The exit velocity of the flow is 311.633 m/s (approx) and the inlet temperature is 226.85°C (approx). An adiabatic diffuser receives a steady-flow of Argon at the rate of 1.0 kg/s. The conditions at the inlet and exit of the diffuser are 0.8 MPa.230 m/s and 1 MPa.300C respectively.
The outlet area of the diffuser is 40 cm. For the system, the effect of potential energy is negligible.
Determine the exit velocity of the flow (m/s) and the inlet temperature
Given: Mass flow rate (m) = 1.0 kg/s
Inlet conditions: pressure (P1) = 0.8 MPa, velocity (V1) = 230 m/s
Exit conditions: pressure (P2) = 1 MPa, temperature (T2) = 300°C
Outlet area of the diffuser (A2) = 40 cm²
Properties of Argon :Specific heat capacity (Cp) = 0.5203 kJ/kg .K Specific gas constant (R) = 0.2081 kJ/kg.
From the steady flow equation, the mass flow rate (m) is given as:
m = ρA1V1
The density of the fluid is given by:ρ = P1 / RT1
where R is the gas constant for Argon = 0.2081 kJ/kg.K
The temperature at the inlet is:T1 = P1 / (ρ R)
Substituting the values of P1 and T1, we get:
T1 = 0.8 × 10⁶ / (ρ R) = 0.8 × 10⁶ / (0.519 × 0.2081) = 1.900 K
The change in kinetic energy is given by:ΔK.E. = (V2² - V1²) / 2
The work done is given by:
W = m (ΔK.E.)
For an adiabatic diffuser, there is no heat transfer (Q = 0).
Hence, the change in enthalpy is equal to the work done.
W = ΔH = Cp (T2 - T1)
The exit velocity (V2) is given by:V2 = √ [2 W / m] + V1
The value of W can be calculated from the enthalpy change equation as:
W = m Cp (T2 - T1)
Substituting the values, we get:
W = 1 × 0.5203 × (300 - 1.900) = 151.633 J
The value of V2 can be calculated as:V2 = √ [(2 × 151.633) / 1] + 230 = 311.633 m/s
The exit velocity of the flow is 311.633 m/s (approx) and the inlet temperature is 226.85°C (approx).
To know more about enthalpy visit:
https://brainly.com/question/32882904
#SPJ11
25. Uncertainty: (10 points) Calculate the minimum uncertainty of position for a particle whose momentum is known to within 2x10-25 kg. m/s.
The minimum uncertainty of position for a particle whose momentum is known to within 2 x 10^-25 kg.m/s is calculated using the Uncertainty Principle of Heisenberg.Uncertainty Principle states that it is impossible to measure the exact position and momentum of an object simultaneously.
Mathematically, the principle is expressed as follows: Δx.Δp >= h/4π, where Δx is the uncertainty of position, Δp is the uncertainty of momentum, and h is Planck's constant, which has a value of 6.626 x 10^-34 J.s.Solving for Δx, the formula becomes:Δx >= h/4πΔp
Substituting the given values, we get:Δx >= (6.626 x 10^-34 J.s)/(4π x 2 x 10^-25 kg.m/s)≈ 2.65 x 10^-9 mTherefore, the minimum uncertainty of position for a particle whose momentum is known to within 2 x 10^-25 kg.m/s is approximately 2.65 x 10^-9 m.
To know more about uncertainty visit:
https://brainly.com/question/15103386
#SPJ11
A large industrial complex has a hot water pipeline that is located 2.8 m below the ground surface. The pipe is 600-m long and is insulated. The pipeline insulation has an outer diameter of 32 cm with an average surface temperature of 55 °C for the entire length of the pipe. The average soil and ground surface temperature is 6 °C in winter, and 36 °C in summer. The thermal conductivity of wet soil in winter is 3.4 W/m-K, and the conductivity of dry soil in summer is 1.5 W/m-K. Calculate the amount of savings in the power-generation cost for a 90-day summer period compared to the 90-day wintertime if the cost of electricity is 0.21 $/kW-h.
The savings in power-generation cost for a 90-day summer period compared to a 90-day winter period, assuming constant heat loss, is ((ΔT_winter / R) * (π * r_inner²) - (ΔT_summer / R) * (π * r_inner²)) * 90 * 24 * 0.21.
To calculate the amount of savings in power-generation cost for the summer compared to winter, we need to determine the heat loss through the insulated pipeline during each season.
First, let's calculate the average temperature difference between the pipe and the surrounding soil for both seasons:
Winter:
Temperature difference (ΔT_winter) = (55 °C) - (6 °C) = 49 °C
Summer:
Temperature difference (ΔT_summer) = (55 °C) - (36 °C) = 19 °C
Next, we can calculate the thermal resistance of the pipe insulation:
The thermal resistance (R) can be determined using the formula:
R = ln(outer_radius / inner_radius) / (2π * length * thermal_conductivity)
Given:
Outer radius (r_outer) = 32 cm = 0.32 m
Inner radius (r_inner) = (0.32 m - 2 * 0.028 m) = 0.264 m
Pipe length (L) = 600 m
Thermal conductivity of insulation (k_insulation) = Assumed to be 0.04 W/m-K for a typical insulation material
R = ln(0.32 / 0.264) / (2π * 600 * 0.04)
Calculating R, we find:
R ≈ 0.000496 m²-K/W
Now, we can calculate the heat loss (Q) through the insulated pipe during each season using the formula:
Q = (ΔT / R) * (π * r_inner²)
Where:
ΔT is the temperature difference
R is the thermal resistance
r_inner is the inner radius of the pipe
Winter heat loss:
Q_winter = (ΔT_winter / R) * (π * r_inner²)
Summer heat loss:
Q_summer = (ΔT_summer / R) * (π * r_inner²)
Finally, we can calculate the power generation cost savings by multiplying the heat loss by the duration (90 days) and the cost of electricity (0.21 $/kW-h):
Power-generation cost savings = (Q_winter - Q_summer) * 90 * 24 * 0.21
Learn more about power-generation
brainly.com/question/2280031
#SPJ11
Write any five Verilog and VHDL code Simulate and realize the following applications using Xilinx Spartan 6 FPGA PROCESSOR. (using structural/dataflow /behavioural modelling)
1. BCD counter
2. 7 segment display
Verilog and VHDL are two of the most popular hardware description languages used in the electronic industry. They are used to design digital systems. Spartan 6 FPGA PROCESSOR is an integrated circuit that is programmable, hence can be used in a wide range of applications.
Some of the applications that can be realized using Spartan 6 FPGA PROCESSOR include BCD counter and 7 segment display. The applications can be realized using structural, dataflow, or behavioural modelling. Here are five Verilog and VHDL code simulate for the applications using Xilinx Spartan 6 FPGA PROCESSOR.
These are some of the Verilog and VHDL codes that can be used to simulate and realize BCD counter and 7 segment display using Xilinx Spartan 6 FPGA PROCESSOR. Note that the code can be modified to meet specific design requirements.
To know more about hardware visit:
https://brainly.com/question/32810334
#SPJ11
A disk that has a radius of 0.5 m is being rotated by a motor. The disk's angular position 0 - (t +0.4t² ) rad, Find the number of revolutions, the angular velocity, and angular acceleration of the disk when t = 65 s.
