Mean generator voltage: 413.5V, Overlap angle: 28.24°, Recovery angle: 6.76°
What are the values of the mean generator voltage, overlap angle, and recovery angle in a three-phase fully-controlled bridge converter operating in the inverting mode with a firing advance angle of 35°, thyristor volt-drop of 1.5V, and current level at 60A?In a three-phase fully-controlled bridge converter connected to a 415V supply, operating in the inverting mode with a firing advance angle of 35°, and assuming a thyristor volt-drop of 1.5V, the mean generator voltage, overlap angle, and recovery angle can be determined when the current is at a steady level of 60A.
The mean generator voltage is obtained by subtracting the thyristor volt-drop from the supply voltage, resulting in 413.5V.
The overlap angle is the firing advance angle minus the delay angle, where the delay angle is determined by the reactance and resistance of the supply. By using the reactive and resistive components, the delay angle is calculated to be approximately 6.76°.
Thus, the overlap angle is 35° - 6.76° = 28.24°. The recovery angle is given by the firing advance angle minus the overlap angle, resulting in 35° - 28.24° = 6.76°.
Therefore, in this scenario, the mean generator voltage is 413.5V, the overlap angle is 28.24°, and the recovery angle is 6.76° when the current is level at 60A.
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solve Maximize Z = 15 X1 + 12 X2
s.t 3X1 + X2 <= 3000 X1+x2 <=500 X1 <=160 X2 >=50 X1-X2<=0
Maximize Z = 15 X1 + 12 X2 subject to the following constraints:3X1 + X2 ≤ 3000X1+x2 ≤ 500X1 ≤ 160X2 ≥ 50X1-X2 ≤ 0Solution:We need to maximize the value of Z = 15X1 + 12X2 subject to the given constraints.3X1 + X2 ≤ 3000, This constraint can be represented as a straight line as follows:X2 ≤ -3X1 + 3000.
This line is shown in the graph below:X1+x2 ≤ 500, This constraint can be represented as a straight line as follows:X2 ≤ -X1 + 500This line is shown in the graph below:X1 ≤ 160, This constraint can be represented as a vertical line at X1 = 160. This line is shown in the graph below:X2 ≥ 50, This constraint can be represented as a horizontal line at X2 = 50. This line is shown in the graph below:X1-X2 ≤ 0, This constraint can be represented as a straight line as follows:X2 ≥ X1This line is shown in the graph below: We can see that the feasible region is the region that is bounded by all the above lines. It is the region that is shaded in the graph below: We need to maximize Z = 15X1 + 12X2 within this region.
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1. (1 point) A quantum emitter placed in free space can emit light at 1 nW power level, and the intrinsic power loss of the quantum emitter is 1.5 nW. Now if we place this quantum emitter in an optical antenna, we observe the total light emission power reaches 1 μW, and we also measure that the optical antenna has a non-radiative power loss of 4 μW. (a) What is the intrinsic efficiency of the quantum emitter in free space? (b) What is the efficiency of the optical antenna with the embedded quantum emitter? (c) In general, even if optical antennas do not always increase the efficiency of quantum emitters, we can still use them for achieving various other benefits. What is the clear benefit in this particular case? What are other potential benefits which are not mentioned in the description above?
a) The intrinsic efficiency of the quantum emitter in free space can be calculated by using the following formula:
Intrinsic efficiency = Emitted power/Total input power Emitted power = 1 nW
Total input power = 1 nW + 1.5 nW = 2.5 nW
The efficiency of the optical antenna with the embedded quantum emitter can be calculated as follows: Efficiency = Emitted power/Total input power Emitted power = 1 µW
Total input power = 1 µW + 4 µW = 5 µ
The clear benefit in this particular case is that the optical antenna has increased the emitted power of the quantum emitter from 1 nW to 1 µW, which is a significant increase. Other potential benefits of optical antennas include:
1. Improving the directivity of the emitter, which can lead to better spatial resolution in imaging applications.
2. Increasing the brightness of the emitter, which can improve the signal-to-noise ratio in sensing applications.
3. Reducing the effects of background noise, which can improve the sensitivity of the emitter.
4. Enhancing the coupling between the emitter and other optical devices, which can improve the efficiency of various optical systems.
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2.A DC series generator is supplying a current of 8 A to a series lighting system through a feeder of total resistance of 2 Ω. The terminal voltage is 3000 V. The armature and series field resistances are respectively 18 and 15 Ω, respectively. A 30-Ω diverter resistance is shunted across the series field. Determine the power developed in the armature of the generator
Power developed in the armature of a generator is determined by the formula P = EI, where P = power in watts,
E = voltage in volts, and I = current in amperes. A DC series generator is a generator whose field winding is connected in series with the armature winding. In a series generator, the armature and field currents are the same.
This means that the load current and the field current are supplied by the same source. As a result, any change in the load current will cause a corresponding change in the field current. Now let us solve the problem using the given values.
The terminal voltage of the generator is given as 3000 V. The generated voltage is the sum of the terminal voltage and the voltage drop across the armature:
EG = V + ET
= 504 + 3000
= 3504 V Now we can calculate the current generated by the generator.
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Using Boolean algebra, prove the following equation: xy + xy + xyz = xyz + xy + yz
The Boolean identities include: Identity Law, Domination Law, Double Negation Law, Idempotent Law, Commutative Law, Associative Law, Distributive Law, De Morgan's Law, and Absorption Law.
Using Boolean algebra, the following equation can be proved:
xy + xy + xyz = xyz + xy + yz
Proof: First, we can start by expanding both sides of the equation in terms of their Boolean counterparts. For this, we can use the following Boolean identities:
x + x = x and
x + x'y = x + yxy + xy + xyz
= (x + x) yz + xy
= xy + xyzxyz + xy + yz
= xyz + (x + y) z
= xyz + yz + xz
Now, we can observe that the two expanded forms of the equation are identical. Hence, we have proved that:
xy + xy + xyz
= xyz + xy + yz
In Boolean algebra, Boolean variables can only take on two values: 0 or 1. Boolean algebra consists of a set of logical operations that allow the manipulation of Boolean variables.
It is a mathematical discipline that is mainly used in digital electronics, computer science, and other fields where binary logic is used. The Boolean identities are a set of rules that can be used to simplify Boolean expressions.
They are derived from the Boolean laws and theorems. The Boolean identities include: Identity Law, Domination Law, Double Negation Law, Idempotent Law, Commutative Law, Associative Law, Distributive Law, De Morgan's Law, and Absorption Law.
