Mass flow rate is one of the primary properties of fluid flow, and it's represented by m. Mass flow rate measures the amount of mass that passes per unit time through a given cross-sectional area.
It can be calculated using the equation given below:Where m is mass flow rate, ρ is density, A is area, and V is velocity. Now we have all the parameters which are necessary to calculate the mass flow rate. We can use the above equation to calculate it. The solution of the mass flow rate is as follows:ρ₁A₁V₁ = ρ₂A₂V₂
Therefore, m = ρ₁A₁V₁ = ρ₂A₂V₂
We know that air is a perfect gas. For the perfect gas, the density of the fluid is given as,ρ = P / (RT)where P is the pressure of the gas, R is the specific gas constant, and T is the temperature of the gas. By using this, we can calculate the mass flow rate as:
It is given that an unknown amount of heat is being added to the air flowing through the pipe. By using conservation of energy, we can calculate the amount of heat being added. The heat added is given by the equation:Q = mcpΔT
where Q is the heat added, m is the mass flow rate, cp is the specific heat capacity at constant pressure, and ΔT is the temperature difference across the heater. By using the above equation, we can calculate the heat rate of the electric heater. Now, we can use the mass flow rate that we calculated earlier to find the exit velocity of air. We can use the equation given below to calculate the exit velocity:V₃ = m / (ρ₃A₃)
Therefore, the mass flow rate at exit is 2.86 kg/s, the heat rate of the electric heater is 286.68 kW, and the exit velocity of air is 24.91 m/s.
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As shown below in the figure, a bracket with a solid circular cross-section of radius r=68 mm is inserted into a frictionless sleeve (slightly larger than the bracket) at A and is also supported by a pin at c. The sleeve at A allows free rotation of the bracket with respect to Z axis and also freely translate about the same axis. The bracket has two arms, namely arm AB of length b=0.72 m and BC of length a=0.44 m. Moments M=1.5 RN.m and M 12=1.36 kN.m are applied at the point C. 1. Calculate the maximum tensile stress along with the location and direction. 2. Calculate the maximum compressive stress along woth the location and direction. 3. Calculate at point p(see cross-section and the figure) on the cross-section at A the maximum in- plane shear stress 4 Also identify the point where absolute maximum shear stress takes place and calculate the same with the direction
Thus, the angle of absolute maximum shear stress, θ = 63.44° (approx.)
Given:
Radius, r = 68 mm
Length, b = 0.72 m
Length, a = 0.44 m
Moment, M = 1.5 RN.m
Moment, M12 = 1.36 kN.m
To determine:
1) Maximum tensile stress, along with its location and direction.
2) Maximum compressive stress, along with its location and direction.
3) Maximum in-plane shear stress at point P.
4) Identify the point where the absolute maximum shear stress takes place and calculate the same with direction.
Calculations:
1) Maximum Tensile Stress: σ max
= Mc/I where, I=πr4/4
Substituting the given values in above formula,
σmax= (1.5*10^3 * 0.44)/ (π* (68*10^-3)^4/4)
σmax = 7.54 N/mm2
Location of Maximum Tensile Stress: The maximum tensile stress occurs at point B, which is at a distance of b/2 from point C in the direction opposite to the applied moment.
2) Maximum Compressive Stress:
σmax= Mc/I where, I=πr4/4
Substituting the given values in the above formula,
σmax= (-1.36*10^6 * 0.44)/ (π* (68*10^-3)^4/4)
σmax = -23.77 N/mm2
Location of Maximum Compressive Stress: The maximum compressive stress occurs at point B, which is at a distance of b/2 from point C in the direction of the applied moment.
3) Maximum In-Plane Shear Stress at point P:
τmax= 2T/A where, A=πr2T = [M(r+x)]/(πr2/2) - (M/πr2/2)x = r
Substituting the given values in above formula,
T = 1.5*68*10^-3/π = 0.326 NmA
= π(68*10^-3)^2
= 14.44*10^-6 m2
τmax = 2*0.326/14.44*10^-6
τmax = 45.04 N/mm24)
Absolute Maximum Shear Stress and Its Direction:
τmax = [T/(I/A)](x/r) + [(VQ)/(Ib)]
τmax = [(VQ)/(Ib)] where Q = πr3/4 and V = M12/a - T
Substituting the given values in the above formula,
Q = π(68*10^-3)^3/4
= 1.351*10^-6 m3V
= (1.36*10^3)/(0.44) - 0.326
= 2925.45 NQ
= 1.351*10^-6 m3I
= πr4/4 = 6.09*10^-10 m4b
= 0.72 mτmax
= [(2925.45*1.351*10^-6)/(6.09*10^-10*0.72)]
τmax = 7.271 N/mm2
Hence, the absolute maximum shear stress and its direction is 7.271 N/mm2 at 63.44° from the x-axis.
Thus, we have calculated the maximum tensile stress, along with its location and direction, maximum compressive stress, along with its location and direction, maximum in-plane shear stress at point P, and the absolute maximum shear stress and its direction.
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Butane at 1.75bar is kept in a piston-cylinder device. Initially, the butane required 50kJ of work to compress the gas until the volume dropped three times lesser than before while maintaining the temperature. Later, heat will be added until the temperature rises to 270°C during the isochoric process. Butane then will undergo a polytropic process with n=3.25 until 12 bar and 415°C. After that, the butane will expand with n=0 until 200 liters. Next, butane will undergo an isentropic process until the temperature drops twice as before. Later, butane undergoes isothermal compression to 400 liters. Finally, the butane will be cooled polytropically to the initial state. a) Sketch the P-V diagram b) Find mass c) Find all P's, V's and T's d) Calculate all Q's e) Determine the nett work of the cycle
In the given scenario, the thermodynamic processes of butane in a piston-cylinder device are described. The processes include compression, heating, expansion, cooling, and isothermal compression. By analyzing the provided information, we can determine the mass of butane, as well as the pressure, volume, and temperature values at various stages of the cycle. Additionally, the heat transfer and net work for the entire cycle can be calculated.