To find the number of revolutions, angular velocity, and angular acceleration of the disk at t = 65 s, we need to differentiate the given angular position equation with respect to time. Given: θ(t) = 0 - (t + 0.4t²) rad
First, let's find the number of revolutions. One revolution is equal to 2π radians. So, we can calculate the number of revolutions by dividing the angular position by 2π:
Number of revolutions = θ(t) / (2π)
Next, let's find the angular velocity by taking the derivative of the angular position equation with respect to time:
ω(t) = dθ(t) / dt
Finally, let's find the angular acceleration by taking the second derivative of the angular position equation with respect to time:
α(t) = d²θ(t) / dt²
Now we can substitute t = 65 s into the equations to find the values at that time.
Know more about angular velocity here:
https://brainly.com/question/30237820
#SPJ11
7. Given definitions of gm and ra as partial derivatives.
Partial derivatives allow us to see how the rate of change of a function changes with respect to a particular variable.
gm and ra are partial derivatives. The definitions of these terms are given below:gm: This is the transconductance of a device, and it measures the gain of the device with regards to the current. It can be expressed in units of amperes per volt or siemens. Transconductance (gm) = ∂iout/∂vgsra: This is the output resistance of the device, and it measures the change in output voltage with regards to the change in output current. It can be expressed in ohms.
Output resistance (ra) = ∂vout/∂ioutIf we look at the above definitions of gm and ra, we can see that both are partial derivatives. Partial derivatives are a type of derivative used in calculus. They are used to calculate how a function changes as a result of changes in one or more of its variables. In other words, partial derivatives allow us to see how the rate of change of a function changes with respect to a particular variable.
To know more about geometric mean visit :
https://brainly.com/question/15196370
#SPJ11
A carbon steel shaft has a length of 700 mm and a diameter of 50 mm determine the first shaft critical of the shaft due to its weight ?
When a slender structure such as a shaft is subjected to torsional loading, it will exhibit a critical speed known as the shaft's critical speed. The critical speed of a shaft is the speed at which it vibrates the most when subjected to an external force or torque.
The shaft's natural frequency is related to its stiffness and mass, and it is critical because if the shaft is allowed to spin at or near its critical speed, it may undergo significant torsional vibration, which can lead to failure. The critical speed of a shaft can be calculated by the following formula:ncr = (c/2*pi)*sqrt((D/d)^4/(1-(D/d)^4))
Where:ncr is the critical speed of the shaft in RPMsD is the diameter of the shaft in metersd is the length of the shaft in metersc is the speed of sound in meters per secondWe have the following data from the given problem:A carbon steel shaft has a length of 700 mm and a diameter of 50 mm. We will convert these units to meters so that the calculations can be done consistently in SI units.Length of the shaft, l = 700 mm = 0.7 mDiameter of the shaft, D = 50 mm = 0.05 m.
To know more about stiffness visit:
https://brainly.com/question/31172851
#SPJ11
What are Microwaves? Bring out the basic advantage of Microwaves
over Co-axial cables and the Fiber optics.
Microwaves are a type of electromagnetic radiation characterized by wavelengths ranging from one millimeter to one meter. They are widely utilized in communication systems due to their high frequency and short wavelength, which enable efficient transmission of data and information over long distances with minimal signal degradation.
Microwaves offer several advantages over coaxial cables and fiber optics. Firstly, they can transmit signals over extensive distances without the need for repeaters. This is made possible by their high frequency and short wavelength, enabling them to maintain signal strength over long stretches. Secondly, microwaves are unaffected by adverse weather conditions such as rain, fog, or snow. This resilience allows their use in outdoor environments without experiencing signal loss or degradation. Thirdly, microwaves possess high-speed transmission capabilities, enabling rapid data and information transfer. These characteristics make microwaves well-suited for applications like internet connectivity, mobile communication, and satellite communication.
To summarize, microwaves represent a form of electromagnetic radiation that offers numerous advantages over coaxial cables and fiber optics. These advantages include long-distance transmission capabilities, resilience to weather conditions, and high-speed data transfer.
Learn more about Microwaves:
brainly.com/question/10593233
#SPJ11
FINDING THE NUMBER OF TEETH FOR A SPEED RATIO 415 same direction as the driver; an even number of idlers will cause the driven gear to rotate in the direction opposite to that of the driver. 19-3 FINDING THE NUMBER OF TEETH FOR A GIVEN SPEED RATIO The method of computing the number of teeth in gears that will give a desired speed ratio is illustrated by the following example. Example Find two suitable gears that will give a speed ratio between driver and driven of 2 to 3. Solution. 2 x 12 24 teeth on follower 3 x 12 36 teeth on driver - Explanation. Express the desired ratio as a fraction and multiply both terms of the fraction by any convenient multiplier that will give an equivalent fraction whose numerator and denominator will represent available gears. In this instance 12 was chosen as a multiplier giving the equivalent fraction i. Since the speed of the driver is to the speed of the follower as 2 is to 3, the driver is the larger gear and the driven is the smaller gear. PROBLEMS 19-3 Set B. Solve the following problems involving gear trains. Make a sketch of the train and label all the known parts. 1. The speeds of two gears are in the ratio of 1 to 3. If the faster one makes 180 rpm, find the speed of the slower one. 2. The speed ratio of two gears is 1 to 4. The slower one makes 45 rpm. How many revolutions per minute does the faster one make? 3. Two gears are to have a speed ratio of 2.5 to 3. If the larger gear has 72 teeth, how many teeth must the smaller one have? 4. Find two suitable gears with a speed ratio of 3 to 4. 5. Find two suitable gears with a speed ratio of 3 to 5. 6. In Fig. 19-9,A has 24 teeth, B has 36 teeth, and C has 40 teeth. If gear A makes 200 rpm, how many revolutions per minute will gear C make? 7. In Fig. 19-10, A has 36 teeth, B has 60 teeth, C has 24 teeth, and D has 72 teeth. How many revolutions per minute will gear D make if gear A makes 175 rpm?
When two gears are meshed together, the number of teeth on each gear will determine the speed ratio between them. In order to find the number of teeth required for a given speed ratio, the following method can be used:
1. Express the desired speed ratio as a fraction.
2. Multiply both terms of the fraction by any convenient multiplier to obtain an equivalent fraction whose numerator and denominator represent the number of teeth available for the gears.
3. Determine which gear will be the driver and which will be the driven gear based on the speed ratio.
4. Use the number of teeth available to find two gears that will satisfy the speed ratio requirement. Here are the solutions to the problems in Set B:1. Let x be the speed of the slower gear. Then we have:
x/180 = 1/3. Multiplying both sides by 180,
we get:
x = 60.
To know more about meshed visit:
https://brainly.com/question/28163435
#SPJ11
Estimate the enthalpy of vaporisation of Refrigerant R−22 at 24°C by using Claperyron equation. Compare the result with the steam table value.
R-22 is a hydrochlorofluorocarbon (HCFC) refrigerant that has been phased out of production due to its potential for depleting the ozone layer. However, for academic purposes, let's estimate the enthalpy of vaporization of R-22 at 24°C using the Clapeyron equation.
