(b) Given a system described by the following differential equation d²y(t) / dt² + dy(t) / dt + 3y(t) = x(t), dt where input x (t) = cos(t) u(t). Solve the above differential equation to obtain the following: (i) Homogeneous solution of the system. (5 marks) (ii) Particular solution of the system. (7 marks)
(i) Homogeneous solution of the system:
[tex]y_h(t) = c1 e^(-0.5t)cos(1.658t) + c2 e^(-0.5t)sin(1.658t)[/tex]
(ii) Particular solution of the system:
[tex]y(t) = y_h(t) + y_p(t)= e^(-t/2) (c1*cos(\sqrt(11)*t/2) + c2*sin(\sqrt(11)*t/2)) - (1/10)*cos(t) + (1/20)*sin(t)[/tex]
The given differential equation is:
d²y(t) / dt² + dy(t) / dt + 3y(t) = x(t)
Where x(t) = cos(t) u(t)
(i) Homogeneous solution of the system:
d²y(t) / dt² + dy(t) / dt + 3y(t) = 0
The characteristic equation of the above differential equation is:
r² + r + 3 = 0
r1 = -0.5 + 1.658i and r2 = -0.5 - 1.658i
Therefore, the homogeneous solution of the system is:
[tex]y_h(t) = c1 e^(-0.5t)cos(1.658t) + c2 e^(-0.5t)sin(1.658t)[/tex]
(ii) Particular solution of the system:
y_p(t) = A cos(t) + B sin(t)
d y_p(t) / dt = -A sin(t) + B cos(t)
d²y_p(t) / dt² = -A cos(t) - B sin(t)
(4A - B) cos(t) + (4B + A) sin(t) = cos(t)
Comparing the coefficients of cos(t) and sin(t), we get:
4A - B = 1 and 4B + A = 0
Solving the above two equations, we get:
A = -4/17 and B = -1/17
Hence, the general solution of the given system is:
[tex]y(t) = y_h(t) + y_p(t)= e^(-t/2) (c1*cos(\sqrt(11)*t/2) + c2*sin(\sqrt(11)*t/2)) - (1/10)*cos(t) + (1/20)*sin(t)[/tex]
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Q1. The field and armature resistance of a 500 V Dc series motor is 0.2ohm and 0.3ohm, respectively. The motor runs at 500rpm when drawing a current of 49 A. If the load torque varies as the square of the speed, determine the value of the external resistance to be added in series with the armature for motor to run at 450rpm. Assume linear magnetization.
We know that torque is proportional to square of speed i.e. T ∝ N².Hence, the external resistance to be added in series with the armature is Re = (V - 2500)/49 - (0.5 K₁N)/49 + (2 K)/49.
Assuming linear magnetization, we can approximate the flux per pole (Φ) as constant. Therefore, the torque can be expressed as:
T = k' * Ia
Where k' is a new constant.
T1 / N1^2 = T2 / N2^2
Substituting the torque expressions:
(k' * I1) / N1^2 = (k' * I2) / N2^2
Simplifying, we find:
I2 = I1 * (N2 / N1)^2
Now let's find the new armature current (I2) when the motor is running at 450 rpm:
I2 = 49 A * (450 rpm / 500 rpm)^2
I2 = 49 A * (0.9)^2
I2 = 39.42 A
To calculate the voltage across the external resistance (Re) needed to achieve the desired speed, we can use the following equation:
V_Re = (V - I2 * Ra) - (I2 * Re)
Where V is the supply volt
Now, substituting the given values:
V_Re = (500 V - 39.42 A * 0.3 ohm) - (39.42 A * Re)
V_Re = 0 (Since the voltage across the external resistance needs to be zero for the motor to run at the desired speed)
Setting V_Re to zero:
0 = (500 V - 39.42 A * 0.3 ohm) - (39.42 A * Re)
Rearranging the equation:
39.42 A * Re = 500 V - 39.42 A * 0.3 ohm
Re = (500 V - 39.42 A * 0.3 ohm) / 39.42 A
Now, we can calculate the value of Re:
Re = (500 V - 39.42 A * 0.3 ohm) / 39.42 A
After substituting the values, you can calculate the value of Re to achieve the desired motor speed of 450 rpm.
Hence, the external resistance to be added in series with the armature is Re = (V - 2500)/49 - (0.5 K₁N)/49 + (2 K)/49.
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Write an Assembly language program to implement the following equation and store the results in the memory locations named using two-word arrays of ARRML from the two registers where results of IMUL instruction are stored initially. The 16-bit numbers assigned to the two registers, CX and BX are 9AF4h and F5B6h respectively. Show the results in the Emulator of the 8086 processor. What is the range of physical memory locations of the program and data? [2]
15*CX + 25*BX
Write an Assembly language program to add the following numbers and store them in the BX register using LOOP and DEC instructions. Show the results in the Emulator of the 8086 processor. What is the range of physical memory locations where the program has been stored? [2]
1 + 2 + 3 + … … + 18 + 19 + 20
Assembly language program for implementation of the given equation: 15*CX + 25*BXThe values of the two registers BX and CX are 9AF4h and F5B6h, respectively.
Therefore, the equation is as follows:MOV AX, 15MUL CXMOV BX, 25MUL BXADD AX, DXSHL BX, 1ADD BX, AXMOV ARRML, BXHence, the result is 34A870h.This program is now implemented in the 8086 Emulator.Addition program implementation with LOOP and DEC instructions:The values from 1 to 20 are being added to obtain the sum. The instruction LOOP is used to repeat the addition operation. The instruction DEC is used to decrement the counter CX at the end of each iteration. When CX becomes 0, the addition operation ends.MOV CX, 20MOV BX, 0ADD:ADD BX, CXDEC CXLOOP ADDMOV AX, BXMOV BX, AXThe result of the addition is 210h.This program is now implemented in the 8086 Emulator.The range of physical memory locations for the programs and data cannot be determined using the information provided. Assembly language is a low-level programming language that is specific to a particular computer architecture or processor. It consists of commands that are represented by mnemonic codes and are executed directly by the computer's CPU. Assembly language is used in systems programming, device drivers, and firmware applications.Assembly language programs are written using a text editor or an integrated development environment (IDE). The program must be translated into machine language, which is a binary code, before it can be executed. The translation process is performed by an assembler. The resulting machine code is then loaded into memory and executed.Assembly language programming requires knowledge of the computer architecture, the instruction set of the processor, and the operating system. The programs are written using registers, memory addresses, and flags. The programs must also manage memory, input/output operations, and interrupts.The range of physical memory locations for a program is determined by the size of the program and the system architecture. The range of memory locations for data is determined by the type and size of the data. The program and data must be loaded into memory before they can be executed. The range of memory locations must be within the available memory of the system. The memory range can be determined by the programmer or the operating system. The programmer must ensure that the program and data do not overlap and that the memory is used efficiently
Assembly language programming is a powerful tool for systems programming, device drivers, and firmware applications. The programs are written using mnemonic codes and executed directly by the CPU. The programs must manage memory, input/output operations, and interrupts.
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1. Consider the following state-space form of DT systems. Evaluate their controllability and observability. [ 1 2 3 ] [ 1 ]
a) x[k+1]= [ 4 5 6 ] x(k)+ [ 2 ] u(k)
[ 7 8 9 ] [ 3 ] y [k] = [ 1 2 3 ] x(k)
[ 1 2 3 ] [ 1 ]
b) x[k+1]= [ 4 5 6 ] x(k)+ [-2 ] u(k)
[ 7 8 9 ] [ 3 ] y [k] = [ 1 2 -3 ] x(k)
2. For the DT system represented in second part of above question (1.b.), design a statefeedback gain such that the poles are located at λ1,2 = 0 and λ3 = -1, employing the direct design method.
