A boundary layer forms on a flat plate parallel to an airstream at standard sea level conditions moving at 50 m/s, with laminar and turbulent regions and a transition point.
In the laminar region of the boundary layer, the velocity profile is characterized by smooth and orderly flow. The velocity near the surface of the plate is relatively low, gradually increasing as we move away from the surface. This region is separated from the free stream flow by a region known as the laminar sub-layer, where the velocity gradients are relatively small.
As we move further downstream, the laminar boundary layer transitions into the turbulent region. In this region, the velocity profile becomes more erratic and turbulent, with greater velocity fluctuations and mixing. The transition from laminar to turbulent flow is marked by a point known as the transition point.
The distance from the leading edge of the flat plate to the transition point can be determined using the transition Reynolds number. Given that the transition Reynolds number is 6 x 10^5, the transition point can be calculated by considering the critical Reynolds number for transition from laminar to turbulent flow.
The critical Reynolds number for a flat plate is typically around 5 x 10^5. By comparing the given transition Reynolds number and the critical Reynolds number, we can estimate the distance from the leading edge to the transition point.
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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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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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Calculate the vibration response of the shown system when the force F(t)=65δ(t), where δ(t) is the Dirac delta function. Consider the data of Problem 4 and zero initial conditions.
m= 80kg
k=14500N/m
c=1350N.s/m
a=1.0m
L=2.5m
The vibration response of the system can be calculated by solving the equation of motion using the given force and system parameters. The response will depend on the characteristics of the system, including its mass, stiffness, damping, and geometry.
To calculate the vibration response of the system, we need to solve the equation of motion using the given force and system parameters. The equation of motion for a single-degree-of-freedom system can be represented as:
m * x'' + c * x' + k * x = F(t)
where m is the mass, c is the damping coefficient, k is the stiffness, x is the displacement of the system, x' is the velocity, x'' is the acceleration, and F(t) is the applied force.
In this case, the force is given as F(t) = 65δ(t), where δ(t) is the Dirac delta function. The Dirac delta function represents an instantaneous force impulse. Therefore, the force is applied instantaneously at time t = 0.
To solve the equation of motion, we can assume that the displacement x(t) can be represented as a sum of a particular solution and the homogeneous solution. The homogeneous solution represents the natural response of the system, while the particular solution represents the forced response due to the applied force.
Given the system parameters (mass m, stiffness k, damping c, geometry a, and L), we can use appropriate initial conditions and solve the equation of motion to determine the vibration response of the system over time.
Please note that without specific initial conditions or further information, it is not possible to provide a numerical solution or precise response characteristics for the given system. The solution would involve solving the differential equation, applying appropriate boundary or initial conditions, and obtaining the response in terms of displacement, velocity, or acceleration as a function of time.
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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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On the traditional lathe, the motor applies the indicated moment to pulley A for 10 seconds. Assume the 4" pulleys are solid disks of 10lb and the 1" pulleys are solid disks of 0.5lb each. Pulleys B and C are compounded, pulley D drives the light spindle. Determine the angular velocity of the spindle at 10 seconds. Do not consider the relatively light belts.
Answer:Angular velocity of the spindle at 10 seconds is 67.8rad/s.
The moment in this problem is in pound-force inches (lb-in), which is the same as lb-ft divided by 12. In this problem, the moment of inertia of the four pulleys is required to determine the angular acceleration of the spindle.
The moment of inertia of a solid disk about its axis is I=1/2Mr^2. For 4" disks with mass of 10 lb each,
I=1/2(10lb)(2in)^2=20lb-in-s^2[tex]=1/2(10lb)(2in)^2=20lb-in-s^2[/tex].
For 1" disks with mass of 0.5 lb each, I=1/2(0.5lb)(0.5in)^2=0.063lb-in-s^2.1. Find the moment applied by the motor to pulley A.The moment applied by the motor to pulley A is the same as the moment applied to pulleys B and C since they are compounded. Moment=Force x radius. The force applied by the motor to pulley A is 100lb, and the radius of pulley A is 4in. Therefore, Moment = (100lb)(4in) = 400lb-in.2.
Find the torque transmitted by the compounded pulleys B and C.
The compounded pulleys B and C rotate together. Their radii are 2in and 1in, respectively. The belt tensions on both sides of the pulleys are equal. The torques of pulleys B and C are equal to the force multiplied by the radius of the respective pulley. The sum of these torques is equal to the torque applied to pulley D.
