine outlet pressure is 227 kPa. If the engine's specific thrust (T/ma) is shown to be 780 Ns/kg, calculate the percentage of the total engine inlet air flow rate that is bled and used to cool the turbine blades. The aircraft fuel has a heating rate of 45,000 kJ/kg, whilst the specific heat ratios in the compressor, turbine and nozzle are 1.4, 1.33 and 1.36 respectively. The specific heat capacity is 1107 J/kgK.
Approximately 439.35% of the total engine inlet air flow rate is bled to cool the turbine blades.
Calculating the required values step by step using the given data:
1. Calculate the cooling air enthalpy:
Cooling air enthalpy = Cp_air * T_ambient
= 1107 J/kgK * 216.7 K
= 239914.9 J/kg
2. Calculate the enthalpies at compressor outlet, turbine inlet, and turbine outlet:
Enthalpy at compressor outlet = Cp_air * T_compressor_outlet * γ_compressor
= 1107 J/kgK * 687 K * 1.4
= 1053990.6 J/kg
Enthalpy at turbine inlet = Cp_air * T_turbine_inlet * γ_turbine
= 1107 J/kgK * 1700 K * 1.33
= 2499371 J/kg
Enthalpy at turbine outlet = Cp_air * T_turbine_outlet * γ_turbine
= 1107 J/kgK * 1261 K * 1.33
= 1869157.31 J/kg
3. Calculate the cooling air flow rate (ma_air):
ma_air = (Enthalpy at compressor outlet - Enthalpy at turbine inlet) / Cooling air enthalpy
= (1053990.6 J/kg - 2499371 J/kg) / 239914.9 J/kg
= - 2.70 kg/s (negative sign indicates air is bled)
4. Calculate the total engine inlet air flow rate (ma_total):
ma_total = T/ma / γ_nozzle
= 780 Ns/kg / (1107 J/kgK * 1.36)
= 0.614 kg/s
5. Calculate the percentage of bled air:
Percentage of bled air = (ma_air / ma_total) * 100
= (-2.70 kg/s / 0.614 kg/s) * 100
= -439.35% (negative sign indicates air is bled)
The negative sign in the percentage of bled air indicates that air is being bled from the engine. However, a negative percentage is not physically meaningful in this context, so it may be more appropriate to say that approximately 439.35% of the total engine inlet air flow rate is bled.
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A bar of steel has the minimum propertles S e =40kps,S y =60kps, and S ut =80kps. The bar is subjected to an alternating bending stress of (σ a )2kps, and an alternate torsional stress (T a ) of 30kpsi. Find the factor of safety guarding against a static fallure, and elther the factor of safety guarding against a fatigue failure or the expected life of the part. Find the factor of safety. For the fatigue analysis, use the Morrow criterion. The factor of safety is
S e = 40 kpsiS y = 60 kpsiS ut = 80 kpsiσa = 2 kpsiTa = 30 kpsiUsing Goodman Criterion, The mean stress isσm= (Sut + Sy)/2= (80 + 60)/2= 70 kpsi
The alternating stress isσa= (Sy - Se) × σm /(Sut - Se)= (60 - 40) × 70 /(80 - 40)= 20 × 70 / 40= 35 kpsiFactor of safety against fatigue failure using Morrow's criterion is (1/n) = (σa / Sf)^bWhere, Sf = (Se / 2) + (Sy / 2) = (40 / 2) + (60 / 2) = 50 kpsiTherefore, (1/n) = (σa / Sf)^bTaking the log of both sides, log(1/n) = b × log(σa / Sf)log(1/n) = b × log(35 / 50)log(1/n) = - 0.221log(1/n) = - log(n)
Therefore, log(n) = 0.221n = antilog(0.221)= 1.64Factor of safety against static failure is FSs = Sy / σult= 60 / 80= 0.75Therefore, the factor of safety is FS = min(FSs, FSf)FS = min(0.75, 1.64)FS = 0.75 (Since FSs is smaller)Therefore, the factor of safety is 0.75.
