At sea-level conditions, the Brake Power of the gasoline engine is 252.56 kW.
To determine the engine parameters at sea-level conditions, we need to account for the change in temperature and pressure.
Given:
Temperature at the location: 15.0 °C
Pressure at the location: 99.4 kPa
Temperature difference: 18.0 °C
To convert the temperature to Kelvin, we add 273.15 to the given temperature:
Temperature at the location in Kelvin = 15.0 + 273.15 = 288.15 K
To convert the pressure to absolute pressure, we add 101.3 kPa (standard atmospheric pressure at sea level):
Pressure at the location in kPa = 99.4 + 101.3 = 200.7 kPa
Next, we can calculate the Brake Power at sea-level conditions.
Brake Power = Rated Power - Mechanical Energy Loss
Rated Power = 308 kW (given)
Mechanical Energy Loss = 18% of Rated Power = 0.18 * 308 kW = 55.44 kW
Brake Power = 308 kW - 55.44 kW = 252.56 kW
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Draw the Bode Diagram (magnitude plot) for the transfer function H(s) = 100(8+4)(s+20) / s(s+8)(8+100)
We have the transfer function
H(s) = 100(8+4)(s+20) / s(s+8)(8+100)
and we can draw the Bode Diagram (magnitude plot) using the above steps.
Given the transfer function,
H(s) = 100(8+4)(s+20) / s(s+8)(8+100)
To draw the Bode Diagram (magnitude plot) for the transfer function
H(s) = 100(8+4)(s+20) / s(s+8)(8+100),
First, we need to find the magnitude of the transfer function.
We know that the magnitude of a transfer function can be found by substituting s = jω and taking the modulus.
Thus,
H(jω) = 100(8+4)(jω+20) / jω(jω+8)(8+100)
Here,
|H(jω)| = |100(8+4)(jω+20) / jω(jω+8)(8+100)|
Let, K = 100(8+4) = 1200
|H(jω)| = |K(jω+20) / jω(jω+8)(8+100)|
|H(jω)| = K |(jω+20) / jω||1 / (jω+8)(8+100)|
|H(jω)| = K |(1+20/jω) / (1+jω/8)(1+jω/100)|
|H(jω)| = K |(1+20/jω) / (1+ jω/8)(1+ jω/100)|
Taking log on both sides,
log |H(jω)| = log K + log |(1+20/jω) / (1+ jω/8)(1+ jω/100)|
log |H(jω)| = log K + log |1+20/jω| - log |1+jω/8| - log |1+jω/100|
Now we will find the values of
|1+20/jω|, |1+jω/8|, and |1+jω/100|
for different values of ω and plot the graph.
The magnitude plot will be in decibels (dB).
So, we need to convert the values into dB.
The magnitude in dB is given by,
20 log |H(jω)| dB = 20 log K + 20 log |1+20/jω| - 20 log |1+jω/8| - 20 log |1+jω/100|
Thus, we have the transfer function
H(s) = 100(8+4)(s+20) / s(s+8)(8+100)
and we can draw the Bode Diagram (magnitude plot) using the above steps.
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Make a schematic diagram for a pcb of pid controller connected with first order RC circuit. Please explain each step for implementation of pid on PCB.
To make a schematic diagram for a PCB of a PID controller connected with the first-order RC circuit and explain the implementation steps of the PID on PCB as shown.
PID stands for proportional-integral-derivative. It is a type of feedback controller that has three main components: the proportional, the integral, and the derivative components. The RC circuit is an electronic circuit composed of a resistor and a capacitor. It is used in low-pass and high-pass filters, oscillators, and other electronic applications.
Step 1: Gather components and tools: The first step is to gather all the necessary components and tools needed to make the PCB. These include the circuit board, components, resistors, capacitors, ICs, and so on.Step 2: Design the circuit board : The next step is to design the circuit board. This can be done using specialized software such as Eagle, KiCAD, or Altium Designer. The design should include the PID controller and the first-order RC circuit. The software will generate a schematic diagram, which can be used to create a layout for the PCB.Step 3: Create the PCB layout: Once the schematic is designed, the next step is to create the PCB layout. This involves placing the components on the board and routing the connections between them. The layout should be optimized to reduce noise and other interference.Step 4: Etch the PCB: When the layout is complete, the next step is to etch the PCB. This is done by transferring the layout onto a copper-clad board and using a chemical solution to remove the unwanted copper. This leaves behind the circuit traces and pads needed to mount the components.Step 5: Solder the components: The final step is to solder the components onto the PCB. This requires a steady hand and a soldering iron.Know more about the PID controller
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Part-B (Fluid mechanics) Question 4 (a) A steady, two-dimensional, incompressible flow field in the xy-plane has a stream function given by = ax3 + by + cx, where a, b, and c are constants: a = 0.5(m.s)-1, b = -2.0 m/s, and c = -1.5 m/s. = == (i) Obtain expressions for velocity components u and v. (4 marks) (ii) Verify that the flow field satisfies the incompressible continuity equation. (4 marks) (iii) The velocity potential (o). (4 marks)
(i) Velocity components u and v:It is known that the velocity components u and v can be determined from the stream function as follows: u = ∂Ψ / ∂y; v = - ∂Ψ / ∂x
Where Ψ = ax3 + by + cx, we have the following:
u = ∂Ψ / ∂y
= b
= -2.0 m/s
(since there is no y-term in Ψ)andv = - ∂Ψ / ∂x = -3ax2 + c= -3(0.5)(x)2 - 1.5 m/s
(ii) Incompressible continuity equation verification:The incompressible continuity equation states that the sum of partial derivatives of u, v, and w with respect to x, y, and z, respectively is zero: ∂u / ∂x + ∂v / ∂y + ∂w / ∂z = 0Since there is no z component and the flow is two-dimensional, the above equation can be written as follows: ∂u / ∂x + ∂v / ∂y = 0
Substituting the expressions for u and v we get: ∂u / ∂x + ∂v / ∂y = ∂(-3ax2 + c) / ∂x + ∂b / ∂y
= 0 + 0
= 0
Hence the flow satisfies the incompressible continuity equation.(iii) The velocity potential o:In an irrotational flow, the velocity components can be derived from a velocity potential function such that u = ∂φ / ∂x and
v = ∂φ / ∂y.