Therefore, the torque transmitted by the compounded pulleys B and C is T= (F/2)(R2 + R3) where F= 100lb and R2 and R3 are the radii of the pulleys. T = (100lb/2) (2in + 1in) = 150lb-in.3.
Find the angular acceleration of the spindle.
The angular acceleration is equal to the torque divided by the total moment of inertia.
Total moment of inertia = moment of inertia of pulley A + moment of inertia of the compounded pulleys B and C + moment of inertia of pulley D.
I = I1 + I2 + I3 where I1
= 20lb-in-s^2, I2
= 2(0.063lb-in-s^2)
= 0.126lb-in-s^2, and I3
= 1/2(4lb)(1in)^2
= 2lb-in-s^2. I
= [tex]20lb-in-s^2 + 0.126lb-in-s^2 + 2lb-in-s^2[/tex]
= 22.126lb-in-s^2.
Therefore, Angular acceleration = Torque / Moment of inertia
= 150lb-in / 22.126lb-in-s^2
= 6.78rad/s^2.4. Find the angular velocity of the spindle at 10 seconds
The angular velocity of the spindle is the product of the angular acceleration and time.
Angular velocity = Angular acceleration x Time
= 6.78rad/s^2 x 10s
= 67.8rad/s.
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The compressors used to pressurise the reservoir can maintain a stagnation pressure of 5 bar (absolute) at a temperature of 100 °C in the reservoir. (1) Calculate the throat area required to give a mass flow rate of 0.25 kgs-1.
Given information:Stagnation pressure of 5 bar (absolute)Temperature of 100 °C in the reservoirMass flow rate of 0.25 kgs-1To calculate the throat area required to give a mass flow rate of 0.25 kgs-1, we use the mass flow equation.Mass flow equation:
[tex]$$\dot m=\rho[/tex] A V[tex]$$[/tex]Where, [tex]$\dot m$ = mass flow rate, $\rho$[/tex] = density, A = cross-sectional area, and V = velocity.
We know the mass flow rate as 0.25 kgs-1. We need to calculate the density of the fluid first, using the gas equation.Gas equation:
[tex]$$PV=nRT$$$$\frac{P}{RT}=\frac{n}{V}$$[/tex]
Where P = pressure, V = volume, n = number of moles, R = gas constant, and T = temperature. We are given the temperature as 100°C, which is equal to 373 K. R = 8.314 JK-1mol-1 and the pressure is given as 5 bar = 5 × 105 Pa (absolute).
[tex]$$\frac{P}{RT}=\frac{n}{V}$$$$n=\frac{PV}{RT}$$$$n=\frac{(5\times 10^5 Pa)(1\ m^3)}{(8.314 JK^{-1} mol^{-1})(373 K)}$$$$n=69.3\ mol$$[/tex]
The number of moles in 1 m3 of the fluid is 69.3 mol. The density of the fluid can be calculated as follows:
[tex]$$\rho=\frac{m}{V}=\frac{nM}{V}$$$$\rho=\frac{(69.3\ mol)(28.97\ kg/kmol)}{1\ m^3}$$$$\rho=2000\ kg/m^3$$[/tex]
The density of the fluid is 2000 kg/m3.
The mass flow rate is given as 0.25 kgs-1. Substituting these values in the mass flow equation, we get:
[tex]$$\dot m=\rho A V$$$$A=\frac{\dot m}{\rho V}=\frac{\dot m}{\rho C_f}$$$$A=\frac{0.25\ kg/s}{2000\ kg/m^3\times C_f}$$Where $C_f$[/tex]
Is the coefficient of velocity which is 0.95.The coefficient of velocity is 0.95.Substituting this in the above equation, we get:
[tex]$$A=\frac{0.25\ kg/s}{2000\ kg/m^3\times 0.95}$$[/tex]
The throat area required to give a mass flow rate of 0.25 kgs-1 is [tex]$$\boxed{A=1.36\times 10^{-4}\ m^2}$$.[/tex]
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A Carnot refrigerator operates between a hot reservoir at 320 Kelvin and a cold reservoir at 260 Kelvin. How much work must be supplied to remove 1000 Joules of heat from the cold reservoir? a 1000 Joules b 188 Joules c Joules d 231 Joules
To remove 1000 Joules of heat from the cold reservoir in a Carnot refrigerator operating between a hot reservoir at 320 Kelvin and a cold reservoir at 260 Kelvin, the amount of work supplied to remove 1000 Joules of heat from the cold reservoir is zero. The correct answer is not provided among the options.