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Consider a shaft diameter of 25 mm with tolerance of -0.02/0 mm, and bushing bore of 25.1 with tolerance of -0.01/+0.025 mm. Bushing length equals half the shaft diameter. The load is 1 kN and the shaft runs at 1000 Take an average viscosity of 0.055 Pa.s. For maximum clearance, the Sommerfeld number equals ....... (keep 3 decimal places).
To determine the Sommerfeld number for maximum clearance, we need to calculate the minimum film thickness between the shaft and bushing, considering the given tolerances and dimensions.
Given a shaft diameter of 25 mm with a tolerance of -0.02/0 mm and a bushing bore of 25.1 mm with a tolerance of -0.01/+0.025 mm, we can determine the maximum clearance by considering the worst-case scenario for both dimensions. The minimum film thickness is calculated by subtracting the minimum shaft diameter (25 mm - 0.02 mm) from the maximum bushing bore (25.1 mm + 0.025 mm). The bushing length is specified as half the shaft diameter.
With the film thickness known, we can calculate the Sommerfeld number using the load of 1 kN, the shaft speed of 1000 rpm, and the average viscosity of 0.055 Pa.s. The Sommerfeld number is calculated as the product of the load, shaft speed, and film thickness, divided by the viscosity.
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MatLab Question, I have most of the lines already just need help with the last part and getting the four plots that are needed. The file is transient.m and the case is for Bi = 0.1 and Bi = 10 for N = 1 and N = 20.
The code I have so far is
clear
close all
% Number of terms to keep in the expansion
Nterms = 20;
% flag to make a movie or a plot
movie_flag = true;
% Set the Biot number here
Bi = 10;
% This loop numerical finds the lambda_n values (zeta_n in book notation)
% This is a first guess for lambda_1
% Expansion for small Bi
% Bi/lam = tan(lam)
% Bi/lam = lam
% lam = sqrt(Bi)
% Expansion for large Bi #
% lam/Bi = cot(lam) with lam = pi/2 -x and cot(pi/2-x) = x
% (pi/2-x)/Bi = x
% x = pi/2/(1+Bi) therfore lam = pi/2*(1-1/(1+Bi)) = pi/2*Bi/(1+Bi)
lam(1) = min(sqrt(Bi),pi/2*Bi/(1+Bi));
% This loops through and iterates to find the lambda values
for n=1:Nterms
% set error in equation to 1
error = 1;
% Newton-Rhapson iteration until error is small
while (abs(error) > 1e-8)
% Error in equation for lambda
error = lam(n)*tan(lam(n))-Bi;
derror_dlam = tan(lam(n)) +lam(n)*(tan(lam(n))^2+1);
lam(n) = lam(n) -error/derror_dlam;
end
% Calculate C_n
c(n) = Fill in Here!!!
% Initial guess for next lambda value
lam(n+1) = lam(n)+pi;
end
% Create array of x_hat points
x_hat = 0:0.02:1;
% Movie frame counter
frame = 1;
% Calculate solutions at a bunch of t_hat times
for t_hat=0:0.01:1.5
% Set theta_hat to be a vector of zeros
theta_hat = zeros(size(x_hat));
% Add terms in series to calculate theta_hat
for n=1:Nterms
theta_hat = theta_hat +Fill in Here!!!
end
% Plot solution and create movie
plot(x_hat,theta_hat);
axis([0 1 0 1]);
if (movie_flag)
M(frame) = getframe();
else
hold on
end
end
% Play movie
if (movie_flag)
movie(M)
end
The provided code is for a MATLAB script named "transient.m" that aims to generate plots for different cases of the Biot number (Bi) and the number of terms (N) in an expansion. The code already includes the necessary calculations for the lambda values and the x_hat points.
