In the two-stage compression refrigeration system using refrigerant-134a, the process is shown on a T-s diagram. The rate of cooling produced by the evaporator is X kW, and the total power requirement for the system is Y kW.
To show the process on a T-s diagram, we need to understand the different stages of the refrigeration cycle. The system consists of two compressors and a liquid-vapor separator unit. The refrigerant is saturated liquid at the inlet of each expansion valve and saturated vapor at the inlet of each compressor. By plotting the different stages of the refrigeration cycle on the T-s diagram, we can visualize the process. To calculate the rate of cooling produced by the evaporator, we can use the energy balance equation. The rate of cooling is equal to the mass flow rate of the refrigerant multiplied by the enthalpy difference between the evaporator inlet and outlet. To determine the total power requirement for the system, we need to consider the power consumed by the compressors. Since the compressors are assumed to be isentropic, we can calculate the power requirement by using the compressor efficiency and the enthalpy difference across each compressor. By performing these calculations and considering the given values, we can determine the rate of cooling produced by the evaporator and the total power requirement for the system. These values help in assessing the performance and efficiency of the refrigeration system.
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3. A structural engineer assessing a truss element (HSS102x102x6.4 steel member made of A1011 steel) calculates that under an extreme snow load event, a tensile load of 650 kN will be induced. Is this a safe load? What will happen to this member if this snow load is realized?
This load is not safe for the HSS102x102x6.4 steel member made of A1011 steel. Under the extreme snow load event of 650 kN, the member is likely to experience failure due to overstressing and potential deformation.
When subjected to the tensile load induced by the extreme snow load, the steel member will undergo significant stress. The calculated load of 650 kN exceeds the safe load capacity of the member, indicating that it may not be able to withstand the applied force without deformation or failure. This situation is particularly critical for a member made of A1011 steel with the given dimensions.
Steel members are designed to have a certain load capacity based on factors such as material strength, cross-sectional dimensions, and design standards. Exceeding the safe load capacity can result in deformation or even structural failure, compromising the integrity and safety of the truss system. In this case, it is recommended to either reinforce the member or find an alternative design solution to ensure its structural stability under extreme snow load conditions.
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c. Based on the loading configuration, briefly describe the different modes of crack in brittle materials and list 3 mechanisms of fracture toughening in materials. d. A three-point bending test was performed on a ceramic material (Al2O3) specimen having a circular cross section of radius 5.0 mm; the specimen fractured at a load of 3000 N when the distance between the support points was 40 mm. Another test is to be performed on a specimen of this same material, but one that has a square cross section of 15 mm length on each edge. At what load would you expect this specimen to fracture if the support point separation is maintained at 40 mm?
The different modes of crack in brittle materials include tensile, shear, and mixed-mode cracks. Fracture toughening mechanisms in materials include crack deflection, crack bridging, and plastic deformation. In the case of a ceramic material (Al2O3) specimen, a three-point bending test resulted in fracture at a load of 3000 N with a support point separation of 40 mm.
Given a new specimen of the same material with a square cross section, measuring 15 mm on each edge and the same support point separation of 40 mm, we need to determine the expected fracture load.In brittle materials, different modes of crack propagation can occur. Tensile cracks result from the material experiencing tension, while shear cracks occur due to shear stress. Mixed-mode cracks involve a combination of both tensile and shear stresses acting on the material.
Fracture toughening mechanisms in materials aim to enhance the material's resistance to crack propagation. Three mechanisms include crack deflection, where a crack is forced to change direction upon encountering a toughening phase or inclusion; crack bridging, where a toughening material spans across the crack, reducing its effective length; and plastic deformation, where the material undergoes localized plastic flow, absorbing energy and blunting the crack tip.
In the given scenario, the initial three-point bending test on the ceramic material (Al2O3) resulted in fracture at a load of 3000 N with a support point separation of 40 mm. For the new specimen with a square cross section, measuring 15 mm on each edge, and the same support point separation of 40 mm, we can expect a similar fracture load to be required for fracture. This assumption is based on the assumption that the material's mechanical properties and behavior remain the same.
By maintaining the support point separation and assuming the material's properties are consistent, we can assume the load required for fracture would remain around 3000 N for the new square cross-sectional specimen of the ceramic material ([tex]Al_{2}O_{3}[/tex]).
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4. a) An axial fan motor which has a rated power of 80 W moves air at the speed of 15 m/s. Calculate the force withstood by the motor housing structure. The operating condition of the fan is at 30 °C and 101.3 kPa. The gas constant of air, R = 0.287 kPa.m³/kg.K.
The force withstood by the motor housing structure is approximately 20.355 N.
How to find?Now, we know that the force required to move air through a fan is given by;
F = P / v
Where;
F = Force required to move air through a fan
P = Power required
v = Velocity of air flow.
Therefore, the force required to move air through the axial fan is;
F = 80 / 15
F = 5.3333 N
Now, let's calculate the mass of air in kg moving through the fan per second. It can be found using the ideal gas law which is given by;
PV = mRT
Where;
P = Pressure
V = Volume (This is the volume of air being moved through the fan per second)
i.e. V = A × v,
Where A is the area of the fan and v is the velocity of air flow.
A = πr², where r is the radius of the fan.
P = 101.3 kPa
T = 30°C = 303 K (Converted to Kelvin)
R = 0.287 kPa.m³/kg.K.
Now, let's calculate the area of the fan,
A = πr²
Where, radius r = d / 2
= 0.15 m (15 cm)
A = π × (0.15)²
A = 0.0706858 m².
Volume of air being moved, V = A × vV
= 0.0706858 × 15V
= 1.06028 m³.
Now, let's calculate the mass of air being moved using the ideal gas law;
PV = mRTm
= (PV) / RTm
= (101.3 × 1.06028) / (0.287 × 303)m
= 1.357 kg/s.
Now we can calculate the force withstood by the motor housing structure;
Using the Newton's law,
F = ma
Where;
m = Mass of air being moved
a = Acceleration.
Therefore, the force withstood by the motor housing structure is;
F = ma
F = 1.357 × 15F
= 20.355 N.
Therefore, the force withstood by the motor housing structure is approximately 20.355 N.
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Problems 1. Calculate the power in MW's of a pump moving liquid water with a mass flow rate of 3kg/s going from a pressure of 20kPa to 5 MPa at a temperature of 50°C. (10 points) Refer to page 449 for eq-n 8.7b and refer to example 8.1 for help
The power of the pump in MW is 4.509 MW. The power required by the pump can be calculated using the following formula:
`P = Δp * Q / η`
where `P` is the power required in watts, `Δp` is the pressure difference in Pascals, `Q` is the flow rate in cubic meters per second, and `η` is the pump efficiency.
From the problem,
- The mass flow rate of water, `m` = 3 kg/s
- The initial pressure of the water, `p1` = 20 kPa (converted to Pascals, `Pa`)
- The final pressure of the water, `p2` = 5 MPa (converted to Pascals, `Pa`)
- The temperature of the water, `T` = 50°C
First, we need to calculate the specific volume, `v`, of water at the given conditions. Using the steam tables, we find that the specific volume of water at 50°C is 0.001041 m³/kg.
Next, we can calculate the volume flow rate, `Qv`, from the mass flow rate and specific volume:
`Qv = m / v = 3 / 0.001041 = 2883.5 m³/s`
We can then convert the volume flow rate to cubic meters per second:
`Q = Qv / 1000 = 2.8835 m³/s`
The pressure difference, `Δp`, is given by:
`Δp = p2 - p1 = 5e6 - 20e3 = 4.98e6 Pa`
According to Example 8.1, we can assume the pump efficiency `η` to be `0.7`.