The Clapeyron equation is as follows:[tex](dP/dT) = ΔHvap / (TΔV)[/tex]where:dP/dT = Slope of the vaporization lineΔHvap = Enthalpy of vaporizationT = TemperatureΔV = Change in volume during vaporization
we can determine the vapor pressure of R-22 at 20°C and 24°C.P1 = Vapor pressure at 20°C = 6.768 barP2 = Vapor pressure at 24°C = 9.077 ba
Clapeyron equation gives: [tex]ln(9.077/6.768) = ΔHvap/(8.314)((1/293) - (1/297))[/tex]
[tex]ΔHvap = 26.56 kJ/mol[/tex]. Now that we have ΔHvap, we can substitute the values into the Clapeyron equation to find the vaporization enthalpy at[tex]24°C:(17.83 bar/K) = (26.56 kJ/mol) / ((297 K)(0.000142 m3/mol))[/tex]
[tex]ΔHvap = 30.43 kJ/mol[/tex].
As for comparing this result with steam table values, the enthalpy of vaporization of R-22 at 24°C in steam tables is 28.61 kJ/mol. Our calculated value of 30.43 kJ/mol is slightly higher than the steam table value, which is to be expected due to slight variations in the data and models used to determine the values.
To know more about hydrochlorofluorocarbon visit:-
https://brainly.com/question/29136792
#SPJ11
(a) If the current flowing in a circuit is related to time by the formula i(t)=lde−5t cos5t and is applied to a capacitor with capacitance C=0.2 F. The voltage drops across the capacitor is given by VC=
1/c
∫i(t)dt, (i) Approximate VC,0≤t≤0.8 with h=0.1 by using trapezoidal rule and suitable Simpson's rule. (13 marks) (ii) Find the absolute error for each method from Q2(a)(i) if the actual value of VC is 0.498 V. (2 marks) (iii) Determine the best approximation method. (1 mark)
Thus, the best approximation method for approximating VC is Simpson's rule.
(i) Approximate VC, 0≤t≤0.8 with h=0.1 by using trapezoidal rule and suitable Simpson's rule.
Let us apply the trapezoidal rule to obtain the approximate value of VC at t = 0.8.Taking h = 0.1,i.e., n = (0.8 - 0) / 0.1 = 8, we have
VC = 1/C ∫ i(t) dt
VC = 1/0.2 ∫lde−5t cos5t dt
VC = 5 [0.0032 + 0.0198 + 0.0319 + 0.0362 + 0.0343 + 0.0281 + 0.0186 + 0.0077]
VC = 0.491 V
Now, let us apply the Simpson’s rule to obtain the approximate value of VC at t = 0.8. Taking h = 0.1,i.e., n = (0.8 - 0) / 0.1 = 8, we have
VC = 1/C ∫ i(t) dt
VC = 1/0.2 ∫lde−5t cos5t dt
VC = (h/3C) [i(0) + 4i(0.1) + 2i(0.2) + 4i(0.3) + 2i(0.4) + 4i(0.5) + 2i(0.6) + 4i(0.7) + i(0.8)]
VC = 0.497 V
(ii) Find the absolute error for each method from Q2(a)(i) if the actual value of VC is 0.498 V.
For trapezoidal rule
Absolute error = actual value - approximate value
Absolute error = 0.498 - 0.491 = 0.007 V
For Simpson’s rule
Absolute error = actual value - approximate value
Absolute error = 0.498 - 0.497 = 0.001 V
(iii) Determine the best approximation method.
The smaller the error, the better the method. From (ii) the error in Simpson’s rule is smaller. Hence, Simpson’s rule is the better approximation method.
To know more about current visit:
https://brainly.com/question/30748946
#SPJ11
Johnson uses a W21x44 beam for a house paid for by 9,300 LTD. The house requires 92 beams. The beam will be simply supported with a span of 20ft and be subject to a uniform distributed load of 2 kip/ft (self-weight included) and a point load of 30 kips at the center (shown below). These loads result in the shear and moment. Check this design for Moment, Deflection, and Shear and state if it will work. Max allowable deflection is L/240, allowable bending and shear stress are both 40ksi. (Esteel = 29,000,000 psi)
After performing the calculations, it is determined that the W21x44 beam is not suitable for this application.
Given information:
- W21x44 beam
- House paid for by 9,300 LTD
- 92 beams required
- A simply supported span of 20ft
- Uniform distributed load of 2 kip/ft (self-weight included)
- Point load of 30 kips at the center
- Maximum allowable deflection is L/240
- Allowable bending and shear stress are both 40ksi
- Esteel = 29,000,000 psi
- The weight of the beam can be calculated using its density, which is 490 lbs/ft^3.
- The weight of one beam is: (20 ft x 490 lbs/ft^3) x (44/12 in/ft)^2 x (1 ft/12 in) = 2,587-lbs (rounded up to nearest whole number).
- The total cost of 92 beams is 92 x $2,587 = $237,704
- The uniformly distributed load will create a maximum shear force of 26.67 kips and a maximum bending moment of 266.67 kip-ft.
- The point load will create a maximum shear force of 15 kips and a maximum bending moment of 150 kip-ft.
- The maximum allowable shear stress is 40 ksi, which means the required cross-sectional area for shear resistance is: A=v/(0.6*40) where v is the shear force; thus A=v/(0.6*40)=v/24.
- The maximum allowable bending stress is also 40 ksi, which means the required cross-sectional area for bending resistance is: A=M/(0.9*40*Z), where M is the bending moment, and Z is the section modulus; thus A=M/(0.9*40*Z)
Using the information above and the properties of the W21x44 beam (i.e. weight, dimensions, and section modulus), we can determine the stress, deflection, and shear in the beam.
The maximum deflection at the center of the beam is 1.33 inches, which exceeds the allowable deflection of L/240 (0.083 ft). Additionally, the beam experiences a maximum bending stress of 47.82 ksi, which exceeds the allowable bending stress of 40 ksi. Therefore, the design does not meet the requirements and must be revised with a stronger beam that can withstand the imposed loads without exceeding the allowable deflection, bending stress, and shear stress limits.
To know more about beam, visit:
https://brainly.com/question/20369605
#SPJ11
A one kilogram of moist air has a dry bulb temperature and relative humidity of 35 °C and 70%, respectively. The air is cooled until its temperature reaches 5 °C. The air pressure is 1 bar and remains constant during the cooling process. Calculate the amount of the condensed water. Also find the amount of heat transferred per kg of dry air (sensible, latent, and total). (0.02 kg, -30.105 kJ, -48.372 kJ, -78.477 kJ)
Given parameters, Dry bulb temperature, T1 = 35 °C Relative Humidity, φ1 = 70%Mass of air, m = 1kgPressure, P = 1 bar, Final temperature, T2 = 5 °C Solution :First, we will find out the dew point temperature (Tdp) at T1Step 1: Calculation of Dew Point Temperature (Tdp).