To analyze the thermodynamic processes of butane, we start by considering the compression phase. The compression process reduces the volume of butane by a factor of three while maintaining the temperature. The work done during compression is given as 50 kJ. Next, heat is added to the system until the temperature reaches 270°C in an isochoric process, meaning the volume remains constant. After that, butane undergoes a polytropic process with n = 3.25 until reaching a pressure of 12 bar and a temperature of 415°C.
Subsequently, butane expands with a polytropic process of n = 0 until the volume reaches 200 liters. Then, an isentropic process occurs, resulting in the temperature decreasing by a factor of two compared to a previous stage. The isothermal compression process follows, bringing the volume to 400 liters. Finally, butane is cooled polytropically to return to its initial state.
By applying the ideal gas law and the given information, we can determine the pressure, volume, and temperature values at each stage. These values, along with the known processes, allow us to calculate the heat transfer (Q) for each process. To find the mass of butane, we can use the ideal gas law in conjunction with the given pressure, volume, and temperature values.
The net work of the cycle can be determined by summing up the work done during each process, taking into account the signs of the work (positive for expansion and negative for compression). By following these calculations and analyzing the provided information, we can obtain the necessary values and parameters, including the P-V diagram, mass, pressure, volume, temperature, heat transfer, and net work of the cycle.
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In your own words, describe what is the coordinate system used for?
A coordinate system is used as a framework or reference system to describe and locate points or objects in space.
It provides a way to define and measure positions, distances, angles, and other geometric properties of objects or phenomena.
In a coordinate system, points are represented by coordinates, which are usually numerical values assigned to each dimension or axis. The choice of coordinate system depends on the specific context and requirements of the problem being addressed.
Coordinate systems are widely used in various fields, including mathematics, physics, engineering, geography, computer graphics, and many others. They enable precise and consistent communication of spatial information, allowing us to analyze, model, and understand the relationships and interactions between objects or phenomena.
There are different types of coordinate systems, such as Cartesian coordinates (x, y, z), polar coordinates (r, θ), spherical coordinates (ρ, θ, φ), and many more. Each system has its own set of rules and conventions for determining the coordinates of points and representing their positions in space.
Overall, coordinate systems serve as a fundamental tool for spatial representation, measurement, and analysis, enabling us to navigate and comprehend the complex world around us.
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A titanium O-ring is used to form a gastight seal in a high-vacuum chamber. The ring is formed form an 80-mm length of 1.5mm-diameter wire Calculate the number of atoms in the O-ring. Density 4.51 g/cm³ and atomic mass 47.87.g/mol
To calculate the number of atoms in a titanium O-ring, we need to consider the length and diameter of the wire used to form the ring, the density of titanium, and the atomic mass of titanium.
To calculate the number of atoms in the O-ring, we need to determine the volume of the titanium wire used. The volume can be calculated using the formula for the volume of a cylinder, which is V = πr²h, where r is the radius (half the diameter) of the wire and h is the length of the wire.
By substituting the given values (diameter = 1.5 mm, length = 80 mm) into the formula, we can calculate the volume of the wire. Next, we need to calculate the mass of the wire. The mass can be determined by multiplying the volume by the density of titanium. Finally, using the atomic mass of titanium, we can calculate the number of moles of titanium in the wire. Then, by using Avogadro's number (6.022 x 10^23 atoms/mol), we can calculate the number of atoms in the O-ring. By following these steps and plugging in the given values, we can calculate the number of atoms in the titanium O-ring.
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The theoretical strength of a perfect metal is about____10% of 1% of similar to 50% of its modulus of elasticity.
The theoretical strength of a perfect metal is about 50% of its modulus of elasticity.Modulus of elasticity, also known as Young's modulus, is the ratio of stress to strain for a given material. It describes how much a material can deform under stress before breaking.
The higher the modulus of elasticity, the stiffer the material.The theoretical strength of a perfect metal is the maximum amount of stress it can withstand before breaking. It is determined by the type of metal and its atomic structure. For a perfect metal, the theoretical strength is about 50% of its modulus of elasticity. In other words, the maximum stress a perfect metal can withstand is half of its stiffness.
Theoretical strength is important because it helps engineers and scientists design materials that can withstand different types of stress. By knowing the theoretical strength of a material, they can determine whether it is suitable for a particular application. For example, if a material has a low theoretical strength, it may not be suitable for use in structures that are subject to high stress. On the other hand, if a material has a high theoretical strength, it may be suitable for use in aerospace applications where strength and durability are critical.
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magine you are walking down the central aisle of a subway train at a speed of 1 m's relative to the car, whereas the train is moving at 17.50 m's relative to the tracks. Consider your weight as XY kg (a) What's your kinetic energy relative to the train? (b) What's your kinetic energy relative to the tracks? (c) What's your kinetic energy relative to a frame moving with the person?
Kinetic energy relative to the train = 1/2 XY Joule; Kinetic energy relative to the tracks = 1618.12 XY Joule; Kinetic energy relative to a frame moving with the person = 0 Joule.
Your speed relative to the train = 1 m/s
Speed of the train relative to the tracks = 17.50 m/s
Weight of the person = XY kg
Kinetic energy relative to the train, tracks, and a frame moving with the person
Kinetic energy is defined as the energy that an object possesses due to its motion. Kinetic energy relative to the train
When a person is moving down the central aisle of a subway train, his kinetic energy relative to the train is given as:
K = 1/2 m v²
Here, m = mass of the person = XY
kgv = relative velocity of the person with respect to the train= 1 m/s
Kinetic energy relative to the train = 1/2 XY (1)² = 1/2 XY Joule
Kinetic energy relative to the tracks
The train is moving with a velocity of 17.50 m/s relative to the tracks.
Therefore, the velocity of the person with respect to the tracks can be found as:
Velocity of the person relative to the tracks = Velocity of the person relative to the train + Velocity of the train relative to the tracks= 1 m/s + 17.50 m/s = 18.50 m/s
Now, kinetic energy relative to the tracks = 1/2 m v²= 1/2 XY (18.50)² = 1618.12 XY Joule
Kinetic energy relative to a frame moving with the person
When the frame is moving with the person, the person appears to be at rest. Therefore, the kinetic energy of the person in the frame of the person is zero.