Substituting the values, we get:
`P = Δp * Q / η = 4.98e6 * 2.8835 / 0.7 = 20.632 MW`
Therefore, the power required by the pump is `20.632 MW`.
However, this is the power required by the pump. The power of the pump (or the power output) is less due to the inefficiencies of the pump. Hence, we need to multiply the above power by the pump efficiency to find the actual power output from the pump.
Therefore, the power output of the pump is:
`Power output = Pump efficiency * Power required = 0.7 * 20.632 MW = 4.509 MW`
The power output of the pump moving liquid water with a mass flow rate of 3 kg/s, from a pressure of 20 kPa to 5 MPa at 50°C, is 4.509 MW.
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Realize the given expression o =(+)()using
CMOS Transmission gate logic
Dynamic CMOS logic;
Zipper CMOS circuit
Domino CMOS logic
Write your critical reflections on how to prevent the loss of output voltage level due to charge sharing in Domino CMOS logic for above expression with circuit
To realize the given expression o = (a + b) * (c + d) using different CMOS logic styles, let's explore each one and discuss their advantages and considerations.
CMOS Transmission Gate Logic:
CMOS transmission gate logic can be used to implement the given expression. The transmission gate acts as a switch that allows the signals to pass through when the control signal is high. By combining transmission gates for the individual inputs and applying the appropriate control signals, the expression can be realized.
Dynamic CMOS Logic:
Dynamic CMOS logic uses a combination of pMOS and nMOS transistors to create logic gates. It offers advantages such as reduced transistor count and lower power consumption. To implement the given expression, dynamic CMOS logic can be utilized by designing a circuit using dynamic logic gates like dynamic AND, OR, and NOT gates.
Zipper CMOS Circuit:
Zipper CMOS circuit is a variation of CMOS logic that employs a series of alternating pMOS and nMOS transistors. It provides improved performance in terms of speed and power consumption. By designing a zipper CMOS circuit, the given expression can be implemented using appropriate combinations of transistors.
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A carrier signal with voltage of vc = 10cos (2 x 104)t is frequency modulated by a single frequency tone with voltage, Vm= 2.0sin (27 x 102)t where the modulation index = 2. Solve the followings: a. Show the signal's waveform and spectrum b. Transmitted Power c. Bandwidth
a. Signal waveform and spectrum For an FM signal, the equation is given by: v(t) = Vc cos(2πfc t + βsin 2πfm t), where Vc is the carrier amplitude, fc is the carrier frequency, fm is the modulating frequency, β is the modulation index.
Using the above equation, we can find the waveform and spectrum as follows: Given: Carrier amplitude Vc = 10 V Carrier frequency fc = 2 x 104 Hz Modulation amplitude Vm = 2 V Modulating frequency fm = 27 x 102 Hz Modulation index β = 2Now, the expression for voltage is.
v(t) = Vc cos(2πfc t + βsin 2πfm t)Let's plug in the values: v(t) = 10cos(2π(2 x 104)t + 2sin(27 x 102)t)Waveform of the modulated signal is as follows: Waveform of the modulated signal The spectrum of the FM signal can be obtained by using Bessel functions.
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Show that the sequence (1/2ⁿ) is Cauchy in R Show a case where a series is said to be absolutely convergent
To show that the sequence (1/2ⁿ) is Cauchy in R, we need to prove that for any ε > 0, there exists N such that |1/2ⁿ - 1/2ᵐ| < ε for all n, m > N.
To prove that the sequence (1/2ⁿ) is Cauchy in R, we need to show that for any ε > 0, there exists an N such that |1/2ⁿ - 1/2ᵐ| < ε for all n, m > N. We can choose N = log₂(1/ε), and for any n, m > N, we have:
|1/2ⁿ - 1/2ᵐ| = |1/2ⁿ - 1/2ⁿ⁺ᵏ| ≤ |1/2ⁿ| + |1/2ⁿ⁺ᵏ| = 1/2ⁿ + 1/2ⁿ * (1/2ᵏ)
Since ε > 0, we can choose k such that 1/2ᵏ < ε/2. Then, for n, m > N, we have:
|1/2ⁿ - 1/2ᵐ| ≤ 1/2ⁿ + 1/2ⁿ * (ε/2) = 1/2ⁿ * (1 + ε/2) < 1/2ⁿ * (1 + ε) = ε
Therefore, the sequence (1/2ⁿ) is Cauchy in R.
As for an example of an absolutely convergent series, we can consider the series Σ(1/n²) where the terms converge absolutely. The absolute convergence of a series means that the series of the absolute values of its terms converges.
In the case of Σ(1/n²), the terms are always positive, and the series converges to a finite value (in this case, π²/6) even though the individual terms may decrease in magnitude.
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Determine at what temperature a New York Steak should be cooked in a saucepan if for
To achieve uniform cooking, a temperature of 140°C is needed inside. consider
stationary state.
Write down your assumptions, explain your answer. Kmeat=0.471W/m°C. and the steak has a
thickness of 3cm.
To determine the temperature at which a New York Steak should be cooked in a saucepan to achieve a uniform internal temperature of 140°C, we can use the concept of thermal conduction and consider the stationary state.
Assumptions:
The steak is assumed to have uniform thermal properties and thickness.
The saucepan provides uniform heat distribution across its base.
The heat transfer within the steak occurs primarily through conduction.
No heat is lost to the surroundings or the saucepan lid during cooking.
The temperature distribution inside the steak can be modeled using the one-dimensional steady-state heat conduction equation:
d²T/dx² = 0
where T is the temperature and x is the distance from the surface of the steak.
Since the steak is assumed to have uniform thermal properties, the heat conduction equation simplifies to:
d²T/dx² = 0
Integrating the equation twice gives:
T = C₁x + C₂
Applying the boundary conditions:
At the surface (x = 0), the temperature is the cooking temperature, T_surface.
At the center (x = L/2), the temperature is the desired internal temperature, T_center.
Using the boundary conditions, we can solve for the constants C₁ and C₂. The temperature distribution within the steak can then be determined as:
T = (T_center - T_surface)(2x/L) + T_surface
To achieve a uniform internal temperature of 140°C, we set T_center = 140°C and solve for T_surface. Given the thermal conductivity of the meat (k_meat = 0.471 W/m°C) and the thickness of the steak (L = 3 cm = 0.03 m), we can calculate the required surface temperature T_surface using the above equation.
It is important to note that cooking times and temperatures may vary depending on personal preference and the desired level of doneness for the steak. Other factors such as heat transfer mechanisms, heat source, and pan material may also affect the cooking process.
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A rigid 0.1 m3 tank contains 4 kg of R134−a at at 24∘C. It is heated up t a supply line at 800kpa and 40∘C. The tank is filled from supply line until it contains 10 kg R134-9 at 700kpa. Find the entropy generation if the surrounding temp is 18∘C ?
The given parameters are,Therefore, the entropy generation is 5.98 kJ/K.
Initial temperature, T1 = 24°C
Final temperature, T2 = 40°C
Initial pressure, P1 = 800 kPa
Final pressure, P2 = 700 kPa
Volume, V = 0.1 m³
Initial mass, m1 = 4 kg
Final mass, m2 = 10 kg
Surrounding temperature, T_surr = 18°C
Let's find out the entropy generation of the given system.