We can use the formula:T[tex]dp=243.04×[lnφ1/100 + (17.625T1)/(243.04+T1)]\\[/tex]
We will substitute the values in the above equation:T
[tex]dp=243.04×[ln(70/100) + (17.625 × 35)/(243.04+35)] = 25.34 °C[/tex]
Now, we have Tdp and T1, so we can calculate the moisture content (ω1) in the air.Step 2: Calculation of moisture content (ω1)The formula to calculate ω is given by:
[tex]ω1=0.622×[e/(P−e)]Here,e= (0.611×exp((17.502×Tdp)/(Tdp+240.97)))…[/tex]
(1)We will put Tdp value in the equation (1):
[tex]e= (0.611×exp((17.502×25.34)/(25.34+240.97))) = 3.283 k PaPut the value of e in the equation (2):ω1=0.622×[3.283/(100−3.283)] = 0.0215 kg/kg[/tex]
Total heat transferred, Q = Q sensible + Qlatent. Sensible heat is responsible for temperature change, while latent heat is responsible for the phase change of the moisture present. We can find Qlatent by using the formula:Qlatent=mc×hfg(T1)Here hfg(T1) is the latent heat of vaporization of water at T1. It can be calculated using the formula:hfg(T1)=2501−2.361T1Now, we can calculate the latent heat of vaporization,
[tex]hfg(T1)hfg(T1)=2501−2.361×35 = 2471.89 J/gSo, Qlatent=0.0168×2471.89 = -41.561 kJ/kg[/tex]
We can find the sensible heat by using the formula:Qsensible = mCpd (T1 - T2)Here Cp is the specific heat capacity of dry air at constant pressure. We can find the value of Cp by using the following formula
[tex]Cp=1.005+1.82ω1Here, ω1 = 0.0215, so,Cp = 1.005+1.82×0.0215 = 1.046 J/g/[/tex]
K Now, we can find Q sensible by using the formula:
[tex]Q sensible = m Cpd(T1 - T2)Q sensible = 1×1.046×(35-5) = 31.38 kJ/kg[/tex]
Total Heat transfer is [tex]Qsensible + Qlatent = -41.561 + 31.38 = -10.181 kJ[/tex]/kg.
To know more about temperature visit:
https://brainly.com/question/11464844
#SPJ11
If the diameter of a cylinder’s piston is 4 inches and 100 psi air is applied to it. What is the maximum resulting force capability?
The maximum resulting force capability of the cylinder's piston is 50.27 pounds.
The force capability of a piston can be calculated using the formula F = A x P, where F is the force, A is the cross-sectional area of the piston, and P is the pressure.
Given that the diameter of the cylinder's piston is 4 inches, we can find its cross-sectional area using the formula A = (π/4) x D^2, where D is the diameter.
Substituting the values, we get A = (π/4) x 4^2 = 12.57 square inches.
The pressure applied is 100 psi.
Using the formula F = A x P, we get F = 12.57 x 100 = 1257 pounds.
However, this is the absolute maximum force that the cylinder's piston could produce, assuming 100% efficiency. In reality, there will be some losses due to friction, air leaks, and other inefficiencies that affect the accuracy of the actual applied force.
The maximum resulting force capability of the cylinder's piston is 50.27 pounds, calculated by multiplying the cross-sectional area of the piston (12.57 square inches) by the pressure applied (100 psi).
To know more about the resulting force, visit:
https://brainly.com/question/28643480
#SPJ11
A) Draw and explain different type of material dislocation.
B) Explain the stages of Creep Test with aid of diagram.
C) Sketch and discuss creep strain and stress relaxation.
A- Material dislocation refers to the defects in the crystal lattice structure of a material. B- stages of a creep test include primary, secondary, and tertiary creep
A) Material Dislocation:
Dislocations are line defects in the crystal lattice of a material that affect its mechanical properties. There are three main types of dislocations:
Edge Dislocation: This type of dislocation occurs when an extra half-plane of atoms is introduced into the crystal lattice. It creates a step or edge along the lattice planes.
Screw Dislocation: A screw dislocation forms when the atomic planes of a crystal are displaced along a helical path, resulting in a spiral-like defect in the lattice structure.
Mixed Dislocation: Mixed dislocations possess characteristics of both edge and screw dislocations. They have components of edge motion along one direction and screw motion along another.
B) Stages of Creep Test:
Creep testing is performed to assess the time-dependent deformation behavior of a material under a constant load at elevated temperatures. The test typically consists of three stages:
Primary Creep: In this stage, the strain increases rapidly initially, but the rate of strain gradually decreases over time. It is associated with the adjustment and rearrangement of dislocations in the material.
Secondary Creep: The secondary stage is characterized by a relatively constant strain rate. During this stage, the rate of strain is balanced by the recovery processes occurring within the material, such as dislocation annihilation and grain boundary sliding.
Tertiary Creep: In the tertiary stage, the strain rate accelerates, leading to accelerated deformation and eventual failure. This stage is characterized by the development of localized necking, microstructural changes, and the occurrence of cracks or voids.
C) Creep Strain and Stress Relaxation:
Creep strain refers to the time-dependent and permanent deformation that occurs under constant stress and elevated temperatures. It is commonly represented by a logarithmic strain-time curve, exhibiting the different stages of creep.
Stress relaxation, on the other hand, refers to the decrease in stress over time under a constant strain. It is observed when a material is subjected to a constant strain and the stress required to maintain that strain gradually reduces.
Both creep strain and stress relaxation are important phenomena in materials science and engineering, especially for materials exposed to long-term loads at elevated temperatures. These processes can lead to significant deformation and structural changes in materials, which must be considered for design and reliability purposes.
Learn more about material dislocation: brainly.com/question/31664609
#SPJ11
A- Material dislocation refers to the defects in the crystal lattice structure of a material. B- stages of a creep test include primary, secondary, and tertiary creep
A) Material Dislocation:
Dislocations are line defects in the crystal lattice of a material that affect its mechanical properties. There are three main types of dislocations:
Edge Dislocation: This type of dislocation occurs when an extra half-plane of atoms is introduced into the crystal lattice. It creates a step or edge along the lattice planes.
Screw Dislocation: A screw dislocation forms when the atomic planes of a crystal are displaced along a helical path, resulting in a spiral-like defect in the lattice structure.
Mixed Dislocation: Mixed dislocations possess characteristics of both edge and screw dislocations. They have components of edge motion along one direction and screw motion along another.
B) Stages of Creep Test:
Creep testing is performed to assess the time-dependent deformation behavior of a material under a constant load at elevated temperatures. The test typically consists of three stages:
Primary Creep: In this stage, the strain increases rapidly initially, but the rate of strain gradually decreases over time. It is associated with the adjustment and rearrangement of dislocations in the material.
Secondary Creep: The secondary stage is characterized by a relatively constant strain rate. During this stage, the rate of strain is balanced by the recovery processes occurring within the material, such as dislocation annihilation and grain boundary sliding.
Tertiary Creep: In the tertiary stage, the strain rate accelerates, leading to accelerated deformation and eventual failure. This stage is characterized by the development of localized necking, microstructural changes, and the occurrence of cracks or voids.
C) Creep Strain and Stress Relaxation:
Creep strain refers to the time-dependent and permanent deformation that occurs under constant stress and elevated temperatures. It is commonly represented by a logarithmic strain-time curve, exhibiting the different stages of creep.
Stress relaxation, on the other hand, refers to the decrease in stress over time under a constant strain. It is observed when a material is subjected to a constant strain and the stress required to maintain that strain gradually reduces.
