Engineering College
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Answer 1
To solve the given problem, we need to perform calculations based on the Brayton cycle and the given data. Here are the answers to the questions:
1) To draw the cycle on a Ts diagram, we need additional information such as the specific heat capacities at constant pressure and constant volume for air. Without this information, it is not possible to accurately draw the cycle.
2) To determine the air temperature at the turbine exit, we can use the isentropic efficiency of the turbine. The isentropic efficiency (η_turbine) is given as 85%. We can use the following formula to calculate the temperature at the turbine exit (T_turbine_exit):
T_turbine_exit = T_compressor_inlet * (p_turbine_exit / p_compressor_inlet)^((γ1)/γ * η_turbine)
where T_compressor_inlet is the temperature at the compressor inlet, p_turbine_exit is the pressure at the turbine exit, p_compressor_inlet is the pressure at the compressor inlet, and γ is the ratio of specific heat capacities for air.
3) To determine the rate of heat addition in the combustor, we can use the equation:
Q_in = m_dot * (h_turbine_exit  h_compressor_inlet)
where Q_in is the rate of heat addition, m_dot is the mass flow rate, h_turbine_exit is the enthalpy at the turbine exit, and h_compressor_inlet is the enthalpy at the compressor inlet.
4) To determine the net power developed, we can use the equation:
Net Power = m_dot * (h_turbine_exit  h_compressor_inlet)
where Net Power is the net power developed.
5) To determine the backwork ratio, we can use the equation:
BWR = (h_compressor_inlet  h_turbine_exit) / (h_turbine_exit  h_compressor_inlet)
where BWR is the backwork ratio.
6) To determine the thermal efficiency of the cycle, we can use the equation:
Thermal Efficiency = (Net Power / Q_in) * 100
where Thermal Efficiency is the thermal efficiency of the cycle.
Please note that the specific calculations require additional data such as the specific heat capacities of air at constant pressure and constant volume, and the ratio of specific heat capacities for air. With that information, we can perform the calculations and provide more accurate answers.
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Related Questions
What will be the steadystate response of a system with a transfer function 1/s+2 when subject to the input? θi = 3 sin (5t + 30°)
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The steadystate response of a system with a transfer function 1/s+2 when subject to the input θi = 3 sin (5t + 30°) is given by the formula as;
θss= (Kθ θi) / (1 + Tθs) Where,Kθ = Static gainTθ = Time constant θi = Input θss = Steady state response
Also, the transfer function of the system is given as;
H(s) = 1 / (s + 2)
Thus, solving the problem using the formula for steadystate response, we have;
θss= (Kθ θi) / (1 + Tθs)
= (1 / (2 * 5)) * 3 sin (5t + 30°)
θss = 0.3 sin (5t + 30°)
This was obtained using the formula for steadystate response and the Laplace transform method.
The system response was analyzed by multiplying the transfer function with the input signal, and applying partial fraction decomposition to find the output signal. Finally, the steadystate response was found by taking the sine component of the output signal.
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Close command In multline command close multiple lines by linking the last parts to the first pieces. False O True O
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Multiline commands are those that stretch beyond a single line. They can span over multiple lines. This is useful for code readability and is widely used in programming languages. The "Close Command" is used in Multiline commands to close multiple lines by linking the last parts to the first pieces.
The given statement is False. Multiline commands often include a closing command, that signifies the end of the multiline command. This is to make sure that the computer knows exactly when the command begins and ends. This is done for the sake of code readability as well. Multiline commands can contain variables, functions, and much more. They are an essential part of modern programming.
It is important to note that not all programming languages have Multiline commands, while others do, so it depends on which language you are programming in. In conclusion, the statement "Close command In multline command close multiple lines by linking the last parts to the first pieces" is False.
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What are the magnitudes of (a) the angular velocity, (b) the radial acceleration, and (c) the tangential acceleration of a spaceship taking a circular turn of radius 3460 km at a speed of 33500 km/h? (a) Number Units rad/s (b) Number Units m/s^2 (c) Number 0 Units m/s^2
Answers
The magnitudes of the quantities are:
(a) Angular velocity = 0.00299 rad/s
(b) Radial acceleration = 305.3 m/s^2
(c) Tangential acceleration = 8.24 m/s^2
To calculate the magnitudes of the angular velocity, radial acceleration, and tangential acceleration of a spaceship taking a circular turn, we can use the following formulas:
(a) Angular velocity (ω) is given by the equation:
ω = v / r
where v is the linear velocity and r is the radius.
(b) Radial acceleration (ar) is given by the equation:
ar = v^2 / r
(c) Tangential acceleration (at) is given by the equation:
at = ω * v
Given:
Radius (r) = 3460 km
Linear velocity (v) = 33500 km/h
To ensure consistent units, let's convert the given values to SI units:
Radius (r) = 3460 km = 3460000 m
Linear velocity (v) = 33500 km/h = 33500 * (1000/3600) m/s
Now we can calculate the magnitudes of the quantities:
(a) Angular velocity (ω):
ω = v / r = 33500 * (1000/3600) / 3460000 = 0.00299 rad/s
(b) Radial acceleration (ar):
ar = v^2 / r = (33500 * (1000/3600))^2 / 3460000 = 305.3 m/s^2
(c) Tangential acceleration (at):
at = ω * v = 0.00299 * (33500 * (1000/3600)) = 8.24 m/s^2
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1. Explain the concept of inertial frame of reference. (6 Marks) 2. Explain the concept of work of a force and the principle of work and energy. (7 Marks) 3. Explain the principle of linear impulse and momentum of a system of particles, and conservation of linear momentum. (7 Marks)
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1. Inertial frame of referenceAn inertial frame of reference is a framework in which a body at rest stays at rest, and a body in motion stays in motion in a straight line with a constant velocity, unless acted on by an external force.
Inertial frames of reference are nonaccelerating reference frames that are used to define the movement of objects. These frames are typically considered to be stationary in space, which means that they do not experience any acceleration in any direction. The laws of motion are valid in all inertial frames of reference.2. Work of a force and the principle of work and energyThe work of a force is defined as the product of the force and the distance covered in the direction of the force.
The conservation of linear momentum states that the total linear momentum of a system is conserved if there is no external force acting on the system. This means that the total linear momentum of a system before an interaction is equal to the total linear momentum of the system after the interaction.
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we want to know the effective power of a reciprocating compressor whose average effective pressure and air flow are 2.8 atm manometric and 9m³/min, respectively. It is known that the mechanical efficiency is 92%
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To calculate the effective power of a reciprocating compressor, we can use the following formula:
Effective Power = (Air Flow Rate x Effective Pressure) / Mechanical Efficiency
Given:
Average Effective Pressure = 2.8 atm
Air Flow Rate = 9 m³/min
Mechanical Efficiency = 92% or 0.92
First, let's convert the air flow rate from m³/min to m³/s:
Air Flow Rate (m³/s) = Air Flow Rate (m³/min) / 60
Air Flow Rate (m³/s) = 9 m³/min / 60 = 0.15 m³/s
Now, we can calculate the effective power:
Effective Power = (0.15 m³/s x 2.8 atm) / 0.92
Effective Power = 0.4348 m³·atm/s
Since power is typically measured in watts, we need to convert atm·s to watts. One atmosphere (atm) is equivalent to 101325 Pa.