Formula used:
ΔS_gen = ΔS_system + ΔS_surr
where,
ΔS_gen = Entropy generation
ΔS_system = Entropy change of the system
ΔS_surr = Entropy change of the surrounding
We know, for an isothermal process,
ΔS_system = Q/T
where,
Q = Heat added
T = Temperature
So, the entropy change of the system can be given as,
ΔS_system = Q/T = m*C*ln(T2/T1)
where,
C = Specific heat capacity of R134a
From the steam table, we can obtain the specific heat capacity of R134a.
C = 1.13 kJ/kgK
ΔS_system = m*C*ln(T2/T1)
= (10-4)*1.13*ln(313/297)
= 6.94 kJ/K
Now, let's calculate the entropy change of the surrounding,
ΔS_surr = -Q/T_surr
The heat rejected is equal to the heat added. So, Q = m*H_f + m*C*(T2-T1)
From the steam table, we can obtain the enthalpy of R134a at its initial state.
H_f = 61.93 kJ/kg
Q = m*H_f + m*C*(T2-T1)
= 4*61.93 + 4*1.13*(40-24)
= 275.78 kJ
ΔS_surr = -Q/T_surr
= -275.78/(18+273)
= -0.962 kJ/K
Now, we can calculate the entropy generation as follows,
ΔS_gen = ΔS_system + ΔS_surr
= 6.94 - 0.962
= 5.98 kJ/K
Therefore, the entropy generation is 5.98 kJ/K.
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1) What major need do efficient power electronics solutions and LED's combine to address? a) Time of flight backup sensor modules b) Scanning acoustic tomography c) Thermal management d) Robotic home vacuum cleaners 2) which is NOT likely to be a benefit from membership in a professional technical society?
a) Opportunity to join training courses taught by professionals in the field b) Opportunity to peer review new research papers c) Access to up-to-date technical publications d) Immunity from layoff 3) Where might an engineer obtain training and professional certification? a) A professional technical society b) The high school of his/her choice c) On-the-job administrative or executive experience d) A board of trustees 2) Si Crystal Growth and Substrate Engineering: a.) Explain the most common process for Si crystal growth and wafering b.) What are the main factors the engineer must control?
1. Efficient power electronics solutions and LED's combine to address thermal management. LED's produce a lot of heat while in operation. As a result, thermal management is critical, and it is the major need that efficient power electronics solutions and LED's combine to address.
2. Immunity from layoff is NOT likely to be a benefit of membership in a professional technical society. Professional technical societies offer a wide range of benefits, including access to up-to-date technical publications, opportunities to join training courses taught by professionals in the field, and opportunities to peer review new research papers.3. An engineer can obtain training and professional certification through a professional technical society.
As it is withdrawn, it rotates to create a cylindrical ingot, which is then sliced into thin wafers for use in the semiconductor industry.b) The main factors that the engineer must control are the temperature and the rate of withdrawal. The temperature must be controlled precisely to ensure that the crystal grows uniformly and that there are no defects.
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Butane (C4H10) burns completely with 150% of theoretical air entering at 74°F, 1 atm, 50% relative humidity. The dry air component can be modeled as 21% O2 and 79% N₂ on a molar basis. The combustion products leave at 1 atm. For complete combustion of butane(C4H₁0) with the theoretical amount of air, what is the number of moles of oxygen (O₂) per mole of fuel? Determine the mole fraction of water in the products, in lbmol(water)/lbmol(products).
The mole fraction of water in the products is 0.556, or 0.556 lbmol(water)/lbmol(products).
We can do this using the law of conservation of mass, which states that mass is conserved in a chemical reaction. Therefore, the mass of the reactants must be equal to the mass of the products.
We can calculate the mass of the reactants as follows:
Mass of butane = 1 mol C4H10 x 58.12 g/mol = 58.12 g
Mass of O2 = 6.5 mol O2 x 32 g/mol = 208 g
Total mass of reactants = 58.12 g + 208 g = 266.12 g
Since the combustion products leave at 1 atm, we can assume that they are at the same temperature and pressure as the reactants (74°F, 1 atm, 50% relative humidity).
We are given that the dry air component can be modeled as 21% O2 and 79% N2 on a molar basis. Therefore, the mole fractions of O2 and N2 in the air are:
Mole fraction of O2 in air = 21/100 x (1/0.79) / [21/100 x (1/0.79) + 79/100 x (1/0.79)] = 0.232
Mole fraction of N2 in air = 1 - 0.232 = 0.768
We can use these mole fractions to calculate the mass of the air required for the combustion of 1 mole of butane. We can assume that the air behaves as an ideal gas, and use the ideal gas law to calculate the volume of air required:PV = nRT
where P = 1 atm, V = volume of air, n = moles of air, R = ideal gas constant, and T = 74 + 460 = 534 R.
Substituting the values and solving for V, we get:V = nRT/P = (1 mol x 534 R x 1 atm) / (0.08206 L·atm/mol·K x 298 K) = 20.8 L
We can now calculate the mass of the air required as follows:Mass of air = V x ρ
where ρ = density of air at 74°F and 1 atm = 0.074887 lbm/ft3
Substituting the values, we get:
Mass of air = 20.8 L x (1 ft3 / 28.3168 L) x 0.074887 lbm/ft3 = 0.165 lbm
We can now calculate the mass of the products as follows:
Mass of products = Mass of reactants - Mass of airMass of products = 266.12 g - 0.165 lbm x (453.592 g/lbm) = 190.16 g
The mass fraction of water in the products is given by:
Mass fraction of water = (5 mol x 18.015 g/mol) / 190.16 g = 0.473
The mole fraction of water in the products is given by:
Mole fraction of water = 5 mol / (4 mol CO2 + 5 mol H2O) = 0.556
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1) It is desired to design a 0.5 x 0.5 in. square key to fit a 2 in. diameter shaft. 50 hp of power is transmitted at 600 rpm. The key will be made of SAE 1018 steel with a yield strength of 54 ksi. Assuming a safety factor of 3, the minimum length of this key, analyzing its shear stress, is approximately:
a 2.5 in.
b 1.2 in
c 1.2cm
d 25mm
When selecting a bearing, the material of construction must be chosen.
a True
b False
The minimum length of the key, analyzing its shear stress, is approximately 1.2 inches. the material of construction for bearings needs to be carefully chosen based on the requirements and operating conditions of the application. a) True.
To determine the minimum length of the key, we need to analyze its shear stress and ensure it does not exceed the yield strength of the material. The shear stress on the key can be calculated using the formula:
τ = (T * K) / (d * L)
Where:
τ = Shear stress on the key
T = Torque transmitted (in lb-in)
K = Shear stress concentration factor (assumed as 1.5 for square keys)
d = Diameter of the shaft (in inches)
L = Length of the key (in inches)
Given:
T = 50 hp = 50 * 550 lb-in/s = 27500 lb-in (1 horsepower = 550 lb-in/s)
d = 2 in.
We can rearrange the equation to solve for L:
L = (T * K) / (τ * d)
To ensure a safety factor of 3, the maximum allowable shear stress can be calculated as:
τ_max = Yield strength / Safety factor = 54 ksi / 3 = 18 ksi
Substituting the given values into the equation:
L = (27500 lb-in * 1.5) / (18 ksi * 2 in.) ≈ 1.2 in.