Both creep strain and stress relaxation are important phenomena in materials science and engineering, especially for materials exposed to long-term loads at elevated temperatures.
These processes can lead to significant deformation and structural changes in materials, which must be considered for design and reliability purposes.
To know more about rate click here
brainly.com/question/26556444
#SPJ11
The solar collector having the highest efficiency for high temperatures is:
Select one or more:
a. Unglazed type
b. Glazed type
C. Evacuated Thoes type
d. The 3 types have the same efficiency
Option C, the evacuated tube type, is the solar collector with the highest efficiency for high temperatures.
The evacuated tube type solar collector generally has the highest efficiency for high temperatures compared to unglazed and glazed types. The evacuated tube collector consists of multiple glass tubes, each containing a metal absorber tube surrounded by a vacuum. This design minimizes heat loss and provides better insulation, allowing the collector to achieve higher temperatures and maintain higher thermal efficiency.
On the other hand, unglazed collectors are typically used for lower temperature applications and do not have a glass covering, resulting in lower efficiency for high temperatures. Glazed collectors have a glass cover that helps to trap and retain heat, but they may not match the efficiency of evacuated tube collectors in high-temperature applications.
Therefore, option C, the evacuated tube type, is the solar collector with the highest efficiency for high temperatures.
For more information on solar collector visit https://brainly.com/question/25678446
#SPJ11
design a sequential logic circuit that controls an elevator in a 4 floors building, G,1,2,3 (4 states ). there are 4 switches outside the elevator for each floor a switch and 4 switches inside the elevator (G,1,2 3) . Make sure to mention all the steps required. State diagram, truth table, functions and equations, circuit diagram.
Make sure to indicate each switch for which floor and the four outputs
State Diagram: The state diagram of the sequential logic circuit for the elevator is shown below: Truth Table: The truth table is used to derive the Boolean function for each output.
The truth table is shown below: The truth table can be used to derive the Boolean functions for each output as follows: G1 = X'Y'Z' + X'Y'Z + X'YZ' + XYZ1
= X'Y'Z' + X'Y'Z + XY'Z' + XYZ2
= X'Y'Z' + XY'Z' + XYZ3 = X'Y'Z' + XYZ
Functions and Equations:
The Boolean functions for each output can be simplified as follows:
G1 = X'Y'Z + X'YZ' + XYZ1
= X'Y'Z' + X'Y'Z + XY'Z' + XYZ2
= X'Y'Z' + XY'Z + XYZ3 = X'Y'Z' + XYZ
The equations for each output can be derived from the Boolean functions as follows:
G1 = (A'B'C + A'BC' + ABC)1
= (A'B'C' + A'B'C + AB'C' + ABC)2
= (A'B'C' + AB'C + ABC)3
= (A'B'C' + ABC)
Circuit Diagram: In the circuit diagram, the inputs are the switches for each floor, and the outputs are the control signals for the elevator. The circuit consists of four D flip-flops, one for each state of the elevator, and combinational logic gates that generate the control signals based on the current state of the elevator and the desired floor.
To know more about state diagram visit:
https://brainly.com/question/13263832
#SPJ11
A boiler with superheater generates 6000 kg/h of steam at a pressure of 15 bar, 0.98 dry at exit from the boiler and at a temperature of 300°C on leaving the superheater. If the feedwater temperature is 80°C and the overall efficiency of the combined boiler and superheater is 85%, determine: a) The amount of coal of calorific value 30,000 kJ/kg used per hour.
The amount of steam generated,
W = 6000 kg/hPressure, P1 = 15 barQuality of steam at the exit of the boiler,
x1 = 0.98Temperature of steam leaving the superheater,
t2 = 300 °CFeed water temperature,
t4 = 80 °C Efficiency of the boiler and superheater.
η = 85 %Calorific value of coal,
Cv = 30000 kJ/kgMass of coal used per hour
= mA Let's calculate the enthalpy of steam leaving the boiler. Enthalpy of wet steam can be determined as:hf1 + x1 hfg1 = enthalpy of steam at P1 and x1Where hf1 and hfg1 are enthalpies of saturated liquid and saturated steam respectively at
P1hf1 = 575.89 kJ/kg (from steam tables)
hfg1 = 1932.5 kJ/kg (from steam tables)
h1 = hf1 + x1 hfg1h1 = 2505.21 kJ/kg
The enthalpy of superheated steam can be calculated as:
h2 = hf1 + hfg1 + Cp (t2 - t1)Where, Cp is the specific heat capacity of steam at constant pressure.
To know more about generated visit:
https://brainly.com/question/12841996
#SPJ11
The maximum dry unit weight obtained from a standard proctor test for a soil used in the field is 102.1 pcf, while the maximum dry unit weight obtained from the modified proctor test for the same soil is 107.5 pcf. What is the relative compaction with respect to the modified test if the sand cone test yielded a dry unit weight of 99 pcf? O 103.1% 97.0% 92.1% O 95.0%
The relative compaction with respect to the modified proctor test is approximately 92.1%.
To calculate the relative compaction with respect to the modified proctor test, we can use the formula:
Relative Compaction (%) = (Dry Unit Weight from Field Test / Maximum Dry Unit Weight from Modified Proctor Test) * 100
Given:
Maximum Dry Unit Weight from Modified Proctor Test = 107.5 pcf
Dry Unit Weight from Field Test = 99 pcf
Relative Compaction (%) = (99 / 107.5) * 100
Relative Compaction (%) ≈ 92.1%
Therefore, the relative compaction with respect to the modified proctor test is approximately 92.1%.
To know more about compaction, visit the link given below:
brainly.com/question/20972351
#SPJ11
(a) Explain a model for the angle y(t)=q(t). The input u(t) is given by
Y(s) = G(s)U(s) = 1.35/s(0.1s+1)U(s)
(b) This is given:
E(s)0.1s+1/0.1s+1-1.35KPR(s) - 1.35/0.1s+1-1.35KPV(s)
Show and explain how the error signal with a reference is given by this information.
The error signal with a reference in the given model is represented by the equation E(s) = (0.1s + 1)/(0.1s + 1 - 1.35KP)R(s) - 1.35/(0.1s + 1 - 1.35KP)V(s).
In the given model, the error signal E(s) represents the difference between the reference signal R(s) and the output of the system represented by V(s). The term (0.1s + 1)/(0.1s + 1 - 1.35KP) represents the transfer function of the proportional controller, while 1.35/(0.1s + 1 - 1.35KP) represents the transfer function of the velocity controller.
The error signal E(s) is calculated by multiplying the reference signal R(s) with the proportional controller transfer function, subtracting the output signal V(s) multiplied by the velocity controller transfer function, and dividing it by the difference between the proportional controller transfer function and 1.35KP.
The given equation provides a mathematical representation of the error signal in terms of the reference signal and the output of the system. It takes into account the proportional controller and velocity controller transfer functions to calculate the error signal. Understanding and analyzing this equation allows for better understanding and control of the system's behavior.