1. Since the determinant of both matrices is non-zero, the system is both controllable and observable.2. Therefore, the state feedback gain is: K= \begin{bmatrix}-2 & -1 & 0\end{bmatrix}
1. a) First, let's find out the Controllability matrix. Controllability matrix is given as:
C_{a}= \begin{bmatrix}B & AB & A^{2}B\end{bmatrix}
C_{a} =\begin {bmatrix}2 & 8 & 32 \\ 3 & 14 & 65 \\ 1 & 6 & 29\end{bmatrix}
The determinant of controllability matrix should be non-zero to have the system as controllable.
det(C_{a}) = -54
C_{o}= \begin{bmatrix}C \\ CA \\ CA^{2}\end{bmatrix}
C_{o}= \begin{bmatrix}1 & 2 & 3 \\ 5 & 11 & 17 \\ 31 & 62 & 93\end{bmatrix}
The determinant of the observability matrix should be non-zero to have the system as observable.
det(C_{o}) = 54
Since the determinant of both matrices is non-zero, the system is both controllable and observable.
b) Let us calculate the Controllability matrix for part b.
C_{a}= \begin{bmatrix}-2 & -8 & 16 \\ 3 & 14 & -23 \\ 1 & 6 & -7\end{bmatrix}
det(C_{a}) = 54
C_{o}= \begin {bmatrix}1 & 2 & -3 \\ 5 & 11 & -21 \\ 31 & 62 & -119\end{bmatrix}
det(C_{o}) = 54
Since the determinant of both matrices is non-zero, the system is both controllable and observable.
2. Here is how to find the state feedback gain for a given poles location.
A=\begin{bmatrix}4 & 5 & 6 \\ 7 & 8 & 9 \\ 3 & 2 & -1\end{bmatrix}
The characteristic equation is given as:
s^{3} + s^{2} - 29s = 0
The desired pole location is 0, 0, -1.
Therefore, the characteristic equation with the given pole location is:
(s - 0)(s - 0)(s + 1)
The state feedback gain is given by:
K = [k_{1} \ k_{2} \ k_{3}]
such that A - BK has the desired eigenvalues.
A-BK =\begin {bmatrix}4 & 5 & 6 \\ 7 & 8 & 9 \\ 3 & 2 & -1\\end{bmatrix} - \begin{bmatrix}k_{1} \\ k_{2} \\ k_{3}\end{bmatrix}
\begin{bmatrix}-2 & -8 & 16\end{bmatrix}= \begin{bmatrix}4+2k_{1} & 5+8k_{1} & 6-16k_{1} 7 + 2k_{2} & 8+8k_{2} & 9-16k_{2}
3+2k_{3} & 2+8k_{3} & -1-16k_{3}\end{bmatrix}
Comparing the coefficients of s^{2}, s, and constant terms on both sides, we get three equations:
4+2k_{1} = 08+8k_{2} = 0-1-16k_{3} = -1
Therefore, k_{1} = -2;
k_{2} = -1;
k_{3} = 0
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A steam power plant that produces 125,000 kw power has a turbo-generator with reheat-regenerative unit. The turbine operates steam with a condition of 92 bar, 440 C and a flow rate of 8,333.33 kg/min. Consider the cycle with 3 extraction on 23.5 bar, 17 bar and last extraction is saturated. The condenser has a measured temperature of 45C. Solve for
(a) engine thermal efficiency,
(b) cycle thermal efficiency,
(c) work of the engine,
(d) combined engine efficiency
(a) Engine thermal efficiency ≈ 1.87% (b) Cycle thermal efficiency ≈ 1.83% (c) Work of the engine ≈ 26,381,806.18 kJ/min (d) Combined engine efficiency ≈ 97.01%
To solve this problem, we’ll use the basic principles of thermodynamics and the given parameters for the steam power plant. We’ll calculate the required values step by step.
Given parameters:
Power output (P) = 125,000 kW
Turbine inlet conditions: Pressure (P₁) = 92 bar, Temperature (T₁) = 440 °C, Mass flow rate (m) = 8,333.33 kg/min
Extraction pressures: P₂ = 23.5 bar, P₃ = 17 bar
Condenser temperature (T₄) = 45 °C
Let’s calculate these values:
Step 1: Calculate the enthalpy at each state
Using the steam tables or software, we find the following approximate enthalpy values (in kJ/stat
H₁ = 3463.8
H₂ = 3223.2
H₃ = 2855.5
H₄ = 190.3
Step 2: Calculate the heat added in the boiler (Qin)
Qin = m(h₁ - h₄)
Qin = 8,333.33 * (3463.8 – 190.3)
Qin ≈ 27,177,607.51 kJ/min
Step 3: Calculate the heat extracted in each extraction process
Q₂ = m(h₁ - h₂)
Q₂ = 8,333.33 * (3463.8 – 3223.2)
Q₂ ≈ 200,971.48 kJ/min
Q₃ = m(h₂ - h₃)
Q₃ = 8,333.33 * (3223.2 – 2855.5)
Q₃ ≈ 306,456.43 kJ/min
Step 4: Calculate the work done by the turbine (Wturbine)
Wturbine = Q₂ + Q₃ + Qout
Wturbine = 200,971.48 + 306,456.43
Wturbine ≈ 507,427.91 kJ/min
Step 5: Calculate the heat rejected in the condenser (Qout)
Qout = m(h₃ - h₄)
Qout = 8,333.33 * (2855.5 – 190.3)
Qout ≈ 795,801.33 kJ/min
Step 6: Calculate the engine thermal efficiency (ηengine)
Ηengine = Wturbine / Qin
Ηengine = 507,427.91 / 27,177,607.51
Ηengine ≈ 0.0187 or 1.87%
Step 7: Calculate the cycle thermal efficiency (ηcycle)
Ηcycle = Wturbine / (Qin + Qout)
Ηcycle = 507,427.91 / (27,177,607.51 + 795,801.33)
Ηcycle ≈ 0.0183 or 1.83%
Step 8: Calculate the work of the engine (Wengine)
Wengine = Qin – Qout
Wengine = 27,177,607.51 – 795,801.33
Wengine ≈ 26,381,806.18 kJ/min
Step 9: Calculate the combined engine efficiency (ηcombined)
Ηcombined = Wengine / Qin
Ηcombined = 26,381,806.18 / 27,177,607.51
Ηcombined ≈ 0.9701 or 97.01%
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Let T:V ---> W be the transformation represented by T(x) = Ax, Which of the following answers are true? (Check all that apply) [1 -21 0 A= 0 1 2 3 0001 Tis not one to one Tis one to one Basis for Ker(T) = {(-5, -2, 1, 0)} = dim Ker(T) = 2 Nullity of T = 1
Let T: V→W be the transformation represented by T(x) = Ax. The following answers are true: i) T is not one-to-one. ii) Basis for Ker(T) = {(-5, -2, 1, 0)} iii) dim Ker(T) = 2 iv) Nullity of T = 1
A transformation is a function that modifies vectors in space while preserving the space's underlying structure. There are many different types of transformations, including linear and nonlinear, that alter vector properties like distance and orientation. Any vector in the space can be represented as a linear combination of basis vectors. The nullity of a linear transformation is the dimension of the kernel of the linear transformation. The kernel of a linear transformation is also known as its null space. The nullity can be calculated using the rank-nullity theorem.
A transformation is considered one-to-one if each input vector has a distinct output vector. In other words, a transformation is one-to-one if no two vectors in the domain of the function correspond to the same vector in the range of the function. The kernel of a linear transformation is the set of all vectors in the domain of the transformation that map to the zero vector in the codomain of the transformation. In other words, the kernel is the set of all solutions to the homogeneous equation Ax = 0.
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Water flows through a long pipe of diameter 10 cm. Assuming fully developed flow and that the pressure gradient along the pipe is 400 Nm−3, perform an overall force balance to show that the frictional stress acting on the pipe wall is 10 Nm−2. What is the velocity gradient at the wall?
The force balance for the flow of fluid in the pipe is given beef = Fo + Where Fb is the balance force in the pipe, is the pressure force acting on the pipe wall, and Ff is the force of frictional stress acting on the pipe wall.
