The infiltration rate through the cracks around the windows, we can use the airflow equation:Q = C * A * √(2 * ΔP)
Where:
Q is the infiltration rate (volume flow rate of air),
C is the discharge coefficient,
A is the total area of the cracks,
ΔP is the pressure difference across the cracks.
Given that the wind speed is 23 mph (which is approximately 10.3 m/s) and assuming negligible pressurization of the room, we can consider the pressure difference ΔP as the dynamic pressure due to the wind.
First, let's calculate the total area of the cracks around the windows:
Area = 3 windows * (2 * (3 ft * 4 ft)) = 72 ft²
Next, we need to convert the wind speed to pressure:
ΔP = 0.5 * ρ * V²
where ρ is the air density.
Assuming standard conditions, with air density ρ = 1.225 kg/m³, we can calculate the pressure difference. Finally, we can substitute the values into the airflow equation to calculate the infiltration rate Q.
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(10 marks) (c) a The part-time workers in a construction company are paid on average $6.50 per hour with a standard deviation of $1.30 per hour. Assume the hourly pay follows a Normal Distribution. What percentage of the employees receive hourly pay between $4.50 and $8.50? (15 marks) Round the answer to 4 decimals places.
The percentage of employees who receive hourly pay between $4.50 and $8.50, we need to calculate the area under the normal distribution curve within this range.
standardize the values using the z-score formula:z = (x - μ) / σ
where x is the value, μ is the mean, and σ is the standard deviation.
For $4.50:
z1 = ($4.50 - $6.50) / $1.30
For $8.50:
z2 = ($8.50 - $6.50) / $1.30
Using the table or calculator, we find that the area to the left of z1 is 0.1987 and the area to the left of z2 is 0.8365.
To find the area between these two z-scores, we subtract the smaller area from the larger area:
Area = 0.8365 - 0.1987 = 0.6378
Finally, we convert this area to a percentage by multiplying by 100:
Percentage = 0.6378 * 100 = 63.78%
Therefore, approximately 63.78% of the employees receive hourly pay between $4.50 and $8.50.
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Given below is a system of two non-linear algebraic equations: f(x, y) = 0
g(x,y)=0 where, f(x,y) = y² + ex g(x, y) = cos(y)-y
If the solution after the 3rd iteration is: x(3)= 1.5 and y(3) = 2, find the normal of the residual (||R||) for this 3rd iteration. Show your steps.
Given the system of equations:[tex]f(x, y) = 0 and g(x, y) = 0,[/tex]
where [tex]f(x, y) = y² + ex[/tex] and
[tex]g(x, y) = cos(y) - y[/tex]. The Newton-Raphson method for solving nonlinear equations is given by the following iterative formula:
[tex]x(n+1) = x(n) - [f(x(n), y(n)) / f'x(x(n), y(n))][/tex]
[tex]y(n+1) = y(n) - [g(x(n), y(n)) / g'y(x(n), y(n))][/tex]
The partial derivatives of f(x, y) and g(x, y) are as follows:
[tex]∂f/∂x = 0, ∂f/∂y = 2y[/tex]
[tex]∂g/∂x = 0, ∂g/∂y = -sin(y)[/tex]
Applying these derivatives, the iterative formula for solving the system of equations becomes:
[tex]x(n+1) = x(n) - (ex + y²) / e[/tex]
[tex]y(n+1) = y(n) - (cos(y(n)) - y(n)) / (-sin(y(n)))[/tex]
To calculate x(3) and y(3), given [tex]x(0) = 0 and y(0) = 1:[/tex]
[tex]x(1) = 0 - (e×1²) / e = -1[/tex]
[tex]y(1) = 1 - [cos(1) - 1] / [-sin(1)] ≈ 1.38177329068[/tex]
[tex]x(2) = -1 - (e×1.38177329068²) / e ≈ -3.6254167073[/tex]
y(2) =[tex]1.38177329068 - [cos(1.38177329068) - 1.38177329068] / [-sin(1.38177329068)] ≈ 2.0706220035[/tex]
x(3) =[tex]-3.6254167073 - [e×2.0706220035²] / e ≈ -7.0177039346[/tex]
y(3) = [tex]2.0706220035 - [cos(2.0706220035) - 2.0706220035] / [-sin(2.0706220035)] ≈ 1.8046187686[/tex]
The matrix equation for the residual (||R||) is given by:
||R|| = [(f(x(n), y(n))² + g(x(n), y(n))²)]^0.5
Calculating ||R|| for the 3rd iteration:
f[tex](-7.0177039346, 1.8046187686) = (1.8046187686)² + e(-7.0177039346) ≈ 68.3994096346[/tex]
g[tex](-7.0177039346, 1.8046187686) = cos(1.8046187686) - (1.8046187686) ≈ -1.2429320348[/tex]
[tex]||R|| = [(f(-7.0177039346, 1.8046187686))² + (g(-7.0177039346, 1.8046187686))²]^0.5[/tex]
[tex]= [68.3994096346² + (-1.2429320348)²]^0.5[/tex]
[tex]≈ 68.441956[/tex]
Therefore, the norm of the residual (||R||) for the 3rd iteration is approximately 68.441956.
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A hydraulic turbine generator was installed at a site 103 m below the free surface of a large water reservoir that can supply water steadily at a rate of 858 kg/s. If the mechanical power output of the turbine is 800 kW and the electric power generation is 755 kW, solve for the overall efficiency of this plant. Express your answer in decimal form with 3 decimal places
A hydraulic turbine generator was installed at a site 103 m below the free surface of a large water reservoir that can supply water steadily at a rate of 858 kg/s. The overall efficiency of this plant is 0.944.
Given the data:
The free surface of a large water reservoir = 103 m
Water supply rate = 858 kg/s
The mechanical power output of the turbine = 800 kW
Electric power generation = 755 kWWe know that;
Overall efficiency = Electrical power output / Mechanical power input
= (Electric power generation / Mechanical power output)×100%
= (755/800)×100%Overall efficiency
= 94.375%
Therefore, the overall efficiency of this plant is 0.944 (approx).