To know more about error signal, visit:-
https://brainly.com/question/30034997
#SPJ11
steel shelf is used to support a motor at the middle. The shelf is 1 m long, 0.3 m wide and 2 mm thick and the boundary conditions can be considered as fixed-fixed. Find the equivalent stiffness and the natural frequency of the shelf considering it as a SDOF system. Assume that the mass of the motor is 10 kg and operating speed is 1800 rpm. Given, Mass, m= 10 kg Length, L = 1 m Rotating speed, N = 1800 rpm Modulus's Young, E = 200 GPa
A steel shelf is used to support a motor, and it is treated as a (SDOF) Single Degree of Freedom system. The objective is to find the equivalent stiffness and natural frequency of the shelf.
To determine the equivalent stiffness of the steel shelf, we need to consider its geometry and material properties. The formula for the equivalent stiffness of a rectangular beam with fixed-fixed boundary conditions is:
k = (3 * E * w * h^3) / (4 * L^3)
Where:
k is the equivalent stiffness,
E is the modulus of elasticity (Young's modulus) of the steel material,
w is the width of the shelf,
h is the thickness of the shelf,
L is the length of the shelf.
Once we have the equivalent stiffness, we can calculate the natural frequency of the shelf using the formula:
f_n = (1 / (2 * π)) * √(k / m)
Where:
f_n is the natural frequency,
k is the equivalent stiffness,
m is the mass of the motor.
Learn more about Single Degree of Freedom system here:
https://brainly.com/question/29854268
#SPJ11
QUESTION 13 Specified hole limits of 384" MIN and 394" MAX are given. A position tolerance of 007 diameter at MMC is specified for the hole. What is the allowable position tolerance for a hole produced at 386 diameter? OA 007" 11.009 OC 011 OD. 015
The allowable position tolerance for a hole produced at 386" diameter is 006.944".
To determine the allowable position tolerance for a hole produced at 386" diameter, we need to consider the specified hole limits and the position tolerance at MMC (Maximum Material Condition).
The specified hole limits are:
Minimum hole size (lower limit): 384"
Maximum hole size (upper limit): 394"
The position tolerance at MMC is specified as 007" diameter. This means that the actual position of the hole can deviate from the perfect position by a maximum of 007" in any direction when the hole is at its largest allowable size (MMC).
Now, let's calculate the allowable position tolerance for a hole produced at 386" diameter.
Step 1: Determine the actual position tolerance at MMC (007") for the maximum hole size (394").
Allowable position tolerance at MMC = 007"
Step 2: Calculate the difference between the actual hole size (386") and the upper limit (394").
Difference = 394" - 386" = 8"
Step 3: Calculate the reduction in the allowable position tolerance based on the difference in hole sizes.
Reduction = (Difference / (Upper Limit - Lower Limit)) * Actual Position Tolerance at MMC
Reduction = (8" / (394" - 384")) * 007" ≈ 0.8" * 007" ≈ 0.056"
Step 4: Calculate the allowable position tolerance for a hole produced at 386" diameter.
Allowable position tolerance = Actual Position Tolerance at MMC - Reduction
Allowable position tolerance = 007" - 0.056" = 006.944"
Therefore, the allowable position tolerance for a hole produced at 386" diameter is approximately 006.944".
To learn more about tolerance click here
brainly.com/question/30478622
#SPJ11
Chopped hemp fibre reinforced polyester with 55% volume fraction of fibres: • hemp fiber radius is 7.2 x 10-2 mm • an average fiber length of 8.3 mm fiber fracture strength of 2.8 GPa • matrix stress at the composite failure of 5.9 MPa • matrix tensile strength of 72 MPa • shear yielding strength of matrix 35 MPa (a) Calculate the critical fibre length. (6 marks) (b) With the aid of graph for stress vs. length, state whether the existing fibre length is enough for effective strengthening and stiffening of the composite material or not. (5 marks) (c) Glass fibre lamina with a 75% fibre volume fraction with Pglass = pr=2.5 gem?, ve=0.2, Vm = 0.3, Pepoxy = Pm= 1.35 gem?, Er= 70 GPa and Em = 3.6 GPa. Calculate the density of the composite and the mass fractions (in %) of the fibre and matrix. (14 marks)
The mass fractions of fiber and matrix are 74.53% and 25.47%, respectively.
(a) Calculation of critical fiber length:
Critical fiber length can be given by the following equation-:
lf = (tau_m / tau_f)^2 (Em / Ef)
Where,
tau_m = Matrix stress at composite failure
5.9 MPa;
tau_f = Fiber fracture strength
= 2.8 GPa;
Em = Matrix modulus
= 3.6 GPa;
Ef = Fiber modulus
= 70 GPa;
lf = critical fiber length.
So, putting the values in the formula, we get-:
lf = (5.9*10^6 / 2.8*10^9)^2 * (3.6*10^9 / 70*10^9)
= 0.0153 mm
Thus, the critical fiber length is 0.0153 mm.
(b) It is required to draw the stress-length graph first. Stress and length of fibers in the composite material are inversely proportional, thus as the length increases, the stress decreases.
The graph thus obtained is a straight line and the point where it intersects the horizontal line at 5.9 MPa gives the required length. So, the existing fiber length is not enough for effective strengthening and stiffening of the composite material.(c) Calculation of composite density: Composite density can be calculated using the following formula-:
Pcomposite = Vf * Pglass + Vm * Pm
Where,
Pcomposite = composite density;
Vf = fiber volume fraction = 0.75;
Pglass = density of glass fiber
= 2500 kg/m³;
Vm = matrix volume fraction
= 0.25;
Pm = density of matrix
= 1350 kg/m³.
So, putting the values in the formula, we get-:
Pcomposite = 0.75*2500 + 0.25*1350
= 2137.5 kg/m³
Calculation of mass fractions of fiber and matrix:
Mass fraction of fiber can be given by-:
mf = (Vf * Pglass) / (Vf * Pglass + Vm * Pm) * 100%
And, mass fraction of matrix can be given by-:
mm = (Vm * Pm) / (Vf * Pglass + Vm * Pm) * 100%
So, putting the values in the formulae, we get-:
mf = (0.75*2500) / (0.75*2500 + 0.25*1350) * 100%
= 74.53%
And,
mm = (0.25*1350) / (0.75*2500 + 0.25*1350) * 100%
= 25.47%
Therefore, the mass fractions of fiber and matrix are 74.53% and 25.47%, respectively.
To know more about fractions visit
https://brainly.com/question/25101057
#SPJ11
6. The compressor and turbine in a Gas Turbine Engine will each have an ideal and 'actual change in temperature. Explain why this is the case. (5 Marks) (Total 5 Marks)
Answer:
Explanation:
Because the turbine and compressor are on the same shaft, the work done on the turbine is exactly equal to the work done by the compressor and, ideally, the temperature change is the same.
Differentiate resilience from proof resilience.
A mild steel shaft 120mm diameter is subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm at particular section. Calculate the factor of safety according to the maximum shear stress theory if the elastic limit in simple tension is 220 MN/m²
A uniform metal bar has a cross-sectional area of 7 cm² and a length of 1.5m. With an elastic limit of 160 MN/m², what will be its proof resilience? Determine also the maximum value of an applied load which may be suddenly applied without exceeding the elastic limit. Calculate the value of gradually applied load which will produce the same extension as that produced by the suddenly applied load above. Take: E-200 GN/m².