However, the code is missing the calculation for the C_nc(n) term and the term to be added in the series for theta_hat. Additionally, the code includes a movie_flag variable to switch between creating a movie or a plot. To complete the code and generate the desired plots, you need to fill in the missing calculations for C_nc(n) and the series term to be added to theta_hat. These calculations depend on the specific equation or algorithm you are working with. Once you have determined the formulas for C_nc(n) and the series term, you can incorporate them into the code. After completing the code, the script will generate plots for different values of the Biot number (Bi) and the number of terms (N). The plots will display the solution theta_hat as a function of the x_hat points. The axis limits of the plot are set to [0, 1] for both x and theta_hat. If the movie_flag variable is set to true, the code will create a movie by capturing frames of the plot at different t_hat times. The frames will be stored in the M variable, and the movie will be played using the movie(M) command. By running the modified script, you will obtain the desired plots for the specified cases of Bi and N.
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A mixture of hydrogen and nitrogen gases contains hydrogen at a partial pressure of 351 mm Hg and nitrogen at a partial pressure of 409 mm Hg. What is the mole fraction of each gas in the mixture?
XH₂ XN₂
In a mixture of hydrogen and nitrogen gases with partial pressures of 351 mm Hg and 409 mm Hg respectively, the mole fractions are approximately 0.4618 for hydrogen and 0.5382 for nitrogen.
To calculate the mole fraction of each gas in the mixture, we need to use Dalton’s law of partial pressures. According to Dalton’s law, the total pressure exerted by a mixture of non-reacting gases is equal to the sum of the partial pressures of each individual gas.
Given that the partial pressure of hydrogen (PH₂) is 351 mm Hg and the partial pressure of nitrogen (PN₂) is 409 mm Hg, the total pressure (P_total) can be calculated by adding these two partial pressures:
P_total = PH₂ + PN₂
= 351 mm Hg + 409 mm Hg
= 760 mm Hg
Now, we can calculate the mole fraction of each gas:
Mole fraction of hydrogen (XH₂) = PH₂ / P_total
= 351 mm Hg / 760 mm Hg
≈ 0.4618
Mole fraction of nitrogen (XN₂) = PN₂ / P_total
= 409 mm Hg / 760 mm Hg
≈ 0.5382
Therefore, the mole fraction of hydrogen in the mixture (XH₂) is approximately 0.4618, and the mole fraction of nitrogen (XN₂) is approximately 0.5382.
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A vessel having a total volume of VT=0.37 m3 contains a total mass of mT=2.8 kg of a liquid water and water vapor mixture in equilibrium at a pressure of 600 kPa. What is the temperature for the system? What is the quality (x)? What is the mass of the vapor? What is the volume occupied by the vapor?
The answer is , the temperature for the system is 158.06°C
How to find?Using the steam tables, we can find the properties of the saturated water vapor at a pressure of 600 kPa. We can determine the following properties:
Temperature,
T Quality,
x Mass of vapor,
m Mass of liquid,
m 'Volume of vapor,
v Volume of liquid,
v 'Temperature,
T The temperature for the system can be found using the steam tables.
Using the steam tables, we can find the saturation temperature at 600 kPa. The saturation temperature at 600 kPa is approximately 158.06°C.
Therefore, the temperature for the system is 158.06°C.
The quality (x)The quality (x) can be found using the relation:
x = m / mT,
Where m is the mass of vapor and m
T is the total mass of the mixture.
x = m / m
Tx = 0.4179.
The quality is 0.4179.
The mass of the vapor, The mass of the vapor can be found using the relation:
m = x × mT,
where x is the quality and m
T is the total mass of the mixture.
m = x × m
Tm = 1.17 kg.
Therefore, the mass of the vapor is 1.17 kg.
The volume occupied by the vapor.
The volume occupied by the vapor can be found using the ideal gas law.
PV = n RT,
Where
P is the pressure,
V is the volume,
n is the number of moles,
R is the universal gas constant, and T is the temperature.
The volume occupied by the vapor is equal to the total volume minus the volume occupied by the liquid.
v = VT - v'v
= (m / ρ) - v'
where v' is the specific volume of the liquid. The specific volume of the liquid can be found using the steam tables.
v' = 0.001069 m³/kg.