Since the flow in this case is incompressible, it is also irrotational. Therefore, we can find the velocity potential φ by integrating the velocity components: u = ∂φ / ∂x
⇒ φ = ∫ u dx + f(y) v
= ∂φ / ∂y
⇒ φ = ∫ v dy + g(x)
Comparing these expressions, we get: ∫ u dx + f(y) = ∫ v dy + g(x)
The left-hand side of this equation can be expressed as follows: ∫ u dx + f(y) = ∫ (-3ax2 + c) dx + f(y)
= -ax3 + cx + f(y)
Similarly, the right-hand side can be expressed as: ∫ v dy + g(x) = ∫ b dy + g(x) = by + g(x)
Comparing the two expressions, we get:-ax3 + cx + f(y) = by + g(x)Differentiating with respect to x, we get: g'(x) = c; Integrating we get g(x) = cx + k1, where k1 is a constant Differentiating with respect to y, we get:f'(y) = b; Integrating we get f(y) = by + k2, where k2 is a constant. Substituting these values in the previous equation, we get:-ax3 + cx + by + k1 = by + cx + k2. Therefore, k1 = k2 = 0The velocity potential is given by: φ = -ax3 / 3 + cx Thus, the velocity potential (o) is -ax3 / 3 + cx.
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The linear burning rate of a solid propellant restricted burning grain is 20 mm/s when the chamber pressure is 80 bar and 40 mm/s when the chamber pressure is 200 bar. determine (i) the chamber pressure that gives a linear burning rate of 30 mm/s (ii) the propellant consumption rate in kg/s if the density of the propellant is 2000 kg/m3, grain diameter is 200 mm and combustion pressure is 100 bar.
(i) To determine the chamber pressure that gives a linear burning rate of 30 mm/s, we can use the concept of proportionality between burning rate and chamber pressure. By setting up a proportion based on the given data, we can find the desired chamber pressure.
(ii) To calculate the propellant consumption rate, we need to consider the burning surface area of the grain, the linear burning rate, and the density of the propellant. By multiplying these values, we can determine the propellant consumption rate in kg/s.
Let's calculate these values:
(i) Using the given data, we can set up a proportion to find the chamber pressure (P) for a linear burning rate (R) of 30 mm/s:
(80 bar) / (20 mm/s) = (P) / (30 mm/s)
Cross-multiplying, we get:
P = (80 bar) * (30 mm/s) / (20 mm/s)
P = 120 bar
Therefore, the chamber pressure that gives a linear burning rate of 30 mm/s is 120 bar.
(ii) The burning surface area (A) of the grain can be calculated using the formula:
A = π * (diameter/2)^2
A = π * (200 mm / 2)^2
A = π * (100 mm)^2
A = 31415.93 mm^2
To calculate the propellant consumption rate (C), we can use the formula:
C = A * R * ρ
where R is the linear burning rate and ρ is the density of the propellant.
C = (31415.93 mm^2) * (30 mm/s) * (2000 kg/m^3)
C = 188,495,800 mm^3/s
C = 0.1885 kg/s
Therefore, the propellant consumption rate is 0.1885 kg/s if the density of the propellant is 2000 kg/m^3, the grain diameter is 200 mm, and the combustion pressure is 100 bar.
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Use Matlab to create the required Bode plots. 1) Design a lead compensator for the system below. The ramp error constant should be K) = 20 and the phase margin should be greater than or equal to 50°. Hand in your uncompensated Bode plot and your compensated Bode plot.
G(s) = 4/s(s+2)
solution
G(s) = 40.16 s+4.39/s+17.64
To make the Bode plots for the given system using MATLAB as well as the design a lead compensator, one can use the code given below
What is the MATLAB?MATLAB is a computer program made for scientists and engineers to study and design things that help make the world better. MATLAB's main component is its language, which is based on matrices and allows for easy expression of mathematical computations.
Therefore, the computer program tends to creates the G_uncompensated transfer function using the special numbers. After that, it creates a graph called the Bode plot using a tool called the bode function. It also gives the graph a name.
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Which of the following devices typically makes greater use of tunneling? (a) Field effect transistor (b) Diode (c) Flash memory
Tunneling is the movement of charged particles or objects through a potential barrier or energy barrier that they would normally be unable to surmount. Tunneling is employed by several electronic devices, especially in solid-state devices such as diodes, flash memories, and field-effect transistors.
It has a tunnel oxide that allows electrons to pass from the channel through the oxide to the floating gate. Diodes, on the other hand, only require a small amount of tunneling in reverse bias. As a result, diodes have a limited tunneling effect.
The flow of electrons across a p-n junction is a significant aspect of diodes. Electrons flow from the n-type region to the p-type region, or vice versa, depending on the polarity. As a result, the correct response is: Flash memory.
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A composite material for a car-repair kit consists of a random mixture of short
glass fibers in a polyester matrix. Estimate the maximum toughness Gc of the
composite. You may assume that the volume fraction of glass is 30%, the fiber
diameter is 15 μm, the fracture strength of the fibers is 1400 MN m^-2, and the
shear strength of the matrix is 30 MN m^-2.
Calculate the critical length 2xc of the fibers above. How would you
expect Gc to change if the fibers were substantially longer than 2xc?
The maximum toughness Gc of the composite material can be estimated by considering the volume fraction of glass, fiber diameter, fracture strength of the fibers, and shear strength of the matrix. To calculate the critical length 2xc of the fibers, we need to determine the aspect ratio of the fibers and its impact on the composite's toughness.
The aspect ratio of the fibers is determined by dividing the fiber length by its diameter.
In this case, the critical length 2xc is the maximum length at which the fibers can still contribute to the toughness of the composite.