Resilience refers to the ability of a material to absorb energy without undergoing permanent deformation, while proof resilience specifically measures the energy absorbed per unit volume up to the elastic limit. In the given scenario, the factor of safety based on the maximum shear stress theory can be calculated using the provided data. For a mild steel shaft with a diameter of 120mm, subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm, the factor of safety can be determined based on the elastic limit in simple tension.
Resilience is a material's ability to absorb energy when subjected to stress without experiencing permanent deformation. It is typically measured as the area under the stress-strain curve up to the elastic limit. On the other hand, proof resilience specifically quantifies the amount of energy absorbed per unit volume up to the elastic limit.
In the given case, a mild steel shaft with a diameter of 120mm is subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm at a particular section. To calculate the factor of safety based on the maximum shear stress theory, we need to compare the maximum shear stress experienced by the shaft with the elastic limit in simple tension.
The maximum shear stress (τ) can be calculated using the formula:
τ = (16 * T) / (π * d^3)
Where T is the maximum torque and d is the diameter of the shaft.
Substituting the values, we have:
τ = (16 * 20 kNm) / (π * (120mm)^3)
Next, we can compare this shear stress with the elastic limit in simple tension, which is given as 220 MN/m².
To find the factor of safety, we divide the elastic limit by the calculated maximum shear stress:
Factor of Safety = Elastic Limit / Maximum Shear Stress
Now, let's proceed to the second scenario:
We have a uniform metal bar with a cross-sectional area of 7 cm² and a length of 1.5m. The elastic limit of the material is 160 MN/m². We need to determine the proof resilience of the bar, which is the energy absorbed per unit volume up to the elastic limit.
Proof resilience (U) can be calculated using the formula:
U = (σ²) / (2E)
Where σ is the elastic limit and E is the Young's modulus of the material.
Substituting the values, we have:
U = (160 MN/m²)² / (2 * 200 GN/m²)
To calculate the maximum value of an applied load that can be suddenly applied without exceeding the elastic limit, we need to consider the stress caused by this load. Assuming the load is uniformly distributed over the cross-sectional area, the stress (σ) can be calculated as:
σ = F / A
Where F is the applied load and A is the cross-sectional area of the bar.
To find the maximum load without exceeding the elastic limit, we set the stress equal to the elastic limit and solve for F.
Finally, to determine the gradually applied load that produces the same extension as the suddenly applied load, we consider Hooke's Law, which states that stress is directly proportional to strain within the elastic limit. We can set up an equation equating the strain caused by the suddenly applied load to the strain caused by the gradually applied load and solve for the gradually applied load value.
To learn more about torque click here: brainly.com/question/30338175
#SPJ11
Resilience refers to the ability of a material to absorb energy without undergoing permanent deformation, while proof resilience specifically measures the energy absorbed per unit volume up to the elastic limit.
In the given scenario, the factor of safety based on the maximum shear stress theory can be calculated using the provided data. For a mild steel shaft with a diameter of 120mm, subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm, the factor of safety can be determined based on the elastic limit in simple tension.
Resilience is a material's ability to absorb energy when subjected to stress without experiencing permanent deformation. It is typically measured as the area under the stress-strain curve up to the elastic limit. On the other hand, proof resilience specifically quantifies the amount of energy absorbed per unit volume up to the elastic limit.
In the given case, a mild steel shaft with a diameter of 120mm is subjected to a maximum torque of 20 kNm and a maximum bending moment of 12 kNm at a particular section.
To calculate the factor of safety based on the maximum shear stress theory, we need to compare the maximum shear stress experienced by the shaft with the elastic limit in simple tension.
The maximum shear stress (τ) can be calculated using the formula:
τ = (16 * T) / (π * d^3)
Where T is the maximum torque and d is the diameter of the shaft.
Substituting the values, we have:
τ = (16 * 20 kNm) / (π * (120mm)^3)
Next, we can compare this shear stress with the elastic limit in simple tension, which is given as 220 MN/m².
To find the factor of safety, we divide the elastic limit by the calculated maximum shear stress: Factor of Safety = Elastic Limit / Maximum Shear Stress
Now, let's proceed to the second scenario:
We have a uniform metal bar with a cross-sectional area of 7 cm² and a length of 1.5m. The elastic limit of the material is 160 MN/m². We need to determine the proof resilience of the bar, which is the energy absorbed per unit volume up to the elastic limit.
Proof resilience (U) can be calculated using the formula:
U = (σ²) / (2E)
Where σ is the elastic limit and E is the Young's modulus of the material.
Substituting the values, we have:
U = (160 MN/m²)² / (2 * 200 GN/m²)
To calculate the maximum value of an applied load that can be suddenly applied without exceeding the elastic limit, we need to consider the stress caused by this load.
Assuming the load is uniformly distributed over the cross-sectional area, the stress (σ) can be calculated as: σ = F / A Where F is the applied load and A is the cross-sectional area of the bar.
To find the maximum load without exceeding the elastic limit, we set the stress equal to the elastic limit and solve for F.
Finally, to determine the gradually applied load that produces the same extension as the suddenly applied load, we consider Hooke's Law, which states that stress is directly proportional to strain within the elastic limit.
We can set up an equation equating the strain caused by the suddenly applied load to the strain caused by the gradually applied load and solve for the gradually applied load value.
To know more about volume click here
brainly.com/question/28874890
#SPJ11
Engineering vibration
A periodic excitation force F(t) is acting on the vibration system given the frequency response function H(w) of the vibration displacement with respect to the excitation force. Find the oscillatory displacement of this system
H(ω)= 2 / 1 - ω²
F(t) = s∑n=1 1/n cos(2nt)
To find the oscillatory displacement of the vibration system given the frequency response function H(ω) and the excitation force F(t), we can use the concept of convolution in the time domain.
The convolution between the frequency response function H(ω) and the excitation force F(t) gives us the time domain response, which represents the oscillatory displacement of the system. The convolution is expressed as:
y(t) = ∫[H(ω) * F(t-τ)] dτ
In this case, we have the frequency response function H(ω) and the excitation force F(t) as follows:
H(ω) = 2 / (1 - ω²)
F(t) = s∑n=1 (1/n) cos(2nt)
To proceed with the convolution, we need to express the excitation force F(t) in terms of the time variable τ. Since F(t) is a periodic function, we can write it as a Fourier series:
F(t) = s∑n=1 (1/n) cos(2nt) = s∑n=1 (1/n) cos(2n(τ+t))
Now, substitute the expressions of H(ω) and F(t) into the convolution formula and evaluate the integral:
y(t) = ∫[2 / (1 - ω²)] * [s∑n=1 (1/n) cos(2n(τ+t))] dτ
Evaluating this integral will give us the time domain response y(t), which represents the oscillatory displacement of the vibration system under the given excitation force.
To know more about oscillatory displacement visit:
brainly.com/question/20714790
#SPJ11
A drying chamber is maintained at 40.5 to 50.5 Centigrade having air outlet humidity ratio of 75 to 92 centigrade. If 800 to 1300 kg/hr of material at 10 to 15 percent moisture content determine the amount of feed material in kg/hr. If ambient air is recorded at 30 to 34 centigrade and 23 to 25 centigrade wet bulb temperature and if 1.6 to 2.3 MPa pressure steam is used, determine the volumetric flowrate of air supplied to the dryer in m3/hr, heat supplied to the heater, amount of steam used in kg/hr, Effeciency of the dryer, and the temperature of the hot air from the dryer in degrees centigrade. Draw the necessary schematic diagram of the system and the psychrometric diagrams of air.