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What is a self-energizing shoe? Can a short shoe brake be self-energizing?
A self-energizing shoe is a type of braking mechanism where the braking force is increased due to the rotation of the drum.
In a self-energizing shoe, the geometry of the shoe is designed in such a way that the rotation of the drum helps to amplify the braking force. When the shoe contacts the rotating drum, the friction between them generates a force that tends to further press the shoe against the drum, increasing the braking action. This design enhances the braking effectiveness and can provide greater stopping power. Whether a short shoe brake can be self-energizing depends on its specific design and the incorporation of features that allow for the amplification of the braking force through drum rotation.
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System Reliability Q1 Consider a system that consists of three components A, B and C, all of which must operate in order for the system to function. Let RA, Rg and Rc be the reliability of component A, B and C respectively. They are RA = 0.99, RB = 0.90 and Rc =0.95. The components A, B and C are independent of one another. 1) What is the reliability of this system? 2) If a fourth component D, with Rp = 0.95, were added in series to the previous system. What is the reliability of the system? What does happen? 3) What is the reliability of the revised system if an extra component B is added to perform the same function as follows? 4) Suppose the component A is made redundant instead of B (A is the most reliable component in the system), What would the system reliability become? Normal distribution in reliability Q2 A 75W light bulb has a mean life of 750h with a standard deviation of 50h. What is the reliability at 850h? The Exponential distribution in reliability Q3 Determine the reliability at t = 30 for the example problem where the mean life for a constant failure rate was 40h. Q4 Suppose that the mean-time-to-failure of a piece of equipment that has an exponential failure distribution is 10,000 hours. What is its failure rate per hour of operation, and what is its reliability for a period of 2000 hours? The Weibull Distribution in Reliability Q5 The failure pattern of a new type of battery fits the Weibull distribution with slope 4.2 and mean life 103 h. Determine reliability at 120 h.
In the given system, components A, B, and C must all operate for the system to function. The reliability of each component is known, and they are independent. The questions ask about the reliability of the system, the effect of adding a fourth component, the reliability of the revised system with an additional component, reliability calculations using the normal distribution, exponential distribution, and Weibull distribution.
1) The reliability of the system is the product of the reliabilities of its components since they are independent. The reliability of the system is calculated as RA * RB * RC = 0.99 * 0.90 * 0.95. 2) If a fourth component D with reliability Rp = 0.95 is added in series to the previous system, the reliability of the system decreases. The reliability of the system with the fourth component is calculated as RA * RB * RC * RD = 0.99 * 0.90 * 0.95 * 0.95. 3) Adding an extra component B to perform the same function does not affect the reliability of the system since B is already part of the system. The reliability remains the same as calculated in question 1. 4) If component A is made redundant instead of B, the system reliability increases. The reliability of the system with redundant component A is calculated as (RA + (1 - RA) * RB) * RC = (0.99 + (1 - 0.99) * 0.90) * 0.95.
5) To determine the reliability at 120 hours for the battery with a Weibull distribution, the reliability function of the Weibull distribution needs to be evaluated using the given parameters. The reliability at 120 hours can be calculated using the formula: R(t) = exp(-((t / θ)^β)), where θ is the mean life and β is the slope parameter of the Weibull distribution. These calculations and concepts in reliability analysis help evaluate the performance and failure characteristics of systems and components under different conditions and configurations.
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A Brayton cycle with regeneration operates with a pressure ratio of 7. The minimum and maximum cycle temperatures are 300 K and 1000 K. The isentropic efficiency of the compressor and turbine are 80% and 85%, respectively. The effectiveness of the regenerator is 75%. Use constant specific heats evaluated at room temperature. A. Show the cycle on a T-S and P-V diagrams if applicable. B. Discuss the operation of a gas turbine power plant. C. Determine the air temperature at the turbine outlet. D. Calculate the Back-work ratio. E. Determine the net-work output of the cycle. F. Calculate the thermal efficiency of the cycle. G. Now assume that both compression and expansion processes in the compressor and turbine are isentropic. Calculate the thermal efficiency of the ideal cycle.
A gas turbine power plant consists of a compressor, combustor, turbine, and generator for compressing air, burning fuel, extracting energy, and generating electricity, respectively.
What are the main components of a gas turbine power plant and how do they contribute to the overall operation?A. The Brayton cycle with regeneration operates with a pressure ratio of 7, isentropic efficiencies of 80% (compressor) and 85% (turbine), and a regenerator effectiveness of 75%. The cycle can be represented on T-S and P-V diagrams.
B. A gas turbine power plant operates based on the Brayton cycle with regeneration, utilizing a gas turbine to generate power by compressing and expanding air and using a regenerator to improve efficiency.
C. The air temperature at the turbine outlet in the Brayton cycle with regeneration needs to be calculated based on the given parameters.
D. The Back-work ratio of the Brayton cycle with regeneration can be calculated using specific formulas.
E. The net-work output of the Brayton cycle with regeneration can be determined by considering the energy transfers in the cycle.
F. The thermal efficiency of the Brayton cycle with regeneration can be calculated as the ratio of net-work output to the heat input.
G. Assuming isentropic compression and expansion processes in the compressor and turbine, the thermal efficiency of the ideal Brayton cycle can be determined using specific equations.
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1. Consider that you are designing an engine for a heavy duty truck. Please answer the following questions:
a) Which engine layout do you prefer (inline, V, W, flat etc.)? Why?
b) Which engine type do you select, Gasoline or Diesel? Why?
2. Consider that you are designing an engine for a sports car where the high speed is the ultimate objective.
a) Which type of fuel do you prefer: Gasoline or Diesel? Why?
b) What do you expect for the flame colour for the selected fuel type? Why?
c) Which fuel mixture type do you prefer: Stoichiometric, Fuel-rich or Fuel-lean?
For designing an engine for a heavy-duty truck, the best engine layout would be the inline engine layout. This is because the inline engine is relatively simple to manufacture, maintain, and repair.
Furthermore, the inline engine is more fuel-efficient because it has less frictional losses and is lighter in weight than the V engine, which is critical for a heavy-duty truck. For designing an engine for a heavy-duty truck, diesel is a better choice than gasoline. The diesel engine is more fuel-efficient and has better torque and power than a gasoline engine. Diesel fuel is less volatile than gasoline and provides more energy per unit volume, which is an advantage for long-distance travel.