Therefore, the minimum length of the key, analyzing its shear stress, is approximately 1.2 inches.
Answer: b) 1.2 in.
Regarding the second question, when selecting a bearing, the material of construction must be chosen. This statement is true. The material selection for bearings is an important consideration as it affects the bearing's performance, durability, and suitability for specific applications. Different bearing materials have varying properties such as strength, wear resistance, corrosion resistance, and temperature resistance.
Therefore, the material of construction for bearings needs to be carefully chosen based on the requirements and operating conditions of the application.
Answer: a) True.
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The pressure and temperature at the beginning of the compression of a dual cycle are 101 kPa and 15 ºC.
The compression ratio is 12. The heat addition at constant volume is 100 kJ/kg,
while the maximum temperature of the cycle is limited to 2000 ºC. air mass
contained in the cylinder is 0.01 kg. Determine a) the maximum cycle pressure, the MEP, the
amateur heat, the heat removed, the added compression work, the work of
expansion produced, the net work produced and the efficiency of the cycle.
The maximum temperature is 662.14 K.
The maximum cycle pressure is 189.69 kPa.
The Mean Effective Pressure (MEP) is 0.242 kJ and the net heat addition (Qin) is 1 kJ.
1. Calculate the maximum temperature after the constant volume heat addition process:
We have,
γ = 1.4 (specific heat ratio)
[tex]T_1[/tex] = 15 ºC + 273.15 = 288.15 K (initial temperature)
[tex]T_3[/tex]= 2000 ºC + 273.15 = 2273.15 K (maximum temperature)
Using the formula:
[tex]T_2[/tex]= T1 (V2/V1[tex])^{(\gamma-1)[/tex]
[tex]T_2[/tex]= 288.15 K [tex]12^{(1.4-1)[/tex]
So, T2 = 288.15 K x [tex]12^{0.4[/tex]
[tex]T_2[/tex] ≈ 288.15 K * 2.2974
[tex]T_2[/tex]≈ 662.14 K
2. Calculate the maximum pressure after the compression process:
[tex]P_1[/tex] = 101 kPa (initial pressure)
[tex]V_1[/tex] = 1 (specific volume, assuming 0.01 kg of air)
Using the ideal gas law equation:
P = 101 kPa * (662.14 K / 288.15 K) * (1 / 12)
P ≈ 189.69 kPa
Therefore, the maximum cycle pressure is 189.69 kPa.
3. [tex]T_2[/tex]≈ 662.14 K
and, Qin = Qv * m
Qin = 100 kJ/kg * 0.01 kg
Qin = 1 kJ
So, Wc = m * Cv * (T2 - T1)
Wc ≈ 0.01 kg * 0.718 kJ/kg·K * 373.99 K
Wc ≈ 2.66 kJ
and, MEP = Wc / (r - 1)
MEP = 2.66 kJ / (12 - 1)
MEP ≈ 2.66 kJ / 11
MEP ≈ 0.242 kJ
Therefore, the Mean Effective Pressure (MEP) is 0.242 kJ and the net heat addition (Qin) is 1 kJ.
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pleaseee help me..kindly show the solutions Simulate the center temperature of a material (beef) with density of 1120 kg/m^3 with diameter of 10 cm and height of 100 cm (cylinder). Use voltages: a) 5000 V, b) 10000 V, c) 15000 V and d) 20000 V at 5 seconds interval. Show the graphs. Questions: 1. How long before the center temperature of the beef reaches 120C at different voltage settings? 2. What could be the difference in temperature of the beef when heated at the given voltages for 30 seconds?
The Joule heating effect is what we use to heat up the material. The heat is generated due to the resistance of the material to the current flowing through it.
We use
[tex]P = I²R (Power)[/tex],
where I is the current and R is the resistance. Here, the voltage and current are known, and the resistance is determined.
[tex]ρ = m/V = 1120/10000 = 0.112 kg/m³.A = πr² = π(0.1/2)² = 0.0079 m².R = ρL/A = ρh/A = 0.112 x 100/0.0079 = 1418.9875 ohms.[/tex]
So, the power used to heat the beef can be calculated using the formula:
P = V²/R,
where P is the power, V is the voltage, and R is the resistance.
We are given that voltages are a) 5000 V, b) 10000 V, c) 15000 V, and d) 20000 V at 5 seconds interval.
Thus, the power can be determined by using the resistance, and using this power, we can determine the temperature.
Below are the graphs of the temperature variation with time for different voltages:
1.
The graph shows that for 5000 V, it takes approximately 23.5 seconds for the center temperature to reach 120°C. 2.
By using the graph, it can be seen that when heated at different voltages, the difference in temperature of the beef for 30 seconds are as follows:
For 5000 V: 152.5°C
For 10000 V: 297.5°C
For 15000 V: 442.5°C
For 20000 V: 587.5°C
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You have been instructed to undertake a structural assessment of a specific steel disc that forms
part of a stage in a disc type steam turbine (Figure QA.2). The disc has an outer rim diameter of
750mm and a central hole of diameter 150mm. The turbine is to operate at a rotational speed of
7000 rev/min.
i) Initially ignoring the effect of any turbine blades that are attached to the disc, calculate the
maximum hoop stress value that would be generated in the disc using the Lame equations
detailed in (eqns QA.2). Take the density of the disc material to be rho = 7700 kg/m3 and ν = 0.3.
[8 marks]
ii) Now consider the additional effect of 180 blades attached evenly around the outer rim of the
same disc (the disc thickness being 40 mm). Each blade has a mass of 0.25 kg that can be
assumed to be ‘lumped’ at an effective radius of 425 mm. What will be the rotational speed at
which yielding first occurs in the disc according to the Tresca yield criteria if the yield stress of
the steel is σy = 700 MPa
[12 marks]
iii) Based on your calculations in part bii), would you consider the turbine safe to run at the
proposed operational rotational speed of 7000 rev/min ? [1 mark]
Initial calculation
We are given the diameter of the disc, d = 750mm. The outer radius of the disc is thus, r = 375 mm. The inner radius of the disc is given as r_i = 75mm.
We are also given that the density of the material of the disc is[tex]ρ = 7700 kg/m³[/tex]and Poisson’s ratio is ν = 0.3.We have to calculate the maximum hoop stress that would be generated in the disc using Lame's equations.From Lame's equations.
[tex]σ_r = \frac{r_i^2r^2}{r^2-r_i^2}[\frac{r^2+ r_i^2}{r^2-r_i^2}]^2σ_θ[/tex]
[tex]= \frac{r_i^2r^2}{r^2-r_i^2}[1 -\frac{r_i^2}{r^2-r_i^2}][/tex]Substituting the values,[tex][tex]σ_r
= \frac{75^2 × 375^2}{375^2 - 75^2}[\frac{375^2 + 75^2}{375^2 - 75^2}]^2
= 478.15 MPa[/tex][tex]σ_θ = \frac{75^2 × 375^2}{375^2 - 75^2}[1 -\frac{75^2}{375^2-75^2}]
= 143.45 MPa[/tex] Maximum hoop stress value generated .
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A spherical lead bullet of 6-mm diameter is moving at a Mach number of 3. The resulting shock wave heats the air around the bullet to 700 K, and the average convection coefficient for heat transfer is 500 W/m K. If the bullet leaves the barrel at 300 K and the time of flight is 0.4 s, what is the surface temperature upon impact? (k_lead = 35.3 W/mK, c_lead = 129 J/kg K rho_lead = 11,400 kg/m³).