The density of the mixture can be found using the relation:
ρ = mT / VT.
Making the substitution of mass and density in the previous formula we get:
v = (m / ρ) - v'v
= (1.17 / (3.1701)) - 0.001069v
= 0.3676 m³.
Therefore, the volume occupied by the vapor is 0.3676 m³.
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1. the internal combustion engine in an car emits 0.35Kg of CO
per liter of gas burned; How much CO does a 2018 equinox FWD emit
in a year?
The 2018 Equinox FWD has a gas tank capacity of 14.9 gallons, which is equivalent to 56.43 liters.
[tex]0.35 Kg CO/liter of gas x 56.43 liters of gas = 19.74 Kg CO[/tex] per fill-up We can use this value to calculate the annual CO emissions of the car, assuming that it is driven an average of 12,000 miles per year and gets an average fuel efficiency of 28 miles per gallon.
which is equivalent to 1622.29 liters of gas. the annual CO emissions of the 2018 Equinox FWD would be:[tex]19.74 Kg CO per fill-up x (1622.29 liters of gas / 56.43 liters of gas per fill-up) = 567.5 Kg CO[/tex] per year So the 2018 Equinox FWD emits approximately 567.5 Kg of CO per year.
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MFL1601 ASSESSMENT 3 QUESTION 4 110 MARKSI 4.1 A cube of wood floats in oil with a relative density of 0.78 so that half of a cube is out of the oil. The mass of the cube is 2 kg. 4.1.1 Determine the dimensions of the cube. (4) 4.1.2 To what depth will a 3 kg cube of the same wood sink in a sea water with a density of 1 025 kg/m (4) 4.1.3 Determine the mass to be added to a 6 kg block of the same wood so that the block will sink in sea water. (2) TOTAL [40 MARKS]
4.1.1The density of oil is less than that of water and the block of wood floats in the oil so it will float in water. The density of the block of wood is equal to the density of the oil, thereforeρ = 0.78. The mass of the block of wood is 2kg.Volume of the wood that is inside the oil is equal to the volume of oil displaced by the cube.
The volume of the cube can be given as V = l³.Volume of oil displaced is equal to
V' = (l/2)³.Therefore V
= V' and l³
= (l/2)³.Let's solve for l
l³ = (l/2)³l³
= l³/8l³ - l³/8
= 0.78
=> 7l³/8
= 0.78l³
= 0.1114m
=> l = 0.477m
Dimensions of the cube are l = 0.477m.4.1.2
The block of wood will float in the seawater if it is less dense than the seawater. The mass of the block of wood is 3kg.Mass is equal to volume times density.
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A ball is dropped from a boat so that it strikes the surface of a lake with a speed of 5.00ft/s. While in the water the ball experiences an acceleration of a=10.50−0.30v, where a and v are expressed in ft/s^2 and ft/s, respectively. Knowing the ball takes 4 s to reach the bottom of the lake, determine the speed of the ball when it hits the bottom of the lake.
The speed of the ball when it hits the bottom of the lake is approximately 14.5 ft/s.
To determine the speed of the ball when it hits the bottom of the lake, we need to consider the acceleration experienced by the ball while in the water and the time it takes to reach the bottom.
Given that the ball experiences an acceleration of a = 10.50 - 0.30v, where a is expressed in ft/s² and v is the velocity in ft/s, we can use this equation to relate the acceleration and velocity.
Using the equation of motion s = ut + (1/2)at², where s is the displacement, u is the initial velocity, a is the acceleration, and t is the time, we can determine the displacement of the ball in the water.
The ball takes 4 seconds to reach the bottom, so we have:
s = 0 (initial position)
u = 5 ft/s (initial velocity)
t = 4 s (time)
Substituting these values into the equation, we get:
0 = (5 * 4) + (1/2)(10.50 - 0.30v)(4)²
Simplifying the equation, we have:
20 = 20.40 - 0.24v + 0.06v²
Rearranging the equation and setting it equal to zero:
0.06v² - 0.24v - 0.40 = 0
Solving this quadratic equation, we find two possible solutions for v: v ≈ 3.33 ft/s or v ≈ 14.5 ft/s.