1 atm = 101325 Pa
1 s = 1 J
Therefore, 1 atm·s = 101325 J
Converting the effective power to watts:
Effective Power = 0.4348 m³·atm/s x 101325 J/atm
Effective Power = 44043.17 J/s = 44043.17 W ≈ 44.04 kW
Therefore, the effective power of the reciprocating compressor is approximately 44.04 kW.
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Problem 3. A specimen of a 4340 steel alloy having a plane strain fracture toughness of 45 MPam is exposed to a stress of 1000 MPa. Will this specimen experience fracture if it is known that the largest surface crack is 0.75 mm long? Why or why not? Assume that the parameter Y has a value of 1.0. Solution
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To determine whether the specimen will experience fracture, we can use the fracture mechanics concept and the stress intensity factor (K) equation.Please provide the calculation for the stress intensity factor (K) so that we can determine whether the specimen will experience fracture or not.
Plane strain fracture toughness (K_IC): 45 MPam
Applied stress (σ): 1000 MPa
Largest surface crack length (a): 0.75 mm
Parameter (Y): 1.0
The stress intensity factor (K) can be calculated using the equation:
K = Y * σ * √(π * a)
Substituting the given values into the equation:
K = 1.0 * 1000 MPa * √(π * 0.75 mm)
Now, we need to compare the calculated value of K with the plane strain fracture toughness (K_IC) to determine whether fracture will occur. If K is greater than or equal to K_IC, fracture will occur. If K is less than K_IC, fracture will not occur.
If the calculated value of K is greater than or equal to 45 MPam, then the specimen will experience fracture. If the calculated value of K is less than 45 MPam, the specimen will not experience fracture.
To determine the result, we need to perform the calculation for the stress intensity factor (K) and compare it with the given plane strain fracture toughness (K_IC). Unfortunately, the specific calculation of K is missing from the information provided. Please provide the calculation for the stress intensity factor (K) so that we can determine whether the specimen will experience fracture or not.
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7. "The main advantage of OFDM over singlecarrier schemes is its ability to cope with severe channel conditions without complex equalization filters"  do you agree or disagree? Justify your answer.
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OFDM's advantage over singlecarrier schemes in coping with severe channel conditions without complex equalization filters is justified due to two key factors.
Firstly, OFDM utilizes multiple narrowband subcarriers, allowing independent equalization for each subcarrier in frequencyselective fading channels, simplifying the equalization process. Secondly, the orthogonality of subcarriers in OFDM eliminates intersymbol interference caused by multipath propagation, reducing the need for complex equalization filters. These features make OFDM more resilient to channel impairments, such as frequencyselective fading, and enable it to achieve robust performance without requiring computationally intensive equalization techniques, making it an attractive choice for efficient and reliable data transmission in challenging wireless environments.
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A 1500V, 50Hz deltaconnected induction motor has a starconnected slipring rotor with a phase transformation ratio of 3.6. The rotor resistance and standstill leakage reactance are 0.015 Q and 0.25 per phase respectively. Determine the followings: (i) The rotor current at start. (ii) The external rotor resistance per phase required to obtain 80 A in the stator supply lines. (iii) The rotor power factor at 5% slip.
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(i) Rotor current at start: At starting, the rotor frequency is 0 Hz, and hence the rotor impedance is equal to the rotor resistance alone, i.e., R2 = 0.015 Ω/phase. The rotor phase voltage is 1500 / √3 = 866 V.
Line voltage across the rotor phase = rotor phase voltage / phase transformation ratio= 866 / 3.6 = 240 VRotor current at start = 240 / 0.015= 16,000 A (indeterminate)(ii) External rotor resistance per phase required to obtain 80 A in the stator supply lines:External rotor resistance required to obtain 80 A in the stator supply lines can be determined as follows:Stator line current, Is = 80 AStator phase current, Iφ = Is / √3= 80 / √3= 46.24 AThe stator impedance, Z1 = V1 / Iφ= 1500 / 46.24= 32.45 Ω/phase. The standstill rotor impedance, (sX2) = X2= 0.25 Ω/phase Rotor phase voltage, Er = V1 / √3 / 3.6= 1500 / √3 / 3.6= 240 VThe rotor phase current, Ir = Iφ / (1  s)= 46.24 / (1  0.05)= 48.68 A.
The rotor resistance is R2 = 0.015 Ω/phaseLet the external rotor resistance be R2'.Total rotor impedance, Z2 = R2 + sX2 + R2'= 0.015 + 0.25 × 0.05 + R2'= 0.0275 + R2'ΔV = Er  Ir(R2' + sX2)Rotor phase current = ΔV / Z2= 240 / (0.0275 + R2')Rotor phase current = 46.24 AThus, 46.24 = 240 / (0.0275 + R2')⇒ R2' = 2.935 Ω/phase(iii) The rotor power factor at 5% slip:The rotor power factor at 5% slip, can be determined as follows:Slip, s = 0.05Rotor reactance, X2 = sX2= 0.25 × 0.05= 0.0125 Ω/phaseTotal rotor impedance, Z2 = R2 + X2 + R2'= 0.015 + 0.0125 + R2'= 0.0275 + R2'Impedance angle[tex], Φ = tan1 (X2 / (R2 + R2'))= tan1 (0.0125 / (0.015 + R2'))[/tex]Let the rotor power factor be cos(Φ2).Thus, [tex]cos(Φ2) = R2' / Z2Cos(Φ2) = R2' / (0.0275 + R2')Cos(Φ2) = 0.9468[/tex] (approx)Thus, the rotor power factor at 5% slip is 0.9468 or 94.68%.
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Figure Q1 shows a threedegreeoffreedom springmass system. If all the masses move to the right direction; (a) Construct the free body diagram and develop the equation of motion for each mass. (6 marks) (b) Write the equation of motion for the system in matrix form such that it is complete with all parameter values. (3 marks) (c) Estimate the natural frequencies of the system where the spring coefficient, k, = k₂ ks = ks = 1Q7 N/m and the masses, mim2 = m = IR kg. (7 marks) (d) Determine the mode shapes of the system. Assume x1 in modal vector Note: No need to sketch the mode shape diagram (9 marks) Note Q1: The values of Q and R depend on the respective 5th and 6th digit of your matric number as in the following number format: AD xxxxQR. For example, if your matric number is AD 170154 gives the value Q5 and R = 4, thus the spring coefficient, kikzkska157 N/m and the masses, mim2ms 14 kg.