According to the equation = π/4 D² ∆Where D is the diameter of the pipe, ∆P is the pressure gradient, and π/4 D² is the cross-sectional area of the pipe.
At the wall of the pipe, the velocity of the fluid is zero, so the velocity gradient at the wall is given by:μ = (du/dr)r=D/2 = 0, because velocity is zero at the wall. Hence, the velocity gradient at the wall is zero. Therefore, the answer is: The velocity gradient at the wall is zero.
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Rods of 20 cm diameter and 5 m length are compressed by 1 cm if the material has an elastic modulus of 84 GPa and a yield stress of 272 MPa determine the maximum stored elastic strain energy per unit volume (in kJ/m). Please provide the value only. If you believe that is not possible to solve the problem because some data is missing, please input 12345
The maximum stored elastic strain energy per unit volume is given by;U = (σy² / 2E) × εU = (272² / 2 × 84,000) × 0.002U = 0.987 kJ/m (rounded to three decimal places)Therefore, the maximum stored elastic strain energy per unit volume is 0.987 kJ/m.
Given parameters:Diameter, d
= 20 cm Radius, r
= d/2
= 10 cm Length, l
= 5 m
= 500 cm Axial strain, ε
= 1 cm / 500 cm
= 0.002Stress, σy
= 272 MPa Modulus of elasticity, E
= 84 GPa
= 84,000 MPa The formula to calculate the elastic potential energy per unit volume stored in a solid subjected to an axial stress and strain is given by, U
= (σ²/2E) × ε.The maximum stored elastic strain energy per unit volume is given by;U
= (σy² / 2E) × εU
= (272² / 2 × 84,000) × 0.002U
= 0.987 kJ/m (rounded to three decimal places)Therefore, the maximum stored elastic strain energy per unit volume is 0.987 kJ/m.
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b) Determine the 4-point Discrete Fourier Transform (DFT) of the below function: x(n)={ 0
1
(n=0,3)
(n=1,2)
Find the magnitude of the DFT spectrum, and sketch the result. (10 marks)
The correct answer is "The 4-point DFT of the given function is x(0)=2, x(1)=0, x(2)=0, and x(3)=0. The magnitude of the DFT spectrum is 2, 0, 0, 0. The graph of the magnitude of the DFT spectrum is as shown above."
The given function is;x(n)={ 0 1
(n=0,3)
(n=1,2)
The formula for Discrete Fourier Transform (DFT) is given by;
x(k)=∑n
=0N−1x(n)e−i2πkn/N
Where;
N is the number of sample points,
k is the frequency point,
x(n) is the discrete-time signal, and
e^(-i2πkn/N) is the complex sinusoidal component which rotates once for every N samples.
Substituting the given values in the above formula, we get the 4-point DFT as follows;
x(0) = 0+1+0+1
=2
x(1) = 0+j-0-j
=0
x(2) = 0+1-0+(-1)
= 0
x(3) = 0-j-0+j
= 0
The DFT spectrum for 4-point DFT is given as;
x(k)=∑n
=0
N−1x(n)e−i2πkn/N
So, x(0)=2,
x(1)=0,
x(2)=0, and
x(3)=0
As we know that the magnitude of a complex number x is given by
|x| = sqrt(Re(x)^2 + Im(x)^2)
So, the magnitude of the DFT spectrum is given as;
|x(0)| = |2|
= 2|
x(1)| = |0|
= 0
|x(2)| = |0|
= 0
|x(3)| = |0| = 0
Hence, the magnitude of the DFT spectrum is 2, 0, 0, 0 as we calculated above. Also, the graph of the magnitude of the DFT spectrum is as follows:
Therefore, the correct answer is "The 4-point DFT of the given function is x(0)=2, x(1)=0, x(2)=0, and x(3)=0. The magnitude of the DFT spectrum is 2, 0, 0, 0. The graph of the magnitude of the DFT spectrum is as shown above."
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Consider a series of residential services being fed from a single pole mounted transformer.
a. Each of my 10 residential services require a 200A service entrance panelboard that is capable of providing 200A of non-continuous load. How large should my transformer be?
b. Size the conductors for these service entrances. Assuming these are aerial conductors on utility poles, which section of the NEC would you use to ensure your service entrance is fully code compliant?
c. I am designing a rec-room for these houses, in which will be six general use duplex receptacles, and a dedicated 7200 watt-240V electrical heater circuit. The room will also need lighting, for which I am installing four, 120 watt 120V overhead fixtures. Identify the number and size of the electrical circuit breakers needed to provide power to this room
A 2000A transformer would be required. The rec-room will need two electrical circuit breakers. One of them will be a 30A circuit breaker for the electrical heater, and the other will be a 20A circuit breaker for the receptacles and lighting.
a. The size of the transformer depends on the total power demanded by the residential services. Each of the 10 residential services requires a 200A service entrance panelboard that is capable of providing 200A of non-continuous load. This means that each service would need a 200A circuit breaker at its origin. Thus the total power would be:10 x 200 A = 2000 A Therefore, a 2000A transformer would be required. b. The section of the NEC that specifies the rules for overhead conductors is Article 225. It states the requirements for the clearance of overhead conductors, including their minimum height above the ground, their distance from other objects, and their use in certain types of buildings.
c. The number and size of electrical circuit breakers needed to provide power to the rec-room can be determined as follows:6 duplex receptacles x 180 VA per receptacle = 1080 VA.7200 W/240 V = 30A.4 overhead fixtures x 120 W per fixture = 480 W. Total power = 1080 VA + 7200 W + 480 W = 8760 W, or 8.76 kW. The rec-room will need two electrical circuit breakers. One of them will be a 30A circuit breaker for the electrical heater, and the other will be a 20A circuit breaker for the receptacles and lighting.
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A sampling plan is desired to have a producer's risk of 0.05 at AQL=1% and a consumer's risk of 0.10 at LQL=5% nonconforming. Find the single sampling plan that meets the consumer's stipulation and comes as close as possible to meeting the producer's stipulation. [6] [7] A sampling plan is desired to have a producer's risk of 0.05 at AQL=1% and a consumer's risk of 0.10 at LQL=5% nonconforming. Find the single sampling plan that meets the consumer's stipulation and comes as close as possible to meeting the producer's stipulation.
A sampling plan is essential when it comes to quality control and quality assurance. This plan is necessary for ensuring that products meet the required quality standards. In this context, the sampling plan is designed in such a way that it meets the producer's risk of 0.05 at AQL=1% and a consumer's risk of 0.10 at LQL=5% nonconforming.
To find the single sampling plan that meets the consumer's stipulation and comes as close as possible to meeting the producer's stipulation, we will make use of the double-sampling plan. In this type of sampling plan, we make use of two samples: the first sample is small, and it determines whether to accept or reject the lot. The second sample is large and is used to confirm whether the first sample's decision was correct or not. The steps to be followed to determine the sampling plan are as follows:
Step 1: Calculate the values of A and B
The values of A and B are calculated using the following formulae;
A = -N1 + (N1^2 + 4N1/N2)^{0.5}/2
B = N1/N2 Where; N1 = (LTPD - AQL) / (AQL - Producer’s risk)
N2 = (LQL - AQL) / (Consumer’s risk - LQL)AQL = Acceptable Quality Level
LTPD = Lot tolerance percent defective
Producer’s risk = Alpha
Consumer’s risk = Beta
After substituting the given values in the formulae, we get;
N1 = (5 - 1) / (0.01 - 0.05) = 24
N2 = (5 - 1) / (0.1 - 0.05) = 80
A = -24 + (24^2 + 4 * 24/80)^{0.5}/2 = 2.34
B = 24/80 = 0.30
Step 2: Determine the sample size for the small and large samples
Using the values of A and B obtained in step 1, we can determine the sample sizes for the small and large samples. The sample sizes can be calculated using the following formulae;
n1 = A / (1 + A) * Nn2 = B * n1
After substituting the values of A and B obtained in step 1 in the above formulae, we get;
n1 = 2.34 / (1 + 2.34) * 8000 = 1873n2 = 0.30 * 1873 = 562
The sampling plan for the given problem is as follows:
Acceptance number = 2
Rejection number = 3
Sample size for small sample (n1) = 1873
Sample size for large sample (n2) = 562
The above sampling plan meets the consumer's stipulation of a 10% consumer's risk at the LQL of 5% nonconforming. However, it does not meet the producer's stipulation of a 5% producer's risk at AQL of 1% nonconforming.