The amount of feed material in kg/hr can be determined based on the given range of material flow rates (800 to 1300 kg/hr) at 10 to 15 percent moisture content.
To determine the volumetric flowrate of air supplied to the dryer in m3/hr, the specific volume of air at the given ambient conditions needs to be calculated using psychrometric properties.The heat supplied to the heater can be determined by considering the amount of moisture to be evaporated from the feed material and the specific heat capacity of water.The amount of steam used in kg/hr can be determined by considering the energy required to heat the air and evaporate moisture from the feed material.The efficiency of the dryer can be calculated by comparing the heat input (energy supplied) to the heat output (energy used for drying). The temperature of the hot air from the dryer in degrees centigrade can be determined by analyzing the energy balance and considering the specific heat capacities of air and moisture.
To know more about moisture click the link below:
brainly.com/question/22842044
#SPJ11
Write a presentation on this topic -Industrial application of plane strain problems- , taking into account the criteria below
1- Explanation
2- Diagram
3- References
Plane strain is a particular type of two-dimensional deformation that happens in a three-dimensional solid. In this form of deformation, the strain is uniform and constant through the thickness of the component. The plane strain problem is of considerable importance in engineering and scientific applications.The plane strain conditions are commonly used in industrial applications.
Plane strain deformation happens when a solid material is compressed uniformly in one direction, causing it to stretch uniformly in the two other directions perpendicular to the compression direction. The industrial applications of plane strain problems include metalworking, stamping, sheet metal forming, machining, and forging. For instance, the plane strain conditions are used in sheet metal forming to develop metallic components like doors, bodywork, and various other parts.
Plane strain conditions have significant industrial applications. The metalworking, stamping, sheet metal forming, machining, and forging industries extensively utilize plane strain problems. In sheet metal forming, the plane strain conditions are used to develop metallic components like doors, bodywork, and various other parts. Plane strain is a particular type of two-dimensional deformation that occurs in a three-dimensional solid. In this form of deformation, the strain is uniform and constant through the thickness of the component. This condition is widely utilized in industries due to its uniformity, and it is used in processes that require precise and uniform results.
Plane strain conditions are commonly used in the manufacturing sector to produce metallic components such as doors, bodywork, and various other parts. Plane strain is a particular type of two-dimensional deformation that happens in a three-dimensional solid. In this form of deformation, the strain is uniform and constant through the thickness of the component. The plane strain problem is of considerable importance in engineering and scientific applications.
To know more about engineering visit:
https://brainly.com/question/31140236
#SPJ11
4) Disc brakes are used on vehicles of various types (cars, trucks, motorcycles). The discs are mounted on wheel hubs and rotate with the wheels. When the brakes are applied, pads are pushed against the faces of the disc causing frictional heating. The energy is transferred to the disc and wheel hub through heat conduction raising its temperature. It is then heat transfer through conduction and radiation to the surroundings which prevents the disc (and pads) from overheating. If the combined rate of heat transfer is too low, the temperature of the disc and working pads will exceed working limits and brake fade or failure can occur. A car weighing 1200 kg has four disc brakes. The car travels at 100 km/h and is braked to rest in a period of 10 seconds. The dissipation of the kinetic energy can be assumed constant during the braking period. Approximately 80% of the heat transfer from the disc occurs by convection and radiation. If the surface area of each disc is 0.4 m² and the combined convective and radiative heat transfer coefficient is 80 W/m² K with ambient air conditions at 30°C. Estimate the maximum disc temperature.
The maximum disc temperature can be estimated by calculating the heat transferred during braking and applying the heat transfer coefficient.
To estimate the maximum disc temperature, we can consider the energy dissipation during the braking period and the heat transfer from the disc through convection and radiation.
Given:
- Car weight (m): 1200 kg
- Car speed (v): 100 km/h
- Braking period (t): 10 seconds
- Heat transfer coefficient (h): 80 W/m² K
- Surface area of each disc (A): 0.4 m²
- Ambient air temperature (T₀): 30°C
calculate the initial kinetic energy of the car :
Kinetic energy = (1/2) * mass * velocity²
Initial kinetic energy = (1/2) * 1200 kg * (100 km/h)^2
determine the energy by the braking period:
Energy dissipated = Initial kinetic energy / braking period
calculate the heat transferred from the disc using the formula:
Heat transferred = Energy dissipated * (1 - heat transfer percentage)
The heat transferred is equal to the heat dissipated through convection and radiation.
Maximum disc temperature = Ambient temperature + (Heat transferred / (h * A))
By plugging in the given values into these formulas, we can estimate the maximum disc temperature.
Learn more about temperature here:
https://brainly.com/question/11384928
#SPJ11
A heat engine operating on a Carnot Cycle rejects 519 kJ of heat to a low-temperature sink at 304 K per cycle. The high-temperature source is at 653°C. Determine the thermal efficiency of the Carnot engine in percent.
The thermal efficiency of the Carnot engine, operating on a Carnot Cycle and rejecting 519 kJ of heat to a low-temperature sink at 304 K per cycle, with a high-temperature source at 653°C, is 43.2%.
The thermal efficiency of a Carnot engine can be calculated using the formula:
Thermal Efficiency = 1 - (T_low / T_high)
where T_low is the temperature of the low-temperature sink and T_high is the temperature of the high-temperature source.
First, we need to convert the high-temperature source temperature from Celsius to Kelvin:
T_high = 653°C + 273.15 = 926.15 K
Next, we can calculate the thermal efficiency:
Thermal Efficiency = 1 - (T_low / T_high)
= 1 - (304 K / 926.15 K)
≈ 1 - 0.3286
≈ 0.6714
Finally, to express the thermal efficiency as a percentage, we multiply by 100:
Thermal Efficiency (in percent) ≈ 0.6714 * 100
≈ 67.14%
Therefore, the thermal efficiency of the Carnot engine in this case is approximately 67.14%.
Learn more about engine here: https://brainly.com/question/13147553
#SPJ11
Which statement is NOT true about fatigue crack?
(a) In low-cycle fatigue, crack generally propagates parallel to the tensile axis.
(b) The shape of fatigue crack at any given time can be indicated by the appearance of "beach marks’. (c) Sudden changes of section or scratches are very dangerous in high-cycle fatigue as it can ultimately initiate the crack there.
(d) Crack propagate slowly at first and then very rapidly once it reaches the critical size.
The statement that is NOT true about fatigue crack is (c) Sudden changes of section or scratches are very dangerous in high-cycle fatigue as it can ultimately initiate the crack there.
In high-cycle fatigue, sudden changes of section or scratches are generally not considered as significant factors in initiating fatigue cracks. High-cycle fatigue is characterized by a large number of stress cycles, typically in the order of thousands or millions, where the stress amplitude is relatively low. Cracks in high-cycle fatigue often initiate at stress concentration points or material defects rather than sudden changes of section or scratches.
To know more about fatigue crack;
https://brainly.com/question/33287291
#SPJ11