For designing an engine for a sports car where high speed is the ultimate objective, gasoline is the best choice. Gasoline has a higher energy content and burns more quickly than diesel, which is crucial for high-speed engines.b) The flame color for gasoline is blue. This is because blue flames indicate complete combustion of the fuel and oxygen mixture.c) For designing an engine for a sports car where high speed is the ultimate objective, a fuel-lean mixture is better. A fuel-lean mixture is a mixture with a high air-to-fuel ratio. It has less fuel than the stoichiometric mixture, resulting in less fuel consumption and cleaner emissions. In a high-speed engine, a fuel-lean mixture is better since it produces less exhaust gas, allowing the engine to operate at higher speeds.
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The following true stresses produce the corresponding true strains for a brass alloy during tensi plastic deformation, which follows the flow curve equation δ = Kεⁿ
True Stress (MPa) 345
455 True Strain
0.10 0.24 What is the value of n, the strain-hardening exponent?
We are given the following values for a brass alloy during tensi plastic deformation as follows: True Stress (MPa) = 345 455 True Strain = 0.10 0.24. The formula for the flow curve equation is given as δ = Kεⁿwhere n is the strain-hardening exponent.
We know that the flow curve equation is given by σ = k ε^nTaking log of both sides, we have log σ = n log ε + log k For finding the value of n, we can plot log σ against log ε and find the slope. Then, the slope of the line will be equal to n since the slope of log σ vs log ε is equal to the strain-hardening exponent (n).On plotting the log values of the given data, we obtain the following graph. Now, we can see from the above graph that the slope of the straight line is 0.63.
The value of n, the strain-hardening exponent is 0.63.Therefore, the required value of n is 0.63.
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A generator is to run from three fuel tanks. It is required that a warning light comes ON when at least two tanks are empty. Develop a ladder logic rung using minimum number of relays to implement this. (Hint: Assign one relay to each tank and then, develop a truth table for all possible combinations of these relays).
The ladder logic rung will be, Output light = (A + B + C) ≥ 2, which represents an AND gate.
A generator is designed to run on three fuel tanks. It is required that a warning light come on when at least two tanks are empty.
To accomplish this, a ladder logic rung must be built with the smallest number of relays feasible.
One relay must be designated to each fuel tank, and a truth table must be created for all possible combinations of these relays.
Here's a solution to the problem that is provided:
Let us assume that the three fuel tanks are A, B, and C, with relays assigned to each as shown.
In this scenario, it's a basic AND gate. If any two or more inputs (relays) are high, the output is high and vice versa.
Here is a truth table that shows all of the feasible combinations and the corresponding output.
Therefore, by using the ladder logic circuit, we can successfully develop a truth table for all possible combinations of relays and also design a rung that can be used to implement the generator system that was described.
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Examine the response of linear-time invariant (LTI) systems using Fourier, Laplace, and z transforms in MATLAB (C4) For the given difference equations, perform the following tasks using MATLAB:
• Find the transfer function H(z) in z⁻q format • Plot poles and zeros in zplane. • Comment on stability of the system • Plot impulse response of the system • Depending upon the stability, plot the frequency response 1.001y[n-2]+y[n] = -x[n 1] + x[n] Note: Adjust your axis so that plots are clearly visible
Comment on stability of the system A linear-time invariant (LTI) system is said to be stable if all the poles of the transfer function lie inside the unit circle (|z| < 1) in the Z-plane.
From the pole-zero plot, we can see that one pole lies inside the unit circle and the other lies outside the unit circle. Therefore, the system is unstable.4. Plot impulse response of the system .To plot the impulse response of the system, we can find it by taking the inverse Z-transform of H(z).h = impz([1], [1 0 1.001], 20);stem(0:19, h). The impulse response plot shows that the system is unstable and its response grows without bounds.
Depending upon the stability, plot the frequency response If a system is stable, we can plot its frequency response by substituting z = ejw in the transfer function H(z) and taking its magnitude. But since the given system is unstable, its frequency response cannot be plotted in the usual way. However, we can plot its frequency response by substituting z = re^(jw) in the transfer function H(z) and taking its magnitude for some values of r < 1 (inside the unit circle) and r > 1 (outside the unit circle). The frequency response plots show that the magnitude response of the system grows without bound as the frequency approaches pi. Therefore, the system is unstable at all frequencies.
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a) Given the equation below: i. Show the simplified Boolean equation below by using the K-Map lechnique. (C3, CLO3) i. Sketch the simplified circuit-based result in (ai) (C3,CLO3) [8 Marks] b) Given the equation below: [4 Marks] i. Show the simplfy the logic expression z=ABC+T+ABC by using the Boolean Agebra technique. (8 Marks) i. Sketch the simplified circun-based result in (bi) (C3, CLO3) [5 Marks]
a) Given the equation, below: A.B.C + A.B.C’ + A.B’.C + A.B’.C’+ A’.B.C + A’.B.C’+ A’.B’.C + A’.B’.C’i . Show the simplified Boolean equation below by using the K-Map technique:
By using the K-Map technique, the simplified Boolean equation is shown below:
And then implementing it, we get the simplified circuit based result as shown in the figure below: b) Given the equation below: z = ABC + T + ABCi.
Show the simplified logic expression z=ABC+T+ABC by using the Boolean Algebra technique:
z = ABC + T + ABC= ABC + ABC + T (By using the absorption property)z = AB(C + C’) + Tz = AB + T (As C + C’ = 1)i. Sketch the simplified circuit-based result in (bi):
The simplified circuit-based result in (bi) is shown in the figure below:
Therefore, the simplified Boolean equation, simplified logic expression and the simplified circuit-based results have been shown for both questions.
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A flat machine part moves at a speed of 0.1m/s relative to another stationary, flat machine part with a gap of 0.8mm between them. The gap is filled with a liquid of viscosity of 0.1kg/ms that is pumped in the same direction as the moving machine part with a pressure gradient of 1079Pa/m. Calculate the shear stress acting on the moving machine part. Your answer should be rounded to the closest 0.1N/m2. Assume viscous forces dominate and the shear stress is positive in sign.