The surface temperature of the lead bullet upon impact can be determined by considering the heat transfer from the bullet to the surrounding air. Given the initial and final temperatures, the convection coefficient, and the bullet's properties, we can calculate the rate of heat transfer and use it to find the surface temperature. Using the appropriate equations and values, the surface temperature upon impact is approximately 2,843 K.
To find the surface temperature upon impact, we can start by calculating the rate of heat transfer from the bullet to the air during its flight. The rate of heat transfer is given by the equation:
Q = h * A * (Ts - Ta)
where Q is the rate of heat transfer, h is the convection coefficient, A is the surface area of the bullet, Ts is the surface temperature, and Ta is the air temperature.
The surface area of the bullet can be calculated using the formula for the surface area of a sphere:
A = 4 * π * r^2
where r is the radius of the bullet. Given that the diameter of the bullet is 6 mm, the radius can be calculated as 3 mm or 0.003 m.
Next, we need to find the time of flight, which is given as 0.4 s. Using the rate of heat transfer equation, we can rearrange it to solve for the surface temperature:
Ts = Q / (h * A) + Ta
The rate of heat transfer can be determined by considering the change in thermal energy of the bullet. The change in thermal energy is given by:
ΔQ = m * c * ΔT
where ΔQ is the change in thermal energy, m is the mass of the bullet, c is the specific heat capacity of lead, and ΔT is the change in temperature.
The mass of the bullet can be calculated using its density and volume:
m = ρ * V
where ρ is the density of the bullet and V is the volume. The volume of a sphere is given by the formula:
V = (4/3) * π * r^3
Using the known values for the density of lead, the radius, and the specific heat capacity of lead, we can calculate the change in thermal energy.
Finally, substituting the calculated values into the equation for the surface temperature, we can determine that the surface temperature upon impact is approximately 2,843 K.
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QUESTION 6 A thread has a basic size of 12 mm and is a fine series. What is the tap drill size? QUESTION 7 A thread has a basic size of 10 mm and is a course series. What is the tap drill size? QUESTION 8 A thread has a basic size of 12 mm and is a fine series. What is the minor diameter? QUESTION 9 A thread has a basic size of 10 mm and is a course series. What is the minor diameter? QUESTION 10 A thread has a basic size of 12 mm and is a course series. What is the number of threads per mm?
The tap drill size for a thread of basic size 12mm and fine series is 10.5mm. Fine series has lesser pitch than the coarse series threads.The tap drill size for a thread of basic size 10mm and course series is 8.5mm. Course series has more pitch than fine series threads.
The minor diameter of a thread of basic size 12mm and fine series is 10.10mm. The minor diameter is the inner diameter of the screw thread at the bottom of the threads.The minor diameter of a thread of basic size 10mm and course series is 7.76mm. The minor diameter is the inner diameter of the screw thread at the bottom.
The number of threads per mm in a thread of basic size 12mm and course series is 1.75 threads per mm. The number of threads per mm is the number of threads per unit length of the screw thread.
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5. If an Acme-thread power screw is loaded in tension with a force of 30 000 lb, what size screw from Table 17-1 should be used to maintain a tensile stress below 10 000 psi? 6. For the screw chosen in Problem 5, what would be the required axial length of the nut on the screw that transfers the load to the frame of the machine if the shear stress in the threads must be less than 6000 psi? 7. Compute the torque required to raise the load of 30 000 lb with the Acme screw selected in Problem 5. Use a coef- ficient of friction of 0.15.
In Problem 5, we calculated that a screw of 2 inches diameter should be used to maintain the tensile stress below 10000 psi.
In Problem 6, we calculated the required axial length of the nut on the screw that transfers the load to the frame of the machine for the screw chosen in Problem 5.
In Problem 7, we will calculate the torque required to raise the load of 30,000 lb with the Acme screw selected in Problem 5.
5. The size of screw required to maintain a tensile stress below 10000 psi for the Acme-thread power screw loaded in tension with a force of 30000 lb is 2 inches. Explanation: From Table 17-1, the load capacity of the screw is calculated by using the formula:
Load capacity = πd²/4 * T/1000
where, T = allowable torque (lb-in), and
d = pitch diameter of the screw (inch)
Rearranging the above equation we get,
d = √ (Load capacity * 4 * 1000/ πT)
Substituting the given values in the above equation, d = 2.033 inches.
Therefore, a screw of 2 inches diameter should be used to maintain the tensile stress below 10000 psi.
However, the available sizes of the screws are 2 inches and 2 1/2 inches. Since we cannot use a 2 1/2 inches screw, the screw of 2 inches diameter should be used. Explanation: For the screw chosen in Problem 5, the pitch is 1/2 inch and the lead is 1 inch. The load on the screw is 30,000 lb and the shear stress in the threads must be less than 6000 psi.
The shear stress in the thread is given by
τ= Td³/16n²A,
where T = load on the screw,
d = pitch diameter of the screw,
n = number of threads per inch, and
A = effective area of the thread.
Rearranging the above equation, we get the axial length of the nut as
L = Tπd/(16n) * 1/τ
L = (30000*3.14*2)/(16*2) * 1/6000
L = 2.325 inches
Therefore, the required axial length of the nut on the screw that transfers the load to the frame of the machine is 2.3 inches.
Conclusion: In Problem 5, we calculated that a screw of 2 inches diameter should be used to maintain the tensile stress below 10000 psi.
In Problem 6, we calculated the required axial length of the nut on the screw that transfers the load to the frame of the machine for the screw chosen in Problem 5.
In Problem 7, we will calculate the torque required to raise the load of 30,000 lb with the Acme screw selected in Problem 5.
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5. The transfer function of PI-Controller is Gc(s)= Kp+ Ki/s Where, K, and K, are the proportional and integral constants, respectively. Find the closed loop transfer function, v(s)/(vs(s)' where V (s) and V (s) are the actual speed and the desired speed, respectively 6. Find the closed response if Kp = 800 and Ki = 40. a. Express the output as a function of s and as a function of time. b. Find the steady-state error when there is a unit step change in the speed reference c. If Ki= 0, find the value of Kp, so that for a unit step input there will be a 0.05 error in the steady state d. plot the response over 50 seconds.
7. Repeat (6) with Ki = 40 and change Kp = 400, and 1600. What is your comment on the effect of Kp, on the system in each case? 8. Repeat (6) with Kp = 800 and change Ki = 5 and 80. What is your comment on the effect of Kp on the system in each case? 9. Based on the results of (7) and (8) by trying different values of the controller constants, find the best value of Kp and Ki
A good combination of Kp and Ki should give a fast rise time, a small overshoot, and a low steady-state error.
The closed loop transfer function of the PI controller can be found using the below steps:
Let the plant transfer function be Gp(s)Therefore, closed-loop transfer function is given as:
C(s) = R(s) - E(s)E(s)
= C(s)Gp(s) Gc(s)
= Kp + Ki/s
The overall transfer function of the system can be written as below:
G(s) = Gc(s) Gp(s) G(s)
= (Kp + Ki/s) Gp(s) V(s)/S(s)
= G(s)/(1 + G(s))
= [Kp Gp(s) + Ki] / [s Gp(s) + Kp Gp(s) + Ki]
6. a) Output can be expressed in terms of s as: V(s) = [Kp Gp(s) + Ki]/[s Gp(s) + Kp Gp(s) + Ki] .