When the fibers are longer than 2xc, they may start to behave as individual fibers rather than reinforcing elements within the matrix.
To estimate Gc, we need to consider the load-carrying capacity and the energy required for crack propagation.
Longer fibers can potentially enhance the load-carrying capacity and toughness of the composite as they can bridge and distribute the applied load more effectively.
However, if the fibers become too long, they may also introduce stress concentration points, leading to reduced toughness.
To assess the change in Gc when the fibers are substantially longer than 2xc, further analysis is required.
It is possible that Gc might increase initially due to improved load transfer, but beyond a certain length, Gc could decrease due to increased stress concentration and reduced interfacial bonding between the fibers and the matrix.
In summary, estimating Gc involves considering the volume fraction of glass, fiber properties, and matrix properties.
The critical length 2xc of the fibers determines the maximum length at which they can contribute to the composite's toughness.
Understanding the relationship between fiber length and Gc is crucial to optimize the composite's performance.
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A steady, incompressible, two dimensional velocity field is given by the following components in the xy plane: u=1.85+2.33x+0.656y v=0.757-2.18x-2.33y According to the given information which term cannot be cancelled during the calculation of the acceleration field?
A. Acceleration in z direction
B. Partial derivative of u with respect to z
C. Partial derivative of v with respect to z
D. Partial derivative of u with respect to time
E. Partial derivative of u with respect to x
According to the given information, the term that cannot be cancelled during the calculation of the acceleration field is the partial derivative of u with respect to time.
The velocity field is given by the components u and v in the xy plane. To calculate the acceleration field, we need to take the derivatives of the velocity components with respect to time and spatial variables.
The acceleration field can be expressed as:
a = (∂u/∂t) + u(∂u/∂x) + v(∂u/∂y) + (∂v/∂t) + u(∂v/∂x) + v(∂v/∂y)
When evaluating this expression, each term can be cancelled if it equals zero or is independent of the variable being differentiated.
In the given information, there is no mention of the z-coordinate or the partial derivatives with respect to z. Therefore, the term involving the acceleration in the z direction (A) and the partial derivatives of u and v with respect to z (B and C) are not relevant and can be cancelled.
However, the partial derivative of u with respect to time (D) is not explicitly given or mentioned in the given information. Since it is not specified that ∂u/∂t equals zero or is independent of time, this term cannot be cancelled during the calculation of the acceleration field. Therefore, the term that cannot be cancelled is the partial derivative of u with respect to time (D. Partial derivative of u with respect to time).
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A steam power plant operates on an ideal reheat-regenerative Rankine cycle and has a net power output of 80 MW. Steam enters the high-pressure turbine at 10 MPa and 550°C and leaves at 0.8 MPa. Some steam is extracted at this pressure to heat the feed water in an open feed water heater. The rest of the steam is reheated to 500°C and is expanded in the low pressure turbine to the condenser pressure of 10 kPa. Show the cycle on a T- s diagram with respect to saturation lines, and determine (a) the mass flow rate of steam through the boiler and (40 Marks) (b) the thermal efficiency of the cycle.
To solve the problem, we need to show the cycle on a T-s diagram using saturation lines and determine the mass flow rate of steam through the boiler and the thermal efficiency of the cycle.
The reheat-regenerative Rankine cycle is commonly used in steam power plants to improve the overall efficiency. In this cycle, steam enters the high-pressure turbine and expands, producing work. After this expansion, some steam is extracted at an intermediate pressure and used to heat the feed water in an open feed water heater. This extraction process helps increase the efficiency of the cycle by utilizing the remaining heat in the extracted steam.
The remaining steam is then reheated to a higher temperature before entering the low-pressure turbine for further expansion. Finally, the steam is condensed in the condenser, and the condensed water is pumped back to the boiler to restart the cycle. By using these processes, the cycle can maximize the utilization of heat and improve the overall efficiency of the power plant.
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a) An amplifier system without feedback has the following specifications: Open loop gain: 90 Input impedance: 25kΩ Output impedance: 5kΩ (i) If the amplifier system employs negative feedback and the close loop gain is 9.5, calculate the system feedback factor, β. (ii) Suppose the negative feedback topology used for the amplifier system in Q3(a)(i) is a current shunt feedback, determine the amplifier, input impedance and output impedance of the amplifier with feedback.
Therefore, the feedback amplifier has an input impedance of 2.09 kΩ and an output impedance of 184.3 Ω.
A feedback amplifier is a type of electronic amplifier that utilizes feedback to regulate the response of the amplifier. This method, also known as negative feedback, entails feeding some of the output back to the input in a phase-reversed form. The fundamental principle is to decrease the gain of the amplifier to a reasonable value while maintaining stability and decreasing distortion.
In an amplifier system without feedback, the open loop gain is 90, input impedance is 25 kΩ, and output impedance is 5 kΩ.
The close loop gain of the amplifier is 9.5.
The feedback factor β of the amplifier can be determined as follows:
β = A / (1 + AB)
Here, A is the open-loop gain, and B is the feedback factor.
β = 90 / (1 + 90 * (9.5 - 1))
= 0.0836 (or 8.36%)
To find out the input impedance of the feedback amplifier, the input impedance of the original amplifier must be multiplied by the feedback factor.
Rin = β * R
= 0.0836 * 25 kΩ
= 2.09 kΩ
The output impedance of the feedback amplifier can be calculated using the following formula:
Rout = R / (1 + AB)
= 5 kΩ / (1 + 90 * (9.5 - 1))
= 184.3 Ω
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Question 1 Not yet answered Marked out of 4.00 A Proportional-Derivative (PD) controller may reduce the stability of the system. Select one: O True O False
Proportional-Derivative (PD) controller is one of the most commonly used types of controllers in control theory. It provides excellent accuracy in controlling the system, but it may reduce the stability of the system when the controller is not set correctly. So, the given statement is True.