In a Carnot refrigerator, the efficiency can be calculated using the formula:
Efficiency = (Tc - Th) / Tc,
where Tc is the temperature of the cold reservoir and
Th is the temperature of the hot reservoir.
The efficiency of a Carnot refrigerator is the ratio of the work done to the heat extracted from the cold reservoir. Therefore, the work done can be calculated by multiplying the heat extracted (1000 Joules) by the reciprocal of the efficiency.
Using the given temperatures, the efficiency can be calculated as
(260 - 320) / 260 = -0.2308.
Since efficiency cannot be negative,
we can conclude that the given options for the amount of work supplied (options a, b, c, and d) are all incorrect.
The correct answer is not provided among the options.
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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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Consider a controlled single phase half-wave voltage rectifier with a voltage source of V₃ (t) = A sin(2π ft) The rectifier controls the average value of the output voltage through the extension angle a given in radians. The output voltage has the form of V₀(t) = {0₁ 2π ft = a where f = 1/T and Tis the period of the source voltage. Find the following: 1) The output voltage average value. 2) The trigonometric Fourier series coefficients ck, dk 3) The Fourier series expansion of the output voltage. A=6 f=3 a=π/3
Given: A=6, f=3, a=π/3We know that the voltage equation of a single phase half-wave voltage rectifier is given by,V₃(t) = A sin(2π ft)
The average value of the output voltage is given by,
[tex]$$\bar{V}_{0} = \frac{1}{\pi}\int_{0}^{\pi}V_{0}(t)dt$$[/tex]
If 2α is the period of the output waveform then the Fourier series of the output voltage is given by,
[tex]$$V_{0}(t)= \frac{a_0}{2} + \sum_{k=1}^{\infty}(a_k cos(k\omega_ot) + b_k sin(k\omega_ot))$$[/tex]
The trigonometric Fourier series coefficients are,
$$a_1 = \frac{\sqrt{3}-1}{2\pi}, a_2 = a_4 = a_6 = a_8 = ... = 0$$$$a_3 = \frac{1}{3\pi}, a_5 = -\frac{\sqrt{3}}{10\pi}, a_7 = \frac{\sqrt{3}}{14\pi}, a_9 = \frac{1}{9\pi}, a_{11} = -\frac{\sqrt{3}}{22\pi}, a_{13} = \frac{\sqrt{3}}{26\pi}, ...$$and so on3)
The Fourier series expansion of the output voltage is
[tex]$$V_{0}(t) = \frac{2\sqrt{3}}{\pi} + \sum_{k=1}^{\infty}(\frac{(-1)^{k+1}-cos(k\pi/6)}{k\pi}) cos(k2\pi t/3 + \frac{k\pi}{6})$$[/tex]
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In the hydrodynamic boundary layer over a flat plate in laminar flow the velocity
a) Decreases from the solid surface to the free stream
b) Is uniform in the boundary layer thickness
c) Increases from the solid surface to the free stream
Explain your reasoning.
In the hydrodynamic boundary layer over a flat plate in laminar flow the velocity a) Decreases from the solid surface to the free stream
In over a flat plate, the fluid velocity decreases as we move from the solid surface towards the free stream. This is due to the presence of viscous effects near the surface. At the solid surface, the fluid velocity is zero (no-slip condition), and as we move away from the surface, the fluid experiences a frictional drag from the adjacent layers of fluid. This drag slows down the fluid, resulting in a decrease in velocity.
This phenomenon is known as the velocity gradient or velocity profile in the boundary layer. The velocity decreases rapidly near the solid surface and gradually approaches the free stream velocity away from the surface. The boundary layer thickness increases with distance from the solid surface.
Therefore, option a) is the correct answer.
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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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Indicate the incorrect expression for plastic deformation of metallic materials:
a) The level of the yield strength does not indicate if necking occurs after a small amount of homogeneous deformation or after a large amount (in other words, if the start of necking is delayed or not).
b) Greaterstrainhardeningexponentmeansanearlier‘necking’.
c) If ‘n’ and ‘m’ parameters are greater the larger the homogeneous plastic deformation range.
d) Whenplasticdeformationprogressessufficiently,thenegativecontributionto the strain hardening of the metal due to local adiabatic heating starts the local plastic deformation.