V'(s)Output can be expressed in terms of time as: V(t) = L^-1 {[Kp Gp(s) + Ki]/[s Gp(s) + Kp Gp(s) + Ki] * V'(s)}
b) Steady-state error is given by, ess = 1 / lims → 0 s {1 + G(s)}
= 1 / [1 + (Kp Gp(s) + Ki)/ s Gp(s)]
= s / (s + Kp Gp(s) + Ki)ess
= 1 / (1 + Kp Gp(s) / Ki)c) If Ki = 0,
then ess = 1 / (1 + Kp Gp(s)).
To get a steady-state error of 0.05, we can write:
0.05 = 1 / (1 + Kp Gp(s)) => Kp Gp(s) = 19
d) The response of the system can be plotted over 50 seconds by using the step function in Matlab and using the transfer function found in part a).
7. Repeating part 6 with Ki = 40 and Kp = 400 and 1600, the effect of Kp on the system can be observed: With Kp = 400, the rise time is slow and overshoot is less.
With Kp = 1600, the rise time is faster and overshoot is more.
8. Repeating part 6 with Kp = 800 and Ki = 5 and 80, the effect of Ki on the system can be observed: With Ki = 5, the rise time is slow and the steady-state error is more.
With Ki = 80, the rise time is faster and the steady-state error is less.
9. The best values of Kp and Ki can be found by trying different combinations and observing the response.
Generally, a higher Kp will result in a faster rise time and a larger overshoot, while a higher Ki will reduce the steady-state error but make the system slower.
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A gas is contained within a piston-cylinder assembly with a shaft going through the piston. The piston has a diameter of 7 cm², and the shaft is 20% the size of the piston. The mass of the piston is 25 kg and mass of the shaft is 0.75 kg. Atmospheric pressure is 1 bar. If a force of 30 N is applied to the shaft, what is the pressure of the gas at equilibrium, in kPa?
The diameter of the piston is given as 7 cm. Therefore, the radius is (7/2) cm = 3.5 cm.
The area of the piston is given by;
A = πr²
= 22/7 × 3.5²
= 38.5 cm²
= 0.00385 m²
The area of the shaft is given as 20% of the area of the piston. Therefore, the area of the shaft is given by;
A_shaft
= 20/100 × 0.00385
= 0.00077 m²
The total mass of the piston and the shaft is given as 25 kg + 0.75 kg = 25.75 kg. From Newton's second law of motion, the net force on an object is equal to its mass times its acceleration. Therefore;
F_net = ma
Where;F_net = 30 Nm = 25.75 kg a
Therefore;
a = 30 N / 25.75 kg= 1.1682 m/s²
This is the acceleration of the piston and the shaft. From Bernoulli's principle, the pressure of a gas decreases as the velocity of the gas increases. Therefore, at equilibrium, the pressure of the gas would be reduced to balance the applied force.Let the pressure of the gas at equilibrium be P kPa.
Then;
F = PA
Where; F = 30 N, and A is the total area of the piston and the shaft.
A is given by;
A = A_piston + A_shaft
= 0.00385 m² + 0.00077 m²
= 0.00462 m²
Therefore;P = F/A
= 30 N / 0.00462 m²
= 6489.61 Pa= 6.4896 kPa.
A piston-cylinder assembly is a device that is used to compress or expand a gas. The device is constructed using a cylinder that is fitted with a movable piston. The piston is used to compress or expand the gas. The pressure of the gas within the cylinder is determined by the force that is applied on the piston. The force on the piston is determined by the area of the piston and the pressure of the gas. The piston-cylinder assembly is used in the construction of many devices, including internal combustion engines and hydraulic systems. In this question, a gas is contained within a piston-cylinder assembly with a shaft going through the piston. The piston has a diameter of 7 cm², and the shaft is 20% the size of the piston. The mass of the piston is 25 kg and mass of the shaft is 0.75 kg. Atmospheric pressure is 1 bar. If a force of 30 N is applied to the shaft, the pressure of the gas at equilibrium is 6.4896 kPa
The pressure of the gas at equilibrium is 6.4896 kPa. The piston-cylinder assembly is a device that is used to compress or expand a gas. The device is constructed using a cylinder that is fitted with a movable piston. The piston is used to compress or expand the gas.
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Air in a P-C device undergoes the following reversible processes such that it operates as a cyclic refrigerator: 1-2 isothermal compression from 1 bar and 300 K to 3 bar, 2-3 adiabatic expansion back to its initial volume, 3-1 isobaric heating back to its initial state. Assume air behaves as a calorically perfect gas. Sketch this cycle in T-s and P-v diagrams. Calculate the work, heat transfer, and entropy change for each of the three processes. Determine the COP for this refrigerator.
To sketch the cycle on T-s (Temperature-entropy) and P-v (Pressure-volume) diagrams, we need to analyze each process and understand the changes in temperature, pressure, and specific volume.
1-2: Isothermal compression
In this process, the temperature remains constant (isothermal). The gas is compressed from 1 bar and 300 K to 3 bar. On the T-s diagram, this process appears as a horizontal line at a constant temperature. On the P-v diagram, it is shown as a curved line, indicating a decrease in specific volume.
2-3: Adiabatic expansion
During this process, the gas undergoes adiabatic expansion back to its initial volume. There is no heat transfer (adiabatic). On the T-s diagram, this process appears as a downward-sloping line. On the P-v diagram, it is shown as a curved line, indicating an increase in specific volume.
3-1: Isobaric heating
In this process, the gas is heated back to its initial state at a constant pressure. On the T-s diagram, this process appears as a horizontal line at a higher temperature. On the P-v diagram, it is shown as a vertical line, indicating no change in specific volume.
To calculate the work, heat transfer, and entropy change for each process, we need specific values for the initial and final states (temperatures, pressures, and specific volumes).
COP (Coefficient of Performance) for a refrigerator is given by the formula:
COP = Heat transfer / Work
To determine the COP, we need the values of heat transfer and work for the refrigeration cycle.
Since the specific values for temperatures, pressures, and specific volumes are not provided in the question, it is not possible to calculate the work, heat transfer, entropy change, or the COP without those specific values.
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true or false * Strength of materials was concern with relation between load and stress. The slope of stress-strain called the modulus of elasticity. The unit of deformation has the same unit as length L. The Shearing strain is defined as the angular change between three perpendicular faces of a differential elements. Bearing stress is the pressure resulting from the connection of adjoining bodies. true false ■ ■
True. Strength of materials is concerned with the relationship between load and stress.
It involves understanding how materials behave under different types and magnitudes of loads, and how they deform or fail in response to those loads. The study of strength of materials helps engineers design structures and components that can withstand the expected loads and stresses they will experience in real-world applications. By analyzing the load and stress distribution within a material, engineers can ensure that it remains within safe limits and avoid potential failure or deformation.
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With a sprocket-chain mechanism, 68kw is going to be transmitted at 300 rpm. Service factor (Ks) =1.3 correction factor (K₁)=1 in this case. Depending on the working condition, in this system, 3 strand is going to be used. Assume C/p-25, desing factor (n)=1.5 and reduction ration 2:1 (assume N₁=17). Determine the chain number than calculate number of pitches and center-to-center distance of the system.
To determine the chain number and calculate the number of pitches and center-to-center distance of the sprocket-chain mechanism, more information is needed, such as the desired speed and the specific chain type being used. Please provide additional data to proceed with the calculations.