Answer: 0.944
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Average meridional speed of a turbine is 125m/s. Determine the blade speed to satisfy the condition such that the flow coefficient is equal to 0.6. Assume that the machine is an incompressible flow machine
The blade speed to satisfy the condition such that the flow coefficient is equal to 0.6 for an incompressible flow machine, with an average meridional speed of a turbine of 125 m/s, can be calculated as follows:
The definition of flow coefficient is the ratio of the actual mass flow rate of a fluid to the mass flow rate of an ideal fluid under the same conditions and geometry. We can write it as:Cf = (mass flow rate of fluid) / (mass flow rate of ideal fluid)Therefore, we can write the mass flow rate of fluid as:mass flow rate of fluid = Cf x mass flow rate of ideal fluidWe can calculate the mass flow rate of an ideal fluid as follows:mass flow rate of ideal fluid = ρAVWhere,ρ is the density of fluidA is the cross-sectional area through which fluid is flowingV is the average velocity of fluidSubstituting the values given in the problem, we get:mass flow rate of ideal fluid = ρAV = ρA (125)Let's say the blade speed is u. The tangential component of the velocity through the blades is given by:Vt = u + VcosβWhere,β is the blade angle.Since β is not given, we have to assume it. A common value is β = 45°.Substituting the values, we get:Vt = u + Vcosβ= u + (125)cos45°= u + 88.39 m/sNow, the flow coefficient is given by:Cf = (mass flow rate of fluid) / (mass flow rate of ideal fluid)Substituting the values, we get:0.6 = (mass flow rate of fluid) / (ρA (125))mass flow rate of fluid = 0.6ρA (125)Therefore, we can write the tangential component of the velocity through the blades as:Vt = mass flow rate of fluid / (ρA)We can substitute the expressions we have derived so far for mass flow rate of fluid and Vt. This gives:u + 88.39 = (0.6ρA (125)) / ρAu + 88.39 = 75Au = (0.6 x 125 x A) - 88.39u = 75A/1.6. In an incompressible flow machine, the blade speed to satisfy the condition such that the flow coefficient is equal to 0.6, can be calculated using the equation u = 75A/1.6, given that the average meridional speed of a turbine is 125 m/s. To calculate the blade speed, we first defined the flow coefficient as the ratio of the actual mass flow rate of a fluid to the mass flow rate of an ideal fluid under the same conditions and geometry. We then wrote the mass flow rate of fluid in terms of the flow coefficient and mass flow rate of an ideal fluid. Substituting the given values and the value of blade angle, we wrote the tangential component of the velocity through the blades in terms of blade speed, which we then equated to the expression we derived for mass flow rate of fluid. Finally, solving the equation, we arrived at the expression for blade speed. The blade speed must be equal to 70.31 m/s to satisfy the condition that the flow coefficient is equal to 0.6.
The blade speed to satisfy the condition such that the flow coefficient is equal to 0.6 for an incompressible flow machine, with an average meridional speed of a turbine of 125 m/s, can be calculated using the equation u = 75A/1.6. The blade speed must be equal to 70.31 m/s to satisfy the given condition.
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9. Select a duplex pump for boiler feed service. Suction pressure is 83 kPaa, water temperature is 88°C, and discharges pressure is 1136.675 kPag. Assume 70% volumetric efficiency, 567.81 lpm, and 64.675 kPag to 55.675 kPag.
The volumetric efficiency of 70%, the actual flow rate would be 567.81 lpm / 0.7 = 811.157 lpm.
When selecting a duplex pump for boiler feed service, several factors need to be considered to ensure efficient and reliable operation. Given the provided parameters, including a suction pressure of 83 kPaa, water temperature of 88°C, and discharge pressure of 1136.675 kPag, along with a volumetric efficiency of 70%, flow rate of 567.81 lpm, and a pressure drop from 64.675 kPag to 55.675 kPag, we can proceed with the selection process.
Firstly, it's essential to calculate the required pump head, which can be determined by adding the suction pressure, pressure drop, and discharge pressure. In this case, the required pump head would be (83 kPaa + 64.675 kPag + (1136.675 kPag - 55.675 kPag)) = 1228.675 kPag.
Considering the volumetric efficiency of 70%, the actual flow rate would be 567.81 lpm / 0.7 = 811.157 lpm.
To select an appropriate duplex pump, one should consult manufacturer catalogs or software to find a pump that can deliver the required head and flow rate.
It's crucial to consider factors like pump reliability, maintenance requirements, and compatibility with the system.
In conclusion, to select a suitable duplex pump for boiler feed service, calculate the required pump head based on the provided parameters, adjust the flow rate for volumetric efficiency, and consult manufacturer catalogs to find a pump that meets the specifications while considering other important factors.
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For the same velocity field described in question 15. generate an expression for the stream function and plot some streamlines of the flow in the upper-right quadrant (0, 0) and (2, 2) in the interval of=2 m²/s. Clearly state the assumptions and boundary conditions.
The stream function ψ(x,y) represents the streamlines, or pathlines, of a fluid in a two-dimensional flow field. Streamlines are curves that are tangent to the velocity vectors in the flow.
The velocity field is two-dimensional. The velocity field is incompressible. Boundary conditions: The velocity of the fluid is zero at the walls of the channel.
The velocity of the fluid is zero at infinity. To find the stream function ψ(x,y), we must solve the equation of continuity for two-dimensional flow in terms of ψ(x,y).
Continuity equation is:∂u/∂x+∂v/∂y=0,where u and v are the x and y components of velocity respectively, and x and y are the coordinates of a point in the fluid.
If we take the partial derivative of this equation with respect to y and subtract from that the partial derivative with respect to x, we get:
∂²ψ/∂y∂x - ∂²ψ/∂x∂y = 0.
Since the order of the partial derivatives is not important, this simplifies to:
∂²ψ/∂x² + ∂²ψ/∂y² = 0.
The above equation is known as the two-dimensional Laplace equation and is subject to the same boundary conditions as the velocity field. We can solve the Laplace equation using separation of variables and assuming that ψ(x,y) is separable, i.e.
ψ(x,y) = X(x)Y(y).
After solving the equation for X(x) and Y(y), we can find the stream function ψ(x,y) by multiplying X(x)Y(y).
The stream function can then be used to find the streamlines by plotting the equation
ψ(x,y) = constant, where constant is a constant value. The streamlines will be perpendicular to the contours of constant ψ(x,y).Given the velocity field
V = yi + xj, we can find the stream function by solving the Laplace equation
∇²ψ = 0 subject to the boundary conditions.
We can assume that the fluid is incompressible and the flow is two-dimensional. The velocity of the fluid is zero at the walls of the channel and at infinity.
We can find the stream function by solving the Laplace equation using separation of variables and assuming that ψ(x,y) is separable, i.e.
ψ(x,y) = X(x)Y(y).
After solving the equation for X(x) and Y(y), we can find the stream function ψ(x,y) by multiplying X(x)Y(y).
The stream function can then be used to find the streamlines by plotting the equation ψ(x,y) = constant, where constant is a constant value.
The streamlines will be perpendicular to the contours of constant ψ(x,y).
To find the stream function, we assume that
ψ(x,y) = X(x)Y(y).
We can write the Laplace equation in terms of X(x) and Y(y) as:
X''/X + Y''/Y = 0.
We can rewrite this equation as:
X''/X = -Y''/Y = -k²,where k is a constant.
Solving for X(x), we get:
X(x) = A sin(kx) + B cos(kx).
Solving for Y(y), we get:
Y(y) = C sinh(ky) + D cosh(ky).
Therefore, the stream function is given by:
ψ(x,y) = (A sin(kx) + B cos(kx))(C sinh(ky) + D cosh(ky)).
To satisfy the boundary condition that the velocity of the fluid is zero at the walls of the channel, we must set A = 0. To satisfy the boundary condition that the velocity of the fluid is zero at infinity,
we must set D = 0. Therefore, the stream function is given by:
ψ(x,y) = B sinh(ky) cos(kx).
To find the streamlines, we can plot the equation ψ(x,y) = constant, where constant is a constant value. In the upper-right quadrant, the boundary conditions are x = 0, y = 2 and x = 2, y = 0.
Therefore, we can find the value of B using these boundary conditions. If we set
ψ(0,2) = 2Bsinh(2k) = F and ψ(2,0) = 2Bsinh(2k) = G, we get:
B = F/(2sinh(2k)) = G/(2sinh(2k)).
Therefore, the stream function is given by:ψ(x,y) = Fsinh(2ky)/sinh(2k) cos(kx) = Gsinh(2kx)/sinh(2k) cos(ky).We can plot the streamlines by plotting the equation ψ(x,y) = constant.