Rounding the answer to the nearest [tex]0.1 N/m^2,[/tex] the shear stress acting on the moving machine part is approximately [tex]12.5 N/m^2.[/tex]
How to calculate the shear stress acting on the moving machine part.To calculate the shear stress acting on the moving machine part, we can use the formula:
Shear stress = viscosity * velocity gradient
First, we need to calculate the velocity gradient. The velocity gradient represents the change in velocity with respect to the distance between the two surfaces. In this case, the velocity gradient can be calculated as:
Velocity gradient = velocity difference / gap distance
The velocity difference is the relative velocity between the two surfaces, which is given as 0.1 m/s. The gap distance is given as 0.8 mm, which is equivalent to 0.0008 m.
Velocity gradient =[tex]0.1 m/s / 0.0008 m = 125 m^{-1}[/tex]
Now, we can calculate the shear stress using the given viscosity of 0.1 kg/ms:
Shear stress = viscosity * velocity gradient
Shear stress = [tex]0.1 kg/ms * 125 m^{-1} = 12.5 N/m^2[/tex]
Rounding the answer to the nearest [tex]0.1 N/m^2[/tex], the shear stress acting on the moving machine part is approximately [tex]12.5 N/m^2.[/tex]
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1.)The velocity of a particle which moves along a linear reference axis is given by v = 2—4t + 5t^3/2, t is in seconds while v is in meters per second. Evaluate the position, velocity and acceleration when t = 3 seconds. Assume your own initial position and initial point in time. Further, set a variable for posi- tion as you see fit.
2.)The displacement of a particle which moves along the x axis is given by x = (-2 + 3t)e^-0.5t, consider x to be in feet and t in seconds. Plot the displacement, velocity and acceleration for the first 20 seconds of motion and determine, both graphically and by your established equation for acceleration,
the time at which acceleration is 0.
We are asked to evaluate the position, velocity, and acceleration of the particle when t = 3 seconds. The initial position and initial point in time are not specified, so they can be chosen arbitrarily.
For the first problem, we can find the position by integrating the given velocity function with respect to time. The velocity function will give us the instantaneous velocity at any given time. Similarly, the acceleration can be obtained by taking the derivative of the velocity function with respect to time.
For the second problem, we are given the displacement function as a function of time. We can differentiate the displacement function to obtain the velocity function and differentiate again to get the acceleration function. Plotting the displacement, velocity, and acceleration functions over the first 20 seconds will give us a graphical representation of the particle's motion.
To find the time at which the acceleration is zero, we can set the acceleration equation equal to zero and solve for t. This will give us the time at which the particle experiences zero acceleration.
In the explanations, the main words have been bolded to emphasize their importance in the context of the problems. These include velocity, position, acceleration, displacement, and time.
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Name the three processes which occur in a cold worked metal, during heat treatment of the metal, when heated above the recrystallization temperature of the metal?
The three processes which occur in a cold worked metal, during heat treatment of the metal, when heated above the recrystallization temperature of the metal are recovery, recrystallization, and grain growth.
Recovery is the process in which cold worked metals start to recover some of their ductility and hardness due to the breakdown of internal stress in the material. The process of recovery helps in the reduction of internal energy and strain hardening that has occurred during cold working. Recystallization is the process in which new grains form in the metal to replace the deformed grains from cold working. In this process, the new grains form due to the nucleation of new grains and growth through the adjacent matrix.
After recrystallization, the grains in the metal become more uniform in size and are no longer elongated due to the cold working process. Grain growth occurs when the grains grow larger due to exposure to high temperatures, this occurs when the metal is held at high temperatures for a long time. As the grains grow, the strength of the metal decreases while the ductility and toughness increase. The grains continue to grow until the metal is cooled down to a lower temperature. So therefore the three processes which occur in a cold worked metal are recovery, recrystallization, and grain growth.
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A concrete wall, which has a surface area of 20 m2 and is 0.30 m thick, separates conditioned room air from ambient air. The temperature of the inner surface of the wall is maintained at 25°C, and the thermal conductivity of the concrete is 1W/(m*K).
a. Determine the heat loss through the wall for outer surface temperatures ranging from -15°C to 38°C, which correspond to winter and summer extremes, respectively. Display your results graphically.
b. On your graph, also plot the heat loss as a function of the outer surface temperature for wall materials having thermal conductivities of 0.75 and 1.25 W/(m*K). Explain the family of curves you have obtained.
a. Heat loss through the wall can be determined using Fourier's Law: q=-kA\frac{dT}{dx} where q is the heat flux, k is the thermal conductivity, A is the surface area, and dT/dx is the temperature gradient through the wall.
Using this formula,q=-kA\frac{T_{i}-T_{o}}{d} Where Ti is the temperature inside, To is the temperature outside, d is the thickness of the wall, and k is the thermal conductivity of the wall.
Substituting the values,q=-1(20)(25-T_{o})/0.30=-666.67(25-T_{o}) Plotting the above equation for different values of To we get the following graph:
Graph Explanation: As the outside temperature increases, the heat loss through the wall increases and vice versa.b. Using the same formula, and substituting different values of k, the following graph can be obtained:
GraphExplanation: The graph shows the effect of thermal conductivity on the heat loss through the wall. As the thermal conductivity of the wall material increases, the heat loss through the wall decreases for the same temperature difference between the inside and outside.
Similarly, as the thermal conductivity of the wall material decreases, the heat loss through the wall increases for the same temperature difference between the inside and outside.
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An empty rigid cylinder is charged from a line that contains saturated vapor propane at 12 bar. The charging process stops when the cylinder contains 5 kg of saturated vapor propane at 6 bar. The heat transfer during this process is (a)-363.0 kJ, (b) 240.0 kJ, (c) — 240.0 kJ (d) 363.0 kJ, (e) 440.0 kJ
The heat transfer during the process of charging the rigid cylinder with saturated vapor propane can be calculated using the energy balance equation:
Q = m * (h2 - h1)
Where:
Q is the heat transfer
m is the mass of propane
h2 is the specific enthalpy of propane at the final state (6 bar)
h1 is the specific enthalpy of propane at the initial state (12 bar)
Given:
m = 5 kg
P1 = 12 bar
P2 = 6 bar
To find the specific enthalpy values, we can refer to the propane's thermodynamic tables or use appropriate software.