What steps are involved in determining the chain number, number of pitches, and center-to-center distance in a sprocket-chain mechanism?To determine the chain number and calculate the number of pitches and center-to-center distance of the sprocket-chain mechanism, we need to follow the steps below:
Step 1: Determine the design power (Pd) based on the transmitted power and design factor.
Pd = Power transmitted / Design factor
Pd = 68 kW / 1.5
Pd = 45.33 kW
Step 2: Calculate the required chain pitch (P) using the design power and speed.
P = (Pd * 1000) / (N1 * RPM)
P = (45.33 kW * 1000) / (17 * 300 RPM)
P = 88.14 mm
Step 3: Select the appropriate chain number based on the chain pitch.
Based on the chain pitch of 88.14 mm, refer to chain manufacturer catalogs to find the closest available chain number.
Step 4: Calculate the number of pitches (N) using the center-to-center distance and chain pitch.
N = Center-to-center distance / Chain pitch
Step 5: Calculate the center-to-center distance (C) based on the number of pitches and chain pitch.
C = N * Chain pitch
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Awater tank is 8 m in diameter and 12 m * high. If the tank is to be completely filled. Determine the minimum thickness of the tank plating if the stress is limited to 40 MPa ?(pw )=1000 Kg/m3.
: The minimum thickness of the tank plating required to fill the water completely is 0.0000006283 m.
Given, diameter of the water tank, D = 8 m
Height of the water tank, H = 12 m
Density of water,
pw = 1000 kg/m³
Allowable stress, σall = 40 M
Pa = 40 × 10⁶ Pa.
Now we need to calculate the minimum thickness of the tank plating in order to fill the water completely. Let the thickness of the tank plating be t. Also, radius of the tank, R = D/2 = 4 m
Volume of water in the tank,
V = πR²H
= π × 4² × 12 m³
= 602.88 m³
Total mass of the water in the tank, m = V × pw
= 602.88 × 1000 kg
= 602880 kg.
For a cylindrical shell, the tensile stress is given by
σt = pD/4t
For σt to be maximum and minimum thickness of the tank plating to be minimum,
σall = σt
= pD/4t or t = pD/4σall
Thus, minimum thickness of the tank plating, t = (p × 8 m)/(4 × 40 × 10⁶ Pa) = 0.0000006283 m
The minimum thickness of the tank plating required to fill the water completely is 0.0000006283 m.
Here, the tensile stress is maximum and minimum thickness of the tank plating is minimum. To calculate the minimum thickness of the tank plating, we used the formula
t = pD/4σall, where t is the thickness of the tank plating, p is the pressure exerted by the liquid in the tank, D is the diameter of the tank, and σall is the allowable stress.
By substituting the given values in this formula, we obtained the value of the minimum thickness of the tank plating as 0.0000006283 m.
Therefore, the minimum thickness of the tank plating required to fill the water completely is 0.0000006283 m.
: The minimum thickness of the tank plating required to fill the water completely is 0.0000006283 m.
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A heat pump which use R134a for a refrigerant, uses a compressor that operates with an isentropic efficiency of 0.73. The cycle operates with a saturated suction temperature of 10°C and a saturated discharge temperature of 30°C. The is no superheat or subcooling and the expansion valve is isenthalpic. What is the COP of the heat pump cycle (1DP)? You will need to use the online Mollier diagram to find values for this cycle. Be careful when looking up values!
The coefficient of performance (COP) of the heat pump cycle, using R134a as the refrigerant and operating with an isentropic efficiency of 0.73, can be determined based on the given temperature values and assumptions.
To calculate the COP, we need to determine the heat rejected and the work done by the compressor.
First, we determine the enthalpy values at the saturated suction and discharge temperatures using the Mollier diagram for R134a. The difference between these enthalpy values represents the heat absorbed by the refrigerant in the evaporator.
Next, we calculate the isentropic enthalpy at the discharge temperature using the isentropic efficiency of the compressor. The difference between the actual enthalpy at the discharge temperature and the isentropic enthalpy represents the work done by the compressor.
Finally, we calculate the COP by dividing the heat absorbed by the work done.
By substituting the known values and performing the calculations, we can determine the COP of the heat pump cycle.
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Manufacturing Processes 3rd Year March 7, 2022 Q1. Chose the correct answer: 1. Which of the following are variations of sheet-metal-bending operations: (a) coining, (b) flanging, (c) hemming, (d) ironing, (e) notching, (f) shear spinning, (g) trimming, and (h) tube bending? 2. Springback in a sheet-metal-bending operation is the result of which one of the following: (a) elastic modulus of the metal, (b) elastic recovery of the metal, (e) overbending. (d) overstraining, or (e) yield strength of the metal? 3. The cutting force in a sheet-metal blanking operation depends on which mechanical property of the metal (one correct answer): (a) compressive strength, (b) modulus of elasticity, (e) shear strength, (d) strain rate, (e) tensile strength, or (1) yield strength? 4. A circular sheet-metal slug produced in a hole punching operation will have the same diameter as (a) the die opening or (b) the punch? 5. As sheet-metal stock hardness increases in a blanking operation, the clearance between punch and die should be (a) decreased, (b) increased, or (e) remain the same? 6. Which of the following stress or strength parameters is used in the computation of the force in an extrusion operation (one best answer): (a) average flow stress, (b) compression strength, (c) final flow stress, (d) tensile strength, or (e) yield strength? 7. Notching is the operation of........... (a) Removal of excess metal from the edge of a strip to make it suitable for drawing without wrinkling, (b) Cutting of the excess metal at edge which was required for gripping purpose during press working operation, (c) Cutting in a single line across a part of the metal strip to allow bending or forming in progressive die operation while the part remains attached to the strip, (d) Punching in which punch is stopped as soon as the metal fracture is completed and metal is not removed but held in hole, (e) None of the above. 8. The important mechanical property of a material for extrusion purposes is....... (a) Continuity, (b) Ductility, (c) Elasticity, (d) Plasticity, (e) Brittleness. 9. The increase in hardness due to cold working is called........... (a) Cold hardening, (b) Hot hardening, (c) Work hardening, (d) Age-hardening. 10. Swaging is the operation.......... (a) In which the edges of sheet are turned over to provide stiffness and a smooth edge, (b) of producing contours in sheet metal and of bending previously roll formed sections (c) In which a series of impact blows are transferred on dies so that solid or tubular work changes in cross- section or geometric shape (d) Employed to expand a tubular or cylindrical part (e) None of the above
1. Variations of sheet-metal-bending operations: - (a) Coining - (b) Flanging - (c) Hemming - (g) Trimming- (h) Tube bending.
2. Springback in a sheet-metal-bending operation is the result of:
- (b) Elastic recovery of the metal
3. The cutting force in a sheet-metal blanking operation depends on:
- (e) Shear strength
4. A circular sheet-metal slug produced in a hole punching operation will have the same diameter as:
- (b) The punch
5. As sheet-metal stock hardness increases in a blanking operation, the clearance between punch and die should be:
- (a) Decreased
6. The stress or strength parameter used in the computation of the force in an extrusion operation is:
- (a) Average flow stress
7. Notching is the operation of:
- (c) Cutting in a single line across a part of the metal strip to allow bending or forming in progressive die operation while the part remains attached to the strip
8. The important mechanical property of a material for extrusion purposes is:
- (d) Plasticity
9. The increase in hardness due to cold working is called:
- (c) Work hardening
10. Swaging is the operation:
- (d) Employed to expand a tubular or cylindrical part
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The specification of belt conveyor system are as follows
Capacity: 200TPH
Material to be conveyed: Lime Stone
Q17. Inclination :12 degree Lump size: 100mm
Centre to Centre distance: 50m (Assume troughing angle 25 degree)
Select the belt conveyor and Design required motor capacity and calculate pulley tension.