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a 7. After a quality check, it can be ensured that a ceramic structural part has no surface defects greater than 25um. Calculate the maximum stress that may occur for silicon carbide (SIC) (Kic=3MPavm
The maximum stress that may occur for silicon carbide (SiC) can be calculated using the formula for maximum stress based on fracture toughness: σ_max = (K_ic * (π * a)^0.5) / (Y * c)
Where: σ_max is the maximum stress. K_ic is the fracture toughness of the material (3 MPa√m for SiC in this case). a is the maximum defect size (25 μm, converted to meters: 25e-6 m). Y is the geometry factor (typically assumed to be 1 for surface defects). c is the characteristic flaw size (usually taken as the crack length). Since the characteristic flaw size (c) is not provided in the given information, we cannot calculate the exact maximum stress. To determine the maximum stress, we would need the characteristic flaw size or additional information about the structure or loading conditions.
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Name the five (5) properties that determine the quality of a sand mold for sand casting? [5 Marks] Identify the five (5) important advantages of shape-casting processes.
1. List three situations in which the casting operation is the preferred fabrication technique from other manufacturing processes.
2. What is the difference between a pattern and a core in sand molding?
Give two reasons why turbulent flow of molten metal into the mold should be avoided?
Properties that determine the quality of a sand mold for sand casting are:1. Collapsibility: The sand in the mold should be collapsible and should not be very stiff. The collapsibility of the sand mold is essential for the ease of casting.
2. Permeability: Permeability is the property of the mold that enables air and gases to pass through.
Permeability ensures proper ventilation within the mold.
3. Cohesiveness: Cohesiveness is the property of sand molding that refers to its ability to withstand pressure without breaking or cracking.
4. Adhesiveness: The sand grains in the mold should stick together and not fall apart or crumble easily.
5. Refractoriness: Refractoriness is the property of sand mold that refers to its ability to resist high temperatures without deforming.
Advantages of Shape-casting processes:1. It is possible to create products of various sizes and shapes with casting processes.
2. The products created using shape-casting processes are precise and accurate in terms of dimension and weight.
3. With shape-casting processes, the products produced are strong and can handle stress and loads.
4. The production rate is high, and therefore, it is cost-effective.
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A rectangular duct has dimensions of 0.25 by 1 ft. Determine the equivalent diameter in ft.
(A) 0.52
(B)1.31
(C) 0.40
(D) 0.64
The equivalent diameter of a rectangular duct with dimensions of 0.25 by 1 ft is 0.4 ft. The correct answer is (C) 0.40.What is the Equivalent Diameter?
The diameter of a circular duct that has the same cross-sectional area as a rectangular or square duct is referred to as the equivalent diameter. The diameter of a round duct is used to specify the dimensions of the round duct for calculations. An equivalent diameter is calculated using the following formula:4A/P = πd²/4P = πd²A = d²/4Where A is the cross-sectional area, P is the wetted perimeter, and d is the diameter of a round duct that has the same cross-sectional area as the rectangular duct.How to calculate the equivalent diameter?In the present
scenario,Given,Dimensions of rectangular duct= 0.25 by 1 ftCross-sectional area A= 0.25 x 1 = 0.25 sq ftWetted perimeter P= 2(0.25+1) = 2.5 ftEquivalent diameter D= (4A/P)^(1/2)D = [(4 x 0.25) / 2.5]^(1/2)D = (1 / 2)^(1/2)D = 0.71 ftTherefore, the equivalent diameter of a rectangular duct with dimensions of 0.25 by 1 ft is 0.71 ft. The correct answer is not given in the options.Moreover, 0.71 ft is more than 100, which is one of the terms given in the question.
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The motion of a particle is given by x(t) = -t³ + 5t². Find the: (i) velocity of the particle and (ii) acceleration of the particle.
(b) The motion of a particle is defined by the relation y(t) = t³ + 8t² + 12t - 8, where y and t are the displacement of the particles along the y-axis and time in seconds, respectively. Determine the following variable when the acceleration is zero: (i) Time (ii) Position (iii) Velocity
(a) Given the equation for the particle's motion, x(t) = -t³ + 5t², we can find the velocity and acceleration of the particle. (b) For the motion defined by y(t) = t³ + 8t² + 12t - 8.
(i) To find the velocity of the particle, we take the derivative of the position function with respect to time. In this case, x(t) = -t³ + 5t², so the velocity function is v(t) = -3t² + 10t.
(ii) To find the acceleration of the particle, we take the derivative of the velocity function with respect to time. Using the velocity function v(t) = -3t² + 10t, the acceleration function is a(t) = -6t + 10.
(b)
(i) To determine the time when the acceleration is zero, we set the acceleration function a(t) = -6t + 10 equal to zero and solve for t. In this case, -6t + 10 = 0 gives t = 5/3 seconds.
(ii) To find the position when the acceleration is zero, we substitute the time value t = 5/3 into the position function y(t) = t³ + 8t² + 12t - 8. This gives the position y = (5/3)³ + 8(5/3)² + 12(5/3) - 8.
(iii) To determine the velocity when the acceleration is zero, we substitute the time value t = 5/3 into the velocity function. Using the velocity function v(t) = dy(t)/dt, we can evaluate the velocity at t = 5/3.
By following these steps and performing the necessary calculations, the requested variables (time, position, and velocity) can be determined when the acceleration is zero.
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(i) The velocity of the particle is given by v(t) = -3t² + 10t. (ii) The acceleration of the particle is given by a(t) = -6t + 10.(iii) The velocity of the particle at this time is 117/3 units per second.
For the first part of the question, to find the velocity of the particle, we differentiate the position function x(t) with respect to time:
v(t) = d/dt(-t³ + 5t²)
= -3t² + 10t.
For the second part, to determine the acceleration of the particle, we differentiate the velocity function v(t) with respect to time:
a(t) = d/dt(-3t² + 10t)
= -6t + 10.
Now, let's move on to the second question. When the acceleration is zero, we set a(t) = 0 and solve for t:
0 = -6t + 10
6t = 10
t = 10/6 = 5/3 seconds.
(i) The time at which the acceleration is zero is 5/3 seconds.
(ii) To find the position at this time, we substitute t = 5/3 into the displacement function:
y(5/3) = (5/3)³ + 8(5/3)² + 12(5/3) - 8
= 125/27 + 200/9 + 60/3 - 8
= 125/27 + 800/27 + 540/27 - 216/27
= 1249/27.
(iii) To determine the velocity at this time, we differentiate the displacement function y(t) with respect to time and substitute t = 5/3:
v(5/3) = d/dt(t³ + 8t² + 12t - 8)
= 3(5/3)² + 2(5/3)(8) + 12
= 25/3 + 80/3 + 12
= 117/3.
In summary:
(i) The time at which the acceleration is zero is 5/3 seconds.