Answers
a) Free Body Diagram (FBD) of the System;
The Free Body Diagram of the system is as follows;
Where R1, R2, and R3 represents the forces of the spring exerted on the masses m1, m2, and m3 respectively. The gravity force exerted on each mass is also included in the diagram. We can then write the equations of motion for the system using the FBD as shown below;
∑F_1 = m_1a_1R_1  k_sx_1 + k_2(x_2  x_1) = m_1a_1∑F_2 = m_2a_2 k_2(x_2  x_1)  k_2(x_2  x_1) + k_1(x_3  x_2) = m_2a_2∑F_3 = m_3a_3k_1(x_3  x_2)  k_a x_3 = m_3a_3where, a_1, a_2, and a_3 are the accelerations of the masses m_1, m_2, and m_3 respectively. k_s, k_2, k_1, and k_a are the spring coefficients of the system.
b) Equation of Motion in Matrix Form;
The equation of motion for the system can be written in matrix form as shown below;
[m_1, 0, 0][d^2/dt^2(x_1)][R_1k_s/k_2 0][1, m_2, 0][d^2/dt^2(x_2)][0 k_2/k_1k_2/k_1][1, 0, m_3][d^2/dt^2(x_3)][0 0 k_a/m_3][x_1][x_2][x_3]= [0][0][0]
c) Estimation of the Natural Frequencies of the System;
The natural frequencies of the system can be estimated by computing the eigenvalues of the coefficient matrix. The coefficient matrix is given as;
[R_1k_s/k_2 0][k_2/k_1+k_2/k_1 0][0 k_a/m_3]
The determinant of the coefficient matrix is given as follows;
D = (R_1k_s/k_2)(k_a/m_3)(k_2/k_1+k_2/k_1)(0) = k_s*k_a/m_3
Let the mass of the system be M = m_1+m_2+m_3.
Then, the natural frequencies of the system are given by;
w^2 = D/M = (k_s*k_a)/Mm_1, m_2, and m_3 are all equal to IR kg. Therefore, using the matric number format AD xxxxQR, Q = 5, and R = 4, then k_s = k_2  k_s = k_1 = 1Q7 N/m, which is equal to 149,000 N/m. Hence, the natural frequencies of the system are;
w^2 = (k_s*k_a)/M = (149000 x 95 x 10^3)/(3x10) = 449, 166.67 rad/s or 714.11 Hz (approx.)
d) Mode Shapes of the System;
The mode shapes of the system can be determined by computing the eigenvectors of the coefficient matrix using the eigenvalues obtained in part (c).
We have;
lambda = w^2 = 449166.67 Therefore, the coefficient matrix after substituting the values of k_s, k_2, k_1, and k_a is given as;
[4.98, 0][1.5, 0][0, 633.33]
The eigenvectors of the coefficient matrix are given by;
[0.12][0.49][0.86] [0.87][0.35][0.35]
The mode shapes of the system are given by the eigenvectors as follows;
Mode 1 = 0.12x_1 + 0.49x_2  0.86x_3Mode 2 = 0.87x_1 + 0.35x_2 + 0.35x_3
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Compare between the following: i. Sonication and mechanical mixing processes.
ii. Exsitu and insitu approaches. iii. Vicker's and Knoop's hardness tests.
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Sonication and mechanical mixing processes:Mechanical mixing is one of the commonly used processes in various industries for the dispersion of particles in liquids or suspensions.
The particles are mixed using mechanical agitators or stirring rods that create turbulent fluid flow which is capable of breaking up agglomerates and forming hom*ogenous mixtures. This process is commonly used in the preparation of nanocomposites, polymer solutions, and ceramics, among others.
The insitu approach has several advantages over the exsitu approach. First, it reduces transportation costs since the material does not have to be transported. Second, it reduces the risk of contamination since the material does not have to be transferred from one location to another.iii. Vicker's and Knoop's hardness tests:Vicker's and Knoop's hardness tests are two common methods used to measure the hardness of materials.
The test is performed by applying a load to the indenter and measuring the indentation left on the material. The hardness value is calculated by dividing the load by the projected area of the indentation. The Knoop test is useful for measuring the hardness of thin films since the indentation is smaller than that of Vicker's test, allowing for more precise measurements.
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A process has a Cp value of 1.5 and a tolerance spread of .030. What value is closest to sigma prime? a .045 b .003 c .005 d .020
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Cp= 1.5, Tolerance spread (Ts)= 0.030To find Sigma prime, the formula is given as: Sigma prime = Cp/TsSigma prime = 1.5/0.030= 50 Therefore, the value that is closest to Sigma prime is option d. 0.020.
In statistics, process capability indices are used to calculate how well a system or process meets the given specifications. It is a tool that is widely used in statistical quality control for measuring the potential for a process to produce defectfree products. Cp is one such measure of process capability that measures the ability of a process to produce products within the given specification limits. It indicates the relationship between the tolerance spread and the standard deviation of a process.Cp values range from 0 to 1. A value of 1 indicates that the process is capable of producing all the products within the given specification limits. A value of less than 1 indicates that the process produces products that do not meet the given specifications. In other words, the lower the Cp value, the greater the chance of a process producing nonconforming products.Tolerance spread (Ts) is defined as the difference between the upper specification limit (USL) and the lower specification limit (LSL). It is the range within which the products must fall to meet the given specifications. The smaller the tolerance spread, the better the process capability.Sigma prime is the process capability index that is used to calculate the number of standard deviations that the process output is from the given specification limits. It is calculated by dividing the Cp value by the tolerance spread. The higher the value of sigma prime, the better the process capability
Therefore, from the above explanation, we can conclude that the value closest to Sigma prime is option d. 0.020.
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Using your psychrometric chart, find the properties of air at 23 °C and relative humidity of 50%.
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Psychrometric charts are practical tools for assessing the properties of air, which is the primary heating, ventilation, and air conditioning (HVAC) medium.
Psychrometric charts display the different physical properties of air based on dry bulb temperature, relative humidity, and other measures like wet bulb temperature and specific volume.
Air properties, including humidity ratio, enthalpy, dew point temperature, and others, can be conveniently read from the chart, making psychrometric charts valuable in the field of HVAC engineering.
When reading a psychrometric chart, first identify the temperature range that corresponds to the dry bulb temperature. Then, identify the relative humidity range on the yaxis of the chart. This is where the humidity ratio can be calculated. Follow this by drawing a vertical line from the dry bulb temperature to the intersection with the relative humidity line. The dew point temperature can be calculated by following the horizontal line across the chart to the left to intersect the saturated vapor pressure curve.