In general, a PD controller is designed to provide faster response to changes in error and to reduce the steady-state error. However, in some cases, a PD controller may be too sensitive to changes in error and produce unstable responses. This instability is caused by the derivative term, which amplifies high-frequency noise in the error signal. As a result, the system may oscillate or even become unstable. To overcome this, it is important to tune the controller gains carefully. A good controller tuning will ensure that the controller responds optimally to changes in error while maintaining stability.
This is usually done using various methods such as Ziegler-Nichols method, Cohen-Coon method, and many more. In conclusion, a PD controller can reduce the stability of the system if not tuned correctly.
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In an orthogonal cutting operation in tuning, the cutting force and thrust force have been measured to be 300 lb and 250 lb, respectively. The rake angle = 10°, width of cut = 0.200 in, the feed is 0.015in/rev, and chip thickness after separation is 0.0375. Determine the shear strength of the work material.
The shear strength of the work material can be determined using the following equation:
Shear strength = Cutting force / (Width of cut × Chip thickness)
By analyzing the forces and using appropriate equations, the shear strength of the work material can be calculated.
In an orthogonal cutting operation, the cutting force and thrust force are measured to be 300 lb and 250 lb, respectively. The rake angle is given as 10°, the width of cut is 0.200 in, the feed rate is 0.015 in/rev, and the chip thickness after separation is 0.0375 in.
Substituting the given values, we have:
Shear strength = 300 lb / (0.200 in × 0.0375 in)
By performing the calculation, the shear strength of the work material can be obtained in the appropriate units. It's important to note that the shear strength of the work material is a measure of its resistance to shear deformation during the cutting process. By determining this value, machinists and engineers can assess the suitability of the material for specific cutting operations and make informed decisions regarding tool selection, cutting parameters, and overall process optimization.
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The Master Productiom Schedule is an aggregated production plan developed during the SOP process O True False
The given statement "The Master Production Schedule is an aggregated production plan developed during the SOP process" is True.
The Master Production Schedule (MPS) is a collection of data that organizes manufacturing plans for a particular period of time. The MPS consists of a list of all of the goods that are planned to be manufactured, as well as the dates on which they are planned to be manufactured.
The MPS is used to guarantee that there are no significant delays in the production process and that manufacturing and inventory costs are minimized. The MPS is essential because it enables planners to adjust their schedules, materials, and resources to suit current market demand and modifications to the supply chain.
The MPS is developed as part of the Sales and Operations Planning (SOP) process.
The SOP is a periodic process that brings together all aspects of the firm, including production, finance, sales, and marketing, to agree on a unified plan for the future.
As a result, the MPS is generated at the conclusion of the SOP procedure and is influenced by the overall business plan, market predictions, and any resource or capacity limitations that were identified throughout the SOP process.
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QUESTION 2 Copy of A power plant was analysed based on Carnot cycle, the evaporator supplie 459 kJ of heat per cycle at 442°C and condenser cooling fluid was maintained at 33°C. Determine the amount of heat rejected in the condenser? Provide answer to no decimal place and insert the unit symbol in kilo.... 1 °C is 273K in this question if required.
A load is mounted on a spring with a spring constant of 324Nm^(-1) and confined to move only vertically, as shown in Figure 3. The wheels which guide the mass can be considered to be frictionless.
The load has a mass, m=4kg, which includes a motor causing the mass to be driven by a force, F = 8 sin wt given in newtons.
Write the inhomogeneous differential equation that describes the system above. Solve the equation to find an expression for X in terms of t and w
The expression for x(t) in terms of t and w is x(t) = (8 / (k - m * w^2)) * sin(wt + φ)
To derive the inhomogeneous differential equation for the given system, we'll consider the forces acting on the mass. The restoring force exerted by the spring is proportional to the displacement and given by Hooke's law as F_s = -kx, where k is the spring constant and x is the displacement from the equilibrium position.
The force due to the motor is given as F = 8 sin(wt).
Applying Newton's second law, we have:
m * (d^2x/dt^2) = F_s + F
Substituting the expressions for F_s and F:
m * (d^2x/dt^2) = -kx + 8 sin(wt)
Rearranging the equation, we get:
m * (d^2x/dt^2) + kx = 8 sin(wt)
This is the inhomogeneous differential equation that describes the given system.
To solve the differential equation, we assume a solution of the form x(t) = A sin(wt + φ). Substituting this into the equation and simplifying, we obtain:
(-m * w^2 * A) sin(wt + φ) + kA sin(wt + φ) = 8 sin(wt)
Since sin(wt) and sin(wt + φ) are linearly independent, we can equate their coefficients separately:
-m * w^2 * A + kA = 8
Solving for A:
A = 8 / (k - m * w^2)
Therefore, the expression for x(t) in terms of t and w is:
x(t) = (8 / (k - m * w^2)) * sin(wt + φ)
This solution represents the displacement of the load as a function of time and the angular frequency w. The phase constant φ depends on the initial conditions of the system.
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Thermal power plants operating on a Rankine Cycle reject considerable quantities of heat to a cooling system via a condenser. If the cooling medium is water in an open loop with the environment it can cause significant thermal pollution of a river or lake at the point of discharge. Consider (0) a CANDU Nuclear Plant, and (ii) a Coal Fired Fossil Plant each of 1000 MW electrical output..
Determine the total rate of heat discharge in the cooling water for each.
A thermal power plant that operates on a Rankine cycle discharges significant amounts of heat to a cooling system through a condenser. If water is used as the cooling medium in an open-loop system with the environment, it may cause substantial thermal pollution of a river or lake at the point of discharge.
The overall rate of heat discharge in the cooling water for each of a CANDU nuclear plant and a coal-fired fossil plant with an electrical output of 1000 MW is given below:CANDU Nuclear PlantIn a CANDU (Canadian Deuterium Uranium) nuclear reactor, the coolant (heavy water) is driven by the heat generated by nuclear fission, and the heat is transferred to water in a separate loop, which generates steam and powers the turbine to generate electricity.The CANDU reactor uses heavy water (deuterium oxide) as a moderator and coolant, which flows through 380 fuel channels in a horizontal pressure tube. The water flows through the core, absorbs heat from the fuel, and then transfers it to a heat exchanger. The heat is then transferred to steam, which drives the turbine to produce electricity.