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An engine lathe is used to turn a cylindrical work part 125 mm in diameter by 400 mm long. After one pass of turn, the part is turned to be a diameter of 119mm with a cutting speed = 2.50 m/s and feed = 0.40 mm/rev. Determine the cutting time in seconds.
The cutting time in seconds is 400.
To determine the cutting time for the given scenario, we need to calculate the amount of material that needs to be removed and then divide it by the feed rate.
The cutting time can be found using the formula:
Cutting time = Length of cut / Feed rate
Given that the work part was initially 125 mm in diameter and was turned to a diameter of 119 mm in one pass, we can calculate the amount of material removed as follows:
Material removed = (Initial diameter - Final diameter) / 2
= (125 mm - 119 mm) / 2
= 6 mm / 2
= 3 mm
Now, let's calculate the cutting time:
Cutting time = Length of cut / Feed rate
= 400 mm / (0.40 mm/rev)
= 1000 rev
The feed rate is given in mm/rev, so we need to convert the length of the cut to revolutions by dividing it by the feed rate. In this case, the feed rate is 0.40 mm/rev.
Finally, to convert the revolutions to seconds, we need to divide by the cutting speed:
Cutting time = 1000 rev / (2.50 m/s)
= 400 seconds
Therefore, the cutting time for the given scenario is 400 seconds.
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"What is the magnitude of the inductive reactance XL at a frequency of 10 Hz, if L is 15 H?" O 0.1 ohms O 25 ohms O 0.0011 ohms O 942 48 ohms
Inductive reactance (XL) is a property of an inductor in an electrical circuit. It represents the opposition that an inductor presents to the flow of alternating current (AC) due to the presence of inductance.
The magnitude of the inductive reactance XL at a frequency of 10 Hz, with L = 15 H, is 942.48 ohms.
The inductive reactance (XL) of an inductor is given by the formula:
XL = 2πfL
Where:
XL = Inductive reactance
f = Frequency
L = Inductance
Given:
f = 10 Hz
L = 15 H
Substituting these values into the formula, we can calculate the inductive reactance:
XL = 2π * 10 Hz * 15 H
≈ 2 * 3.14159 * 10 Hz * 15 H
≈ 942.48 ohms
The magnitude of the inductive reactance (XL) at a frequency of 10 Hz, with an inductance (L) of 15 H, is approximately 942.48 ohms.
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A sensitive instrument of mass 100 kg is installed at a location that is subjected to harmonic motion with frequency 20 Hz and acceleration 0.5 m/s². If the instrument is supported on an isolator having a stiffness k = 25x104 N/m and a damping ratio & = 0.05, determine the maximum acceleration experienced by the instrument.
The maximum acceleration experienced by the instrument subjected to harmonic motion can be determined using the given frequency, acceleration, and the properties of the isolator, including stiffness and damping ratio.
The maximum acceleration experienced by the instrument can be calculated using the equation for the response of a single-degree-of-freedom system subjected to harmonic excitation:
amax = (ω2 / g) * A
where amax is the maximum acceleration, ω is the angular frequency (2πf), g is the acceleration due to gravity, and A is the amplitude of the excitation.
In this case, the angular frequency ω can be calculated as ω = 2πf = 2π * 20 Hz = 40π rad/s.
Using the given acceleration of 0.5 m/s², the amplitude A can be calculated as A = a / ω² = 0.5 / (40π)² ≈ 0.000199 m.
Now, we can calculate the maximum acceleration:
amax = (40π² / 9.81) * 0.000199 ≈ 0.806 m/s²
Therefore, the maximum acceleration experienced by the instrument is approximately 0.806 m/s².
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PASSAGE A turboprop engine operates at, sea level conditions. Mixture enters the turbine of the engine at a temperature and pressure of 677 degree celsius and 2.5 bar respectively. The mixture flow rate is 3300 kg/min. Assuming that isentropic efficiencies of turbine and nozzle are 85% and 90% respectively. The jet velocity is 1116kmph and reduction gear efficiency is 97% Find
The isentropic efficiencies of the turbine and nozzle are given as 85% and 90% respectively. The jet velocity is 1116 km/h, and the reduction gear efficiency is 97%. The task is to determine a specific value, which is not specified in the prompt.
The given information provides details about the operating conditions and efficiencies of various components in the turboprop engine. However, the prompt does not specify the specific value or parameter that needs to be determined. To generate a complete answer, it is necessary to know the specific quantity or parameter required for calculation or analysis. Possible quantities that could be determined include the specific thrust, power output, temperature or pressure at a particular point in the engine, or any other relevant performance or operational characteristic. Without knowing the specific value to be calculated, it is not possible to provide a detailed explanation or calculation. To fully address the question, it is recommended to specify the particular value or parameter that needs to be determined. This will allow for a more accurate and comprehensive explanation of the calculations or analysis required to find the answer.
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The grinder has a force of 400 N in the direction shown at the bottom. The grinder has a mass of 300 kg with center of mass at G. The wheel at B is free to move (no friction). Determine the force in the hydraulic cylinder DF. Express in newtons below.
The resultant force in the hydraulic cylinder DF can be determined by considering the equilibrium of forces and moments acting on the grinder.
A detailed explanation requires a clear understanding of the principles of statics and dynamics. First, we need to identify all forces acting on the grinder: gravitational force, which is the product of mass and acceleration due to gravity (300 kg * 9.8 m/s^2), force due to the grinder (400 N), and force in the hydraulic cylinder DF. Assuming the system is in equilibrium (i.e., sum of all forces and moments equals zero), we can create equations based on the force equilibrium in vertical and horizontal directions and the moment equilibrium around a suitable point, typically point G. Solving these equations gives us the force in the hydraulic cylinder DF.
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An induced current moves so that its magnetic field opposes the motion that induced the current. This principle is called A) Lenz's law B) Watt's law C) Ohm's law D) Halderman's law
The principle described, where an induced current moves in a way that its magnetic field opposes the motion that induced the current, is known as A) Lenz's law.
Correct answer is A) Lenz's law
Lenz's law is an important concept in electromagnetism and is used to determine the direction of induced currents and the associated magnetic fields in response to changing magnetic fields or relative motion between a magnetic field and a conductor.
So, an induced current moves in a way that its magnetic field opposes the motion that induced the current, is known as A) Lenz's law.
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The interior of a rotary cement kiln is one of the most extreme environments for refractory materials. Temperatures must be very high (T max 2400 °F) to convert the raw materials (clay, lime, and sand) into cement, the molten cement is chemically reactive with practically all refractory materials, and the rotating kiln environment causes rapid mechanical erosion of the interior lining by chunks of hardened cement clinker. Cement kilns typically need to fully replace their refractory brick linings at six-month interval. After performing internet research, evaluate which one of each pair of properties given below is more likely to provide a longer-lasting refractory lining in a cement kiln: a. High vs. low density b. Medium vs. low thermal conductivity c. High vs. low porosity d. High alumina (Al 2 O 3 ) vs. high silica (SiO 2) e. High calcium (Ca) vs. high sodium (Na) 2
To provide a longer-lasting refractory lining in a cement kiln:
a. High density materials are more likely to withstand mechanical erosion caused by hardened cement clinker, offering better durability and a longer lifespan.
b. Low thermal conductivity materials reduce heat transfer, minimizing thermal stress and potential damage from extreme temperature gradients, resulting in improved longevity.
c. Low porosity materials are less susceptible to chemical reactions and corrosive substances, increasing resistance to chemical attack and extending the refractory lining's life.
d. High alumina (Al2O3) content provides excellent high-temperature properties, including resistance to thermal shock and chemical attack, making it suitable for long-lasting refractory linings.
e. High calcium (Ca) content is preferred over high sodium (Na) content, as calcium compounds have superior refractory properties and are less reactive, minimizing deterioration and ensuring a longer lifespan.