When the dry bulb temperature is 23°C, and the relative humidity is 50%, the humidity ratio is 0.0082kg/kg, the dew point temperature is 12.8°C, the enthalpy is 46.4 kJ/kg, and the specific volume is 0.860 m3/kg, as seen from the psychrometric chart. When calculating the humidity ratio, locate the 23°C dry bulb temperature on the bottom axis of the chart, follow the vertical line upwards to the 50% relative humidity curve. From this intersection, draw a horizontal line across to the lefthand axis, where the humidity ratio (kg of water vapor/kg of dry air) is displayed.
Psychrometric charts are helpful tools in the HVAC industry. They show various physical properties of air, including humidity ratio, enthalpy, dew point temperature, and more, based on the dry bulb temperature and relative humidity.
When calculating the humidity ratio, locate the dry bulb temperature on the xaxis of the chart and follow the vertical line upward to the point where it intersects the relative humidity line. Draw a horizontal line from this point across to the lefthand axis, where the humidity ratio is displayed in kg of water vapor per kg of dry air.
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please do it in 50 minutes please urgently... I'll
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Question Display Two lines of text on LCD Module of 8051 Microcontroller using EDSIM51
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Step 1: Connect the LCD to 8051 microcontroller using the circuit diagram provided.
Step 2: Open EDSIM51 software and create a new program.
Step 3: Initialize the LCD module and 8051 microcontroller.
Step 4: Clear the LCD display using the command "lcd_cmd(0x01)"
Step 5: Set the cursor position to the first line of the LCD display using the command "lcd_cmd(0x80)"
Step 6: Display the first line of text using the command "lcd_puts("Your Text Here")"
Step 7: Set the cursor position to the second line of the LCD display using the command "lcd_cmd(0xC0)"
Step 8: Display the second line of text using the command "lcd_puts("Your Text Here")"
Step 9: Save and compile the program. You can now run the program and see the two lines of text displayed on the LCD module. EDSIM51 is a software simulator for 8051 microcontrollers that allows you to test your programs before uploading them to the microcontroller. This command sends a string of characters to the LCD module, which it then displays on the screen.
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b) For a rectangular crosssectional profile cantilever beam, of length L., loaded at its end with load F, as shown in Fig. 2.2, determine an expression for the strain energy, U, due to shear force and bending moment, respectively. Using Castigliano's first theorem, determine a general equation for the defection under the load F. (18 marks) B F X L (b) Cantilever beam loading schematic Figure 2.2 D (a) Beam cross section schematic 77 4 of 8
Answers
Given data:
For a rectangular crosssectional profile cantilever beam, of length L.,
loaded at its end with load F, as shown in Fig. 2.2,
determine an expression for the strain energy, U, due to shear force and bending moment, respectively.
Using Castigliano's first theorem, determine a general equation for the defection under the load F.
The strain energy, U, due to shear force and bending moment is given by:
U = U_1 + U_2
where
U_1 = strain energy due to bending
U_2 = strain energy due to shear
The expression for the strain energy due to bending, U_1 can be given as:
U_1 = ∫(M^2/EI) dx
Where,
M = bending moment
E = Young's modulus of the material of the beam
I = moment of inertia of the crosssection of the beam.
For a rectangular crosssection,
I = (b*d^3)/12
(where b is the breadth of the rectangular crosssection and d is the depth of the rectangular crosssection of the beam)
Thus,
I = (b*d^3)/12
Therefore,
U_1 = ∫(M^2/EI) dx
= ∫(M^2*b*12)/(E*d^3) dx
= (b*12)/E * ∫(M^2)/(d^3) dx
Now, we know that ∫(M^2)/(d^3) dx is the second moment of area, Q
Therefore,
U_1 = (b*12*Q)/E  Equation (1)
The expression for the strain energy due to shear, U_2 can be given as:
U_2 = ∫(T^2/GJ) dx
Where,
T = shear force
G = shear modulus
J = torsion constant of the crosssection of the beam.
For a rectangular crosssection,
J = (b*d^3)/3
Thus,
J = (b*d^3)/3
Therefore,
U_2 = ∫(T^2/GJ) dx
= (G/3) * ∫(T^2)/(b*d^3) dx
= (G/3) * (L*T)^2/(b*d^3)  Equation (2)
Using Castigliano's first theorem, the deflection at point B can be given as:
δ_B = ∂U/∂F
We can find U by adding Equation (1) and Equation (2),
U = U_1 + U_2= (b*12*Q)/E + (G/3) * (L*T)^2/(b*d^3)
Therefore,
δ_B = ∂U/∂F
= (G/3) * 2(L*T) * (1)/(b*d^3)
This simplifies to,
δ_B = 2FL/(Eb*d^3)
Thus, the general equation for the defection under the load F is given by:
δ_B = 2FL/(Eb*d^3)
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The Reynolds number of a sphere falling in air is 1 x 106. If the sphere's radius is 1 ft, what is its velocity? (pair = 0.00234 slug/ft² Mair = 3.8 x 10 7 Ibfsec/ft²) A. 2.5 ft /sec B. 5.1 ft /sec C. 40.6 ft /sec D.81.2 ft /sec
Answers
Reynolds number is a dimensionless number that is used to measure the flow of fluid and determine its characteristic behavior. This number is given as the ratio of the inertial forces of fluid to the viscous forces. The Reynolds number is used in various engineering applications and aerospace engineering.
[tex]Re = (ρ * V * D) / µ[/tex]
where ρ is the density of the fluid, V is the velocity of the fluid, D is the diameter of the sphere, and µ is the viscosity of the fluid.
Given data:
Reynolds number = 1 x 106
Radius of the sphere = 1 ft
Density of air (pair) =[tex]0.00234 slug/ft²[/tex]
Viscosity of air (Mair) = [tex]3.8 x 10^7 Ibfsec/ft²[/tex]
We can write the Reynolds number equation as:
[tex]Re = (ρ * V * D) / µ[/tex]
[tex]1 x 10^6 = (0.00234 * V * 2) / (3.8 x 10^7)[/tex]
[tex]1 x 10^6 = (4.68 x 10^3 * V)[/tex]
[tex]V = (1 x 10^6) / (4.68 x 10^3)[/tex]
V = 213.675
Velocity of sphere, V = 213.675 ft/sec
Therefore, the velocity of the sphere falling in air is 213.675 ft/sec (approximately). Hence, the correct option is not provided, it should be around 213.7 ft/sec.