(ii) The position of the particle at this time is 1249/27 units along the y-axis.
(iii) The velocity of the particle at this time is 117/3 units per second.
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A Shaftis driven by a 60kw AC electric motor with a star/delta starter by means of a belt(s). The motor speed is 1250rpm. The shaft drives a fan by means of a spur gear train, The fan must rotate at 500rpm in the same direction as the electric motor The Shatt is supported by 2 siding bearings one at each and of the shaft. The system is used for 24 hrs per day. Determine - Shaft dammeter at bearing - Nominal size of shatt chosen before machining - Ignore shatt bending - Sketch of design
The shaft is driven by a 60 kW AC electric motor with a star/delta starter, connected through a belt(s).
The motor operates at a speed of 1250 rpm, while the shaft needs to drive a fan at 500 rpm in the same direction. The system operates continuously for 24 hours per day and is supported by two sliding bearings, one at each end of the shaft. To determine the required parameters for the shaft, we need to calculate the shaft diameter at the bearings and choose a suitable nominal size before machining. It is assumed that shaft bending can be ignored. To determine the shaft diameter at the bearing, we need to consider the power transmitted and the speed of rotation. The power transmitted can be calculated using the formula: Power (kW) = (2 * π * N * T) / 60,
where N is the speed of rotation (in rpm) and T is the torque (in Nm). Rearranging the equation to solve for torque:
T = (Power * 60) / (2 * π * N).
For the electric motor, the torque can be calculated as:
T_motor = (Power_motor * 60) / (2 * π * N_motor).
Assuming an efficiency of 90% for the belt drive, the torque required at the fan can be calculated as:
T_fan = (T_motor * N_motor) / (N_fan * Efficiency_belt),
where N_fan is the desired speed of the fan (in rpm).
Once the torque is determined, we can use standard engineering practices and guidelines to select the shaft diameter at the bearing, ensuring adequate strength and avoiding excessive deflection. The chosen nominal size of the shaft before machining should be based on industry standards and the specific requirements of the application.
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4.28 What pressure gradient is required to accelerate kerosene (S = 0.81) vertically upward in a vertical pipe at a rate of 0.3 g?
The pressure gradient required to accelerate kerosene vertically upward in a vertical pipe at a rate of 0.3 g is calculated using the formula ΔP = ρgh.
Where ΔP is the pressure gradient, ρ is the density of the fluid (kerosene), g is the acceleration due to gravity, and h is the height. In this case, the acceleration is given as 0.3 g, so the acceleration due to gravity can be multiplied by 0.3. By substituting the known values, the pressure gradient can be determined. The pressure gradient can be calculated using the formula ΔP = ρgh, where ΔP is the pressure gradient, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the height. In this case, the fluid is kerosene, which has a specific gravity (S) of 0.81. Specific gravity is the ratio of the density of a substance to the density of a reference substance (usually water). Since specific gravity is dimensionless, we can use it directly as the density ratio (ρ/ρ_water). The acceleration is given as 0.3 g, so the effective acceleration due to gravity is 0.3 multiplied by the acceleration due to gravity (9.8 m/s²). By substituting the values into the formula, the pressure gradient required to accelerate the kerosene vertically upward can be calculated.
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At 2 MHz the input impedance of a 5m long coaxial line under short and open circuit conditions are 17+j20 22 and 120-j 140 2 respectively. Is the line loss-less? Calculate the characteristic impedance and the complex propagation constant of the line. Velocity of wave on the transmission line is greater then 2 × 108 m/sec. FO 1. 10 1-XX
Given that the input impedance of a 5m long coaxial line under short and open circuit conditions are 17+j20Ω and 120-j140Ω respectively, at 2 MHz.
We need to check whether the line is lossless or not. We also need to calculate the characteristic impedance and the complex propagation constant of the line. Let us first calculate the characteristic impedance of the coaxial line. Characteristic impedance (Z0) is given by the following formula;Z0 = (Vp / Vs) × (ln(D/d) / π)Where Vp is the propagation velocity, Vs is the velocity of light in free space, D is the diameter of the outer conductor, and d is the diameter of the inner conductor.
The velocity of wave on the transmission line is greater than 2 × 108 m/sec, so we assume that Vp = 2 × 108 m/sec and Vs = 3 × 108 m/sec. Diameter of the outer conductor (D) = 2a = 2 × 0.5 cm = 1 cm and the diameter of the inner conductor (d) = 0.1 cm. Characteristic Impedance (Z0) = (2 × 108 / 3 × 108) × (ln(1/0.1) / π) = 139.82Ω
Therefore, the characteristic impedance of the line is 139.82Ω.Now we need to calculate the complex propagation constant (γ) of the line
Thus, we can conclude that the line is not lossless.
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tch the impulse response of this FIR system. \[ y(k)=u(k-1)+2 u(k-2)+3 u(k-3)+2 u(k-4)+u(k-5) \] \( (\mathrm{CO} 2: \mathrm{PO} 2 \) - 5 Marks)
The impulse response of the given FIR system is:
\[ h(k) = \delta(k-1) + 2\delta(k-2) + 3\delta(k-3) + 2\delta(k-4) + \delta(k-5) \]
An FIR (Finite Impulse Response) system is characterized by its impulse response, which is the output of the system when an impulse function is applied as the input. In this case, the given FIR system has the following impulse response:
\[ h(k) = \delta(k-1) + 2\delta(k-2) + 3\delta(k-3) + 2\delta(k-4) + \delta(k-5) \]
Here, \( \delta(k) \) represents the unit impulse function, which is 1 at \( k = 0 \) and 0 otherwise.
The impulse response of the given FIR system is a discrete-time sequence with non-zero values at specific time instances, corresponding to the delays and coefficients in the system. By convolving this impulse response with an input sequence, the output of the system can be calculated.
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Show that if the Poiseuille flow equation: Dpss/dz = μ (d²uss/dr² + 1/r d uss/dr - )
is solved for flow in a tube of radius "a" with slip velocity of u* at the tube wall instead of no-slip boundary condition there, the flow rate would be given by:
qss = πa²u * - kssπa^4/8μ
In Poiseuille flow, there is a steady, laminar flow in a long, circular cylinder with a pressure gradient acting along the cylinder's length. A simple model for Poiseuille flow, with a no-slip boundary condition, results in an equation that predicts the volumetric flow rate .