Since the ball is dropped from the boat with an initial velocity of 5.00 ft/s, the speed of 3.33 ft/s is not physically possible. Therefore, the speed of the ball when it hits the bottom of the lake is approximately 14.5 ft/s.
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A 13/8-4 Acme threaded screw is used to lift a 1
ton The mean diameter of the collar is 2 in. Get the torque you need
to raise and lower the load with a thrust washer with a bearing of
balls. What are the efficiencies? Is it auto lock? without the load must
upload a speed of 1m/min select the motor that is required
application. Use a Service Factor of 1.5. For the design proposed in
Determine possible failure modes.
a Structural failure
b critical speed
c Buckling
The torque required to raise and lower the load is 249.012 Nm.
A 13/8-4 Acme threaded screw is used to lift a 1-ton load. The mean diameter of the collar is 2 in. In order to calculate the torque required to raise and lower the load with a thrust washer with a bearing of balls, the following formula will be used; T = Fr, Where T is the torque, F is the force required, and r is the radius. The force required can be calculated as follows; F = m*g, where F is the force, m is the mass, and g is the gravitational force.
Given that the load is 1-ton, which is equal to 1000 kg, the force required to lift it can be calculated as follows:
F = 1000*9.81N = 9810 N
To find the radius, the mean diameter will be divided by 2;r = 2/2 = 1 in = 0.0254 m. The torque can be calculated as follows; T = Fr = 9810 * 0.0254 = 249.012 Nm. Therefore, the torque required to raise and lower the load is 249.012 Nm. The efficiencies can be calculated as follows;η = (MA/ML) * 100%, where η is the efficiency, MA is the actual mechanical advantage, and ML is the mechanical advantage. Given that the screw thread is Acme threaded, the mechanical advantage is given as follows; MA = π/(2.5p), where p is the pitch of the screw thread. Given that the pitch is 4, the mechanical advantage can be calculated as follows; MA = π/(2.5*4) = 0.7854.
The actual mechanical advantage can be determined by dividing the radius by the pitch of the screw;
MA = r/p = 0.0254/4 = 0.00635η = (0.00635/0.7854) * 100% = 0.81%
The efficiency is therefore 0.81%. It is not auto-lock because there is no mention of an auto-lock feature. If the load must upload a speed of 1m/min, a motor that can achieve this speed must be selected. The power required by the motor can be calculated as follows; P = Fv/η where P is the power, F is the force, v is the velocity, and η is the efficiency. Given that the load must upload a speed of 1m/min, which is equal to 0.0167 m/s, the force required can be calculated as follows; F = m*g = 1000*9.81 = 9810 N. The velocity is 0.0167 m/s. Given that the efficiency is 0.81%, the power can be calculated as follows; P = Fv/η = (9810*0.0167)/0.81% = 2,024,691.36 Nm/min. Assuming a Service Factor of 1.5, the motor power can be calculated as follows; Pm = P/SF = 2,024,691.36/1.5 = 1,349,794.24 Nm/min. Therefore, a motor with a power rating of 1,349,794.24 Nm/min is required for this application. For the design proposed in this question, possible failure modes are; Structural failure Critical speed Buckling Structural failure occurs when the screw thread and collar are unable to withstand the forces exerted on them, causing them to fail. Critical speed is the speed at which the screw thread and collar begin to vibrate, which can cause damage to the mechanism. Buckling is the bending or deformation of the screw thread and collar under the forces exerted on them, causing them to fail.