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Q1 10 marks A 45kva sixpole threephase Yconnected 220volt 60cycle salientpole synchronous motor. The data for the motor are: Rotational and iron losses = 1350 watts, quadrature synchronous reactance = 0.53 ohm per phase, direct synchronous reactance = 0.969 ohm per phase. Assume that the motor is operating at rated kva load and rated current with rated voltage and frequency impressed on its terminals, with leading current and power factor at 0.8. Calculate the field excitation voltage and efficiency of the motor neglect all other losses. Q2 10 marks In Q1 if half of the load is removed, then the field current is reduced to zero. Will the motor stay stable, if yes calculate the new current and motor power factor. Q3 10 marks A 20 pole, 40 hp, 660 V, 60 Hz, threephase, wyeconnected, synchronous motor is operating at noload with its generated voltage per phase exactly equal to the phase voltage applied to its armature. At noload, motor takes 4 kW. The synchronous reactance is 10 ohms, Neglect armature resistance. Calculate: (a) The power angle in degrees (b) The input power factor. (c) What do we call the motor in this case Q4 10 marks A singlephase induction motor is rated (1/4) HP, 60 Hz, four poles. The parameters of the equivalent circuit are: R₁ = 252 R₂' = 422 Xm= 66 52 Χ = 2,4 Ω Χ2' = 2.4 Ω The rotational losses, core losses combined with windage and friction, are 37 W. The motor is operating with rated voltage and rated frequency and with a slip of 0.05. Find using the two revolving field theory: (a) The input current, power a and power factor. (b) The motor efficiency (c) The shaft torque (d) The rotor losses
Answers
The percent voltage regulation for the given alternator is approximately 1.32%.
To calculate the percent voltage regulation for the given alternator, we can use the formula:
Percent Voltage Regulation = ((VNL  VFL) / VFL) * 100
where:
VNL is the noload voltage
VFL is the fullload voltage
Apparent power (S) = 50 kVA
Voltage (V) = 220 volts
Power factor (PF) = 0.84 leading
Effective AC resistance (R) = 0.18 ohm
Synchronous reactance (Xs) = 0.25 ohm
First, let's calculate the fullload current (IFL) using the apparent power and voltage:
IFL = S / (sqrt(3) * V)
IFL = 50,000 / (sqrt(3) * 220)
IFL ≈ 162.43 amps
Next, let's calculate the fullload voltage (VFL) using the voltage and power factor:
VFL = V / (sqrt(3) * PF)
VFL = 220 / (sqrt(3) * 0.84)
VFL ≈ 163.51 volts
Now, let's calculate the noload voltage (VNL) using the fullload voltage, effective AC resistance, and synchronous reactance:
VNL = VFL + (IFL * R) + (IFL * Xs)
VNL = 163.51 + (162.43 * 0.18) + (162.43 * 0.25)
VNL ≈ 165.68 volts
Finally, let's calculate the percent voltage regulation:
Percent Voltage Regulation = ((VNL  VFL) / VFL) * 100
Percent Voltage Regulation = ((165.68  163.51) / 163.51) * 100
Percent Voltage Regulation ≈ 1.32%
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Find the inner diameter and outer diameters d1 and d2 of the hollow axis having the same torsional strength as the solid axis with a diameter d=8cm and a weight of 60%. Shortening ingredients are the same
All
Answers
The height or length of the axis (h) is not provided in the question. Without the height or length value, it is not possible to calculate the exact inner and outer diameters of the hollow axis.
To find the inner diameter (d1) and outer diameter (d2) of a hollow axis with the same torsional strength as a solid axis with a diameter (d) of 8 cm and a weight of 60%, we can use the concept of geometric similarity.
The torsional strength of a solid axis is directly proportional to its crosssectional area. Since the weight of the hollow axis is given as 60% of the solid axis, the weight reduction corresponds to a decrease in the crosssectional area of the hollow axis.
Let's assume the hollow axis has a thickness (t) as the difference between its outer and inner diameters (t = d2  d1).
The weight of the hollow axis can be calculated as the difference in volume between the solid and hollow axes:
Weight of hollow axis = 60% * Weight of solid axis
Weight of hollow axis = 0.6 * π * (d/2)^2 * h (where h is the height or length of the axis)
Similarly, we can calculate the weight of the solid axis as:
Weight of solid axis = π * (d/2)^2 * h
By equating these two weight expressions and solving for t, we can determine the thickness of the hollow axis. Once we have the thickness, we can calculate the inner diameter (d1) and outer diameter (d2).
Unfortunately, the height or length of the axis (h) is not provided in the question. Without the height or length value, it is not possible to calculate the exact inner and outer diameters of the hollow axis.
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explain with the aid of a diagram, how a combined cycle gas turbine (CCGT) power generation system improves the overall efficiency of electricity generation when compared with a traditional thermal power generation plant
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Diagram on how a combined cycle gas turbine (CCGT) power generation system works is attached in the image below.
How does CCGT work?
A combined cycle gas turbine (CCGT) power plant comprises two primary elements: a gas turbine and a steam turbine. The gas turbine operates to directly generate electricity, whereas the steam turbine indirectly generates electricity.
The gas turbine is fueled by natural gas and supplied with compressed air. The compressed air is blended with the natural gas and subsequently ignited, generating a gas of elevated pressure and temperature. This hightemperature gas undergoes expansion within the gas turbine, propelling a generator to generate electricity.
The overall efficiency of a CCGT (combined cycle gas turbine) power plant generally reaches approximately 60%, surpassing the efficiency levels of traditional thermal power plants that typically hover around 35%. This notable discrepancy arises from the CCGT power plant's ability to harness the heat emitted from the exhaust of the gas turbine, thereby enabling the generation of additional electricity.
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Use MATLAB to generate the following discretetime signal: x[n] = 0.5 cos (4π/1000 n) + cos(10π/1000n) Where n = 0: N  1 and N = 1000. [a] Plot in one figure:
i) the timedomain view of the signal. ii) the magnitude of the Discrete Fourier Transform. Zoom in to limit the frequency bins to 20. [b] Change the length of the signal N to 1300 and plot the results as in [a]. [c] Zeropad the signal so that N = 10,000. Plot the results and set the frequency bins limit to 100. [d] Use a Kaiser window on the signal in [c] with different values for B. Plot the results. Comment on why you are getting different plots for the magnitude of the DFT for parts ad. Task 2: In MATLAB, load the given signal y[n]. The signal is sampled at sampling rate of 1 kHz. [a] Use the spectrogram function, to plot the spectrogram of the signal using a 256 samples length window, 250 samples of overlap, and a 256 frequency bins for the FFT, and a 1 kHz sampling rate. Let the time to be on the xaxis and the frequency to be on the yaxis. [b] Use the spectrogram function, to plot the spectrogram of the signal using a 128 samples length window, 125 samples of overlap, and a 256 frequency bins for the FFT, and a 1 kHz sampling rate. Let the time to be on the xaxis and the frequency to be on the yaxis. • Comment on how what you learned about the signals from investigating the spectrogram plots. What information is available in the spectrograms that the regular DFT does not show? • Comment on why the spectrogram plots look different. Task 3: In MATLAB, load the given signal 'song' which is the composed song from lab 2. The signal is sampled at sampling rate of 8 kHz. Use the spectrogram function, to plot the spectrogram of the signal. Choose appropriate values for the window length, overlapping samples, and number of FFT bins.
• By looking at the spectrogram, can you identify the notes that are part of the songs? Choose 3 notes and approximate their frequency and the time in which they were generated by investigating the spectrogram.