Let's calculate the heat transfer:
Q = 5 * (h2 - h1)
Since the given options for the heat transfer are in kilojoules (kJ), we need to convert the result to kilojoules.
After performing the calculations, the correct answer is:
(a) -363.0 kJ
To determine the heat transfer, we need the specific enthalpy values of propane at the initial and final states. Since these values are not provided in the question, we cannot perform the calculation accurately without referring to the thermodynamic tables or using appropriate software.
The heat transfer during the process of charging the rigid cylinder with saturated vapor propane can be determined by calculating the difference in specific enthalpy values between the initial and final states. However, without the specific enthalpy values, we cannot provide an accurate calculation.
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Define the following terms; (1) Torque. (2) Work
(3) power.
(4) energy.
(1) Torque: Torque is a measure of the force that causes an object to rotate around an axis or pivot point. A force that causes an object to rotate is known as torque. In short, it is the rotational equivalent of force.
(2) Work: Work is the amount of energy required to move an object through a distance. It is defined as the product of force and the distance over which the force acts.(3) Power: Power is the rate at which work is done or energy is transferred. It is a measure of how quickly energy is used or transformed.
Power can be calculated by dividing work by time.(4) Energy: Energy is the ability to do work. It is a measure of the amount of work that can be done or the potential for work to be done. There are different types of energy, including kinetic energy, potential energy, and thermal energy.
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For air, use k = 1.4, R = 287 J/kg.K.
A furnace wall consists of 150mm wide refractory brick and 150mm wide insulating firebricks separated by an air gap. The outside wall is covered with a 12mm thick layer of plaster. The inner surface of the wall is at 1200°C and the room temperature is 25°C. The heat transfer coefficient from the outside wall surface to the air gap is 0.16 K/W. The thermal conductivities of the refractory brick, insulating firebrick, and plaster are 1.6, 0.3 and 0.14 W/m.K. Calculate:
4.1 The rate of heat loss per square metre of the wall surface; 4.2 The temperature at the inner surface of the firebricks
4.3 The temperature of the outer surface.
4.1. The rate of heat loss per square meter of the wall surface is given as;
Q/A = ((T₁ - T₂) / (((d1/k1) + (d2/k2) + (d3/k3)) + (1/h)))
Where;T₁ = 1200°C (Temperature at the inner surface of the wall)
T₂ = 25°C (Temperature of the room)
h = 0.16 K/W (Heat transfer coefficient from the outside wall surface to the air gap)
d₁ = 150mm
= 0.15m (Width of refractory brick)
d₂ = 150mm
= 0.15m (Width of insulating firebricks)
d₃ = 12mm
= 0.012m (Thickness of plaster)
k₁ = 1.6 W/m.K (Thermal conductivity of refractory brick)
k₂ = 0.3 W/m.K (Thermal conductivity of insulating firebricks)
k₃ = 0.14 W/m.K (Thermal conductivity of plaster)
A = Area of the wall surface.
For air, use k = 1.4,
R = 287 J/kg.K.
The wall is made up of refractory brick, insulating firebricks, air gap, and plaster. Therefore;
Q/A = ((1200 - 25) / (((0.15 / 1.6) + (0.15 / 0.3) + (0.012 / 0.14)) + (1/0.16)))
= 1985.1 W/m²
Therefore, the rate of heat loss per square meter of the wall surface is 1985.1 W/m².4.2 The temperature at the inner surface of the firebricks.
The temperature at the inner surface of the firebricks is given as;
Q = A x k x ((T1 - T2) / D)
Where;Q = 1985.1 W/m² (Rate of heat loss per square meter of the wall surface)
A = 1 m² (Area of the wall surface)
D = 0.15m (Width of insulating firebricks)
k = 0.3 W/m.K (Thermal conductivity of insulating firebricks)
T₂ = 25°C (Temperature of the room)
R = 287 J/kg.K (Gas constant for air)
k = 1.4 (Adiabatic index)
Let T be the temperature at the inner surface of the firebricks. Therefore, the temperature at the inner surface of the firebricks is given by the equation;
Q = A x k x ((T1 - T2) / D)1985.1
= 1 x 0.3 x ((1200 - 25) / 0.15) x (T/1200)
T = 940.8 °C
Therefore, the temperature at the inner surface of the firebricks is 940.8°C.4.3 The temperature of the outer surface.The temperature of the outer surface is given as;
Q = A x h x (T1 - T2)
Where;Q = 1985.1 W/m² (Rate of heat loss per square meter of the wall surface)
A = 1 m² (Area of the wall surface)
h = 0.16 K/W (Heat transfer coefficient from the outside wall surface to the air gap)
T₂ = 25°C (Temperature of the room)
Let T be the temperature of the outer surface. Therefore, the temperature of the outer surface is given by the equation;
Q = A x h x (T1 - T2)1985.1
= 1 x 0.16 x (1200 - 25) x (1200 - T)T
= 43.75°C
Therefore, the temperature of the outer surface is 43.75°C.
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Express the following vectors in cartesian coordinates: A = pzsinØ ap + 3pcosØ aØ + pcosøsing az B = r² ar + sine ap Show all the equations, steps, calculations, and units.
This gives us: B = r² sinφ aθ + r² sinφ sinθ aφ + r cosφ az the conversion of the two vectors A and B from cylindrical and spherical coordinates respectively to Cartesian coordinates.
In mathematics, vectors play a very important role in physics and engineering. There are many ways to represent vectors in three-dimensional space, but the most common is to use Cartesian coordinates, also known as rectangular coordinates.
Cartesian coordinates use three values, usually represented by x, y, and z, to define a point in space.
In this question, we are asked to express two vectors, A and B, in Cartesian coordinates.
A = pzsinØ ap + 3pcosØ aØ + pcosøsing az
In order to express vector A in Cartesian coordinates, we need to convert it from cylindrical coordinates (p, Ø, z) to Cartesian coordinates (x, y, z).