A 1000 MW electrical output CANDU nuclear plant has a total rate of heat discharge of 2.5 x 10¹³ J/h in the cooling water. Coal-Fired Fossil Plant A coal-fired power plant generates electricity by burning pulverized coal to heat a water-filled boiler to produce steam, which then drives a turbine to generate electricity. The flue gases are discharged to the atmosphere via a stack. Water is used to cool the steam in the condenser. The water used for cooling is discharged into the environment after the heat from the steam is extracted .A 1000 MW electrical output coal-fired fossil plant has a total rate of heat discharge of 2.7 x 10¹⁴ J/h in the cooling water.
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A three-phase 440-V, 51-kW, 60-kVA inductive load operates at 60 Hz and is Y-connected. It is desired to correct the power factor to 0.95 lagging. What value of capacitor should be placed in parallel with each load impedance?
To correct the power factor to 0.95 lagging, a capacitor of approximately 18.75 kVAR should be placed in parallel with each load impedance.
To correct the power factor of the inductive load, we need to add a capacitor in parallel to provide reactive power to offset the reactive power of the load. The reactive power (Q) can be calculated using the power factor (PF) and the apparent power (S).
Given:
Voltage (V) = 440 V
Power (P) = 51 kW
Apparent power (S) = 60 kVA
Power factor (PF) = 0.95 lagging
The reactive power can be calculated using the formula:
Q = S * sqrt(1 - PF^2)
Q = 60 kVA * sqrt(1 - 0.95^2)
Q = 60 kVA * sqrt(1 - 0.9025)
Q = 60 kVA * sqrt(0.0975)
Q = 60 kVA * 0.3125
Q = 18.75 kVAR
Now, we can calculate the required capacitance (C) using the formula:
C = Q / (2 * π * f * V^2)
Where:
f = Frequency = 60 Hz
V = Voltage = 440 V
C = 18.75 kVAR / (2 * π * 60 Hz * (440 V)^2)
C ≈ 18.75 * 10^3 / (2 * π * 60 * (440)^2) Farads
Calculating this value will give you the required capacitance in Farads that should be placed in parallel with each load impedance to correct the power factor to 0.95 lagging.
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) A symmetrical compound curve consists of left transition (L-120m), circular transition (R=340m), and right transition curve. Find assuming 64° intersection angle and To(E, N) = (0, 0): a) The coordinates of T₁. b) The deflection angle and distance needed to set T2 from T1. c) The coordinates of T2. (4%) (6%) (4%) 3) Given: a mass diagram as shown below with 0.85 grading factor applied to cut
A symmetrical compound curve is made up of a left transition curve, a circular transition curve, and a right transition curve. Given the intersection angle of 64 degrees and a point To(E,N)=(0,0), the coordinates of T1, the deflection angle, and distance needed to set T2 from T1, as well as the coordinates of T2, are to be found
To find the coordinates of T1, we first need to calculate the length of the circular curve and the lengths of both the transition curves. Lt = 120 m (length of left transition curve)
To find the deflection angle and distance needed to set T2 from T1, we first need to calculate the length of the right transition curve. Lt = 120 m (length of left transition curve)
Lr = 5.94 m (length of the circular curve)
Ln = Lt + Lr (total length of left transition curve and circular curve)
Ln = 120 + 5.94
= 125.94 mRr
= 340 m (radius of the circular curve)γ
= 74.34 degrees (central angle of the circular curve)y
= 223.4 m (ordinate of the circular curve).
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A small orifice 0.0003 cubic meter in area is in the vertical side of a rectangular tank. The horizontal sectional area of the tank is 0.40 cubic meter. At a given instant the head on the orifice is 1.2 m and 312 seconds later it ia 0.60m. Using C=0.60, ehat will be the height of water from 1.2m after a lapse of 240 seconds?
The height of water after 240 seconds will be 2.91 meters.
The formula for discharging water through an orifice is given by;
Q= CdA √2gh
Where, Q= flow of water
C= co-efficient of discharge
A= area of orifice
g= acceleration due to gravity
h= height of water above the orifice
Height of water above the orifice (h1) = 1.2 m
Height of water above the orifice (h2) = 0.6 m
Time taken (t) = 312 s
Coefficient of discharge (C) = 0.6
Area of orifice (A) = 0.0003 cubic meter
Sectional area of the tank (a) = 0.4 cubic meter
Time after which we need to find the height of water (T) = 240 seconds
Now, Let’s calculate the flow rate of water through the orifice
Initial flow rate, Q1 = CdA √2gh1Q1 = 0.6 × 0.0003 × √2 × 9.81 × 1.2
Q1 = 0.00191 cubic meters per second
Final flow rate,
Q2 = CdA √2gh2Q2 = 0.6 × 0.0003 × √2 × 9.81 × 0.6
Q2 = 0.00116 cubic meters per second
Let the height of water after time T be h3
Therefore, the final flow rate of water through the orifice is;
Q3 = CdA √2gh3Q3 = 0.6 × 0.0003 × √2 × 9.81 × h3
From the formula of continuity; Q1 = Q2 = Q3
Since Q1 = 0.00191 cubic meters per second and Q2 = 0.00116 cubic meters per second
Q3 = 0.00116 cubic meters per second
h3 = (Q1/Q3)² × h1h3 = (0.00191/0.00116)² × 1.2h3 = 2.91 meters
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The characteristic equation of a system is S⁴ +25³ +25² +3S+ K = 0 Determine the range of the parameter K such that the system is stable.
The range of the parameter K for system stability is K > -125.
To determine the stability of the system, we need to analyze the characteristic equation. The characteristic equation of the system is given as S⁴ + 25³ + 25² + 3S + K = 0. Stability of a system is determined by the roots of its characteristic equation.