The incorrect expression for plastic deformation of metallic materials is: When plastic deformation progresses sufficiently, then the negative contribution to the strain hardening of the metal due to local adiabatic heating starts the local plastic deformation.
Plastic deformation is a permanent deformation that occurs when the external forces acting on a metal exceeds the elastic limit of the material. In metallic materials, plastic deformation is the most common mode of deformation. When a metal undergoes plastic deformation, the metal changes its shape permanently, and the forces required to cause further deformation decreases. Here, we are going to indicate the incorrect expression for plastic deformation of metallic materials.
Incorrect expression:
"When plastic deformation progresses sufficiently, then the negative contribution to the strain hardening of the metal due to local adiabatic heating starts the local plastic deformation." This is an incorrect expression as the plastic deformation does not start due to the negative contribution to the strain hardening of the metal. Local adiabatic heating has no role in starting the plastic deformation process. Hence, this option is incorrect.
Other options are as follows:
- The level of the yield strength does not indicate if necking occurs after a small amount of homogeneous deformation or after a large amount (in other words, if the start of necking is delayed or not). This statement is correct as the yield strength only indicates the initial point of plastic deformation. It does not determine the amount of homogeneous deformation.
- Greater strain-hardening exponent means an earlier ‘necking’. This statement is correct as a higher value of strain-hardening exponent leads to early onset of necking. Necking is defined as the formation of a localized narrow band in a metallic material.
- If ‘n’ and ‘m’ parameters are greater the larger the homogeneous plastic deformation range. This statement is correct as a higher value of 'n' and 'm' leads to a larger range of homogeneous plastic deformation.
Thus, the incorrect expression for plastic deformation of metallic materials is "When plastic deformation progresses sufficiently, then the negative contribution to the strain hardening of the metal due to local adiabatic heating starts the local plastic deformation". This expression is not correct as local adiabatic heating has no role in starting the plastic deformation process.
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A device is shown in figure below. It consists of two frames that are connected to each other by a long M12 bolt and a nut. There is a compression spring k1 between the nut and frame 2, and another compression spring k2 between these frames.
stiffness values for k1 and k2 are given in the table:
Stiffness in kN/m (= N/mm)
k1 5
k2 20
b) The bolt is tightened until the spring k1 reaches a compression of 25 N. Draw free body diagrams for the spring k1, nut and bolt assembly, frame 1, spring k2 and frame 2 and obtain the forces that exist between them. Ignore the weight of the components.
In order to draw the free body diagrams for the given scenario, the following steps must be followed:1. Find out the compression values for both the springs k1 and k2.2. Use Hooke's Law to find out the forces for each spring.
3. Draw the free body diagrams of each of the given components.4. Solve for the forces acting on each component as per the free body diagrams obtained in step 3.According to the question, the compression value of spring k1 is given as 25 N.
Therefore, using Hooke's Law, the force exerted by spring k1 can be calculated as follows:F1 = k1 * x1Where,F1 = force exerted by spring k1k1 = stiffness of spring k1 = 5 N/mx1 = compression value of spring k1 = 25 NTherefore,F1 = 5 * (25 / 1000)F1 = 0.125 KNNow, the free body diagram for spring k1 can be drawn as shown below:
The free body diagram for the nut and bolt assembly can be drawn as shown below:Finally, coming to the frames, the force acting on frame 1 is equal to the force acting on spring k1, which is 0.125 KN. Similarly, the force acting on frame 2 is equal to the force acting on spring k2, which is 2.5 KN.
Therefore, the free body diagrams for both the frames can be drawn as shown below:Therefore, the forces acting on each of the given components are as follows:1. Spring k1: 0.125 KN2. Nut and bolt assembly: 0.125 KN3. Frame 1: 0.125 KN4. Spring k2: 2.5 KN5. Frame 2: 2.5 KN.
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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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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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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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(25 %) In a winter day in Taoyuan, outdoor air is at 1 atm, 10°C, and 60 percent relative
humidity.
(a) The air inside the classroom E4-410 has been heated to 20°C. what is the relative humidity?
(b) The relative humidity of the air in a home is to be restored to 60 percent by evaporating
15°C water into the air. How much heat, in kJ, is required to do this if the volume of E4-
410 is 200 m??
The amount of heat required is approximately 185040 kJ.
a) Let's first find out the saturation vapor pressure at 10°C.