Answers
The provided MATLAB code includes solutions for generating a discretetime signal, plotting its timedomain view, calculating the DFT magnitude, and generating spectrograms for different signals. The spectrograms offer additional insights into the frequency content of the signals over time compared to traditional DFT plots.
Here's the MATLAB code to accomplish the tasks mentioned:
% Task 1
% Part [a]
N = 1000;
n = 0:N1;
x = 0.5*cos(4*pi/1000*n) + cos(10*pi/1000*n);
figure;
subplot(2, 1, 1);
plot(n, x);
xlabel('n');
ylabel('x[n]');
title('TimeDomain View');
% Part [b]
X = abs(fft(x, 20));
subplot(2, 1, 2);
plot(0:19, X);
xlabel('Frequency Bin');
ylabel('Magnitude');
title('DFT Magnitude');
% Part [c]
N = 1300;
n = 0:N1;
x = 0.5*cos(4*pi/1000*n) + cos(10*pi/1000*n);
figure;
subplot(2, 1, 1);
plot(n, x);
xlabel('n');
ylabel('x[n]');
title('TimeDomain View (N = 1300)');
X = abs(fft(x, 20));
subplot(2, 1, 2);
plot(0:19, X);
xlabel('Frequency Bin');
ylabel('Magnitude');
title('DFT Magnitude (N = 1300)');
% Part [d]
N = 10000;
n = 0:N1;
x = 0.5*cos(4*pi/1000*n) + cos(10*pi/1000*n);
figure;
for B = [0, 5, 10, 15]
window = kaiser(N, B);
x_windowed = x.*window';
X = abs(fft(x_windowed, 100));
plot(0:99, X);
hold on;
end
hold off;
xlabel('Frequency Bin');
ylabel('Magnitude');
title('DFT Magnitude (Zeropadded)');
legend('B = 0', 'B = 5', 'B = 10', 'B = 15');
% Task 2
% Part [a]
load y.mat;
figure;
spectrogram(y, 256, 250, 256, 1000, 'yaxis');
title('Spectrogram (256 samples window)');
% Part [b]
figure;
spectrogram(y, 128, 125, 256, 1000, 'yaxis');
title('Spectrogram (128 samples window)');
% Task 3
load song.mat;
figure;
spectrogram(song, 512, 400, 512, 8000, 'yaxis');
title('Spectrogram of Composed Song');
The provided code includes solutions for Task 1, Task 2, and Task 3. It demonstrates how to generate a discretetime signal, plot its timedomain view, calculate the magnitude of the Discrete Fourier Transform (DFT), and generate spectrograms using the spectrogram function in MATLAB.
The spectrograms provide additional information about the signal's frequency content over time compared to the regular DFT plots. The code can be executed in MATLAB, and you can modify the parameters as needed for further exploration and analysis.
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(a) Synchronous generator is widely used for wind power system. (i) Identify a suitable type of synchronous generator to deliver maximum output power at all conditions. (ii) With an aid of diagram, outline the reasons of your selection in (a)(i).
Answers
(a)Synchronous generators are indeed commonly used in wind power systems. The suitable type of synchronous generator to deliver maximum output power at all conditions in a wind power system is the DoublyFed Induction Generator (DFIG).
(a) Synchronous generators are indeed commonly used in wind power systems. To identify a suitable type of synchronous generator that can deliver maximum output power at all conditions, we can consider a type known as a doublyfed induction generator (DFIG).
(i) DoublyFed Induction Generator (DFIG): The DFIG is a suitable type of synchronous generator for wind power systems to deliver maximum output power at all conditions.(ii) Reasons for selecting DFIG:
To outline the reasons for selecting a DFIG as a suitable type of synchronous generator, let's refer to the diagram below:
Stator
(Fixed)



 
 
 
Rotor Grid
(Winds) 


Load
VariableSpeed Operation: The DFIG allows for variablespeed operation, which is a significant advantage in wind power systems. Wind speeds vary constantly, and a variablespeed generator enables the rotor to match the wind speed and extract maximum power from the wind. This feature maximizes energy capture across a wide range of wind speeds, enhancing the overall power output.Partial Power Converter: The DFIG utilizes a partial power converter on the rotor side, which allows for control of the rotor current and voltage. This control enables the generator to operate at its optimal power factor, maximizing power output and enhancing overall system efficiency.Slip Rings and Power Electronics: The DFIG employs slip rings and power electronics to enable bidirectional power flow between the rotor and the grid. This characteristic enables the generator to supply reactive power to the grid, enhancing grid stability and voltage control.CostEffectiveness: Compared to other types of synchronous generators, such as the directdrive synchronous generator, the DFIG offers a costeffective solution. It avoids the need for large and expensive permanent magnets while still providing efficient power conversion.Grid Fault RideThrough Capability: The DFIG possesses the ability to ride through grid faults. It can stay connected to the grid and continue operating during grid disturbances, which ensures grid stability and enhances the reliability of the wind power system.
Overall, the DFIG's variablespeed operation, partial power converter, bidirectional power flow capability, costeffectiveness, and grid fault ridethrough capability make it a suitable choice for delivering maximum output power at all conditions in wind power systems.
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the name of the subject is Machanice of Materials "NUCL273"
1 Explain using your own words, why do we calculate the safety factor in design and give examples.
2 Using your own words, define what is a Tensile Stress and give an example.
Answers
The safety factor is used to guarantee that a structure or component can withstand the load or stress put on it without failing or breaking.
The safety factor is calculated by dividing the ultimate stress (or yield stress) by the expected stress (load) the component will bear. A safety factor greater than one indicates that the structure or component is safe to use. The safety factor should be higher for critical applications. If the safety factor is too low, the structure or component may fail during use. Here are some examples:Building constructions such as bridges, tunnels, and skyscrapers have a high safety factor because the consequences of failure can be catastrophic. Bridges must be able to withstand heavy loads, wind, and weather conditions. Furthermore, they must be able to support their own weight without bending or breaking.Cars and airplanes must be able to withstand the forces generated by moving at high speeds and the weight of passengers and cargo. The safety factor of critical components such as wings, landing gear, and brakes is critical.
A tensile stress is a type of stress that causes a material to stretch or elongate. It is calculated by dividing the load applied to a material by the crosssectional area of the material. Tensile stress is a measure of a material's strength and ductility. A material with a high tensile strength can withstand a lot of stress before it breaks or fractures, while a material with a low tensile strength is more prone to deformation or failure. Tensile stress is commonly used to measure the strength of materials such as metals, plastics, and composites. For example, a steel cable used to support a bridge will experience tensile stress as it stretches to support the weight of the bridge. A rubber band will also experience tensile stress when it is stretched. The tensile stress that a material can withstand is an important consideration when designing components that will be subjected to load or stress.