For the system to be stable, all the roots of the characteristic equation must have negative real parts. In this case, the system has a quartic characteristic equation, and we need to consider the coefficients and the parameter K.
The coefficient of the S⁴ term is 1, which implies that the system has a leading coefficient of 1, indicating the presence of a stable pole at the origin. The coefficient of the S³ term is 25³, the coefficient of the S² term is 25², and the coefficient of the S term is 3. These coefficients alone do not affect the stability of the system.
The parameter K plays a crucial role in determining stability. For the system to be stable, the values of K should be such that all the roots of the characteristic equation have negative real parts. To achieve this, we can analyze the value of K in relation to the other coefficients.
Since K is a constant term, it does not affect the real parts of the roots. However, to maintain stability, the value of K should be chosen in such a way that it does not cause any roots to have positive real parts. Therefore, for stability, K must be greater than the sum of the coefficients of the S term and the constant term, which is -125. Hence, the range of the parameter K for system stability is K > -125.
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A heat pump is operating based on a thermodynamic cycle with processes following the sequence of: i. Isothermal compression followed by, ii. Adiabatic expansion and next, iii. Isobaric expansion to return to the initial state. a. Sketch the PV diagram of this thermodynamic cycle. In your diagram, indicate the direction of the cycle, total work and the sign of the total work. Does the cycle absorb or release heat? b. Redraw the cycle in a TV-diagram. Indicate its direction and name all processes. The slope and curvature of all process curves must be quantitatively correct. To achieve this, write the TV relationship for the individual processes. c. Determine the coefficient of performance of this heat pump, given that: COP=IQout/lWinI The isothermal process has pressure ratio of 5 and the working fluid can be treated as monoatomic ideal gas. (Note that this is not a Carnot cycle. Note further that the solution of this problem requires you to first develop and simplify an equation for the COP before you can proceed with any calculation of values). d. The heat pump is used to keep a house at a temperature of 20°C using water river (5°C) as the heat source. The heat pump requires 10kW of energy to achieve the heating requirement. Find the total entropy change of this heating process. Determine if this process reversible or irreversible? e. Demonstrate the heat exchange between the cycle and the thermal reservoirs in a TS diagram. Briefly explain your arguments to support your findings in part(d).
a) The PV diagram of the thermodynamic cycle is sketched, indicating the direction of the cycle, total work, and the sign of the total work. The cycle absorbs or releases heat.
b) The TV diagram of the cycle is redrawn, indicating the direction and naming all processes. The process curves have quantitatively correct slopes and curvatures with the TV relationship for each process.
c) The coefficient of performance (COP) of the heat pump is determined using a simplified equation. The isothermal process pressure ratio and the assumption of a monoatomic ideal gas are considered.
a) The PV diagram of the thermodynamic cycle consists of three processes: isothermal compression, adiabatic expansion, and isobaric expansion. The cycle is shown in a clockwise direction. The total work is represented by the area enclosed by the cycle, and its sign depends on whether the work is done by the system or on the system. The cycle either absorbs or releases heat, depending on the direction of heat transfer during each process.
b) The TV diagram is redrawn to illustrate the cycle. The processes are named according to their characteristics. The isothermal compression process is represented by a horizontal line, the adiabatic expansion process by a steep curve, and the isobaric expansion process by a vertical line. The slopes and curvatures of the process curves are quantitatively correct, reflecting the specific relationships between temperature and volume for each process.
c) To determine the coefficient of performance (COP) of the heat pump, the equation COP = Q_out / W_in is used. However, an equation for COP must be developed and simplified before any calculations can be made. The given information specifies a pressure ratio for the isothermal process and assumes an ideal monoatomic gas as the working fluid.
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Air is compressed steadily by a 15-kW compressor from 100 kPa and 17°C to a final state of 600 kPa and 57°C at a rate of 5 kg/min. During this process, some heat transfer takes place between the compressor and the surrounding medium at 17°C. Determine the entropy generated during this process per unit mass in kJ/kg K. Assume constant specific heats at an average temperature.
To determine the entropy generated per unit mass during the compression process, we need to calculate the change in entropy using the given information of the compressor's operation.
The change in entropy (ΔS) can be calculated using the equation ΔS = ∫(δQ / T), where δQ is the heat transfer and T is the temperature. Since the process is assumed to be steady and the specific heats are assumed constant, we can simplify the equation to ΔS = cp * ln(T2/T1) - R * ln(P2/P1), where cp is the specific heat at constant pressure and R is the specific gas constant.
Given:
Initial pressure (P1) = 100 kPa
Initial temperature (T1) = 17°C = 17 + 273 = 290 K
Final pressure (P2) = 600 kPa
Final temperature (T2) = 57°C = 57 + 273 = 330 K
Power input to the compressor (W) = 15 kW
Mass flow rate (m_dot) = 5 kg/min
First, we need to calculate the change in specific entropy (Δs) using the equation Δs = cp * ln(T2/T1) - R * ln(P2/P1). The specific heat cp can be determined using the average temperature, which is (T1 + T2) / 2. Next, we calculate the total entropy generated (ΔS_total) by multiplying the change in specific entropy (Δs) by the mass flow rate (m_dot) and the specific heat (cp). Finally, we divide the total entropy generated by the mass flow rate (m_dot) to obtain the entropy generated per unit mass. By performing these calculations, we can determine the entropy generated during the compression process per unit mass in kJ/kg K.