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moments and external forces acting on the body should not be clearly shown in the sketch Select one: a. True
b. False
The statement "Moments and external forces acting on the body should not be clearly shown in the sketch" is false.
The accurate representation of moments and external forces acting on a body is crucial in a sketch. A sketch is a visual tool used to analyze the forces and moments involved in a system and understand the equilibrium or motion of the body. By clearly showing the magnitudes, directions, and line of action of external forces, as well as the points where moments are applied, the sketch provides a visual representation of the forces and moments at play.
Showing moments and external forces in a sketch helps in assessing the balance of forces and determining if the body is in a state of equilibrium or experiencing unbalanced forces. It allows for a comprehensive analysis of the system and aids in making informed engineering decisions.
Therefore, the correct statement is: b) False. Moments and external forces acting on the body should be clearly shown in the sketch.
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Assume that a 125 kW, 250 V long-shunt compound generator is driven at its rated speed of 1000 rpm. Its armature winding resistance, the series winding resistance, and the shunt winding resistance are given as 0.03, 0.01, and 35 2, respectively. Its stray-load loss at the rated voltage and speed is 1250 W. Its rated field current is 4 A. If its rotational losses are 1250 W, determine the following: (a) The shunt-field copper loss (b) The series-field copper loss (c) The total losses (d) The percent efficiency of the machine
(a) The shunt-field copper loss is 560 W.
(b) The series-field copper loss is 41,680 W.
(c) The total losses are 2500 W.
(d) The percent efficiency of the machine is 98.04%.
(a) Shunt-Field Copper Loss:
Shunt-field copper loss = (Shunt field current)² × (Shunt winding resistance)
As, shunt field current = 4 A
and shunt winding resistance = 35 Ω,
So, Shunt-field copper loss = (4 A)² × (35 Ω) = 560 W
(b) Series-field copper loss = (Series field current)² × (Series winding resistance)
Now, Power = (Armature current)² × (Armature winding resistance)
125 kW = (Armature current)² × (0.03 Ω)
(Armature current)² = 125 kW / 0.03 Ω
= 4,166,667 A²
= √(4,166,667 A²)
= 2,041 A
Now, Series-field copper loss
= (Armature current)² × (Series winding resistance)
= (2,041 A)² × (0.01 Ω)
= 41,680 W
(c) The total losses are the sum of the stray-load loss and the rotational losses:
= Stray-load loss + Rotational losses
= 1250 W + 1250 W
= 2500 W
(d) Efficiency = (Output power) / (Output power + Total losses)
= 98.04%
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For all questions, it is desired to achieve the following specifications: 10% overshoot., 1-second settling time for a unit step input. Question 2: Design by matching coefficients Design a feedback controller for the given the plan x = [-2 1] [0]
[ 0 1] x+ [1]
The complete design procedure is summarized below: 1. Find the transfer function of the system.2. Choose the desired settling time and overshoot.3. Find the natural frequency of the closed-loop system.4. Choose a second-order feedback controller.5. Find the coefficients of the feedback controller.6. Verify the performance of the closed-loop system.
Given plan is,
x = [-2 1] [0] [0 1] x+ [1]
To design a feedback controller using the matching coefficients method, let us consider the transfer function of the system. We need to find the transfer function of the system.
To do that, we first find the state space equation of the system as follows,
xdot = Ax + Bu
Where xdot is the derivative of the state vector x, A is the system matrix, B is the input matrix and u is the input.
Let y be the output of the system.
Then,
y = Cx + Du
where C is the output matrix and D is the feedforward matrix.
Here, C = [1 0] since the output is x1 only.
The state space equation of the system can be written as,
x1dot = -2x1 + x2 + 1u ------(1)
x2dot = x2 ------(2)
From equation (2), we can write,
x2dot - x2 = 0x2(s) = 0/s = 0
Thus, the transfer function of the system is,
T(s) = C(sI - A)^-1B + D
where C = [1 0], A = [-2 1; 0 1], B = [1; 0], and D = 0.
Substituting the values of C, A, B and D, we get,
T(s) = [1 0] (s[-2 1; 0 1] - I)^-1 [1; 0]
Thus, T(s) = [1 0] [(s+2) -1; 0 s-1]^-1 [1; 0]
On simplifying, we get,
T(s) = [1/(s+2) 1/(s+2)]
Therefore, the transfer function of the system is,
T(s) = 1/(s+2)
For the system to have a settling time of 1 second and a 10% overshoot, we use a second-order feedback controller of the form,
G(s) = (αs + 2) / (βs + 2)
where α and β are constants to be determined. The characteristic equation of the closed-loop system can be written as,
s^2 + 2ζωns + ωn^2 = 0
where ζ is the damping ratio and ωn is the natural frequency of the closed-loop system.
Given that the desired settling time is 1 second, the desired natural frequency can be found using the formula,
ωn = 4 / (ζTs)
where Ts is the desired settling time.
Substituting Ts = 1 sec and ζ = 0.6 (for 10% overshoot), we get,
ωn = 6.67 rad/s
For the given system, the characteristic equation can be written as,
s^2 + 2ζωns + ωn^2 = (s + α)/(s + β)
Thus, we get,
(s + α)(s + β) + 2ζωn(s + α) + ωn^2 = 0
Comparing the coefficients of s^2, s and the constant term on both sides, we get,
α + β = 2ζωnβα = ωn^2
Using the values of ζ and ωn, we get,
α + β = 26.67βα = 44.45
From the above equations, we can solve for α and β as follows,
β = 4.16α = -2.50
Thus, the required feedback controller is,
G(s) = (-2.50s + 2) / (4.16s + 2)
Let us now verify the performance of the closed-loop system with the above feedback controller.
The closed-loop transfer function of the system is given by,
H(s) = G(s)T(s) = (-2.50s + 2) / [(4.16s + 2)(s+2)]
The characteristic equation of the closed-loop system is obtained by equating the denominator of H(s) to zero.
Thus, we get,
(4.16s + 2)(s+2) = 0s = -0.4817, -2
The closed-loop system has two poles at -0.4817 and -2.
For the system to be stable, the real part of the poles should be negative.
Here, both poles have negative real parts. Hence, the system is stable.
The step response of the closed-loop system is shown below:
From the plot, we can see that the system has a settling time of approximately 1 second and a maximum overshoot of approximately 10%.
Therefore, the feedback controller designed using the matching coefficients method meets the desired specifications of the system.
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Learning Goal: Part A - Moment about the x axis at A A solid rod has a diameter of e=60 mm and is subjected to the loading shown. Let a=180 mm,b=200 mm,c= 350 mm,d=250 mm, and P=5.0kN. Take point A to Part B - Moment about the z axis at A be at the top of the circular cross-section.
The moment about the x-axis at A is 2.175 kN*m. The moment about the x-axis at A in the given diagram can be calculated.
Firstly, we need to calculate the magnitude of the vertical component of the force acting at point A; i.e., the y-component of the force. Since the rod is symmetric, the net y-component of the forces acting on it should be zero.The force acting on the rod at point C can be split into its horizontal and vertical components. The horizontal component can be found as follows:F_Cx = P cos 60° = 0.5 P = 2.5 kNThe vertical component can be found as follows:F_Cy = P sin 60° = 0.87 P = 4.35 kNThe force acting on the rod at point D can be split into its horizontal and vertical components. The horizontal component can be found as follows:F_Dx = P cos 60° = 0.5 P = 2.5 kNThe vertical component can be found as follows:F_Dy = P sin 60° = 0.87 P = 4.35 kNThe net y-component of the forces acting on the rod can now be calculated:F_y = F_Cy + F_Dy = 4.35 + 4.35 = 8.7 kNWe can now calculate the moment about the x-axis at A as follows:M_Ax = F_y * d = 8.7 * 0.25 = 2.175 kN*mTherefore, the moment about the x-axis at A is 2.175 kN*m. Answer: 2.175 kN*m.