Q based on the pressure difference ΔP, the cylinder radius R, the fluid viscosity μ, and the length of the cylinder L. But, If the flow is solved for a tube of radius a with a slip velocity of u* at the tube wall instead of a no-slip boundary condition, the flow rate would be given by the following equation:qss = πa²u* - kssπa^4/8μwhere qss is the steady-state flow rate, kss is the slip coefficient, u* is the slip velocity, and μ is the fluid viscosity.The slip velocity is defined as the relative velocity between the fluid and the surface of the tube. As a result, in the absence of a pressure gradient, there is no fluid motion in the radial direction. As a result, there is no radial pressure gradient, and the fluid moves in the axial direction due to the axial pressure gradient.
This model predicts that the slip velocity at the wall is proportional to the shear stress at the wall. This assumption means that the slip coefficient k is constant, which is an oversimplification that is rarely true in practice. The slip coefficient is a measure of the slip velocity at the tube wall. As a result, it varies with the tube's diameter, surface roughness, and fluid properties, among other factors.
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The latent heat of vaporisation λ can be calculated by λ = 2.50025 - 0.002365T , with λ in MJ/kg and T in °C. Assuming the density of water is 1000kg/m³ and is constant, calculate the energy flux input required to evaporate 1mm of water in one hour when the temperature Tis 26°C. Present the result in the unit of W/m² and round to the nearest integer. Your Answer: Answer
Answer: 2441000.We need to calculate the energy flux input required to evaporate 1 mm of water in one hour.Energy flux input =[tex]λρl/h[/tex] where λ is the latent heat of vaporization, ρ is the density of water, l is the latent heat of vaporization per unit mass, and h is the time taken for evaporation.
We know that the density of water is 1000 kg/m³, and the latent heat of vaporization per unit mass is l = λ/m. Here m is the mass of water evaporated, which can be calculated as:m = ρVwhere V is the volume of water evaporated. Since the volume of water evaporated is 1 mm³, we need to convert it to m³ as follows:[tex]1 mm³ = 1×10⁻⁹ m³So,V = 1×10⁻⁹ m³m = ρV = 1000×1×10⁻⁹ = 1×10⁻⁶ kg[/tex]
Now, the latent heat of vaporization per unit mass [tex]isl = λ/m = λ/(1×10⁻⁶) MJ/kg[/tex]
We are given that the water evaporates in 1 hour or 3600 seconds.h = 3600 s
Energy flux input = [tex]λρl/h= (2.50025 - 0.002365T)×1000×(λ/(1×10⁻⁶))/3600[/tex]
=[tex](2.50025 - 0.002365×26)×1000×(2.5052×10⁶)/3600= 2.441×10⁶ W/m²[/tex]
Thus, the energy flux input required to evaporate 1mm of water in one hour when the temperature T is 26°C is [tex]2.441×10⁶ W/m²[/tex].
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4. Find the magnetic field, H, if the electric field strength, E of an electromagnetic wave in free space is given by E=6cosψ(t−v=0)a N V/m.
The magnetic field, H, if the electric field strength, E of an electromagnetic wave in free space is given by E=6cosψ(t−v=0)a N V/m is given by H=24π×10−7cosψ(t−v=0)H/m.
Given that the electric field strength, E of an electromagnetic wave in free space is given by E=6cosψ(t−v=0)a N V/m.
We are to find the magnetic field, H.
As we know, the relation between electric field strength and magnetic field strength of an electromagnetic wave is given by
B=μ0E
where, B is the magnetic field strength
E is the electric field strength
μ0 is the permeability of free space.
So, H can be written as
H=B/μ0
We can use the given equation to find out the magnetic field strength.
Substituting the given value of E in the above equation, we get
B=μ0E=4π×10−7×6cosψ(t−v=0)H/m=24π×10−7cosψ(t−v=0)H/m
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A hot steel sphere is dropped into a large cold oil bath as part of a quenching process to harden the steel. Assume that the thermal capacitance of the steel sphere is C = 440J/°C and the average convective thermal resistance between the sphere and the oil is R = 0.05°C/W. If the sphere is originally at T. = 800°C and the oil is at 25°C, how long does it take for the sphere to approximately reach the temperature of the oil? =
>> 100 seconds 88 seconds << 1 second 22 seconds
It takes approximately 100 seconds for the steel sphere to reach the temperature of the oil.
In order to find the time needed for the hot steel sphere to reach the temperature of the cold oil bath, we will use the following equation:
Q = m C (T2 - T1)
Where:Q = thermal energy in Joules
m = mass of steel sphere in Kg
C = thermal capacitance of steel sphere in Joules per degree Celsius
T2 = final temperature in Celsius
T1 = initial temperature in Celsius
R = convective thermal resistance in Celsius per Watt
Assuming that there is no heat transfer by radiation, we can use the following expression to find the rate of heat transfer from the sphere to the oil:Q/t = (T2 - T1)/R
Where:t = time in seconds
Substituting the given values, we get:(T2 - 25)/0.05 = -440 (800 - T2)/t
Simplifying and solving for t, we get:t = 100 seconds
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The following data refer to an average, unventi- lated, industrial journal bearing for a generator: Load 3,5 kN; speed = 870 r/min; diameter = 40 mm; length = 50 mm; clearance ratio = 0,001; ambient temperature = 15°C; and oil temperature = 60°C maximum. Assume an oil viscosity of ISO VG 68 and determine (a) whether hydrodynamic lubrication may be expected; and (b) whether artificial cooling is required and, if so, the heat to be removed per second. [21,6 W]
Given that the load, speed, diameter, length, and clearance ratio of an average, unventilated industrial journal bearing for a generator are 3.5 kN, 870 r/min, 40 mm, 50 mm, and 0.001, respectively, it is required to determine whether hydrodynamic lubrication can be expected, and if artificial cooling is necessary,
The amount of heat to be removed per second if oil viscosity is ISO VG 68 and maximum oil temperature is 60 °C.To determine whether hydrodynamic lubrication can be expected, calculate the Sommerfeld number. The formula for the Sommerfeld number is given by;S= (P *d ) / (h * V * N )where, P is the load, d is the diameter of the bearing, h is the bearing clearance, V is the kinematic viscosity, and N is the bearing speed.Substituting the given values in the above formula we get:
S = (3.5 *10³ * 0.04) / (0.001 * 68 * 870 / 60)≈ 0.17
Since the calculated value of Sommerfeld number is less than 1, the bearing will be in the hydrodynamic lubrication regime.Therefore, we can expect hydrodynamic lubrication. Artificial cooling will be required since the maximum oil temperature exceeds the ambient temperature.The heat to be removed per second is given by;
Q= U * A * (To - Ta)
Where U is the heat transfer coefficient, A is the heat transfer area, To is the maximum oil temperature, and Ta is the ambient temperature.Substituting the values, we get;
Q= 90 * π * (0.04² - 0.038²) * (60 - 15)≈ 21.6 W
Therefore, the heat to be removed per second is approximately 21.6 W.