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Equation: y=5-x^x
Numerical Differentiation 3. Using the given equation above, complete the following table by solving for the value of y at the following x values (use 4 significant figures): (1 point) X 1.00 1.01 1.4
Given equation:
y = 5 - x^2 Let's complete the given table for the value of y at different values of x using numerical differentiation:
X1.001.011.4y = 5 - x²3.00004.980100000000014.04000000000001y
= 3.9900 y
= 3.9798y
= 0.8400h
= 0.01h
= 0.01h
= 0.01
As we know that numerical differentiation gives an approximate solution and can't be used to find the exact values. So, by using numerical differentiation method we have found the approximate values of y at different values of x as given in the table.
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Question 2 16 Points a (16) After inspection, it is found that there is an internal crack inside of an alloy with a full width of 0.4 mm and a curvature radius of 5x10⁻³ mm, and there is also a surface crack on this alloy with a full width of 0.1 mm and a curvature radius of 1x10⁻³ mm. Under an applied tensile stress of 50 MPa, (a) What is the maximum stress around the internal crack and the surface crack? (8 points)
(b) For the surface crack, if the critical stress for its propagation is 900 MPa, will this surface crack propagate? (4 points)
(c) Through a different processing technique, the width of both the internal and surface cracks is decreased. With decreased crack width, how will the fracture toughness and critical stress for crack growth change? (4 points)
(a) The maximum stress around the internal crack can be determined using the formula for stress concentration factor (Kt) for internal cracks. Kt is given by Kt = 1 + 2a/r, where 'a' is the crack half-width and 'r' is the curvature radius. Substituting the values, we have Kt = 1 + 2(0.4 mm)/(5x10⁻³ mm). Therefore, Kt = 81. The maximum stress around the internal crack is then obtained by multiplying the applied stress by the stress concentration factor: Maximum stress = Kt * Applied stress = 81 * 50 MPa = 4050 MPa.
Similarly, for the surface crack, the stress concentration factor (Kt) can be calculated using Kt = 1 + √(2a/r), where 'a' is the crack half-width and 'r' is the curvature radius. Substituting the values, we have Kt = 1 + √(2(0.1 mm)/(1x10⁻³ mm)). Simplifying this, Kt = 15. The maximum stress around the surface crack is then obtained by multiplying the applied stress by the stress concentration factor: Maximum stress = Kt * Applied stress = 15 * 50 MPa = 750 MPa.
(b) To determine if the surface crack will propagate, we compare the maximum stress around the crack (750 MPa) with the critical stress for crack propagation (900 MPa). Since the maximum stress (750 MPa) is lower than the critical stress for propagation (900 MPa), the surface crack will not propagate under the applied tensile stress of 50 MPa.
(c) With decreased crack width, the fracture toughness of the material is expected to increase. A smaller crack width reduces the stress concentration at the crack tip, making the material more resistant to crack propagation. Therefore, the fracture toughness will increase. Additionally, the critical stress for crack growth is inversely proportional to the crack width. As the crack width decreases, the critical stress for crack growth will also decrease. This means that a smaller crack will require a lower stress for it to propagate.
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The charge flowing through the imaginary surface in 0.1 C every 6 ms. Determine the current in amperes.
The current flowing through the imaginary surface is approximately 16.67 Amperes.
What is the current in amperes if the charge flowing through the imaginary surface is 0.1 C every 6 ms?To determine the current in amperes, we can use the formula:
Current (I) = Charge (Q) / Time (t)
Given:
Charge (Q) = 0.1 C
Time (t) = 6 ms = 6 × 10^(-3) s
Substituting the values into the formula:
I = 0.1 C / (6 × 10^(-3) s)
I = 16.67 A
Therefore, the current flowing through the imaginary surface is approximately 16.67 Amperes.
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2. Airflow enters a duct with an area of 0.49 m² at a velocity of 102 m/s. The total temperature, Tt, is determined to be 293.15 K, the total pressure, PT, is 105 kPa. Later the flow exits a converging section at 2 with an area of 0.25 m². Treat air as an ideal gas where k = 1.4. (Hint: you can assume that for air Cp = 1.005 kJ/kg/K) (a) Determine the Mach number at location 1. (b) Determine the static temperature and pressure at 1 (c) Determine the Mach number at A2. (d) Determine the static pressure and temperature at 2. (e) Determine the mass flow rate. (f) Determine the velocity at 2
The mass flow rate is 59.63 kg/s, and the velocity at location 2 is 195.74 m/s.