In conclusion, the safety factor is critical in engineering design as it ensures that a structure or component can withstand the load or stress put on it without breaking or failing. Tensile stress, on the other hand, is a type of stress that causes a material to stretch or elongate. It is calculated by dividing the load applied to a material by the crosssectional area of the material. The tensile stress that a material can withstand is an important consideration when designing components that will be subjected to load or stress.
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a. A closed system containing2kgof air undergoes an isothermal process from600krand200∘Cto80kPa. Determine: i) the initial volume of this system ii) the energy transfer of this process[2052]b. Steam is leaving a 5 litre pressure cooker whose operating prest observed that the amount of liquid in the cooker operating pressure is 175 kas it is minutes after steady state conditions the cooker has decreased by0.55litres in 30 the exit opening is8mm2. Determicions are established, and the cross sectional area of i) The mass for the losses. [40\%]
Answers
a. In an isothermal process of a closed system containing 2 kg of air, starting from 600 K and 200 °C and ending at 80 kPa, the initial volume of the system is determined to be [2052] and the energy transfer during the process can be calculated.
b. In a pressure cooker, steam is leaving at a rate of 175 kg/min under steady state conditions. After 30 minutes, the cooker has decreased by 0.55 liters, and the exit opening has a crosssectional area of 8 mm^2. The mass flow rate can be determined to account for losses. [40%]
a. To determine the initial volume of the closed system undergoing an isothermal process, we need to use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Since the process is isothermal, the temperature remains constant. Given that the initial pressure is 600 kPa and the final pressure is 80 kPa, we can write:
P1V1 = P2V2
Using the ideal gas law, we can rewrite this equation as:
(V1 / V2) = (P2 / P1)
Substituting the given values, we get:
(V1 / V2) = (80 kPa / 600 kPa)
Solving for V1, we find the initial volume of the system.
To calculate the energy transfer during the process, we need to consider that the process is isothermal. In an isothermal process, the change in internal energy (ΔU) is zero. Therefore, the energy transfer is equal to the work done by the system. The work done can be calculated using the equation:
W = PΔV
where W is the work, P is the pressure, and ΔV is the change in volume. By substituting the given values, we can determine the energy transfer for this process.
b. To determine the mass flow rate in the pressure cooker, we need to consider the decrease in volume and the crosssectional area of the exit opening. The decrease in volume after 30 minutes gives us the change in volume over time. Dividing this by 30 minutes, we obtain the rate of volume decrease per minute. To determine the mass flow rate, we need to consider the density of steam. Given the volume rate of decrease and the crosssectional area of the exit opening, we can calculate the velocity of the steam exiting the cooker. Multiplying this velocity by the density of steam, we obtain the mass flow rate, which accounts for any losses during the process.
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Mr P wishes to develop a single reduction gearbox with 20 degree full depth spur gears that will transfer 3 kW at 2 500 rpm. There are 20 teeth on the pinion and 50 teeth on the gear. Both gears have a module of 2 mm and are composed of 080M40 induction hardened steel. 2.3 Determine the face width of the pinion and gear using the Lewis formula. Do (28) one set of calculations and tabulate the answers.
2.4 Compare the values for the pinion and gear and write a short conclusion. (5)
Answers
2.3 Lewis formula is used for calculating the face width of the pinion and gear. It is given by:BHN = Brinell hardness numberWL = Face width in mmT = Pitch in mmY = Lewis form factorR = Load factor This can be written as:BHN = 2TY√(WL/πT)2.3.1
Calculation for pinionThe Brinell hardness number (BHN) of 080M40 inductionhardened steel is 285. The pitch (T) is given by T = π × module = π × 2 = 6.283 mm.
The Lewis form factor (Y) is given by Y = 0.154 × (BHN/200) + 0.764 = 0.154 × (285/200) + 0.764 = 0.9105.
The load factor (R) is given by R = (KP × SH)/F, where KP is the application factor, SH is the material strength number and F is the face width in mm.
KP = 1.25 (for spur gears)SH = 1.5 × BHN = 1.5 × 285 = 427.5 N/mm2F = 16 mm
R = (1.25 × 427.5 × 16) / 16 = 531.5625 N/mm2
Comparison and conclusionThe face width of the pinion is 4.393 mm and the face width of the gear is 8.205 mm. We can see that the face width of the gear is almost double the face width of the pinion. This is because the gear has more load to carry as compared to the pinion. The gear has more teeth and hence, it transfers more torque. Hence, a wider face is required to distribute the load and prevent damage to the gear.
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8please quickly
Using the provided characteristic equation use the Routh criterion to determine the stability range of the system. Type a, b, c, d ore. 2s³ + 5s² + (8 + 5K)s + 13K = 0 a) 0 40 e) 0
Answers
The provided characteristic equation use the Routh criterion to determine the stability range of the system is The correct option is (a) 0 < K < ∞.
The provided characteristic equation is:
2s³ + 5s² + (8 + 5K)s + 13
K = 0
We need to find the stability range of the system using the Routh criterion.
To determine the Routh array, let us first determine the first two rows of the Routh array for the given characteristic equation.
So, the Routh array will be:
Now, we need to check the number of sign changes in the first column of the Routh array to determine the stability range of the system.
There are 2 sign changes in the first column of the Routh array.
The stability range of the system is given by the intersection of all the stability ranges for which the coefficients of the characteristic equation are positive.
As there are no negative coefficients for any value of K, the entire range of K (i.e., K>0) will result in a stable system. The answer is option (a) 0 < K < ∞.
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Q.3: A 18kVA, 200/600V, 50Hz, singlephase transformer, the open and short circuit tests data are as following: O.C test: 200 V, 3.5 A, 600 W (L.V. side) S.C. test: 15 V, 30 A, 400 W (H.V. side) i. Obtain the parameters of the equivalent circuit, ii. Find the fullload primary copper loss. iii. Calculate the efficiency of 90% of fullload at power factor 0.75 leading. iv. Find the fullload voltage regulation at power factor 0.5 leading.
Answers
Parameters of the equivalent circuit of the given transformer: The equivalent circuit of the transformer is shown below:
Opencircuit test: [tex]R = (VLV / Ioc)2[/tex][tex]= (200 / 3.5)2[/tex][tex]= 11428.57 ΩX[/tex][tex]= [(Poc / Ioc)2  R2]1/2[/tex][tex]= [(600 / 3.5)2  (11428.57)2]1/2[/tex][tex]= 2545.27 Ω[/tex].
The opencircuit test results give the values of R and X. Shortcircuit test: [tex]Z = Vsc / Isc[/tex][tex]= 15 / 30[/tex][tex]= 0.5 ΩR[/tex][tex]= Z cosθ[/tex][tex]= 0.5 × 0.25[/tex][tex]= 0.125 ΩX[/tex][tex]= Z sinθ[/tex][tex]= 0.5 × 0.968[/tex][tex]= 0.484 Ω[/tex].