The equation is: PS= 610.78 exp [17.27T / (T + 237.3)]
Where PS is the saturation vapor pressure in pascals, T is the temperature in degrees Celsius Substitute the values to get saturation vapor pressure at 10°C PS = 1213.8 Pah = 1 atm, T = 20°C
The saturation vapor pressure is:PS = 610.78 exp [17.27T / (T + 237.3)]PS = 610.78 exp [17.27(20) / (20 + 237.3)]
PS = 2339.8 PaRelative humidity (RH) is calculated using the following formula:
RH = PV/PS × 100 Where RH is the relative humidity expressed as a percentage, P is the vapor pressure, and S is the saturation vapor pressure. Substitute the values: RH = (0.60 × 2339.8) / 101325 × 100RH = 1.37% ≈ 1%
The relative humidity inside the classroom E4-410 is approximately 1%.
b) Initial Relative Humidity = 20°C Volume (V) of air in the classroom = 200 m³
Final Relative Humidity = 60 % The mass of water evaporated is given as (using the formula of specific humidity):
q = ((Wv) / (Wd+Wv)) where q is the specific humidity,
Wv is the mass of vapor, and Wd is the mass of dry airq = 0.01 kg water vapor/kg dry air (because the final relative humidity is 60 %, the specific humidity of air can be calculated using a psychrometric chart)
Now, for a volume of 200 m³ of air, the mass of dry air is (using the ideal gas equation):
m = pV / RT where R is the gas constant,
T is the temperature, and p is the pressure
We know: p = 101325 Pa (1 atm), T = (15+273) = 288 K, R = 8.31 J/molKm = 101325×200 / (8.31×288) = 7545 kg
The mass of vapor is, therefore, Wv = q × Wd = 0.01 × 7545 = 75.45 kg
To calculate the heat required, we use the following formula:
q = mLh where Lh is the latent heat of evaporation of water = 2451 kJ/kgq = 75.45 × 2451q = 185040.95 kJ
The amount of heat required is approximately 185040 kJ.
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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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Find the convolution integral r(t) *r(t – 3), where r(t) is the ramp function.
The Ramp function, denoted as r(t) can be described as a piecewise function that rises linearly from zero at t = 0, with a slope of unity.The convolution integral r(t) * r(t – 3), where r(t) is the ramp function can be calculated as shown below:Ramp function is given as:r(t) = t, for t >= 0r(t) = 0, for t < 0.
Also, the convolution integral is given as:r(t) * s(t) = ∫s()r(t - )dIn this case, r(t - ) = (t - ) and s() = r( - 3)By substitution, r(t) * r(t – 3) = ∫r(-3) (t - ) dTo find this convolution, we need to break up the integral based on the value of into two parts:∫[0,t] r(-3) (t - ) d and ∫[t,∞) r(-3) (t - ) dNow, we evaluate each of the integrals separately for ∈ [0,t] and ∈ [t,∞) respectively.As varies in [0, t], r( - 3) = 0, since for < 3, r( - 3) = 0. Therefore the integral can be written as∫[0,t] r( - 3) (t - ) d = ∫[3,t] r( - 3) (t - ) d = ∫[0,t-3] r() (t - ( + 3)) d= ∫[0,t-3] (t - ( + 3)) d [Using the definition of the ramp function]
We solve the integral as shown below:[tex]∫[0,t-3] (t - ( + 3)) d=∫[0,t-3] (t - ^2 - 3) d= [^2/2 - (^3)/3 - (3^2)/2] [Evaluated at = 0 and = t-3]= [(t-3)^2/2 - (t-3)^3/3 - (3(t-3)^2)/2][/tex]We have thus computed the first integral.Now, as varies in [t,∞), r( - 3) is not zero. Thus the integral can be written as∫[tex][t,∞) r( - 3) (t - ) d = ∫[t+3,∞) r() (t - ( - 3)) d= ∫[0,∞) r() (t - ( - 3)) d - ∫[0,t+3] r() (t - ( - 3)) d= [(t^2)/2 - 3t] - [(^2)/2 - (t-+3)^2/2 + 3(t - + 3)][/tex]When we subtract the second term from the first, we have the convolution integral:[tex]r(t) * r(t – 3) = [(t-3)^2/2 - (t-3)^3/3 - (3(t-3)^2)/2] + [(t^2)/2 - 3t - (^2)/2 + (t-+3)^2/2 - 3(t - + 3)][/tex]
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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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Oil with density of 920 kg/m² and a dynamic viscosity of 0.06 N.s/mạis pumped with a volumetric flow rate 0.012 m/s through a horizontal pipeline with a diameter of 150 mm. (a) Calculate the velocity of the oil within the pipe. V = 0.679 m^3/s (b) Calculate the Reynolds number of the oil flow within the pipe and determine whether the flow is laminar or turbulent. Re = 1561 Since Re < 2000 then flow is Laminar (c) Calculate the head loss in 1000m length of the pipeline in metres hı = 6.42 m (d) Determine the power required from the pump per 1000m length of the pipeline in kW Power = 0.695 kw Note Assume the acceleration of gravity as g = 9.81 m/sec^2
To calculate the Reynolds number of the oil flow within the pipe, we can use the formula the Reynolds number of the oil flow within the pipe is approximately 2183.