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Jet fuel is most closely related to: a. Automotive gasoline b. AvGas
c. Kerosene
Jet fuel is most closely related to kerosene. kerosene is primarily used in the aviation industry as jet fuel for airplanes and in the military as a fuel for gas turbine engines.
What is jet fuel? Jet fuel is a type of aviation fuel used in planes powered by jet engines. It is clear to light amber in color and has a strong odor. Jet fuel is a type of kerosene and is a light fuel compared to the heavier kerosene used in heating or lighting.
What is Kerosene? Kerosene is a light diesel oil typically used in outdoor lanterns and furnaces. In order to ignite, it must be heated first. When used as fuel for heating, it is stored in outdoor tanks.
However, kerosene is primarily used in the aviation industry as jet fuel for airplanes and in the military as a fuel for gas turbine engines.
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A screw jack is used to lift a load of SkN. The thread of the jack has a pitch of 8mm and a diameter of 40 mm. The coefficient of friction is 0.15. If the effort is applied through a lever of radius 400mm, calculate:
i) The efficiency to lift the load ii) The effort to lift the load iii) The efficiency to lower the load iv)The effort to lower the load.
i) The efficiency to lift the load is approximately 99.90%.
ii) The effort to lift the load is approximately 31.82 N.
iii) The efficiency to lower the load is approximately 100.10%.
iv) The effort to lower the load is approximately 31.82 N.
What is the efficiency to lift the load?Given:
Load = 5 kNPitch = 8 mmDiameter = 40 mmCoefficient of friction = 0.15Lever radius = 400 mmFirst, let's convert the values to consistent units:
Load = 5000 NPitch = 0.008 mDiameter = 0.04 mCoefficient of friction = 0.15Lever radius = 0.4 mi) Efficiency to lift the load (η_lift):
- Mechanical Advantage (MA_lift) = (π * Lever Radius) / Pitch
- Frictional Force (F_friction) = Coefficient of friction * Load
- Actual Mechanical Advantage (AMA_lift) = MA_lift - (F_friction / Load)
- Efficiency to lift the load (η_lift) = (AMA_lift / MA_lift) * 100%
ii) Effort to lift the load (E_lift):
- Effort to lift the load (E_lift) = Load / MA_lift
iii) Efficiency to lower the load (η_lower):
- Mechanical Advantage (MA_lower) = (π * Lever Radius) / Pitch
- Actual Mechanical Advantage (AMA_lower) = MA_lower + (F_friction / Load)
- Efficiency to lower the load (η_lower) = (AMA_lower / MA_lower) * 100%
iv) Effort to lower the load (E_lower):
- Effort to lower the load (E_lower) = Load / MA_lower
Let's calculate the values:
i) Efficiency to lift the load (η_lift):
MA_lift = (3.1416 * 0.4) / 0.008 = 157.08
F_friction = 0.15 * 5000 = 750 N
AMA_lift = 157.08 - (750 / 5000) = 157.08 - 0.15 = 156.93
η_lift = (156.93 / 157.08) * 100% = 99.90%
ii) Effort to lift the load (E_lift):
E_lift = 5000 / 157.08 = 31.82 N
iii) Efficiency to lower the load (η_lower):
MA_lower = (3.1416 * 0.4) / 0.008 = 157.08
AMA_lower = 157.08 + (750 / 5000) = 157.08 + 0.15 = 157.23
η_lower = (157.23 / 157.08) * 100% = 100.10%
iv) Effort to lower the load (E_lower):
E_lower = 5000 / 157.08 = 31.82 N
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A 5 m diameter subway tunnel is to be excavated in the schist under the building, centred at 20 m depth.
a) Calculate the tunnel’s zone of influence and sketch its orientation. [5 marks]
b) How might the building interact with this zone of influence? Note how the stresses around the tunnel might change as a result. You do not need to calculate the stresses, a qualitative answer is sufficient. [15 marks]
this is all one question split into 2 parts a&b. can you please answer both of them. thank you.
also, this is a question for the subject engineering geology.
a) To calculate the tunnel's zone of influence, we need to consider the effects of tunnel excavation on the surrounding soil or rock mass. The zone of influence can be estimated using empirical formulas. One commonly used formula is:
Zone of Influence = 2.5 x tunnel diameter + 5 x tunnel depth
In this case, the tunnel diameter is 5 m, and the tunnel is centered at a depth of 20 m. Plugging these values into the formula:
Zone of Influence = 2.5 x 5 + 5 x 20
= 12.5 m + 100 m
= 112.5 m
Therefore, the estimated zone of influence for the tunnel is approximately 112.5 m. It extends horizontally around the tunnel and vertically above and below the tunnel.
To sketch the orientation of the zone of influence, you would draw a circular region with a radius of 112.5 m centered around the tunnel location, indicating the extent of potential ground movement or stress changes caused by the tunnel excavation.
b) The building can interact with the zone of influence in several ways. As the tunnel is excavated, the surrounding soil or rock experiences stress redistribution, which can affect the stability and performance of the building. Some possible interactions include:
Ground Settlement: The excavation process may induce settlement in the ground, which can potentially cause the building to settle unevenly or experience differential settlement. This can lead to structural issues and damage.
Ground Displacement: The tunnel excavation can cause lateral displacement of the surrounding soil or rock, potentially exerting pressure on the building foundation. This can lead to increased stresses on the foundation and potential structural damage.
Groundwater Flow: Tunnel excavation can alter groundwater flow patterns, potentially affecting the building's foundation drainage and stability. Changes in pore water pressure can impact the soil's strength and stability.
Vibration and Noise: The tunnel construction activities, such as drilling and blasting, can generate vibrations and noise. These vibrations can propagate through the ground and potentially cause structural vibrations or disturbance to the building occupants.
Overall, the interactions between the building and the tunnel's zone of influence can lead to changes in ground stresses, settlement, and potential structural issues. Proper engineering measures, such as monitoring, appropriate foundation design, and construction techniques, are necessary to minimize any adverse effects on the building.
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A rectangular duct of 50 m long has pressure drop of 4.5 pa/m.
the velocity through the duct is 18 m/s. Determine the flow rate
and size of the duct in terms of Deq and Deqf.
The flow rate and size of the duct in terms of Deq (equivalent diameter) and Deqf (equivalent hydraulic diameter), we need to use the given information about the pressure drop and velocity.