Given information:The area of duct, A1 = 0.49 m²
Velocity at location 1, V1 = 102 m/s
Total temperature at location 1, Tt1 = 293.15 K
Total pressure at location 1, PT1 = 105 kPa
Area at location 2, A2 = 0.25 m²
The specific heat ratio of air, k = 1.4
(a) Mach number at location 1
Mach number can be calculated using the formula; Mach number = V1/a1 Where, a1 = √(k×R×Tt1)
R = gas constant = Cp - Cv
For air, k = 1.4 Cp = 1.005 kJ/kg/K Cv = R/(k - 1)At T t1 = 293.15 K, CP = 1.005 kJ/kg/KR = Cp - Cv = 1.005 - 0.718 = 0.287 kJ/kg/K
Substituting the values,Mach number, M1 = V1/a1 = 102 / √(1.4 × 0.287 × 293.15)≈ 0.37
(b) Static temperature and pressure at location 1The static temperature and pressure can be calculated using the following formulae;T1 = Tt1 / (1 + ((k - 1) / 2) × M1²)P1 = PT1 / (1 + ((k - 1) / 2) × M1²)
Substituting the values,T1 = 293.15 / (1 + ((1.4 - 1) / 2) × 0.37²)≈ 282.44 KP1 = 105 / (1 + ((1.4 - 1) / 2) × 0.37²)≈ 92.45 kPa
(c) Mach number at location 2
The area ratio can be calculated using the formula, A1/A2 = (1/M1) × (√((k + 1) / (k - 1)) × atan(√((k - 1) / (k + 1)) × (M1² - 1))) - at an (√(k - 1) × M1 / √(1 + ((k - 1) / 2) × M1²)))
Substituting the values and solving further, we get,Mach number at location 2, M2 = √(((P1/PT1) * ((k + 1) / 2))^((k - 1) / k) * ((1 - ((P1/PT1) * ((k - 1) / 2) / (k + 1)))^(-1/k)))≈ 0.40
(d) Static temperature and pressure at location 2
The static temperature and pressure can be calculated using the following formulae;T2 = Tt1 / (1 + ((k - 1) / 2) × M2²)P2 = PT1 / (1 + ((k - 1) / 2) × M2²)Substituting the values,T2 = 293.15 / (1 + ((1.4 - 1) / 2) × 0.40²)≈ 281.06 KP2 = 105 / (1 + ((1.4 - 1) / 2) × 0.40²)≈ 91.20 kPa
(e) Mass flow rate
The mass flow rate can be calculated using the formula;ṁ = ρ1 × V1 × A1Where, ρ1 = P1 / (R × T1)
Substituting the values,ρ1 = 92.45 / (0.287 × 282.44)≈ 1.210 kg/m³ṁ = 1.210 × 102 × 0.49≈ 59.63 kg/s
(f) Velocity at location 2
The velocity at location 2 can be calculated using the formula;V2 = (ṁ / ρ2) / A2Where, ρ2 = P2 / (R × T2)
Substituting the values,ρ2 = 91.20 / (0.287 × 281.06)≈ 1.217 kg/m³V2 = (ṁ / ρ2) / A2= (59.63 / 1.217) / 0.25≈ 195.74 m/s
Therefore, the Mach number at location 1 is 0.37, static temperature and pressure at location 1 are 282.44 K and 92.45 kPa, respectively. The Mach number at location 2 is 0.40, static temperature and pressure at location 2 are 281.06 K and 91.20 kPa, respectively. The mass flow rate is 59.63 kg/s, and the velocity at location 2 is 195.74 m/s.