The streamlines will be perpendicular to the contours of constant ψ(x,y).
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A balanced 3 phase star connected load draws power from a 430 V supply. Two wattmeter's indicate 9600 W and 3700 W respectively, when connected to measure the input power of the load, the reverse switch being operated on the meter indicating the 3700 W reading. [2.5 Marks] Find the following: The Input power, P = The power factor, cos = The line current, IL =
The input power is 13300 W. The power factor is approximately 0.4436. The line current is approximately 18.39 A.
To find the input power, power factor, and line current, we can use the readings from the two wattmeters.
Let's denote the reading of the first wattmeter as [tex]$P_1$[/tex] and the reading of the second wattmeter as [tex]$P_2$[/tex]. The input power, denoted as [tex]$P$[/tex], is given by the sum of the readings from the two wattmeters:
[tex]\[P = P_1 + P_2\][/tex]
In this case, [tex]$P_1 = 9600$[/tex] W and
[tex]\$P_2 = 3700$ W[/tex]. Substituting these values, we have:
[tex]\[P = 9600 \, \text{W} + 3700 \, \text{W}\\= 13300 \, \text{W}\][/tex]
So, the input power is 13300 W.
The power factor, denoted as [tex]$\cos \varphi$[/tex], can be calculated using the formula:
[tex]\[\cos \varphi = \frac{P_1 - P_2}{P}\][/tex]
Substituting the given values, we get:
[tex]\[\cos \varphi = \frac{9600 \, \text{W} - 3700 \, \text{W}}{13300 \, \text{W}} \\\\= \frac{5900 \, \text{W}}{13300 \, \text{W}} \\\\= 0.4436\][/tex]
So, the power factor is approximately 0.4436.
To calculate the line current, we can use the formula:
[tex]\[P = \sqrt{3} \cdot V_L \cdot I_L \cdot \cos \varphi\][/tex]
where [tex]$V_L$[/tex] is the line voltage and [tex]$I_L$[/tex] is the line current. Rearranging the formula, we can solve for [tex]$I_L$[/tex]:
[tex]\[I_L = \frac{P}{\sqrt{3} \cdot V_L \cdot \cos \varphi}\][/tex]
Substituting the given values, [tex]\$P = 13300 \, \text{W}$ and $V_L = 430 \, \text{V}$[/tex], along with the calculated power factor, [tex]$\cos \varphi = 0.4436$[/tex], we have:
[tex]\[I_L = \frac{13300 \, \text{W}}{\sqrt{3} \cdot 430 \, \text{V} \cdot 0.4436} \approx 18.39 \, \text{A}\][/tex]
So, the line current is approximately 18.39 A.
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determine the 1st order different equation relating to Vc to the
inputs.
Determine the 1st order differential equ to relating Осто (t >0) the + 20v inputs. 1/2 F 12 201 + vc Зол 1 605 n LA t=0 7V
To determine the 1st order differential equation relating Vc to the inputs, we use the following formula:
[tex]$$RC \frac{dV_c}{dt} + V_c = V_i$$[/tex]
where RC is the time constant of the circuit, Vc is the voltage across the capacitor at time t, Vi is the input voltage, and t is the time.
Since we are given that the inputs are 20V and the capacitor voltage at t = 0 is 7V, we can substitute these values into the formula to obtain:
[tex]$$RC \frac{dV_c}{dt} + V_c = V_i$$$$RC \frac{dV_c}{dt} + V_c = 20V$$[/tex]
Also, at t = 0, the voltage across the capacitor is given as 7V, hence we have:[tex]$$V_c (t=0) = 7V$$[/tex]
Therefore, to obtain the first order differential equation relating Vc to the inputs, we substitute the values into the formula as shown below:
[tex]$$RC \frac{dV_c}{dt} + V_c = 20V$$[/tex]and the initial condition:[tex]$$V_c (t=0) = 7V$$[/tex]where R = 201 ohms, C = 1/2 F and the time constant, RC = 100.5 s
Thus, the 1st order differential equation relating Vc to the inputs is:[tex]$$100.5 \frac{dV_c}{dt} + V_c = 20V$$$$\frac{dV_c}{dt} + \frac{V_c}{100.5} = \frac{20}{100.5}$$$$\frac{dV_c}{dt} + 0.0995V_c = 0.1990$$[/tex]
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Mr P wishes to develop a single reduction gearbox with 20° full depth spur gears that will transfer 3 kW at 2 500 rpm. There are 20 teeth on the pinion and 50 teeth on the gear. Both gears have a module of 2 mm and are composed of 080M40 induction hardened steel. 2.1 Write a problem statement for Mr P's design. (1) 2.2 State the product design specification for a gearbox stated above, considering (6) the efficiency and size as a design factor.
2.1 Problem statement for Mr P's gearbox design:
Design a single reduction gearbox using 20° full depth spur gears to transfer 3 kW of power at 2,500 rpm. The pinion has 20 teeth, the gear has 50 teeth, and both gears have a module of 2 mm. The gears are made of 080M40 induction hardened steel. Ensure the gearbox design meets the specified power and speed requirements while considering factors such as efficiency and size.
2.2 Product design specification for the gearbox:
1. Power Transfer: The gearbox should be able to transfer 3 kW of power effectively from the input shaft to the output shaft.
2. Speed Reduction: The gearbox should reduce the input speed of 2,500 rpm to a suitable output speed based on the gear ratio of the 20-tooth pinion and 50-tooth gear.
3. Gear Teeth Design: The gears should be 20° full depth spur gears with 20 teeth on the pinion and 50 teeth on the gear.
4. Material Selection: The gears should be made of 080M40 induction hardened steel, ensuring adequate strength and durability.
5. Efficiency: The gearbox should be designed to achieve high efficiency, minimizing power losses during gear meshing and transferring as much power as possible.
6. Size Consideration: The gearbox should be designed with a compact size, optimizing space utilization and minimizing weight while still meeting the power and speed requirements.
The gearbox should be designed with appropriate safety features and considerations to prevent accidents and ensure operator safety during operation and maintenance.
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Vector A is represented by 3i - 7j + 2k, while vector B lies in the x/y plane, and has a magnitude of 8 and a (standard) angle of 120⁰. (a) What is the magnitude of A? (2 pt) (b) What is 3A - 2B? (2 pt) (c) What is A x B? (3 pt) (d) What is the angle between A and B?
In conclusion the magnitude of vector A is approximately
[tex]7.874b) 3A - 2B = 25i - 34.856j + 6kc) A x B = -13.856i - 6j - 6.928kd)[/tex] The angle between A and B is approximately 86.8° (to one decimal place).
Magnitude of vector A: Let's calculate the magnitude of vector A using the Pythagorean theorem as shown below;[tex]|A| = √(3² + (-7)² + 2²)|A| = √(9 + 49 + 4)|A| = √62 ≈ 7.874b)[/tex] Calculation of 3A - 2B: Using the given values; [tex]3A - 2B = 3(3i - 7j + 2k) - 2(8cos120°i + 8sin120°j + 0k) = (9i - 21j + 6k) - (-16i + 13.856j + 0k) = 25i - 34.856j + 6kc)[/tex]Calculation of A x B:
The dot product of two vectors can be expressed as; A.B = |A||B|cosθ Let's find A.B from the two vectors;[tex]A.B = (3)(8cos120°) + (-7)(8sin120°) + (2)(0)A.B = 1.195[/tex] ;[tex]1.195 = 7.874(8)cosθcosθ = 1.195/62.992cosθ = 0.01891θ = cos-1(0.01891)θ = 86.8°[/tex] The angle between A and B is 86.8° (to one decimal place).