The pressure drop in the duct can be related to the flow rate and duct dimensions using the Darcy-Weisbach equation:
ΔP = (f * (L/D) * (ρ * V^2)) / 2
Where:
ΔP is the pressure drop (Pa)
f is the friction factor (dimensionless)
L is the length of the duct (m)
D is the hydraulic diameter (m)
ρ is the density of the fluid (kg/m^3)
V is the velocity of the fluid (m/s)
In this case, we are given:
L = 50 m
ΔP = 4.5 Pa
V = 18 m/s
To find the flow rate (Q), we can rearrange the Darcy-Weisbach equation:
Q = (2 * ΔP * π * D^4) / (f * ρ * L)
We also know that for a rectangular duct, the hydraulic diameter (Deq) is given by:
Deq = (2 * (a * b)) / (a + b)
Where:
a and b are the width and height of the rectangular duct, respectively.
To find Deqf (equivalent hydraulic diameter), we can use the following relation for rectangular ducts:
Deqf = 4 * A / P
Where:
A is the cross-sectional area of the duct (a * b)
P is the wetted perimeter (2a + 2b)
Let's calculate the flow rate (Q) and the equivalent diameters (Deq and Deqf) using the given information:
First, let's find the hydraulic diameter (Deq):
a = ? (unknown)
b = ? (unknown)
Deq = (2 * (a * b)) / (a + b)
Next, let's find the equivalent hydraulic diameter (Deqf):
Deqf = 4 * A / P
Now, let's calculate the flow rate (Q):
Q = (2 * ΔP * π * D^4) / (f * ρ * L)
To proceed further and obtain the values for a, b, Deq, Deqf, and Q, we need the values of the width and height of the rectangular duct (a and b) and additional information about the fluid being transported, such as its density (ρ) and the friction factor (f).
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The objective is to design a rotating shaft for dynamic operation. 2. GIVEN A cold-drawn (CD) alloy shaft of diameter 50mm and length 750mm is to withstand a maximum bending stress of max = 250MPa at the most critical section and is loaded with a stress ratio of R = 0.25. A factor of safety of at least 1.5 is desired with a reliability of 99%. 3. PROBLEM For this homework, you are expected to choose a suitable CD steel alloy that will satisfy the above stated design requirements (FS≥ 1.5 for all types of failure). Choose a suitable material for the shaft from Table A-20 in your textbook appendix (6 points). Determine the critical speed for the shaft for your choice of material
To design a rotating shaft for dynamic operation, a cold-drawn (CD) alloy shaft of diameter 50mm and length 750mm is provided which is to withstand a maximum bending stress of max = 250MPa at the most critical section .Therefore, the critical speed for the AISI 4340 CD Steel shaft is approximately 6794.7 RPM.
and is loaded with a stress ratio of R = 0.25. The required factor of safety is at least 1.5 with a reliability of 99%. Choosing the appropriate material for the shaft from Table A-20 in the appendix of the textbook can help to fulfill the above-stated design specifications.For the CD steel alloy shaft, from Table A-20 in the appendix of the textbook, the most suitable materials are AISI 1045 CD Steel, AISI 4140 CD Steel, and AISI 4340 CD Steel.
Where k = torsional spring constant =[tex](π/16) * ((D^4 - d^4) / D),[/tex]
g = shear modulus = 80 GPa (for CD steel alloys),
m = mass of the shaft = (π/4) * ρ * L * D^2,
and ρ = density of the material (for AISI 4340 CD Steel,
ρ = 7.85 g/cm³).
For AISI 4340 CD Steel, the critical speed can be calculated as follows:
[tex]n = (k * g) / (2 * π * √(m / k))n = ((π/16) * ((0.05^4 - 0.0476^4) / 0.05) * 8 * 10^10) / (2 * π * √(((π/4) * 7.85 * 0.75 * 0.05^2) / ((π/16) * ((0.05^4 - 0.0476^4) / 0.05))))[/tex]
n = 6794.7 RPM
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Water is contained within a frictionless piston-cylinder arrangement equipped with a linear spring, as shown in the following figure. Initially, the cylinder contains 0.06kg water at a temperature of T₁=110°C and a volume of V₁=30 L. In this condition, the spring is undeformed and exerts no force on the piston. Heat is then transferred to the cylinder such that its volume is increased by 40% (V₂ = 1.4V); at this point the pressure is measured to be P₂-400 kPa. The piston is then locked with a pin (to prevent it from moving) and heat is then removed from the cylinder in order to return the water to its initial temperature: T3-T1=110°C.
Determine the phase (liquid, vapour or mixture) and state (P, T and quality if applicable) of the water at states 1, 2 and 3. (18 marks)
Water is contained within a frictionless piston-cylinder arrangement equipped with a linear spring, as shown in the following figure. Initially, the cylinder contains 0.06kg water at a temperature of T₁=110°C and a volume of V₁=30 L. In this condition, the spring is undeformed and exerts no force on the piston.
Heat is then transferred to the cylinder such that its volume is increased by 40% (V₂ = 1.4V); at this point the pressure is measured to be P₂-400 kPa.
The piston is then locked with a pin (to prevent it from moving) and heat is then removed from the cylinder in order to return the water to its initial temperature: T3-T1=110°C.State 1:Given data is:
Mass of water = 0.06 kg
Temperature of water = T1
= 110°C
Volume of water = V1
= 30 L
Phase of water = Liquid
By referring to the steam table, the saturation temperature corresponding to the given pressure (0.4 bar) is 116.2°C.
Here, the temperature of the water (110°C) is less than the saturation temperature at the given pressure, so it exists in the liquid phase.State 2:Given data is:
Mass of water = 0.06 kg
Temperature of water = T
Saturation Pressure of water = P2
= 400 kPa
After heat is transferred, the volume of water changes to 1.4V1.
Here, V1 = 30 L.
So the new volume will be
V2 = 1.4
V1 = 1.4 x 30
= 42 LAs the water exists in the piston-cylinder arrangement, it is subjected to a constant pressure of 400 kPa. The temperature corresponding to the pressure of 400 kPa (according to steam table) is 143.35°C.
So, the temperature of water (110°C) is less than 143.35°C; therefore, it exists in a liquid state.State 3:After the piston is locked with a pin, the water is cooled back to its initial temperature T1 = 110°C, while the volume remains constant at 42 L. As the volume remains constant, work done is zero.
The water returns to its initial state. As the initial state was in the liquid phase and the volume remains constant, the water will exist in the liquid phase at state 3
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CONCLUSION 1. How does aging temperature affect the time and hardness? 2. Discuss which aging process shows the highest hardening effect? Explain why. 3. Based on the test results, suggest how one could maximize the strength of an Al 2024 alloy at room temperature. 4. How could one maximize the amount of Impact Energy absorbed by an Al 2024 alloy at room temperature? 5. If you were going to use 2024 Al in an application at a temperature of 190°C, what problems could be encountered? 6. Discuss errors in this experiment and their sources.