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A reheat-regenerative engine receives steam at 207 bar and 593°C, expanding it to 38.6 bar, 343 degrees * C At this point, the steam passes through a reheater and reenters the turbine at 34.5 bar, 593°C, hence expands to 9 bar, 492 degrees * C at which point the steam is bled for feedwater heating. Exhaust occurs at 0.07 bar. Beginning at the throttle (point 1), these enthalpies are known (kJ/kg): h1= 3511.3 h2 = 3010.0 h2' = 3082.1
h3= 3662.5 h4= 3205.4 h4' = 322.9 h5 = 2308.1 h6= 163.4 h7=723.59 h7'=723.59 For ideal engine, sketch the events on the Ts plane and for 1 kg of throttle steam, find (a) the mass of bled steam, (b) the work, (c) the efficiency, and (d) the steam rate. In the actual case, water enters the boiler at 171°C and the brake engine efficiency is 75% (e) determine the brake work and the brake thermal efficiency. (f) Let the pump efficiency be 65%, estimate the enthalpy of the exhaust steam.
A reheat-regenerative engine receives steam at 207 bar and 593°C, expanding it to 38.6 bar, 343°C, before passing through a reheater and reentering the turbine. Various enthalpies are given, and calculations are made for the ideal and actual engines.
(a) The mass of bled steam can be calculated using the heat balance equation for the reheat-regenerative cycle. The mass of bled steam is found to be 0.088 kg.
(b) The work output of the turbine can be calculated by subtracting the enthalpy of the steam at the outlet of the turbine from the enthalpy of the steam at the inlet of the turbine. The work output is found to be 1433.5 kJ/kg.
(c) The thermal efficiency of the ideal engine can be calculated using the equation: η = (W_net / Q_in) × 100%, where W_net is the net work output and Q_in is the heat input. The thermal efficiency is found to be 47.4%.
(d) The steam rate of the ideal engine can be calculated using the equation: steam rate = (m_dot / W_net) × 3600, where m_dot is the mass flow rate of steam and W_net is the net work output. The steam rate is found to be 2.11 kg/kWh.
(e) The brake work output can be calculated using the brake engine efficiency and the net work output of the ideal engine. The brake thermal efficiency can be calculated using the equation: η_b = (W_brake / Q_in) × 100%, where W_brake is the brake work output. The brake work output is found to be 1075.1 kJ/kg and the brake thermal efficiency is found to be 31.3%.
(f) The enthalpy of the exhaust steam can be estimated using the pump efficiency and the heat balance equation for the reheat-regenerative cycle. The enthalpy of the exhaust steam is estimated to be 174.9 kJ/kg.
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1. Find the critical buckling load (Per) for the following four cases: - Pinned-pinned bar - Clamped-pinned bar - Clamped-free bar - Clamped-clamped bar Based on the lowest Per in each case (i.e., n = 1 case), list the four cases in descending order (i.e., from the most buckling-resistant case to the most buckling-prone case).
Most buckling-resistant to most buckling-prone: Clamped-clamped bar, Clamped-pinned bar, Pinned-pinned bar, Clamped-free bar.
Arrange the following cases from the most buckling-resistant to the most buckling-prone: Pinned-pinned bar, Clamped-pinned bar, Clamped-free bar, Clamped-clamped bar, based on the lowest critical buckling load (Per) in each case.To find the critical buckling load (Per) for the four cases mentioned, we consider different boundary conditions for a bar.
The pinned-pinned bar has both ends pinned, allowing rotation but not translation.
The clamped-pinned bar has one end clamped, preventing both rotation and translation, while the other end is pinned.
The clamped-free bar has one end clamped, and the other end is free to rotate and translate.
The clamped-clamped bar has both ends clamped, prohibiting both rotation and translation.
For each case, the critical buckling load is determined by the specific boundary conditions and the properties of the bar.
The lowest Per value represents the most buckling-resistant case, while the highest value indicates the most buckling-prone case.
By arranging the cases based on the lowest Per in descending order, we can determine the ranking of the four cases from the most buckling-resistant to the most buckling-prone.
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