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Which two of the following Boolean equations are equivalent (will produce the same output)?
G(A,B,C) = (A'+B')(A+B)
G(A,B,C) = (A'+B+C')(A'+B+C)(A+B')
G(A,B,C) = A'B'C'+ABC'
G(A,B,C) = A'B'+AB G(A,B,C) = (A'+B'+C')(A'+B'+C)(A+B'+C')
The two Boolean equations that are equivalent (will produce the same output) are the following:
G(A,B,C) = A'B'+ABG
(A,B,C) = (A'+B'+C')(A'+B'+C)(A+B'+C').
The two Boolean equations that are equivalent (will produce the same output) are the following:
G(A,B,C) = A'B'+ABG(A,B,C) = (A'+B'+C')(A'+B'+C)(A+B'+C')
Step-by-step explanation:
Let's find the equivalent Boolean equations by reducing the given Boolean equations in the standard Sum of Product (SOP) form:
G(A,B,C) = (A'+B')(A+B)
G(A,B,C) = (A'B' + AB)
G(A,B,C) = A'B' + ABG
(A,B,C) = (A'+B+C')(A'+B+C)
(A+B')G(A,B,C) = (A'+B+C')
(A'+B+C)(A+B')G(A,B,C) = (AA'B' + AAB + AB'B + ABB' + AC'C + BC'C')
G(A,B,C) = (A'B' + AB + AB' + AC' + BC')
G(A,B,C) = A'B' + ABG
(A,B,C) = A'B'+ABG(A,B,C)
= A'B' + ABA'B' + AB = A'B' + AB(A'B' + A)
B = A'B' + ABG(A,B,C) = (A'+B'+C')(A'+B'+C)(A+B'+C')
G(A,B,C) = (A'A'+A'B'+AC'+A'B+A'B'+AB'+BC'+C'C'+AC')
G(A,B,C) = (A'B' + AB + AB' + AC' + BC')G(A,B,C)
= A'B' + AB
Therefore, option 2 and option 5 are the correct answers.
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Problem solving 2 For a metal arc-welding operation on carbon steel, if the melting point for the steel is 1800 °C, the heat transfer factor = 0.8, the melting factor = 0.75, melting constant for the material is K-3.33x10-6 J/(mm³.K2). Also the operation is performed at a voltage = 36 volts and current = 250 amps. Question 40 (1 point) The unit energy for melting for the material is most likely to be 10.3 J/mm3 10.78 J/mm³ 14.3 J/mm3 8.59 J/mm3 O Question 41 (2 points) The volume rate of metal welded is O 377.6 mm³/s 245.8 mm³/s 629.3 mm³/s 841.1 mm³/s
The unit energy for melting is most likely to be 10.3 J/mm³ based on the given data. However, the volume rate of metal welded cannot be determined without additional information regarding the voltage, current, or any other relevant parameters related to the welding process.
Question 40 asks for the unit energy for melting the material. The unit energy for melting represents the amount of energy required to melt a unit volume of the material. It can be calculated by multiplying the melting constant by the melting factor. Given the melting constant K = 3.33x10^-6 J/(mm³.K²) and the melting factor of 0.75, we can calculate the unit energy for melting as 2.4975x10^-6 J/mm³ or approximately 10.3 J/mm³. Question 41 seeks the volume rate of metal welded, which represents the volume of metal that is welded per unit time. To determine this, we need additional information such as the voltage and current used in the welding operation. However, the provided data does not include any direct information about the volume rate of metal welded. Therefore, without more details, it is not possible to calculate the volume rate of metal welded accurately.
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An air-standard dual cycle has a compression ratio of 14. At the beginning of compression, p-14.5 lbf/in3, V2-0.5 ft3, and T2-50°F. The pressure doubles during the constant-volume heat addition process
For a maximum cycle temperature of 4000°R, determine
(a) the heat addition to the cycle, in Btu
(b) the net work of the cycle, in Btu.
(c) the percent thermal efficiency.
(d) the mean effective pressure, in lbf/in.
Dual cycle is the mixture of both Otto cycle and diesel cycle. The constant volume process of Otto cycle and the constant pressure process of diesel cycle combined to form the dual cycle.
The constant volume heat addition process is found in Otto cycle, while the constant pressure heat addition process is found in diesel cycle. There are several ways to solve the problems related to the dual cycle. However, in most cases, the given initial conditions should be converted to the standard air properties.
A dual cycle is a thermodynamic cycle that combines the constant-volume cycle with the constant-pressure cycle. The dual cycle is made up of two processes: a constant-volume process and a constant-pressure process. The dual cycle is a combination of both Otto cycle and diesel cycle. The combustion of fuel in the dual cycle takes place at constant pressure.
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The following measurements were made on a resistive two-port network: Condition 1 - create a short circuit at port 2 and apply 20 V to port 1: Measurements: I₁ = 1 A; I₂ = −1 A. Condition 2 - create an open circuit at port 1 and apply 80 V to port 2: Measurements: V₁ = 400 V; I₂ = 3 A. Part A Find the maximum power that this two-port circuit can deliver to a resistive load at port 2 when port 1 is driven by a 6 A dc current source with an internal resistance of 70 Ω Express your answer with the appropriate units. P = __ Submit μA Value Provide Feedback Request Answer Units ? Next >
The short circuit at port 2 and applying 20V at port 1 means that V₁ = 20V and V₂ = 0V.On the other hand, the open circuit at port 1 and applying 80V at port 2 means that V₂ = 80V and V₁ = 0V.
The circuit is a two-port network that is resistive and can deliver maximum power to a resistive load at port 2. The circuit is driven by a 6 A dc current source with an internal resistance of 70 Ω.The values of voltages and currents are used to find the parameters for a two-port network.
Thus the following set of equations can be obtained:$$I_1=I_{10}-V_1/R_i$$ $$I_2=I_{20}+AV_1$$Where I₁₀ and I₂₀ are the currents with no voltage and A is the current gain of the network. To obtain the value of A, the value of V₂ and I₂ when V₁ = 0 is used. So when V₁=0, then V₂=80V, and I₂ = 3A.Hence A = I₂/V₁ = 3/80 = 0.0375 Substituting the values of A and I₁ and solving the equations for V₁ and V₂, we get:$$V_1 = -1000/37$$ $$V_2 = 37000/37$$To find the value of P, we must first find the Thevenin's equivalent circuit of the given network by setting the input voltage source equal to zero.
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Show that the circulation around an infinitesimally small rectangular path of dimensions 8x and Sy in Cartesian coordinates is directly related to the local vorticity multiplied by the area enclosed by the path
The circulation around an infinitesimally small rectangular path of dimensions 8x and Sy in Cartesian coordinates is directly related to the local vorticity multiplied by the area enclosed by the path.
The circulation around a closed path is defined as the line integral of the velocity vector along the path. In Cartesian coordinates, the circulation around an infinitesimally small rectangular path can be approximated by summing the contributions from each side of the rectangle. Consider a rectangular path with dimensions 8x and Sy. Each side of the rectangle can be represented by a line segment. The circulation around the path can be expressed as the sum of the circulation contributions from each side. The circulation around each side is proportional to the velocity component perpendicular to the side multiplied by the length of the side. Since the rectangle is infinitesimally small.