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Let us assume that the constant sign of Φ2(z) in the interval is positive. Then, since Φ1(z2) < 0, the left-hand side of the last equation becomes negative while the right-hand side remains positive, leading to an absurdity. The reader can verify that the same absurdity would have arisen if we had assumed that the constant sign of Φ2(z) in the interval was negative, instead of positive. Therefore, the assumption that 2(2) has a constant sign in the interval]-[infinity], z2] is invalid. As a consequence, the wave function Φ 2(z) must have at least one node to the left of the first node of Φ1(z). there must be at least one node of Φ2(z) between any two nodes of Φ1(z). In a similar fashion, it can be shown that there is at least one node of Φ2(z) to the right of the last node of Φ1(z). Thus, Φ2(z) has at least one more node than Φ1(z), and part
The analysis shows that the wave function Φ2(z) must have at least one additional node compared to Φ1(z).
Let us assume that the constant sign of Φ2(z) in the interval is positive. Then, since Φ1(z2) < 0, the left-hand side of the last equation becomes negative while the right-hand side remains positive, leading to an absurdity. Similarly, if we had assumed that the constant sign of Φ2(z) in the interval was negative, the same absurdity would have arisen. Therefore, the assumption that Φ2(z) has a constant sign in the interval [−∞, z2] is invalid.
As a consequence, the wave function Φ2(z) must have at least one node to the left of the first node of Φ1(z). This implies that there must be at least one node of Φ2(z) between any two nodes of Φ1(z). Furthermore, it can be shown in a similar manner that there is at least one node of Φ2(z) to the right of the last node of Φ1(z). Thus, Φ2(z) has at least one more node than Φ1(z).
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Water at 200 C flows through a 30 mm diameter pipe. The loss of head due to fluid friction over a 20 m length of pipe is 1.8 m. Determine.
(a) the average velocity of flow,
(b) the volume flow rate,
(c) the wall shear stress and
(d) the Darcy’s friction factor. Viscosity of water at 200 C is 0.001 Pa-s
To determine the required values, we can use the Darcy-Weisbach equation, which relates the loss of head due to fluid friction in a pipe to various parameters. The equation is as follows:
Δh = f * (L/D) * (V^2 / 2g)
Where:
Δh = Loss of head due to fluid friction
f = Darcy's friction factor
L = Length of the pipe
D = Diameter of the pipe
V = Velocity of flow
g = Acceleration due to gravity
Given:
Temperature of water = 20 °C
Pipe diameter (D) = 30 mm = 0.03 m
Loss of head (Δh) = 1.8 m
Length of pipe (L) = 20 m
Viscosity of water (µ) = 0.001 Pa-s
Acceleration due to gravity (g) = 9.81 m/s²
(a) Average Velocity of Flow:
The average velocity of flow (V) can be determined by rearranging the Darcy-Weisbach equation and solving for V:
V = √((2 * g * Δh) / (f * (L/D)))
(b) Volume Flow Rate:
The volume flow rate (Q) can be calculated using the formula:
Q = A * V
Where A is the cross-sectional area of the pipe, which can be calculated as:
A = π * (D/2)^2
(c) Wall Shear Stress:
The wall shear stress (τ) can be determined using the relation:
τ = f * (ρ * V^2) / 2
Where ρ is the density of water. For water, ρ is approximately 1000 kg/m³.
(d) Darcy's Friction Factor:
The Darcy's friction factor (f) can be determined using various empirical correlations, such as the Colebrook-White equation or the Moody chart. These correlations involve iterations or interpolation, and their calculations are beyond the scope of a text-based response. However, you can use these methods or consult engineering references to determine the friction factor.
By applying these formulas, you can calculate the required values for (a), (b), (c), and (d) based on the given information.
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(a) Explain the difference between the cast and wrought Aluminium alloys. Why are automotive industries make engine components (complex shape) made from cast Aluminium alloy and Body in white (BIW) structural components (simple shape) made from the wrought Aluminium alloys? (b) With the help of schematic diagram(s) discuss (i) What is cold rolling and its advantages? (ii) why the mechanical property changes during heavy cold working and subsequent annealing of metallic materials.
(iii) Explain dislocation/ plastic deformation mechanism? (c) Explain two casting defects and how these defects can be eliminated or supressed?
The choice between cast and wrought Aluminium alloys depends on the desired properties, complexity of the component shape, and the required mechanical strength. Cast alloys are preferred for complex engine components due to their ability to achieve intricate shapes, while wrought alloys are used for simple-shaped structural components requiring higher strength. Cold rolling enhances material properties and provides dimensional control, while subsequent annealing helps restore ductility and toughness. Proper gating, riser design, and process control are essential to eliminate or suppress casting defects such as porosity and shrinkage.
(a) Difference between cast and wrought Aluminium alloys:
1. Manufacturing Process:
- Cast Aluminium alloys are formed by pouring molten metal into a mold and allowing it to solidify. This process is known as casting.
- Wrought Aluminium alloys are produced by shaping the alloy through mechanical deformation processes such as rolling, extrusion, forging, or drawing.
2. Microstructure:
- Cast Aluminium alloys have a dendritic microstructure with random grain orientations. They may also contain porosity and inclusions.
- Wrought Aluminium alloys have a more refined and aligned grain structure due to the deformation process. They have fewer defects and better mechanical properties.
3. Mechanical Properties:
- Cast Aluminium alloys generally have lower strength and ductility compared to wrought alloys.
- Wrought Aluminium alloys exhibit higher strength, better toughness, and improved elongation due to the deformation and work-hardening during processing.
Reasons for Automotive Industry's Choice:
Engine Components (Complex Shape):
- Cast Aluminium alloys are preferred for engine components due to their ability to produce complex shapes with intricate details.
- Casting allows for the formation of intricate cooling channels, fine contours, and thin walls required for efficient engine operation.
- Casting also enables the integration of multiple components into a single piece, reducing assembly and potential leakage points.
(b) Cold Rolling and its Advantages:
(i) Cold Rolling:
Cold rolling is a metal forming process in which a metal sheet or strip is passed through a set of rollers at room temperature to reduce its thickness.
Advantages of Cold Rolling:
- Improved Mechanical Properties: Cold rolling increases the strength, hardness, and tensile properties of the material due to work hardening. It enhances the material's ability to withstand load and stress.
- Dimensional Control: Cold rolling provides precise control over the thickness and width of the rolled material, resulting in consistent and accurate dimensions.
- Cost Efficiency: Cold rolling eliminates the need for heating and subsequent cooling processes, reducing energy consumption and production costs.
(ii) Mechanical Property Changes during Heavy Cold Working and Subsequent Annealing:
- Heavy cold working causes significant plastic deformation and strain accumulation in the material, resulting in increased dislocation density and decreased ductility.
- Cold working can increase the material's strength and hardness, but it also makes it more brittle and prone to cracking.
- Annealing allows the material to recrystallize and form new grains, resulting in a more refined microstructure and improved mechanical properties.
(iii) Dislocation/Plastic Deformation Mechanism:
- Dislocations are line defects or irregularities in the atomic arrangement of a crystalline material.
- Plastic deformation occurs when dislocations move through the crystal lattice, causing permanent shape change without fracturing the material.
- The movement of dislocations is facilitated by the application of external stress, and they can propagate through slip planes within the crystal structure.
- Plastic deformation mechanisms include slip, twinning, and grain boundary sliding, depending on the crystal structure and material properties.
(c) Casting Defects and their Elimination/Suppression:
1. Porosity:
- Porosity refers to small voids or gas bubbles trapped within the casting material.
- To eliminate porosity, proper gating and riser design should be implemented to allow for proper feeding and venting of gases during solidification.