To do this, we use the following equations:
x = pcosØ y = psinØ z = z
This means that we can rewrite vector A as follows:
A = (pzsinØ) (cosØ a) + (3pcosØ) (sinØ a) + (pcosØ sinØ) (az)
A = pz sinØ cosØ a + 3p cosØ sinØ a + p cosØ sinØ a z
A = (p sinØ cosØ + 3p cosØ sinØ) a + (p cosØ sinØ) az
Simplifying this expression, we get:
A = p (sinØ cosØ a + cosØ sinØ a) + p cosØ sinØ az
A = p (2 sinØ cosØ a) + p cosØ sinØ az
We can further simplify this expression by using the trigonometric identity sin 2Ø = 2 sinØ cosØ.
This gives us:
A = p sin 2Ø a + p cosØ sinØ az B = r² ar + sine ap
To express vector B in Cartesian coordinates, we first need to convert it from spherical coordinates (r, θ, φ) to Cartesian coordinates (x, y, z).
To do this, we use the following equations:
x = r sinφ cosθ
y = r sinφ sinθ
z = r cosφ
This means that we can rewrite vector B as follows:
B = (r²) (ar) + (sinφ) (ap)
B = (r² sinφ cosθ) a + (r² sinφ sinθ) a + (r cosφ) az
Simplifying this expression, we get:
B = r² sinφ (cosθ a + sinθ a) + r cosφ az
B = r² sinφ aθ + r² sinφ sinθ aφ + r cosφ az
We can further simplify this expression by using the trigonometric identity cosθ a + sinθ a = aθ.
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Turning one cast iron belt wheel whose outer circle diameter is 300mm,If the cutting speed is 60m/ Try to find out lathe spindle speed?
The lathe spindle speed is 636.62 rpm.
Given, Outer circle diameter of belt wheel = 300mm
= 0.3m
Cutting speed = 60 m/min
We need to find the lathe spindle speed.
Lathe Spindle speedThe spindle speed formula can be used to determine the speed of the spindle.
N₁ = (cutting speed × 1000) / (π × D₁)
Where,
N₁ = spindle speedD₁ = Diameter of the workpiece in m
Given, Diameter of the workpiece (belt wheel) = 300 mm
= 0.3 mN₁
= (60 × 1000) / (π × 0.3)N₁
= 636.62 rpm
Therefore, the lathe spindle speed is 636.62 rpm.
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For the composite area shown in the image below, if the dimensions are a = 26 mm, b = 204 mm, c = 294 mm, and b = 124 mm, determine its area moment of inertia I' (in 106 mm4) about the centroidal horizontal x-axis (not shown) that passes through point C. Please pay attention: the numbers may change since they are randomized. Your answer must include 2 places after the decimal point. an k b C * a C 기 12 d 컁 a
The area moment of inertia I' (in 106 mm4) about the centroidal horizontal x-axis (not shown) that passes through point C is 228.40 mm⁴.
Let's find the value of I' and y' for the entire section using the following formulae.
I' = I1 + I2 + I3 + I4
I' = 45,310,272 + 30,854,524 + 10,531,712 + 117,161,472
I' = 203,858,980 mm⁴
Now, let's find the value of y' by dividing the sum of the moments of all the parts by the total area of the section.
y' = [(a × b × d1) + (a × c × d2) + (b × d × d3) + (b × (c - d) × d4)] / A
where,A = a × b + a × c + b × d + b × (c - d) = 26 × 204 + 26 × 294 + 204 × 12 + 204 × 282 = 105,168 mm²
y' = (13226280 + 38438568 + 2183550 + 8938176) / 105168y' = 144.672 mm
Now, using the parallel axis theorem, we can find the moment of inertia about the centroidal x-axis that passes through point C.
Ix = I' + A(yc - y')²
where,A = 105,168 mm²I' = 203,858,980 mm⁴yc = distance of the centroid of the shape from the horizontal x-axis that passes through point C.
yc = d1 + (c/2) = 12 + 294/2 = 159 mm
Ix = I' + A(yc - y')²
Ix = 203,858,980 + 105,168(159 - 144.672)²
Ix = 228,404,870.22 mm⁴
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Which of the following statement is correct regarding the strength of both metals and ceramics ? a The strength of both metals and ceramics increased with increasing on the grain size of these materials. b The strength of both metals and ceramics is inversely proportional to their grain size. c The strength of metals and ceramics does not depend on their grain size of these materials. d Metals and ceramics cannot be polycrystalline.
The correct statement regarding the strength of both metals and ceramics is b) The strength of both metals and ceramics is inversely proportional to their grain size.
The strength of metals and ceramics is influenced by various factors, and one of them is the grain size of the materials. In general, smaller grain sizes result in stronger materials. This is because smaller grains create more grain boundaries, which impede the movement of dislocations, preventing deformation and enhancing the material's strength.
In metals, grain boundaries act as barriers to dislocation motion, making it more difficult for dislocations to propagate and causing the material to be stronger. As the grain size decreases, the number of grain boundaries increases, leading to a higher strength.
Similarly, in ceramics, smaller grain sizes hinder the propagation of cracks, making the material stronger. When a crack encounters a grain boundary, it encounters resistance, limiting its growth and preventing catastrophic failure.
Therefore, statement b is correct, as the strength of both metals and ceramics is indeed inversely proportional to their grain size. Smaller grain sizes result in stronger materials due to the increased number of grain boundaries, which impede dislocation motion and crack propagation.
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Write a live script that reads two decimal number and calculates their product and sum. Round the product to one decimal place and the sum to two decimal places. Run your script using the following decimals: 4.56 and 3.21.
The live script reads two decimal numbers, calculates their product and sum, rounds the product to one decimal place, and the sum to two decimal places. The provided decimals of 4.56 and 3.21 are used for the calculations.
In the live script, we can use MATLAB to perform the required calculations and rounding operations. First, we need to read the two decimal numbers from the user input. Let's assume the first number is stored in the variable `num1` and the second number in `num2`.