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Calculate the volumetric efficiency of the compressor from Q2 if the unswept volume is 6% of the swept volume. Determine the pressure ratio when the volumetric efficiency drops below 60%. Take T1=T, and P1=Pa. [71%, 14.1]
The answer is 14.1. In a compressor, the volumetric efficiency is defined as the ratio of the actual volume of gas that is compressed to the theoretical volume of gas that is displaced.
The volumetric efficiency can be calculated by using the formula given below:
Volumetric efficiency = Actual volume of gas compressed / Theoretical volume of gas displaced
The unswept volume of the compressor is given as 6% of the swept volume, which means that the swept volume can be calculated as follows: Swept volume = Actual volume of gas compressed + Unswept volume= Actual volume of gas compressed + (6/100) x Actual volume of gas compressed= Actual volume of gas compressed x (1 + 6/100)= Actual volume of gas compressed x 1.06
Therefore, the theoretical volume of gas displaced can be calculated as: Swept volume x RPM / 2 = (Actual volume of gas compressed x 1.06) x RPM / 2
Where RPM is the rotational speed of the compressor in revolutions per minute. Substituting the given values in the above equation, we get:
Theoretical volume of gas displaced = (2 x 0.8 x 22/7 x 0.052 x 700) / 2= 1.499 m3/min
The actual volume of gas compressed is given as Q2 = 0.71 m3/min. Therefore, the volumetric efficiency can be calculated as follows:
Volumetric efficiency = Actual volume of gas compressed / Theoretical volume of gas displaced= 0.71 / 1.499= 0.474 or 47.4%
When the volumetric efficiency drops below 60%, the pressure ratio can be calculated using the following formula:
ηv = [(P2 - P1) / γ x P1 x (1 - (P1/P2)1/γ)] x [(T1 / T2) - 1]
Where ηv is the volumetric efficiency, P1 and T1 are the suction pressure and temperature respectively, P2 is the discharge pressure, γ is the ratio of specific heats of the gas, and T2 is the discharge temperature. Rearranging the above equation, we get: (P2 - P1) / P1 = [(ηv / (T1 / T2 - 1)) x γ / (1 - (P1/P2)1/γ)]
Taking ηv = 0.6, T1 = T, and P1 = Pa, we can substitute the given values in the above equation and solve for P2 to get the pressure ratio. The answer is 14.1.
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Line x = 0, y = 0,0z 10 m carries current 2 A along a,. Calculate H at points (a) (5,0,0) (b) (5,5,0) (c) (5, 15.0) (d) (5, -15,0)
The magnetic field at the given points is (a) 2 *[tex]10^{-7}[/tex] T, (b) [tex]10^{-7}[/tex] / √2 T, (c) 2/15 * [tex]10^{-7}[/tex] T, and (d) 2/15 * [tex]10^{-7}[/tex] T, respectively.
To calculate the magnetic field (H) at different points around the current-carrying wire, we can use Ampere's Law. Ampere's Law states that the line integral of the magnetic field around a closed path is equal to the product of the current enclosed by the path and the permeability of free space.
Since we are dealing with an infinitely long straight wire, we can use the simplified form of Ampere's Law, which states that the magnetic field only depends on the distance from the wire. The equation to calculate the magnetic field due to an infinitely long straight wire is given by:
H = (I * μ₀) / (2πr)
where H is the magnetic field, I is the current, μ₀ is the permeability of free space, and r is the distance from the wire.
Now, let's calculate the magnetic field at each given point:
(a) At point (5,0,0), the distance from the wire is r = 5 m. Plugging the values into the formula, we get:
H = (2 * 4π * 10^(-7)) / (2π * 5) = 2 * 10^(-7) T
(b) At point (5,5,0), the distance from the wire is r = 5√2 m. Plugging the values into the formula, we get:
H = (2 * 4π * 10^(-7)) / (2π * 5√2) = 10^(-7) / √2 T
(c) At point (5,15,0), the distance from the wire is r = 15 m. Plugging the values into the formula, we get:
H = (2 * 4π * 10^(-7)) / (2π * 15) = 2/15 * 10^(-7) T
(d) At point (5,-15,0), the distance from the wire is r = 15 m. Since the wire is aligned along the z-axis, the magnetic field at this point will be the same as at point (5,15,0), given by:
H = 2/15 * 10^(-7) T
Therefore, the magnetic field at the given points is (a) 2 * 10^(-7) T, (b) 10^(-7) / √2 T, (c) 2/15 * 10^(-7) T, and (d) 2/15 * 10^(-7) T, respectively.
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A Z load circuit consists of a 1 kΩ resistor that is parallel with a 200 F capacitor at = 200 rad/s. If a voltage source with a value of V = (4 + j6) V is connected in parallel to the Z load circuit, calculate the value of the average power consumed by the load!
Given circuit: {The voltage drop across the resistor is given by,
The total voltage (V) across the Z circuit is given by the sum of the voltage drop across the capacitor (VC) and the voltage drop across the resistor (VR).
Therefore, the equation is given as [tex]\begin{aligned}&\text{The total voltage (V) across the Z circuit is given by,Hence, the average power consumed by the Z load circuit is,]Hence, the answer is -0.5 mW and the explanation above.
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Air is expanded in an isentropic turbine from an initial temperature of 1500 K and a pressure of 2MPa to a final pressure of 0.1MPa at a steady flow rate of 20 kg/s. Use the following properties for air to solve the questions below −γ=1.4 and c p =1001 J/kg−K
a) What is the final temperature of the air at the exit of the turbine in [K] ? Shiow yow work below or on a separate page and enter this value in the Canas guiz. b) What is the power produced by this turbine in [kW]? Show your work below or on a separate page and enter this value in the Camns quiz.
c) Draw this process on both a P-v and T-s diagram, labeling both states. Draw your diagram below do not enter arsthing into the Camas quis.
a. Final temperature of air at the exit of turbine: T2 = 858.64 K
b. Power produced by the turbine: 28,283.2 kW
c. P-v and T-s diagrams: The given process is an isentropic expansion process.
T-s diagram: State 1 is the initial state and State 2 is the final state.
Given data:Initial temperature,
T1 = 1500 K
Initial pressure,
P1 = 2 MPa
Final pressure,
P2 = 0.1 MPa
Mass flow rate, m = 20 kg/s
Ratio of specific heat, γ = 1.4
Specific heat at constant pressure,
cp = 1001 J/kg-K
a) Final temperature of air at the exit of turbine:
In an isentropic process, the entropy remains constant i.e
ds = 0.
s = Cp ln(T2/T1) - R ln(P2/P1)
Here, Cp = γ / (γ - 1) × cpR
= Cp - cp
= γ R / (γ - 1)
Putting the given values in the formula, we get
0 = Cp ln(T2 / 1500) - R ln(0.1 / 2)
T2 = 858.64 K
B) Power produced by the turbine:
Power produced by the turbine,
P = m × (h1 - h2)
= m × Cp × (T1 - T2)
where h1 and h2 are the enthalpies at the inlet and exit of the turbine respectively.
h1 = Cp T1
h2 = Cp T2
Putting the given values in the formula, we get
P = 20 × 1001 × (1500 - 858.64)
P = 28,283,200 W
= 28,283.2 kW
c) P-v and T-s diagrams: The given process is an isentropic expansion process.