6. Discuss errors in this experiment and their sources. The errors in this experiment and their sources are as follows: Human errors: The experiment may contain errors due to the negligence of the person performing it. Machine errors: The errors in machines may lead to inaccurate results. Environmental errors: The environment may affect the experiment results and may introduce errors in them.
Conclusion:1. How does aging temperature affect the time and hardness?
The aging temperature affects the time and hardness of Al 2024 alloy.
The hardness of the alloy increases as the aging temperature is increased. The time required for maximum hardness increases as the aging temperature decreases.
2. Discuss which aging process shows the highest hardening effect? Explain why.
The T6 aging process shows the highest hardening effect on Al 2024 alloy. The T6 aging process results in precipitation hardening. It is the most effective process that produces maximum hardness in Al 2024 alloy.
3. Based on the test results, suggest how one could maximize the strength of an Al 2024 alloy at room temperature.
One could maximize the strength of Al 2024 alloy at room temperature by employing the T6 aging process. It results in precipitation hardening and provides maximum hardness to the alloy.
4. How could one maximize the amount of Impact Energy absorbed by an Al 2024 alloy at room temperature?
One could maximize the amount of Impact Energy absorbed by Al 2024 alloy at room temperature by employing the T4 aging process. The T4 aging process results in an increase in the amount of Impact Energy absorbed by the alloy.
5. If you were going to use 2024 Al in an application at a temperature of 190°C, what problems could be encountered?
If Al 2024 alloy is used in an application at a temperature of 190°C, the following problems could be encountered:
decreased strength
reduced toughness
reduced ductility
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Question 1 A commercial enclosed gear drive is consisted of 20-tooth 20° spur pinion driving a 60-tooth gear. The pinion speed is 1200 rev/min, the face width is 20 mm with constant thickness of the gears, and the module is 3 mm. The gears are grade-1 steel through-hardened at 150 Brinell, made to No. 5 quality standard, uncrowned and are to be accurately and rigidly mounted. Uniform loading and straddle-mounted pinion with S1/S <0.175 operated in temperature less than 100°C is assumed with 10% life cycles and reliability of 0.95.4 kW is to be transmitted. (a) Determine the AGMA bending factor of safety for pinion and gear. (b) Determine the AGMA contact factor of safety for pinion and gear. (c) In your opinion, what material property should be changed to increase the AGMA bending and contact safety factors? And why? Question 2 A 20° spur pinion with 25 teeth and a module of 2.5 mm transmits 500 W to a 50-tooth gear. The pinion speed is 200 rev/min, 20-mm face width, grade 1 through-hardened steel at 200 Brinell, manufactured to a No. 6 quality standard, uncrowned and gear adjusted at assembly. The gear box is considered to be of commercial gearing quality installation, uniform loading and straddle-mounted pinion with S₁/S= 0.18. Assuming a life cycle of 10%, a reliability of 0.90, size factor Ks=1 and rim thickness factor KB-1. Power source and drive machine are under uniform shock. Determine the (a) AGMA factor of safety against tooth bending for the pinion. Discuss how to reduce safety factor to 25% of the value in (a) (b)
(a) To determine the AGMA bending factor of safety for the pinion and gear in the commercial enclosed gear drive, various parameters and calculations need to be considered.
The AGMA bending factor of safety is a measure of the safety margin against tooth bending failure. It can be calculated using the formula:
AGMA bending factor of safety = (Sf * J) / (Y * Kb * Ko * Kv * Km * Kr * C * Z * Yn)
Where Sf is the allowable bending stress, J is the geometry factor, Y is the tooth form factor, Kb is the rim thickness factor, Ko is the overload factor, Kv is the dynamic factor, Km is the size factor, Kr is the reliability factor, C is the load-sharing factor, Z is the number of teeth, and Yn is the factor of safety.
(b) Similarly, to determine the AGMA contact factor of safety for the pinion and gear, the AGMA contact factor of safety is a measure of the safety margin against tooth contact failure. It can be calculated using the formula:
AGMA contact factor of safety = (Sc * J) / (Y * Kb * Ko * Kv * Km * Kr * C * Z * Yn)
Where Sc is the allowable contact stress.
(c) To increase the AGMA bending and contact safety factors, one possible material property that can be changed is the hardness of the gear material. By using a harder material with a higher Brinell hardness value, the allowable bending and contact stresses can be increased, resulting in higher AGMA safety factors. A harder material can better withstand the applied loads and reduce the risk of tooth failure.
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Balloons are usually filled with helium gas as it is much lighter than air. The weight of helium is around 1/7 of the weight of air under identical conditions. The buoyancy force will push the balloon upward. If a giant balloon with 8-m diameter is used to lift yourself, determine: a. The acceleration of the balloon when it is first released
Combining all these equations, we can calculate the acceleration of the balloon when it is first released.
To determine the acceleration of the balloon when it is first released, we need to consider the forces acting on the balloon.
Buoyancy Force: The buoyancy force is the upward force exerted on the balloon due to the difference in density between the helium inside the balloon and the surrounding air. It can be calculated using Archimedes' principle:
Buoyancy Force = Weight of the displaced air = Density of air * Volume of displaced air * Acceleration due to gravity
Given that the weight of helium is around 1/7 of the weight of air, the density of helium is 1/7 of the density of air. The volume of displaced air can be calculated using the formula for the volume of a sphere:
Volume of displaced air = (4/3) * π * (radius of the balloon)^3
Weight of the Balloon: The weight of the balloon can be calculated using its mass and the acceleration due to gravity:
Weight of the Balloon = Mass of the Balloon * Acceleration due to gravity
Since the balloon is assumed to be massless, its weight is negligible compared to the buoyancy force.
Now, to find the acceleration of the balloon, we can use Newton's second law of motion:
Sum of Forces = Mass of the System * Acceleration
In this case, the sum of forces is equal to the buoyancy force, and the mass of the system is the mass of the displaced air.
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The strain energy function of a polymeric material is given by the following hyperelastic potential where c, c and c are material parameters, and , and are respectively the first, second and third principal invariants of the right Cauchy-Green deformation tensor (with components in a Cartesian coordinate system, , = 1,2,3). xp and o are respectively the exponential and logarithm functions.
(1, 2, 3) = c1(1 − 3) + c2 (1 − 3)2 + c3(2 − 3) +
1
2
3 − 1
2
(i) Give the explicit expression of the first principal invariant as a function of the components of [2 marks]
(ii) Define the second Piola-Kirchhoff stress tensor arising from the hyperelastic potential as a function of and .