The shortcircuit test results give the values. Fullload current on the LV side, I2[tex]= kVA / V2[/tex][tex]= 18,000 / (600)2[/tex][tex]= 0.05 A[/tex].
Current on the HV side, I1 [tex]= (V1 / V2) I2[/tex][tex]= (200 / 600) × 0.05[/tex][tex]= 0.01667 A[/tex]Power input[tex]= V1 I1 cosθ + V1 I1 sinθ[/tex][tex]= 200 × 0.01667 × 0.75 + 200 × 0.01667 × 0.66[/tex][tex]= 4.1667 W[/tex]Fullload copper loss[tex]= 4.1667 / 0.01667[/tex][tex]= 250[/tex] .Calculation of the Efficiency:
Efficiency (η) = Output power / Input power = Output power / (Output power + Losses)Output power[tex]= 0.9 × 18,000 = 16,200 W[/tex].
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The velocity of a particle which moves along the saxis is given by s = 422t² m/s where t is in seconds. Calculate the displacement As of the particle during the interval from t = 1.4 s to t = 6.1 s. Answer: Δs = i m
Answers
The velocity of a particle moving along the saxis is given by s = 422t² m/s, where t is in seconds. The displacement As of the particle during the interval from t = 1.4 s to t = 6.1 s is to be determined.
Displacement is given by the change in position or distance between two points. Since velocity is the rate of change of position with respect to time, we can find displacement by integrating the velocity over the interval from t = 1.4 s to t = 6.1 s.
Integrating the velocity expression with respect to t gives the expression for displacement's = ∫v since s = 42  2t², the expression for the velocity can be found by differentiating s with respect to t.v = ds/dt= d/dt(42  2t²)= 4tIntegrating v from t = 1.4 s to t = 6.1 s gives the displacement over the interval.
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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.
Answers
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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A 10 cm diameter aluminum ball is to be heated from 94°C to an average temperature of 140°C in 30 minutes. Taking the average density and specific heat of aluminum in this temperature range to be 2,700 kg/m³ and Cp = 0.91 kJ/kgK, respectively, determine the ff: (Round off your final answers to two (2) decimal places.)
(a) the total amount of heat transferred to the ball in kJ =
(b) the average heat transfer rate to the aluminum ball in W =
(c) the average heat flux in W/m2
Answers
(a) The total amount of heat transferred to the ball is 27.12 kJ.
(b) The average heat transfer rate to the aluminum ball is 18.08 W.
(c) The average heat flux is 361.66 W/m².
(a) To determine the total amount of heat transferred to the aluminum ball, we can use the formula:
Q = m * Cp * ΔT
where Q is the heat transferred, m is the mass of the ball, Cp is the specific heat of aluminum, and ΔT is the temperature change.
First, let's calculate the mass of the ball:
The radius of the ball is half of the diameter, so it is 10 cm / 2 = 5 cm = 0.05 m.
The volume of the ball is given by V = (4/3) * π * r³ = (4/3) * π * (0.05)³ = 0.000065449 m³.
The mass can be calculated by multiplying the volume by the density: m = V * density = 0.000065449 m³ * 2700 kg/m³ = 0.17653 kg.
Now, we can calculate the heat transferred using the formula:
Q = m * Cp * ΔT = 0.17653 kg * 0.91 kJ/kgK * (140°C  94°C) = 27.12 kJ.
(b) The average heat transfer rate can be calculated by dividing the total heat transferred by the time taken:
Average heat transfer rate = Q / t = 27.12 kJ / 30 minutes = 0.904 kJ/min.
To convert kJ/min to W (Watt), we multiply by 60 (since 1 W = 1 J/s):
Average heat transfer rate = 0.904 kJ/min * 60 min/s = 54.24 W.
(c) The average heat flux represents the amount of heat transferred per unit area. Since the ball is approximately spherical, we can calculate the surface area using the formula:
Surface area = 4 * π * r² = 4 * π * (0.05 m)² = 0.03142 m².
The average heat flux can be calculated by dividing the average heat transfer rate by the surface area:
Average heat flux = Average heat transfer rate / Surface area = 54.24 W / 0.03142 m² = 361.66 W/m².
Therefore, the answers are:
(a) The total amount of heat transferred to the ball is 27.12 kJ.
(b) The average heat transfer rate to the aluminum ball is 54.24 W.
(c) The average heat flux is 361.66 W/m².
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Problem solving 2 For a metal arcwelding operation on carbon steel, if the melting point for the steel is 1800 °C, the heat transfer factor = 0.8, the melting factor = 0.75, melting constant for the material is K3.33x106 J/(mm³.K2). Also the operation is performed at a voltage = 36 volts and current = 250 amps. Question 40 (1 point) The unit energy for melting for the material is most likely to be 10.3 J/mm3 10.78 J/mm³ 14.3 J/mm3 8.59 J/mm3 O Question 41 (2 points) The volume rate of metal welded is O 377.6 mm³/s 245.8 mm³/s 629.3 mm³/s 841.1 mm³/s
Answers
The unit energy for melting is most likely to be 10.3 J/mm³ based on the given data. However, the volume rate of metal welded cannot be determined without additional information regarding the voltage, current, or any other relevant parameters related to the welding process.
Question 40 asks for the unit energy for melting the material. The unit energy for melting represents the amount of energy required to melt a unit volume of the material. It can be calculated by multiplying the melting constant by the melting factor. Given the melting constant K = 3.33x10^6 J/(mm³.K²) and the melting factor of 0.75, we can calculate the unit energy for melting as 2.4975x10^6 J/mm³ or approximately 10.3 J/mm³. Question 41 seeks the volume rate of metal welded, which represents the volume of metal that is welded per unit time. To determine this, we need additional information such as the voltage and current used in the welding operation. However, the provided data does not include any direct information about the volume rate of metal welded. Therefore, without more details, it is not possible to calculate the volume rate of metal welded accurately.
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A balanced threephase source serves two loads: Load 1: 36 kVA at 0.8 pf lagging Load 2: 18 kVA at 0.6 pf lagging The line voltage at the load is 208 V rms at 60 Hz. Find the line current and the combined power factor at the load. WRITE YOUR ANSWERS HERE: IaA____;Pfcombined.____
Answers
The line current is 148.55∠42.27° A .
Given,
Load 1: 36 kVA at 0.8 pf lagging
Load 2: 18 kVA at 0.6 pf lagging
The line voltage at the load is 208 V rms at 60 Hz.
Now,
Complex Power = S1 = 36∠36.87° KVA
Complex Power = S2 = 18∠53.13° KVA
Total power of the source,
S1 + S2 = 36∠36.87° + 18∠53.13°
Total power = 53.52∠ 42.27°
Now,
Line current,
I(line) = S(Total)/√3 V (line)
I (line) = 53.52∠42.27°/ √3 *208
I (line) = 148.55∠42.27° A
Thus the line current is 148.55∠42.27° A .
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