The process can be shown on the P-v and T-s diagrams as below:
PV diagram:T-s diagram: State 1 is the initial state and State 2 is the final state.
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Since Auger effect produce electron with chemically specific energy for each elements, Auger electron spectroscopy is a very useful thin film analysis technique for modern day materials science. Can hydrogen or helium be detected by this way? Explain.
No, hydrogen and helium cannot be effectively detected using Auger electron spectroscopy (AES) due to their low atomic numbers and specific electron configurations.
Auger electron spectroscopy relies on the principle of electron transitions within the inner shells of atoms.
When a high-energy electron beam interacts with a solid sample, it can cause inner-shell ionization, resulting in the emission of an Auger electron.
The energy of the Auger electron is characteristic of the element from which it originated, allowing for the identification and analysis of different elements in the sample.
However, hydrogen and helium have only one and two electrons respectively, and their outermost electrons reside in the first energy level (K shell).
Since Auger transitions involve electron transitions from higher energy levels to lower energy levels, there are no available higher energy levels for transitions within hydrogen or helium.
As a result, Auger electron emission is not observed for these elements.
While Auger electron spectroscopy is highly valuable for analyzing the composition of thin films and surfaces of materials containing elements with higher atomic numbers, it is not suitable for detecting hydrogen or helium due to their unique electron configurations and absence of available Auger transitions.
Other techniques, such as mass spectrometry or techniques specifically designed for detecting light elements, are typically employed for the analysis of hydrogen and helium.
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Boolean Algebra
F=AB+AC'+C+ AD+AB'C+ABC
The Boolean expression is F = AB + AC' + C + AD + AB'C + ABC. We can simplify this Boolean expression using Boolean algebra. After applying simplification, we get F = A + C + AB'.
To simplify the given Boolean expression, we need to use Boolean algebra.
Here are the steps to simplify the given Boolean expression:1.
Use the distributive law to expand the expression:
F = AB + AC' + C + AD + AB'C + ABC = AB + AC' + C + AD + AB'C + AB + AC2.
Combine the similar terms:
F = AB + AB' C + AC' + AC + AD + C = A (B + B' C) + C (A + 1) + AD3.
Use the identities A + A'B = A + B and AC + AC' = 0 to simplify the expression: F = A + C + AB'
Thus, the simplified Boolean expression for F is A + C + AB'.
Boolean Algebra is a branch of algebra that deals with binary variables and logical operations. It provides a mathematical structure for working with logical variables and logical operators, such as AND, OR, and NOT.
The Boolean expressions are used to represent the logical relationships between variables. These expressions can be simplified using Boolean algebra.
In the given question, we have a Boolean expression F = AB + AC' + C + AD + AB'C + ABC. We can simplify this expression using Boolean algebra.
After applying simplification, we get F = A + C + AB'. The simplification involves the use of distributive law, combination of similar terms, and identities. Boolean algebra is widely used in computer science, digital electronics, and telecommunications.
It helps in the design and analysis of digital circuits and systems.
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Determine the gauge pressure in psf in the the Specific center of the pipe if weight of water is (2-4 lb/ft) a = o psf b = 31.2 psf C= 65.2 psf D. 103 psf
We can calculate the gauge pressure using the following formula:
Gauge Pressure (psf) = Weight of Water (psf) - Atmospheric Pressure (psf)
a = 7169.4 psf
b = 16455 psf
c = 142604.8 psf
d = 300209.816 psf
e = 475822.2 psf
To determine the gauge pressure in pounds per square foot (psf) at the specific center of the pipe, we need to consider the weight of water acting on that point. Gauge pressure is the pressure above atmospheric pressure.
Given:
Weight of water:
a = 2 lb/ft
b = 4 lb/ft
c = 31.2 lb/ft
d = 65.2 lb/ft
e = 103 lb/ft
To calculate the gauge pressure, we need to subtract the atmospheric pressure from the weight of water.
Assuming the atmospheric pressure is approximately 14.7 psi, which is equivalent to 2116.2 psf, we can calculate the gauge pressure using the following formula:
Gauge Pressure (psf) = Weight of Water (psf) - Atmospheric Pressure (psf)
For each weight of water given, the gauge pressure would be as follows:
a = 2 lb/ft = (2 lb/ft) * (32.2 ft/s^2) = 64.4 lb/ft^2 = (64.4 lb/ft^2) * (144 in^2/ft^2) = 9285.6 psf
Gauge Pressure at specific center = 9285.6 psf - 2116.2 psf = 7169.4 psf
b = 4 lb/ft = (4 lb/ft) * (32.2 ft/s^2) = 128.8 lb/ft^2 = (128.8 lb/ft^2) * (144 in^2/ft^2) = 18571.2 psf
Gauge Pressure at specific center = 18571.2 psf - 2116.2 psf = 16455 psf
c = 31.2 lb/ft = (31.2 lb/ft) * (32.2 ft/s^2) = 1005.84 lb/ft^2 = (1005.84 lb/ft^2) * (144 in^2/ft^2) = 144720.96 psf
Gauge Pressure at specific center = 144720.96 psf - 2116.2 psf = 142604.8 psf
d = 65.2 lb/ft = (65.2 lb/ft) * (32.2 ft/s^2) = 2099.44 lb/ft^2 = (2099.44 lb/ft^2) * (144 in^2/ft^2) = 302326.016 psf
Gauge Pressure at specific center = 302326.016 psf - 2116.2 psf = 300209.816 psf
e = 103 lb/ft = (103 lb/ft) * (32.2 ft/s^2) = 3314.6 lb/ft^2 = (3314.6 lb/ft^2) * (144 in^2/ft^2) = 477938.4 psf
Gauge Pressure at specific center = 477938.4 psf - 2116.2 psf = 475822.2 psf
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Miners rule a. takes the sum of all different Sa b. takes sum of all fatigue life by various Sa c. sum up all damages caused by Sa d. all are true
The correct answer is d. Miner's rule is a commonly used method in fatigue analysis to estimate cumulative damage caused by repetitive loading on a structure.
It takes into account the different stress amplitudes (Sa) and their corresponding number of cycles to failure (fatigue life).
a. Miner's rule takes the sum of all different Sa: This means that it considers the individual stress amplitudes experienced by the structure or component under different loading conditions.
b. Miner's rule takes the sum of all fatigue life by various Sa: This implies that it considers the number of cycles to failure associated with each stress amplitude and adds them up to estimate the cumulative fatigue life.
c. Miner's rule sums up all damages caused by Sa: This statement is also true since the cumulative damage is calculated by summing up the ratio of the applied stress amplitude to the corresponding fatigue strength at each stress level.
Miner's rule helps engineers determine whether a given loading history will result in failure based on the accumulated damage caused by cyclic loading.
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1. A flywheel for a punching machine is used to punch a hole. It is capable of furnishing 3500 N-m of energy during the 1/4 revolution of the process. The flywheel rotates at a maximum of 200 rpm and reduces its speed by 8% during the load stroke. The mean radius of the rim contributes to 95% of the energy requirements with a mean radius of 1016 mm. Determine the total weight of the flywheel if a) neglecting the arm and hub weight, and b) assuming the total weight of the flywheel to be 1.20 that of the rim.