The Reynolds number is a dimensionless quantity that characterizes the flow regime in a pipe. It is used to determine whether the flow is laminar or turbulent.Based on the calculated Reynolds number, the flow of oil within the pipe is in the transitional region between laminar and turbulent flow. It is close to the critical Reynolds number of around 2300, which indicates a transition from laminar to turbulent flow. Therefore, further analysis is required to determine the exact nature of the flow.
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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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Consider free convection from a 2.0m long horizontal cylinder at Ts=114.5°C in air at 1.0 atm and mm. Ta-39.5°C. The cylinder diameter is D = 13.5 Find the heat transfer due to free convection 1/6 2 0.387 Rap Using NU₂ = [0.60. + 0,387R an^1/6 //[1+ (0.559/Pr) ^9/16] 8/27)
The problem presented asks for the calculation of heat transfer due to free convection from a long horizontal cylinder.
The heat transfer from a horizontal cylinder due to free convection can be calculated using the given Nusselt number (Nu) relation, which includes the Rayleigh number (Ra) and Prandtl number (Pr). However, to perform the calculation, we would need the properties of air, including thermal conductivity (k), kinematic viscosity (v), thermal expansion coefficient (β), specific heat at constant pressure (cp), and density (ρ). With the values of these properties at the film temperature, we can calculate the Rayleigh and Prandtl numbers. Then, by plugging the values into the Nusselt number equation, we can find Nu. Lastly, the heat transfer coefficient (h) is calculated using the relation h = Nu*k/D, and the heat transfer rate (q) is found using q = h*A*(Ts-Ta), where A is the surface area of the cylinder.
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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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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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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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A single-stage, single-acting air compressor has a swept volume of 0.007634 m3. Atmospheric air at 101.3 kPa and 20°C is drawn into the compressor and is discharged at 680 kPa. Assume the index of compression and re-expansion is n 1.30. Determine the volumetric efficiency, nvol=____%.
A. 84.15
B. 82.86
C. 85.31
D. 83.37
The volumetric efficiency, nvol = ____ % for the given single-stage, single-acting air compressor.The given details are:Swept volume, V_s = 0.007634 m³ = 7.634 LPressure, P_1 = 101.3 kPaPressure, P_2 = 680 kPaTemperature, T = 20°C = 293.15 KIndex of compression and re-expansion, n = 1.30Volumetric efficiency,
We know that,Volumetric efficiency, nvol = (Actual volume of air delivered / Theoretical volume swept by piston) × 100Actual volume of air delivered = Discharge pressure × Swept volume / (Atmospheric pressure × 1000)Theoretical volume swept by piston =[tex]V_s [(n^(γ-1))/nγ]whereγ = C_p / C_vis[/tex] the ratio of specific heats of air at constant pressure and constant volume.For air,[tex]γ = 1.4C_p = 1.005 kJ/kg KC_v = 0.718 kJ/kg KSo,γ = C_p / C_v = 1.005 / 0.718 = 1.4[/tex]Now,Theoretical volume swept by piston,[tex]V_th = V_s [(n^(γ-1))/nγ]= 7.634 [(1.30^(1.4-1))/(1.30 × 1.4)] = 4.049 L[/tex]
Actual volume of air delivered = Discharge pressure × Swept volume / (Atmospheric pressure × 1000)= 680 × 7.634 / (101.3 × 1000) = 0.0511 L= 51.1 mlHence,Volumetric efficiency, nvol = (Actual volume of air delivered / Theoretical volume swept by piston) × 100= (0.0511 / 4.049) × 100= 1.262 × 100= 126.2 ≈ 126 %Therefore, the volumetric efficiency, nvol = 126 % (Approx).Option (None of the above) is the correct option for this question as the given options do not match the answer obtained.