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installation time and eliminates the need for mounting on many types of roofs; the financing and installation of industrial large-scale solar (for example the British Motor Company consisting of 1 606 photovoltaic panels), and the installation of electric vehicle charging stations, in partnership with Rabobank. In the first quarter of 2016, the CEO of SolarCity, Lyndon Rive, reported to shareholders that targets for most initiatives have been exceeded. In terms of MW installed, the target has been exceeded by 19%, the revenue grew by 82% year on year, and $728 million in asset financing has been raised year on year. Furthermore, the CEO also reported on a milestone achieved: the opening of a 100 MW cell-producing plant that will produce cells that exceed the average efficiency by 20%. In addition, with the use of the Zep mounting system, the number of component parts needed to install a module has been reduced from eight to three, the installation time of 16 modules (4 KW) has been reduced to 20 minutes, and the risk of damaging roofs has been reduced as well. To accomplish the mission of SolarCity, teams have to constantly work to enhance scale and efficiency, and tirelessly tear down external barriers to solar. SolarCity has been structured to functionally focus on overcoming each potential barrier in the path of higher penetration of distributed solar: the sales/operations focus is on lowering the upfront costs of solar; the finance teams focus on reducing the cost of capital; research and development (R&D) focus on developing new technology that reduces costs; and the government affairs team works to promote a regulatory policy that supports distributed solar. Policy is the most important initiative to drive higher distributed solar, and small wins have been achieved on the policy front despite the efforts of utility groups to undermine the economics of solar. These wins include energy regulators rejecting fixed charges that penalise distributed solar customers in Mexico, Colorado, Minnesota and Kansas (US). In order to meet the long-term projections, SolarCity has to expand its workforce, increase installation efficiency and exceed the current rate of growth in MW installed. However, there are implementation risks that may lead to a difference between what is projected and actually realised. These risks include: the future level of demand for their solar systems; the availability and timely supply of cost-effective solar panels; the ability to integrate new businesses acquired into existing operations; the effect of changes in rebates/taxes; changes in strategic planning decisions; and the reallocation of internal resources. Overall, solar at the scale envisioned by SolarCity will lead to more competition, as opposed to protecting monopolies, and innovative solutions to meet the world's clean energy needs. Question : Comment on how changes in macro-environment forces affected and influenced SolarCity's decision making in the organisatio
SolarCity is a company that has gained significant importance in the solar energy sector. The company operates to overcome each possible obstacle in the path of higher penetration of distributed solar. SolarCity has been positively affected by changes in macro-environment forces.
Below are some of the points that provide insights on how changes in macro-environment forces affected and influenced SolarCity's decision making in the organization: Policy - Policy is the most important initiative to drive higher distributed solar.
The government affairs team works to promote a regulatory policy that supports distributed solar. Policy changes have a direct impact on the financials of the company.
As a result, SolarCity is impacted by changes in policy. Small wins have been achieved on the policy front despite the efforts of utility groups to undermine the economics of solar.
Technology - The company is driven by the continuous development of new technology that reduces costs. The R&D team focuses on developing new technology that reduces costs.
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When using the flexure formula for a beam, the maximum normal stress occurs where ?
Group of answer choices
A. at a point on the cross-sectional area farthest away from the neutral axis
B. at a point on the cross-sectional area closest to the neutral axis
C. right on the neutral axis
D. halfway between the neutral axis and the edge of the beam
The maximum normal stress occurs at a point on the cross-sectional area farthest away from the neutral axis.
Option A is correct. When a beam is subjected to bending, the top fibers of the beam are compressed while the bottom fibers are stretched. The neutral axis is the location within the beam where there is no change in length during bending. As we move away from the neutral axis, the distance between the fibers increases, leading to higher strains and stresses. Therefore, the point on the cross-sectional area farthest away from the neutral axis experiences the maximum normal stress. This is important to consider when analyzing the structural integrity and strength of beams under bending loads.
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A reciprocating compressor draws in 500 cubic feet per minute of air whose density is 0.079lb/cu ft and discharges it with a density of 0.304lb/ cu ft. At suction, p1=15psia; at discharge , p 2 = 80 psia. The increase in the specific internal energy is 33.8 Btu and the heat transferred from the air by cooling is 13Btu/lb. Determine the work on the air in Btu/min and in hp. Neglect change in kinetic energy.
The work on the air is approximately 22.24 Btu/min and 0.037 hp.
To determine the work on the air in Btu/min and in horsepower (hp), we can use the following equations and steps:
1. Calculate the mass flow rate (m_dot) of air using the given volumetric flow rate (Q_dot) and air density (ρ):
m_dot = Q_dot * ρ
Here, Q_dot = 500 cubic feet per minute and ρ = 0.079 lb/cu ft.
Substituting these values, we get:
m_dot = 500 * 0.079 = 39.5 lb/min
2. Determine the change in specific internal energy (Δu) using the given increase in specific internal energy (Δu_in) and mass flow rate (m_dot):
Δu = Δu_in * m_dot
Here, Δu_in = 33.8 Btu and m_dot = 39.5 lb/min.
Substituting these values, we get:
Δu = 33.8 * 39.5 = 1334.3 Btu/min
3. Calculate the work done on the air (W_dot) using the change in specific internal energy (Δu) and mass flow rate (m_dot):
W_dot = Δu / 60
Since the given units are in Btu/min, we divide by 60 to convert it to Btu/s.
Substituting the value of Δu, we get:
W_dot = 1334.3 / 60 = 22.24 Btu/s
4. Convert the work done to horsepower (hp):
1 hp = 550 ft-lbf/s
1 Btu/s = 778 ft-lbf/s
W_hp = W_dot / (778 * 550)
Substituting the value of W_dot, we get:
W_hp = 22.24 / (778 * 550) = 0.037 hp
Therefore, the work on the air is approximately 22.24 Btu/min and 0.037 hp.
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1 22 Problem 4: Revolute-Prismatic Manipulator (25 points). Consider the two-link manipulator shown in 0 0 Fig. 4 with di 0. Link 1 has an inertia tensor given by о ту о and mass mi. Assume that link 2 0 01). has all its mass, m2, located at a point at the end-effector. Derive the dynamic equations for the manipulator. Assume that gravity is directed along –zo. Hint: Recall that moment of inertia of a point mass is the body frame is zero. ] d2 21 02 01 22 21 YY1 22 Y Y2 De di 20 Yo 00 To
The dynamic equations for the given two-link manipulator can be derived by considering the inertia tensors, masses, and the location of the mass at the end-effector of link 2.