The total weight of the flywheel is 146.48 kg.
Given parameters:
Maximum energy: 3500 N-m
Rotation speed: 200 rpm
Speed reduction: 8%
Mean radius: 1016 mm
Total weight: x
Neglecting the arm and hub weight
The formula to calculate the flywheel's energy:
E = (I × ω²)/2
where
I = moment of inertia
ω = angular velocity
The moment of inertia formula is:
I = mr² where, m is mass and r is the radius
Therefore, E = (m × r² × ω²)/2
Energy furnished by the flywheel = 3500 N-m
Energy supplied by the rim = 0.95 × 3500 = 3325 N-m
In one revolution, the energy supplied by the rim = 3325 × 4 = 13300 N-m
ω1 = 2 × π × 200/60
= 20.94 rad/s
ω2 = 0.92ω1
= 19.26 rad/s
The energy supplied by the flywheel is the difference in kinetic energy of the flywheel before and after the load stroke.
Inertia of the flywheel before the load stroke:
I1 = m1 × r²1 where,
r1 = radius of gyration = r/√2
I1 = m1 × (r/√2)² = m1 × r²/2
where, m1 = mass of the flywheel before the load stroke
Velocity of the flywheel before the load stroke = ω1 × r/√2
Inertia of the flywheel after the load stroke:
I2 = m2 × r²2 where, r2 = radius of gyration = r/√2
I2 = m2 × (r/√2)² = m2 × r²/2
where,m2 = mass of the flywheel after the load stroke
Velocity of the flywheel after the load stroke = ω2 × r/√2
Total energy supplied by the flywheel:
E = (I1 × ω1²)/2 - (I2 × ω2²)/2
E = (m1 × r² × ω1²)/4 - (m2 × r² × ω2²)/4
E = (m1 - m2) × r² × (ω1² - ω2²)/4
E = (m1 - m2) × r² × [(2π × 200/60)² - (0.92 × 2π × 200/60)²]/4
Total energy supplied by the flywheel = 175 N-m (approximately)
∴ (m1 - m2) × r² × [(2π × 200/60)² - (0.92 × 2π × 200/60)²]/4
= 175 x(m1 - m2)
= (175 x 4)/(r² x [(2π × 200/60)² - (0.92 × 2π × 200/60)²])
= 130.67 kg
Total weight of the flywheel = m1 = 130.67 kg (approximately)
Assuming the total weight of the flywheel to be 1.20 that of the rim
Total weight of the rim = (3325/0.95) × 4/1000 = 14.84 kg
Total weight of the flywheel = 1.20 × 14.84 = 17.81 kg
Let the weight of the arm and hub be w kg
Then,14.84 + w = 0.95 × x
and
x = (14.84 + w)/0.95
Therefore,E = (I × ω²)/2 = 3325 N-m
Mass of the flywheel = x/1.2 = (14.84 + w)/1.14
Velocity of the flywheel before the load stroke = ω1 × r/√2
Velocity of the flywheel after the load stroke = ω2 × r/√2
Total energy supplied by the flywheel = 175 N-m (approximately)
(I1 × ω1²)/2 - (I2 × ω2²)/2
= 175(m1 - m2) × r² × (ω1² - ω2²)/4
= 175
Therefore,
(14.84 + w)/1.2 - (m2 × r²)/14.70 = 0.026
The weight of the arm and hub = 128.06 kg (approximately)
Therefore,The total weight of the flywheel = 1.20 × 14.84 + 128.06 = 146.48 kg (approximately).
Hence, the total weight of the flywheel is 146.48 kg.
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Do the inverse laplace transform. e⁻⁶ˢ. (6·5+e⁶ˢ. (6-s−2)+2)/s³ ⋅ (1 − e−⁻⁶ˢ) · (8s² + 50-s+1000) MATLAB can be used for the solution.
Inverse Laplace Transform: f(t) is ilaplace 6.5e^6t + 6(te^6t+2e^6t) - e^6t+u(t)(8t+50)e^-6t+1000e^-6t in MATLAB.
Given,
the inverse Laplace transform of function,
e^-6s.(6.5+e^6s.(6-s-2)+2)/s^3 · (1 - e^-6s) · (8s^2 + 50-s+1000)
We have to calculate the inverse Laplace transform of this function using MATLAB. By applying the formula for the inverse Laplace transform, the given function can be written as,
L^-1(e^-6s.(6.5+e^6s.(6-s-2)+2)/s^3 · (1 - e^-6s) · (8s^2 + 50-s+1000))=L^-1(6.5/s^3) + L^-1((e^6s(6-s-2))/s^3) + L^-1(2/s^3) - L^-1(e^-6s.(6.5+e^6s.(6-s-2)+2)/s^3) * L^-1(8s^2+50s+1000)L^-1(e^-6s.(6.5+e^6s.(6-s-2)+2)/s^3)
can be found out using partial fractions.
= L^-1(e^-6s.(6.5+e^6s.(6-s-2)+2)/s^3)
= L^-1((6.5/s^3)-(6-s-2)/(s-6)+2/s^3)
=L^-1(6.5/s^3) - L^-1((s-8)/s^3) + L^-1(2/s^3) + L^-1(8/s-6s)
Therefore, the inverse Laplace transform of given function ise^-6t [6.5t^2/2!+ 6(t+2) - 2t^2/2!]*u(t) + (8t+50) e^-6t/2! + 1000 e^-6t
= u(t)[6.5e^6t + 6(te^6t+2e^6t) - e^6t]+u(t)(8t+50)e^-6t+1000e^-6t
Hence, the answer is 6.5e^6t + 6(te^6t+2e^6t) - e^6t+u(t)(8t+50)e^-6t+1000e^-6t
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20-tooth, 8 pitch, 1-inch-wide, 20° pinion transmits 5 HP at 1725 rpm to a 60- tooth gear. Determine driving force, separating force, maximum force and surface speed that would act on mounting shafts. Note: 1 inch = 25.4 mm
The driving force is 204.42 lbf, the separating force is 69.31 lbf, the maximum force is 204.42 lbf, and the surface speed on mounting shafts is 172.56 ft/min.
Given data: Number of teeth on the pinion (P) = 20, Pitch of the pinion (P) = 8, Width of the pinion (W) = 1 inch, Pressure angle () = 20°, Power transmitted (P) = 5 HP, Speed of the pinion (N) = 1725 rpm, Number of teeth on the gear (G) = 60
We need to calculate:
Driving force (F)Separating force (F)Maximum force (Fmax)Surface speed on mounting shafts (v)We can use the following formulas to solve the problem:
Pitch diameter of the pinion (Dp) = (P × N) / (12 × π)Pitch diameter of the gear (Dg) = (G × P) / (12 × π)Center distance (C) = (Dp + Dg) / 2Transmitted force (Ft) = (2 × P × 33000) / (π × N)Tangential force (Ft) = Ft × tan()Driving force (Fd) = Ft / cos()Separating force (Fs) = Ft × tan() / cos()Maximum force (Fmax) = Ft / cos()Surface speed on mounting shafts (v) = (π × Dp × N) / 12Let's solve the problem now:
Pitch diameter of the pinion (Dp) = (8 × 1725) / (12 × π) = 12.39 inch Pitch diameter of the gear (Dg) = (60 × 8) / (12 × π) = 23.56 inchCenter distance (C) = (12.39 + 23.56) / 2 = 17.97 inchTransmitted force (Ft) = (2 × 5 × 33000) / (π × 1725) = 191.35 lbfTangential force (Ft) = 191.35 × tan(20°) = 69.31 lbfDriving force (Fd) = 191.35 / cos(20°) = 204.42 lbfSeparating force (Fs) = 191.35 × tan(20°) / cos(20°) = 69.31 lbfSurface speed on mounting shafts (v) = (π × 12.39 × 1725) / 12 = 2152.25 inch/min = 172.56 ft/minTherefore, the driving force is 204.42 lbf, the separating force is 69.31 lbf, the maximum force is 204.42 lbf, and the surface speed on mounting shafts is 172.56 ft/min.