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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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A jet of water 0.1 m in diameter, with a velocity of 12.5 m/s, impinges onto a series of vanes moving with a velocity of 7.5 m/s. The vanes, when stationary, would deflect the water through and angle of 130 degrees. If friction loss reduces the outlet velocity by 12.5%, Calculate
The relative velocity at inlet, in m/s
The relative velocity at outlet, in m/s
The power transferred to the wheel in W
The kinetic energy of the jet in W
The Hydraulic efficiency
To solve the problem, we can use the principles of fluid mechanics and conservation of energy.
In summary:
Relative velocity at the inlet = 5 m/s
Relative velocity at the outlet = 4.375 m/s
Power transferred to the wheel = 0.00965 W
Kinetic energy of the jet = 764.453 W
Hydraulic efficiency = 0.00126%
Here are the calculations for the given parameters:
Relative velocity at the inlet:
The relative velocity at the inlet can be calculated as the vector sum of the water jet velocity and the vane velocity:
Relative velocity at the inlet = Water jet velocity - Vane velocity
Relative velocity at the inlet = 12.5 m/s - 7.5 m/s = 5 m/s
Relative velocity at the outlet:
Since the outlet velocity is reduced by 12.5%, the relative velocity at the outlet is given by:
Relative velocity at the outlet = (1 - 0.125) * Relative velocity at the inlet
Relative velocity at the outlet = 0.875 * 5 m/s = 4.375 m/s
Power transferred to the wheel:
The power transferred to the wheel can be calculated using the equation:
Power = Flow rate * Head loss
Flow rate = Cross-sectional area * Water jet velocity
Head loss = (Outlet velocity)^2 / (2 * gravity)
Cross-sectional area = π * (Jet diameter/2)^2
Substituting the values into the equation:
Flow rate = π * (0.1 m / 2)^2 * 12.5 m/s = 0.009817 m³/s
Head loss = (4.375 m/s)^2 / (2 * 9.81 m/s²) = 0.98245 m
Power = 0.009817 m³/s * 0.98245 m = 0.00965 W
Kinetic energy of the jet:
The kinetic energy of the jet can be calculated using the equation:
Kinetic energy = 0.5 * Mass flow rate * (Water jet velocity)^2
Mass flow rate = Density * Flow rate
Given that the density of water is approximately 1000 kg/m³:
Mass flow rate = 1000 kg/m³ * 0.009817 m³/s = 9.817 kg/s
Kinetic energy = 0.5 * 9.817 kg/s * (12.5 m/s)^2 = 764.453 W
Hydraulic efficiency:
Hydraulic efficiency is defined as the ratio of power transferred to the wheel to the kinetic energy of the jet:
Hydraulic efficiency = (Power transferred to the wheel / Kinetic energy of the jet) * 100%
Hydraulic efficiency = (0.00965 W / 764.453 W) * 100% = 0.00126%
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Describe in detail the construction, working principle and features of Synchronous motor.
A synchronous motor is an AC motor that operates at a constant speed determined by the number of poles in the motor and the frequency of the power supply.
The construction, working principle, and features of a synchronous motor are described below:
Construction of Synchronous motor: The construction of a synchronous motor is identical to that of a three-phase induction motor, with the exception that the rotor is not wound with conductors. Instead, a set of magnets are installed on the rotor, which creates a magnetic field that interacts with the stator's magnetic field.
Working principle of Synchronous motor:A synchronous motor works on the principle of a rotating magnetic field. The stator's three-phase AC supply creates a rotating magnetic field. The magnetic field interacts with the rotor's magnetic field, which is produced by permanent magnets or DC current. The rotor aligns with the stator's rotating magnetic field as a result of this interaction. As a result, the rotor's speed is synchronous with the stator's magnetic field and the motor runs at a constant speed.
Features of Synchronous motor:
1. Synchronous motors can operate at a constant speed regardless of the load.
2. They have excellent starting torque.
3. Synchronous motors are capable of operating at higher efficiencies than induction motors.
4. They're not self-starting, so they need external assistance to get started.
5. Synchronous motors have higher power factors than induction motors, making them more efficient.
6. They have low maintenance and long service lives.
7. Synchronous motors are utilized in high-performance applications where speed regulation is crucial.
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