To derive the dynamic equations for the two-link manipulator, we need to consider the kinetic and potential energy of the system. The kinetic energy is determined by the motion of the manipulator, while the potential energy is influenced by the gravitational force.
In this case, we have two links in the manipulator. Link 1 has an inertia tensor given by о ту о and a mass m1. Link 2 has all its mass, m2, located at the end-effector point. To derive the dynamic equations, we need to compute the Lagrangian, which is the difference between the kinetic and potential energy of the system.
The Lagrangian of the system can be expressed as:
L = T - V,
where T represents the total kinetic energy and V represents the total potential energy.
The kinetic energy T can be calculated as the sum of the kinetic energies of each link. For link 1, the kinetic energy is given by:
T1 = 0.5 * m1 * v1^2 + 0.5 * w1^T * о * w1,
where v1 is the linear velocity of link 1 and w1 is the angular velocity of link 1.
Similarly, for link 2, since all its mass is located at the end-effector, the kinetic energy can be simplified as:
T2 = 0.5 * m2 * v2^2 + 0.5 * w2^T * о * w2,
where v2 is the linear velocity of the end-effector and w2 is the angular velocity of the end-effector.
The potential energy V is determined by the gravitational force acting on the system. Assuming gravity is directed along –zo, the potential energy can be written as:
V = (m1 * g * r1z) + (m2 * g * r2z),
where g is the acceleration due to gravity and r1z and r2z are the z-components of the positions of the center of mass of link 1 and the end-effector, respectively.
By calculating the Lagrangian L = T - V and applying the Euler-Lagrange equations, we can derive the dynamic equations for the manipulator.
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In fatigue loading material fails a, below ultimate strength b. none c. above ultimate strength d. below yield strength
In fatigue loading, material fails below the ultimate strength.
Fatigue failure occurs when a material fails under repeated or cyclic loading below its ultimate strength. Fatigue failure is characterized by the accumulation of microcracks and damage, which eventually lead to failure, even though the applied stress levels are below the ultimate strength of the material. Fatigue failure is a time-dependent phenomenon and is influenced by factors such as stress amplitude, stress concentration, and the number of loading cycles.
Certain environmental conditions, such as high temperature, corrosive environments, or exposure to chemicals, can accelerate the fatigue crack growth rate and decrease the fatigue life of materials. Intrinsic material defects such as inclusions, voids, or impurities can act as stress raisers and contribute to fatigue failure. These defects can promote crack initiation and propagation, reducing the fatigue life of the material.
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STATICS OF RIGID BODIES
Please give at least 3 applications of Mechanics to your daily
life and depict the principle being applied in each
application.
Statics of rigid bodies is an integral part of Mechanics. This branch of physics is responsible for analyzing the forces and moments of objects that are at rest.
The importance of mechanics in our daily life can not be overemphasized as it has an endless list of practical applications. Here are three examples of how mechanics is applied in our daily life:
Bridges: Every time you walk on a bridge, you are witnessing an application of Mechanics. Bridges are structures that are designed to withstand forces acting upon them, such as the weight of vehicles and pedestrians that use them. In bridge engineering, the principles of statics of rigid bodies and material strength are utilized to ensure that the bridges are strong enough to support the loads they are subjected to.
This includes the selection of materials, such as concrete, steel, and wood, and the arrangement of structural elements to create a stable and durable structure. In conclusion, Mechanics is an important field that has practical applications in our daily life. Through the use of the principles of statics of rigid bodies and material strength, engineers can design structures and objects that are strong, safe and efficient.
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Using sketches, describe the carburisation process for steel
components?
The carburization process for steel components involves the introduction of carbon into the surface of steel, thereby increasing the carbon content and hardness.
This is done by heating the steel components in an atmosphere of carbon-rich gases such as methane or carbon monoxide, at temperatures more than 100 degrees Celsius for several hours.
Step 1: The steel components are placed in a carburizing furnace.
Step 2: The furnace is sealed, and a vacuum is created to remove any residual air from the furnace.
Step 3: The furnace is then filled with a carbon-rich atmosphere. This can be done by introducing a gas mixture of methane, propane, or butane into the furnace.
Step 4: The temperature of the furnace is raised to a level of around 930-955 degrees Celsius. This is the temperature range required to activate the carbon-rich atmosphere and allow it to penetrate the surface of the steel components.
Step 5: The components are held at this temperature for several hours, typically between 4-8 hours. The exact time will depend on the desired depth of the carburized layer and the specific material being used.
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Steam in a rigid tank is at a pressure of 400psia and a temperature of 600°F. As a result of heat transfer, the temperature decreases to 70°F. Determine the % of the total mass that is liquid in the final state, and the % of volume occupied by the liquid and vapor at the final state.
To determine the percentage of the total mass that is liquid in the final state and the percentage of volume occupied by the liquid and vapor at the final state, we need to use the steam tables to obtain the properties of steam at the given conditions.
First, we look up the properties of steam at the initial state of 400 psia and 600°F. From the steam tables, we find that at these conditions, steam is in a superheated state.
Next, we look up the properties of steam at the final state of 70°F. At this temperature, steam is in a compressed liquid state.
Using the steam tables, we find the specific volume (v) of steam at the initial and final states.
Now, to calculate the percentage of the total mass that is liquid in the final state, we can use the concept of quality (x), which is the mass fraction of the vapor phase.
The quality (x) can be calculated using the equation:
x = (v_final - v_f) / (v_g - v_f)
Where v_final is the specific volume of the final state, v_f is the specific volume of the saturated liquid at the final temperature, and v_g is the specific volume of the saturated vapor at the final temperature.
To calculate the percentage of volume occupied by the liquid and vapor at the final state, we can use the equation:
% Volume Liquid = x * 100
% Volume Vapor = (1 - x) * 100
Please note that the specific volume values and calculations depend on the specific properties of steam at the given conditions. It is recommended to refer to steam tables or use steam property software to obtain accurate values for the calculations.
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