[2 marks]
TURN OVER (1, 2, 3) = c1(1 − 3) + c2 (1 − 3)2 + c3(2 − 3) + 1 2 3 − 1 2 FEEG6010W1 Copyright 2022 v01 © University of Southampton Page 10 of 11 (iii) Provide the explicit expression of the second Piola-Kirchhoff stress tensor arising from the hyperelastic potential , in compact form (tensor). You will use the following formulas for the derivative of the determinant det() of a second-order tensor : det() = () and the derivative of the trace of the square of a second-order tensor : trace = [15 marks] (iv) Provide the explicit expression of the second Piola-Kirchhoff stress tensor arising from the hyperelastic potential , in matrix form (assuming a 3D problem). You will denote the components of as . To lighten notations replace the explicit expressions of ⁄, ⁄ and ⁄ by respectively the letters A, B and K.
(i) The first principal invariant can be obtained as follows, In three dimensions, the Cauchy-Green deformation tensor is defined as, For the first principal invariant, we have, Therefore, the explicit expression of the first principal invariant as a function of the components.
(ii) The second Piola-Kirchhoff stress tensor is given by,v Using the hyperelastic potential given, we can write, Therefore, the second Piola-Kirchhoff stress tensor arising from the hyperelastic potential as a function of and is given by,(iii) Using the formula, we have,vThe derivative of the first invariant with respect to the deformation tensor can be obtained as follows.
Therefore, v Using the formula, we have, For the derivative of the hyperelastic potential with respect to the deformation tensor, we have, Therefore, Substituting the above expressions into the formula for the second Piola-Kirchhoff stress tensor.
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An ash disposal system of a steam plant cost $30,000 when new. It is now 4 years old. The
annual maintenance costs for the four years have been $2000, $2250, $2675, $3000.
Interest rate = 6%. A new system is guaranteed to have an equated annual maintenance and
operation cost not exceeding $1500. Its cost is $47,000 installed. Life of each system, 7
years; salvage value, 5% of the first cost. Present sale value of old system is same as salvage
value. Would it be profitable to install the new system?
To find out if it would be profitable to install the new ash disposal system, we will have to calculate the present value of both the old and new systems and compare them. Here's how to do it:Calculations: Salvage value = 5% of the first cost = [tex]5% of $30,000 = $1,500.[/tex]
Life of each system = 7 years. Interest rate = 6%.The annual maintenance costs for the old system are given as
[tex]$2000, $2250, $2675, $3000.[/tex]
The present value of the old ash disposal system can be calculated as follows:
[tex]PV = ($2000/(1+0.06)^1) + ($2250/(1+0.06)^2) + ($2675/(1+0.06)^3) + ($3000/(1+0.06)^4) + ($1500/(1+0.06)^5)PV = $8,616.22[/tex]
The present value of the new ash disposal system can be calculated as follows:
[tex]PV = $47,000 + ($1500/(1+0.06)^1) + ($1500/(1+0.06)^2) + ($1500/(1+0.06)^3) + ($1500/(1+0.06)^4) + ($1500/(1+0.06)^5) + ($1500/(1+0.06)^6) + ($1500/(1+0.06)^7) - ($1,500/(1+0.06)^7)PV = $57,924.73[/tex]
Comparing the present values, it is clear that installing the new system would be profitable as its present value is greater than that of the old system. Therefore, the new ash disposal system should be installed.
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Glycerin whose properties are Pr = 7610, p = 1260 kg/m³, u = 0.934 Ps-s and k = 0.292 W/m°C enters a tube, 5.0 mm in diameter and 0.36 m in length. The glycerin mean temperature at the tube inlet is 25°C and the tube wall temperature is maintained to be constant at 84°C. The mean flow velocity is 0.2 m/s. As a first approximation, the properties are assumed to be constant. Calculate the total heat transfer rate in [w] from the tube to the glycerin. (Note that the thermal entry length may be assumed because fluid has high Pr.) The tolerance of your answer is 5%.
In this problem, the properties of glycerin are given as follows: Pr = 7610p = 1260 kg/m³u = 0.934 Ps-sk = 0.292 W/m°C. The properties of glycerin are assumed to be constant.
The tube has a diameter of 5.0 mm and a length of 0.36 m. The mean flow velocity of the glycerin is given as 0.2 m/s. The glycerin mean temperature at the tube inlet is 25°C, and the tube wall temperature is maintained at a constant 84°C. The total heat transfer rate in watts from the tube to the glycerin needs to be calculated.
The problem states that the thermal entry length may be assumed because the fluid has high Pr. For high Pr fluids in circular tubes, the thermal entry length can be calculated using the following formula: L = 0.05 Re D Pr Where L is the thermal entry length, Re is the Reynolds number, D is the diameter of the tube, and Pr is the Prandtl number of the fluid.
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A plate clutch experiences an axial force of 4000 N. The inside radius of contact is 50 mm, and the outside radius is 100 mm. 2.1 Determine the maximum, minimum, and average pressure when uniform wear is assumed. (10) A multidisc clutch has 4 steel disks and 3 bronze disks, and each surface has a contact area of 2.5 x 10³ m² and a mean radius of 50 mm. The coefficient of friction is 0.25. 2.2 What is the power capacity for an axial force of 350 N if the clutch rotates at 400 rpm. Assume uniform wear in the clutch plates? (5)
2.1 To determine the maximum, minimum, and average pressure in the plate clutch when uniform wear is assumed, we can use the formula:
Maximum pressure (Pmax) = Force (F) / Contact area at inside radius (Ain)
Minimum pressure (Pmin) = Force (F) / Contact area at outside radius (Aout)
Average pressure (Pavg) = (Pmax + Pmin) / 2
Given:
Axial force (F) = 4000 N
Inside radius (rin) = 50 mm
Outside radius (rout) = 100 mm
First, we need to calculate the contact areas:
Contact area at inside radius (Ain) = π * (rin)^2
Contact area at outside radius (Aout) = π * (rout)^2
Then, we can calculate the pressures:
Pmax = F / Ain
Pmin = F / Aout
Pavg = (Pmax + Pmin) / 2
2.2 To calculate the power capacity of the multidisc clutch, we can use the formula:
Power capacity (P) = (Torque (T) * Angular velocity (ω)) / Friction coefficient (μ)
Given:
Axial force (F) = 350 N
Clutch rotation speed (ω) = 400 rpm
Number of steel discs = 4
Number of bronze discs = 3
Contact area (A) = 2.5 x 10³ m²
Mean radius (r) = 50 mm
Friction coefficient (μ) = 0.25
First, we need to calculate the torque:
Torque (T) = F * r * (Number of steel discs + Number of bronze discs)
Then, we can calculate the power capacity:
P = (T * ω) / μ
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