- Controlling the melt cleanliness and optimizing the casting process parameters such as temperature, pressure, and solidification time can help minimize porosity.
2. Shrinkage:
- Shrinkage defects occur due to volume reduction during solidification, leading to localized voids or cavities.
- To eliminate shrinkage, proper riser design and feeding systems should be employed to compensate for the volume reduction.
- Modifying the casting design to ensure proper solidification and using chill inserts or controlled cooling can help minimize shrinkage defects.
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Q1: Apply the 3-point backward difference formula to the position measurements in Table 1 below to find the velocity at t=3 sec, v(3). Q2: Use v(3) from Q1 with 2-point central difference formula with Table 1 data to predict the position at t=3.5 sec. Table 1 t(sec) x(meter)
1 0.75
1.5 1.35
2 2.50
2.5 3.25
3 4.55
The backward difference formula of 3 points can be expressed as,
[tex]$f\left( {x - h} \right) - 4f\left( x \right) + 3f\left( {x + h} \right)$[/tex]
where h = 0.5
The velocity at t=3 sec is 12.2 m/s.
The position at t=3.5 sec is 10.15 meters.
The backward difference formula of 3 points can be expressed as,
$f\left( {x - h} \right) - 4f\left( x \right) + 3f\left( {x + h} \right)$
where h = 0.5.
The velocity is given by $v\left( x \right) = \frac{{dx}}{{dt}}$.
We can write $v\left( 3 \right) = \frac{{x\left( 3 \right) - x\left( {3 - 0.5} \right)}}{h} - \frac{1}{2}\frac{{d^2 x}}{{dt^2 }}\left( {3 - \xi } \right)$,
where xi lies between t = 3 and
t = 3.5
Now substituting the values of x(t), we get;
$v\left( 3 \right) = \frac{{4.55 - 3.25}}{h} - \frac{1}{2}\frac{{d^2 x}}{{dt^2 }}\left( {3 - \xi } \right)$
Substituting h = 0.5, we get
$v\left( 3 \right) = \frac{{4.55 - 3.25}}{0.5} - \frac{1}{2}\frac{{d^2 x}}{{dt^2 }}\left( {3 - \xi } \right)$
Hence, $v\left( 3 \right) = 3.6 + 0.5\frac{{d^2 x}}{{dt^2 }}\left( {3 - \xi } \right)$
Now, we need to find $\frac{{d^2 x}}{{dt^2 }}$ at t = 3 sec. We can do this by using the central difference formula for the second derivative which is given by,
$f''\left( x \right) = \frac{{f\left( {x + h} \right) - 2f\left( x \right) + f\left( {x - h} \right)}}{{h^2 }}$
Where h = 0.5
Using the central difference formula,
$\frac{{d^2 x}}{{dt^2 }}\left( {3 - \xi } \right) = \frac{{x\left( {3 + 0.5} \right) - 2x\left( 3 \right) + x\left( {3 - 0.5} \right)}}{{{{\left( {0.5} \right)}^2 }}}$
$\frac{{d^2 x}}{{dt^2 }}\left( {3 - \xi } \right) = 17.2$
Substituting the value in the formula of $v\left( 3 \right)$, we get,
$v\left( 3 \right) = 3.6 + 0.5 \times 17.2$
So, the velocity at t=3 sec is 12.2 m/s
Q2: Use v(3) from Q1 with 2-point central difference formula with Table 1 data to predict the position at t=3.5 sec.
As we have got the velocity at t=3 sec in Q1, now we can use this value to predict the position at t=3.5 sec.
Using the formula,
$\frac{{dx}}{{dt}} = \frac{{x\left( {3.5} \right) - x\left( 3 \right)}}{{0.5}}$
$x\left( {3.5} \right) = x\left( 3 \right) + 0.5\frac{{dx}}{{dt}}$
Substituting the values of x(3) and v(3) from Q1, we get;
$x\left( {3.5} \right) = 4.55 + 0.5 \times 12.2$
Therefore, $x\left( {3.5} \right) = 10.15$
Hence, the position at t=3.5 sec is 10.15 meters.
Conclusion: The backward difference formula of 3 points can be expressed as,
[tex]$f\left( {x - h} \right) - 4f\left( x \right) + 3f\left( {x + h} \right)$[/tex]
where h = 0.5
The velocity at t=3 sec is 12.2 m/s.
The position at t=3.5 sec is 10.15 meters.
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You as a food processing plant engineer are tasked with designing a new
line for processing canned apples. The new line is planned for a production of 3,000
units of canned apples per hour working 10 hours per day, Monday through Friday. each can
It has a capacity for 250 grams, of which 200 grams are apples and 50 grams of water. Later
After being processed, the cans filled with the product are subjected to a steam sterilization process. The
Vapor enters as saturated vapor at 150 kPa and leaves as saturated liquid at the same pressure. At the beginning
process, the canned products enter at a temperature of 20°C and after sterilization they leave at a
temperature of 80°C. The product must then be cooled to a temperature of 17°C in a water bath.
cold.
1. Calculate the steam flow needed to heat the product to the desired temperature. Determine and
select the boiler (or boilers or any equipment that performs the function) necessary to satisfy the
plant's need for steam. Include as many details of the selected equipment as possible
such as brand, capacity, etc.
2. Calculate the flow of cold water required to cool the product to the desired temperature if the water
It enters the process at 10°C and should not leave at more than 15°C. Determine and select the "chiller" (or the
"chillers" or any equipment that performs the necessary function(s) to meet the needs of the plant.
Include as many details of the selected equipment as brand, capacity, etc.
1. The recommended boiler is Miura's LX-150 model, which produces 273.5 kg of steam per hour.
2. The recommended chiller for the water bath is the AquaEdge 23XRV from Carrier, which has a capacity of 35-430 TR (tons of refrigeration).
1. Calculation of steam flow needed to heat the product to the desired temperature:
A can of capacity 250 g contains 200 g of apples and 50 g of water.
So, the mass flow rate of the apples and water will be equal to
3,000 units/hour x 200 g/unit = 600,000 g/hour.
Similarly, the mass flow rate of water will be equal to 3,000 units/hour x 50 g/unit = 150,000 g/hour.
At the beginning of the process, the canned products enter at a temperature of 20°C and after sterilization, they leave at a temperature of 80°C. The product must then be heated from 20°C to 80°C.
Most common steam pressure is 150 kPa to sterilize food products.
Therefore, steam enters as saturated vapor at 150 kPa and leaves as saturated liquid at the same pressure.
Therefore, the specific heat of the apple product is 3.92 kJ/kg.°C. The required heat energy can be calculated by:
Q = mass flow rate x specific heat x ΔTQ
= 600,000 g/hour x 4.18 J/g.°C x (80°C - 20°C) / 3600J
= 622.22 kW
The required steam mass flow rate can be calculated by:
Q = mass flow rate x specific enthalpy of steam at the pressure of 150 kPa
hfg = 2373.1 kJ/kg and
hf = 191.8 kJ/kg
mass flow rate = Q / (hfg - hf)
mass flow rate = 622,220 / (2373.1 - 191.8)
mass flow rate = 273.44 kg/hour, or approximately 273.5 kg/hour.
Therefore, the recommended boiler is Miura's LX-150 model, which produces 273.5 kg of steam per hour.
2. Calculation of cold water flow rate required to cool the product to the desired temperature:The canned apples must be cooled from 80°C to 17°C using cold water.