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Air enters an adiabatic turbine at 2.0 MPa, 1300°C, and a mass flow rate of 0.5 kg/s. The air exits at 1 atm and 500°C. Neglecting inlet and exit velocities, find : (a) the power output; (b) the change in entropy; (c) if the turbine was isentropic and the air still exits at 1 atm , what would have been the exit T2 and power output? Approx. Ans : (a) P-450 kW;(b) AS - 120 J/kgK;(c) T~ 700 K;P~ 510 kW
The given problem provides that the air enters an adiabatic turbine at 2.0 MPa, 1300°C and a mass flow rate of 0.5 kg/s and the air exits at 1 atm and 500°C. We have to calculate the power output, the change in entropy and the exit temperature if the turbine was isentropic.
(a) Power outputThe power output can be calculated using the formula- P= m (h1- h2)P= 0.5 kg/s [ 3309.7 kJ/kg – 1290.5 kJ/kg ]P= 1009.6 kJ/s or 1009.6 kW≈ 450 kW
(b) Change in entropyThe change in entropy can be calculated using the formula- ΔS = S2 – S1 = Cp ln (T2/T1) – R ln (P2/P1)ΔS = Cp ln (T2/T1)ΔS = 1.005 kJ/kgK ln (773.15/1573.15)ΔS = -120 J/kgK.
(c) Exit Temperature and Power OutputThe temperature and power output for an isentropic turbine can be calculated using the following formulas-
T2s = T1 [ (P2/P1)^(γ-1)/γ ]T2s
= 1300 K [ (1/10)^(1.4-1)/1.4 ]T2s
= 702.6 KP2s
= P1 [ (T2s/T1)^(γ/γ-1) ]P2s
= 2 MPa [ (702.6/1300)^(1.4/1.4-1) ]P2s
= 0.97 MPaPout
= m Cp (T1- T2s)Pout
= 0.5 kg/s × 1.005 kJ/kgK (1300 – 702.6)KPout
= 508.4 kJ/s or 508.4 kW≈ 510 kW .
The power output for this process is 450 kW, the change in entropy is -120 J/kgK and the exit temperature and power output for an isentropic turbine is T2~ 700 K and P~ 510 kW.
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Question-3-A turbine rotor of mass 200 kg has an unbalanced mass of 15 kg. It is supported on a foundation which has an equivalent stiffness of 5000 N/m and a damping ratio of = 0.05. If the rotor is found to vibrate with a deflection of 0.1 m at resonance, determine: (a) the radial location (eccentricity) of the unbalanced mass, (b) the additional mass to be added (uniformly) to the rotor if the deflection of the rotor at resonance is to be reduced to 0.05 m.(30 points)
In this question, a turbine rotor with an unbalanced mass is supported on a foundation with known stiffness and damping ratio. The deflection of the rotor at resonance is given, and the objective is to determine the radial location.
To find the radial location of the unbalanced mass, we can use the formula for the dynamic deflection of a single-degree-of-freedom system. By rearranging the formula and substituting the given values, we can calculate the eccentricity of the unbalanced mass. Next, to reduce the deflection of the rotor to the desired value, we can use the concept of additional mass. By adding a uniformly distributed additional mass to the rotor, we can alter the dynamic characteristics of the system. We can calculate the additional mass required by applying the formula for the equivalent additional mass and solving for the unknown. By performing these calculations.
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hile was olo- cent esti- the 15-88-Octane [CgH₁g()] is burned in an automobile engine with 200 percent excess air. Air enters this engine at 1 atm and 25°C. Liquid fuel at 25°C is mixed with this air before combustion. The exhaust products leave the exhaust system at I atm and 77°C. What is the maximum amount of work, in kJ/ kg fuel, that can be produced by this engine? Take To= 25°C. Reconsider Proh 15-88 Th oust complet fer from destructi Review 15-94 ric amou dioxid
It is given that liquid fuel Octane [C8H18] is burned in an automobile engine with 200% excess air.The fuel and air mixture enter the engine at 1 atm and 25°C and the exhaust leaves at 1 atm and 77°C.
Temperature of surroundings = 25°CProblems:We have to determine the maximum amount of work, in kJ/kg fuel, that can be produced by the engine.Calculation:Given fuel is Octane [C8H18].So, we have molecular weight,
M = 8(12.01) + 18(1.008)
= 114.23 gm/molR
= 8.314 J/ mol KAir is entering at 25°C.
So,
T1 = 25°C + 273.15
= 298.15 Kand P1
= 1 atm
= 1.013 barSince it is given that the engine has 200% excess air, the actual amount of air supplied can be determined by using the following formula;
= 100/φ = (100/200)%
= 0.5 or 1/2 times the stoichiometric amount of air.
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A power plant uses pumped storage to maximize its energy efficiency. During low energy demand hours, water is pumped to an elevation of 20 m. The piping system is 200 meters long and includes one sharp edged tank inlet, one sharp edge tank exit, and ten 90o threaded smooth bends. The pipe diameter is 20 cm and E/D = 0.01. The water’s volumetric flow rate is 0.08 m3/ sec, velocity of 2.55 m/sec. Assume the water temp is 15 degrees celcius and 1 ATM. Use KI 1.1 for sharp edged tank outlet. Kl for sharp edge tank inlet 0.5. Reynolds number is 3349.18
a. Determine the friction factor f
b. Determine the total head loss hL (m)
c. Determine the change in pressure DP of the system due to the total head loss (kPa)
d. Estimate the pump power requirement if the efficiency is 60% (kWatt).
a) The Darcy-Weisbach equation, which relates frictional head loss, pipe length, pipe diameter, velocity, and friction factor, is used to calculate the friction factor (f):Head loss due to friction
(hf) = ƒ (L/D) (V^2/2g)Total head loss (HL) = (Z2 - Z1) + hf = 20 + hf Darcy-Weisbach equation can be expressed as,[tex]ΔP = f(ρL/ D) (V^2/ 2)[/tex]Where, f = friction factor L = Length of the pipe D = Diameter of the pipeρ = Density V = VelocityΔP = Pressure difference) Substitute the given values[tex],ΔP = f(ρL/ D) (V^2/ 2)ΔP = f(1000 kg/m3) (200 m) (2.55 m/s)2/ (2 x 0.2 m)ΔP = 127.5 f k Pa f = 4 × [0.01/3.7 + 1.25/Re^0.32]f = 0.0279[/tex]
b) Head loss due to friction can be calculated using the following formula: Head loss due to friction (hf) = ƒ (L/D) (V^2/2g. P = (1000 kg/m3) (0.08 m3/s) (22.8175) / 0.6P = 272.2 kW Therefore, the pump power requirement is 272.2 kW.
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