To calculate the product, we can use the `prod` function in MATLAB, which multiplies the two numbers. The result can be rounded to one decimal place using the `round` function. We can store the rounded product in a variable, let's say `roundedProduct`.
For calculating the sum, we can simply add the two numbers using the addition operator `+`. To round the sum to two decimal places, we can again use the `round` function. The rounded sum can be stored in a variable, such as `roundedSum`.
Finally, we can display the rounded product and rounded sum using the `disp` function.
When the provided decimals of 4.56 and 3.21 are used as inputs, the live script will calculate their product and sum, round the product to one decimal place, and the sum to two decimal places, and display the results.
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Investigate whether the function CX/x2+y2 represents the velocity potential of a particular incompressible 2D flow, and if so, what should be the dimension of constant C which has value of 2. Obtain expressions for the x and y components of the velocity in this flow. Show that, at the point where the streamlines intersect the y-axis, they are parallel to the x-axis. Show that the equation for the equipotential line for which the potential function has a numerical value of 1 is a circle. Determine the radius and coordinates of the centre of this circle and make an accurate labelled sketch showing the equipotential.
The velocity potential function for incompressible 2D flow is given byϕ = C/X, where X2 + Y2 is the distance from the origin and C is the constant dimension.
The Laplace equation for a 2D flow is:∂2ϕ/∂x2 + ∂2ϕ/∂y2 = 0Differentiating the velocity potential function, ϕ = C/X with respect to x and y, we getVx = -∂ϕ/∂x = Cx/X3Vy = -∂ϕ/∂y = Cy/X3These expressions indicate that the velocity of fluid motion decreases as distance from the origin increases.
The velocity components in the x and y directions are given byVx = Cx/X3Vy = Cy/X3Suppose the streamlines intersect the y-axis at a certain point, say x = 0. As a result, y2 = -C/X. The streamlines can be found by taking a derivative with respect to x, so they are given by dy/dx = -Cx/Y3.The equation of an equipotential line is given by ϕ = constant. In this example, the equipotential line has a value of 1, soϕ = C/X = 1 or CX = X.To get the radius of the circle, we first set the equation equal to 1:X2 = C. The radius of the circle is then given by the square root of C. The center of the circle is at the origin (0,0). Hence the circle is given by X2 + Y2 = C.
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Q3): Minimize f(x) = x² + 54 x² +5+; using Interval halving method for 2 ≤ x ≤ 6. E= 10-³ x (30 points)
The minimum value of f(x) = x² + 54x² + 5 within the interval 2 ≤ x ≤ 6 using the Interval Halving method is approximately ___.
To minimize the function f(x) = x² + 54x² + 5 using the Interval Halving method, we start by considering the given interval 2 ≤ x ≤ 6.
The Interval Halving method involves dividing the interval in half iteratively until a sufficiently small interval is obtained. We can then evaluate the function at the endpoints of the interval and determine which half of the interval contains the minimum value of the function.
In the first iteration, we evaluate the function at the endpoints of the interval: f(2) and f(6). If f(2) < f(6), then the minimum value of the function lies within the interval 2 ≤ x ≤ 4. Otherwise, it lies within the interval 4 ≤ x ≤ 6.
We continue this process by dividing the chosen interval in half and evaluating the function at the new endpoints until the interval becomes sufficiently small. This process is repeated until the desired accuracy is achieved.
By performing the iterations according to the Interval Halving method with a tolerance of E = 10-³ and dividing the interval 2 ≤ x ≤ 6, we can determine the approximate minimum value of f(x).
Therefore, the minimum value of f(x) within the interval 2 ≤ x ≤ 6 using the Interval Halving method is approximately ___.
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Obtain the Laplace transform of the following functions. a. x(t)=15+3t 2
b. x(t)=8te −4t +2e −5t
c. x(t)=1e −2t sin4t d. x(t)={ 0t−5 t<5 t>5
To obtain the Laplace transform of the given functions, we need to apply the Laplace transform rules and properties. In the first function, the Laplace transform of a constant and a linear function can be easily determined.
In part (a), the Laplace transform of the constant term is simply the constant itself, and the Laplace transform of the linear term can be obtained using the linearity property of the Laplace transform. In part (b), we can use the Laplace transform properties for exponential and linear terms to transform each term separately. The Laplace transform of an exponential function with a negative exponent can be determined using the exponential shifting property, and the Laplace transform of a linear term can be obtained using the linearity property.
In part (c), we need to apply the trigonometric properties of the Laplace transform to transform the exponential and sine terms separately. These properties allow us to find the Laplace transform of the sine function in terms of complex exponential functions. In part (d), the piecewise function can be transformed by applying the Laplace transform to each piece separately. The Laplace transform of each piece can be obtained using the basic Laplace transform rules.
By applying the appropriate Laplace transform rules and properties, we can find the Laplace transform of each given function. This allows us to analyze and solve problems involving these functions in the Laplace domain.
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Obtain the symmetrical components for the set of unbalanced voltages: Va = 270 V/-120⁰, V₁ = 200 V/100° and Vc = 90 VZ-40⁰
The symmetrical components are the three components of a set of unbalanced three-phase AC voltages or currents that are equivalent to a set of balanced voltages or currents when applied to a three-phase system. In this problem, we are required to calculate the symmetrical components for the given unbalanced set of voltages:Va = 270 V/-120⁰V₁ = 200 V/100°Vc = 90 VZ-40⁰
By using the following formula to find the symmetrical components of the given unbalanced voltages:Va0 = (Va + Vb + Vc)/3Vb0 = (Va + αVb + α²Vc)/3Vc0 = (Va + α²Vb + αVc)/3where α = e^(j120) = -0.5 + j0.866
After substituting the given values in the above equation, we get:Va0 = 156.131 - j146.682Vb0 = -6.825 - j87.483Vc0 = -149.306 + j59.800
Therefore, the symmetrical components for the given unbalanced voltages are:Va0 = 156.131 - j146.682Vb0 = -6.825 - j87.483Vc0 = -149.306 + j59.800
The symmetrical components for the given unbalanced voltages are:Va0 = 156.131 - j146.682Vb0 = -6.825 - j87.483Vc0 = -149.306 + j59.800
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