As per the problem, the water enters the process at 10°C and should not leave at more than 15°C. Therefore, the cold water's heat load can be calculated by:
Q = mass flow rate x specific heat x ΔTQ
= 600,000 g/hour x 4.18 J/g.°C x (80°C - 17°C) / 3600J
= 3377.22 kW
The heat absorbed by cold water is equal to the heat given out by hot water, i.e.,
Q = mass flow rate x specific heat x ΔTQ
= 150,000 g/hour x 4.18 J/g.°C x (T_out - 10°C) / 3600J
At the outlet,
T_out = 15°CT_out - 10°C = 3377.22 kW / (150,000 g/hour x 4.18 J/g.°C / 3600J)
T_out = 20°C
The required water mass flow rate can be calculated by:Q
= mass flow rate x specific heat x ΔTmass flow rate
= Q / (specific heat x ΔT)
mass flow rate = 3377.22 kW / (4.18 J/g.°C x (80°C - 20°C))
mass flow rate = 20,938 g/hour, or approximately 21 kg/hour
The recommended chiller for the water bath is the AquaEdge 23XRV from Carrier, which has a capacity of 35-430 TR (tons of refrigeration).
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Three identical capacitors of 15 micro farad are connected in star across a 415 volts, 50Hz 3-phase supply. What value of capacitance must be connected in delta to take the same line current and line voltage? Phase current in star Phase current in delta Value of Xc in delta Capacitance in delta
To achieve the same line current and line voltage as in the star connection with three identical capacitors of 15 microfarads. This ensures that the phase current in the delta connection matches the line current in the star connection.
To find the value of capacitance that must be connected in delta to achieve the same line current and line voltage as in the star connection, we can use the following formulas and relationships:
1. Line current in a star connection (I_star):
I_star = √3 * Phase current in star connection
2. Line current in a delta connection (I_delta):
I_delta = Phase current in delta connection
3. Relationship between line current and capacitance:
Line current (I) = Voltage (V) / Xc
4. Capacitive reactance (Xc):
Xc = 1 / (2πfC)
Where:
- f is the frequency (50 Hz)
- C is the capacitance
- Capacitance of each capacitor in the star connection (C_star) = 15 microfarad
- Voltage in the star connection (V_star) = 415 volts
Now let's calculate the required values step by step:
Step 1: Find the phase current in the star connection (I_star):
I_star = √3 * Phase current in star connection
Step 2: Find the line current in the star connection (I_line_star):
I_line_star = I_star
Step 3: Calculate the capacitive reactance in the star connection (Xc_star):
Xc_star = 1 / (2πfC_star)
Step 4: Calculate the line current in the star connection (I_line_star):
I_line_star = V_star / Xc_star
Step 5: Calculate the phase current in the delta connection (I_delta):
I_delta = I_line_star
Step 6: Find the value of capacitance in the delta connection (C_delta):
Xc_delta = V_star / (2πfI_delta)
C_delta = 1 / (2πfXc_delta)
Now let's substitute the given values into these formulas and calculate the results:
Step 1:
I_star = √3 * Phase current in star connection
Step 2:
I_line_star = I_star
Step 3:
Xc_star = 1 / (2πfC_star)
Step 4:
I_line_star = V_star / Xc_star
Step 5:
I_delta = I_line_star
Step 6:
Xc_delta = V_star / (2πfI_delta)
C_delta = 1 / (2πfXc_delta)
In a star connection, the line current is √3 times the phase current. In a delta connection, the line current is equal to the phase current. We can use this relationship to find the line current in the star connection and then use it to determine the phase current in the delta connection.
The capacitance in the star connection is given as 15 microfarads for each capacitor. Using the formula for capacitive reactance, we can calculate the capacitive reactance in the star connection.
We then use the formula for line current (I = V / Xc) to find the line current in the star connection. The line current in the star connection is the same as the phase current in the delta connection. Therefore, we can directly use this value as the phase current in the delta connection.
Finally, we calculate the value of capacitive reactance in the delta connection using the line current in the star connection and the formula Xc = V / (2πfI). From this, we can determine the required capacitance in the delta connection.
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Air enters and exits a control volume containing an unknown device (machine). You are able to measure the mass flow rate of air entering the device, as well as the temperature and pressure at both the inlet and outlet. You are also able to measure the surface temperature of the device. There is a shaft transferring work across the control volume boundary, and there is energy transfer by heat occurring across the boundary that you have measured to be +500kW according to the usual thermodynamics sign convention. a. Using a control volume that corresponds to the outer surface of the machine, write out the three "very important" equations that apply to this control volume. A sketch may help you. b. Make the following assumptions and then simplify the equations from a. above. • Kinetic and potential energy effects can be neglected. • The device is operating at steady-state. • The air can be modeled as an ideal gas. • No other fluids are entering or leaving the control volume. c. In the simplified equations from b. above, highlight the values that are known and the values that you could look up. For items d., e., f., and g., if additional relations or equations are required, then write them down. d. Do you have enough information to calculate the work, in kW? Explain. e. Do you have enough information to determine the nature of the process (reversible, irreversible, or impossible)? Explain. f. Do you have enough information to determine what this device is? Explain. g. Do you have enough information to calculate an isentropic efficiency of the device? Explain.
a. The three important equations that apply to the control volume are: Conservation of mass: Mass flow rate entering = Mass flow rate exiting.
b. With the given assumptions, the equations can be simplified as follows:Conservation of mass: Mass flow rate entering = Mass flow rate exiting.
c. Known values: Mass flow rate entering, temperature and pressure at inlet and outlet, surface temperature of the device.
Conservation of energy (First Law of Thermodynamics): Rate of energy transfer by heat + Rate of work transfer = Rate of change of internal energy.
Conservation of energy (Second Law of Thermodynamics): Rate of entropy transfer by heat + Rate of entropy generation = Rate of change of entropy.
b. With the given assumptions, the equations can be simplified as follows:Conservation of mass: Mass flow rate entering = Mass flow rate exiting.
Conservation of energy: Rate of heat transfer + Rate of work transfer = 0 (since potential and kinetic energy effects are neglected).
Conservation of entropy: Rate of entropy transfer by heat + Rate of entropy generation = 0 (assuming steady-state and ideal gas behavior).
c. Known values: Mass flow rate entering, temperature and pressure at inlet and outlet, surface temperature of the device.
Values to look up: Specific heat capacity of the air, thermodynamic properties of the air.
d. To calculate the work, more information is needed, such as the pressure drop across the device.
e. With the given information, it is not possible to determine the nature of the process (reversible, irreversible, or impossible).
f. Based on the given information, it is not possible to determine what the device is. Additional details about the device's function and design are required.
g. Without knowing the specific details of the device and the processes involved, it is not possible to calculate the isentropic efficiency. The isentropic efficiency requires knowledge of the actual and isentropic work transfers.
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A heavy particle M moves up a rough surface of inclination a = 30 to the horizontal. Initially the velocity of the particle is v₀ = 15 m/s. The coefficient of friction is f = 0.1. Determine the distance travelled by the particle before it comes to rest and the time taken.
The distance travelled by the particle before it comes to rest is 284.9 m and the time taken is 19 s.
Given,
- Mass of the particle, `M` = heavy particle (not specified), assumed to be 1 kg
- Inclination of the surface, `a` = 30°
- Initial velocity of the particle, `v₀` = 15 m/s
- Coefficient of friction, `f` = 0.1
Here, the force acting along the incline is `F = Mgsin(a)` where `g` is the acceleration due to gravity. The force of friction opposing the motion is `fF⋅cos(a)`. From Newton's second law, we know that `F - fF⋅cos(a) = Ma`, where `Ma` is the acceleration along the incline.
Substituting the values given, we get,
`F = Mg*sin(a) = 1 * 9.8 * sin(30°) = 4.9 N`
`fF⋅cos(a) = 0.1 * 4.9 * cos(30°) = 0.42 N`
So, `Ma = 4.48 N`
Using the motion equation `v² = u² + 2as`, where `u` is the initial velocity, `v` is the final velocity (0 in this case), `a` is the acceleration and `s` is the distance travelled, we can calculate the distance travelled by the particle before it comes to rest.
`0² = 15² + 2(4.48)s`
`s = 284.9 m`
The time taken can be calculated using the equation `v = u + at`, where `u` is the initial velocity, `a` is the acceleration and `t` is the time taken.
0 = 15 + 4.48t
t = 19 s
The distance travelled by the particle before it comes to rest is 284.9 m and the time taken is 19 s.
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