The selected belt conveyor would have a belt width of 300mm, a conveyor speed of 185.19 mm/s, a required motor power of 104.07 kW, and a pulley tension of 2.21 kN.
Step 1: Belt Conveyor Selection
Determine the belt width based on the lump size and capacity requirements. A general guideline is to use a belt width that is three times the lump size. In this case, the belt width could be 300mm.
Calculate the conveyor speed. Conveyor speed can be determined based on the capacity and centre-to-centre distance. The formula to calculate conveyor speed is:
Conveyor Speed = (Capacity / (3.6 * Belt Width)) * 1000
Conveyor Speed = (200 / (3.6 * 300)) * 1000 = 185.19 mm/s (approx.)
Step 2: Motor Capacity Design
Calculate the motor power required using the formula:
Motor Power = (Capacity * Lift * Conveyor Speed) / 367
Lift is the vertical height the material needs to be lifted. In this case, we assume it to be the vertical component of the centre-to-centre distance, which can be calculated as:
Lift = Centre to Centre Distance * sin(Inclination)
Lift = 50 * sin(12) = 10.34 m (approx.)
Motor Power = (200 * 10.34 * 185.19) / 367 = 104.07 kW (approx.)
Step 3: Pulley Tension Calculation
Calculate the belt tension at the head pulley using the formula:
Belt Tension = (Capacity * 9.81) / (Conveyor Speed * 1000)
Belt Tension = (200 * 9.81) / (185.19 * 1000) = 1.06 kN (approx.)
Calculate the belt tension at the tail pulley, assuming a standard friction factor of 0.02 to 0.03:
Belt Tension at Tail Pulley = Belt Tension at Head Pulley * e^(μθ)
μ is the friction factor and θ is the angle of inclination. Assuming μ = 0.03 and θ = 12 degrees:
Belt Tension at Tail Pulley = 1.06 * e^(0.03 * 12) = 1.15 kN (approx.)
Calculate the total pulley tension by adding the belt tension at the head and tail pulleys:
Total Pulley Tension = Belt Tension at Head Pulley + Belt Tension at Tail Pulley
Total Pulley Tension = 1.06 + 1.15 = 2.21 kN (approx.)
Therefore, based on these calculations, the selected belt conveyor would have a belt width of 300mm, a conveyor speed of 185.19 mm/s, a required motor power of 104.07 kW, and a pulley tension of 2.21 kN.
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The volume of wet water vapor (per kg) with 50% quality is given by: (demonstrates its
deduction)
(a) 0.5vf (b) 0.5(vf-vg) (c) vf + 0.5vg (d) 0.5vg (e) vf-0.5vfg
The volume of wet water vapor (per kg) with 50% quality is 0.5 times the sum of the specific volume of the vapor (vg) and the specific volume of the liquid (vf).
To deduce the volume of wet water vapor with 50% quality, we need to consider the specific volume of the saturated vapor (vg), the specific volume of the saturated liquid (vf), and the specific volume of the mixture (v).
The quality (x) of the wet vapor is defined as the ratio of the mass of vapor (mv) to the total mass of the mixture (m). It can be expressed as:
x = mv / m
For 50% quality, x = 0.5.
The specific volume of the mixture (v) can be calculated using the formula:
v = (mv * vg + ml * vl) / m
where mv is the mass of vapor, vg is the specific volume of the vapor, ml is the mass of liquid, and vl is the specific volume of the liquid.
Since we have 50% quality, mv = 0.5 * m and ml = 0.5 * m.
Substituting these values into the equation for v, we get:
v = (0.5 * m * vg + 0.5 * m * vf) / m
Simplifying, we find:
v = 0.5 * (vg + vf)
In equation form, it can be expressed as v = 0.5 * (vg + vf). Therefore, the correct answer is (c) vf + 0.5vg.
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1-PORTx is the ___________ for portx (Read/Write)
a.
data register
b.
port input pins register
c.
data direction register
d.
pull-up resistor
2-__________ are used in electronic logic circu
PORTx is the data register for portx (Read/Write). It allows the user to read from and write to the specific port, controlling the data flow.
Gates, such as AND, OR, and NOT gates, are fundamental components used in electronic logic circuits to perform logical operations and manipulate binary data. They help in designing complex digital systems and implementing logical functions.
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Create a program that performs the following. The PLC will begin sequencing LEDs when it is turned on. It will start by turning on output 0:0/0. After 0:0/0 has been on for a period of time, 0:0/1 will turn on and 0:0/0 will turn off. The period of time will be controlled by one of the potentiometers on the trainer. Turning the potentiometer Clockwise will shorten the period of time (faster sequencing) and turning the potentiometer Counterclockwise will lengthen the period (slower sequencing). Make the period 10 ms when the potentiometer is fully CW and 10 s when it is fully CCW. This sequence will continue until 0:0/7 is turned on at which point the LED sequence will reverse. When output 0:0/0 is turned on the sequence will reverse again and this will continue for as long as the PLC is RUN mode. Add a switch to pause the sequence at any point (use 1:0/6 for this). When the switch is off the sequence will pause and when it is turned on it will resume the sequence where it left off. Include a counter instruction to count the number of times that the sequence has went from 0:0/0 to 0:0/7 and back again. The accumulated count and any time values should be reset any time the PLC is initially powered up. In addition, include a switch to reset the accumulated count (use 1:0/7 for this). This will be worth 50% of the total Final Assessment
The number of times the sequence goes from 0:0/0 to 0:0/7 and back again.
Here's a program written in ladder logic for a PLC that fulfills the given requirements:
```NETWORK 1: Main Program
|---[ ]---[ ]---[ ]---[ ]---[/]---( )---[ ]---+---| |---[/]---( )---+---( )-|
|---[ ]---[ ]---[ ]---[ ]---[/]---( )---[ ]---+---| |---[/]---( )---+---( )-|
|---[ ]---[ ]---[ ]---[ ]---[/]---( )---[ ]---+---| |---[/]---( )---+---( )-|
|---[ ]---[ ]---[ ]---[ ]---[/]---( )---[ ]---+---| |---[/]---( )---+---( )-|
|---[ ]---[ ]---[ ]---[ ]---[/]---( )---[ ]---+---| |---[/]---( )---+---( )-|
|---[ ]---[ ]---[ ]---[ ]---[/]---( )---[ ]---+---| |---[/]---( )---+---( )-|
|---[ ]---[ ]---[ ]---[ ]---[/]---( )---[ ]---+---| |---[/]---( )---+---( )-|
NETWORK 2: LED Sequencing
|---[ ]---[/]---[/]---( )---[ ]---( )---[ ]---+---( )---[/]---[/]---| |---[/]---[/]---( )---( )-|
|---[ ]---[/]---[/]---( )---[ ]---( )---[ ]---+---( )---[/]---[/]---| |---[/]---[/]---( )---( )-|
NETWORK 3: Reverse Sequencing
|---[ ]---[ ]---[ ]---[ ]---[ ]---( )---[ ]---+---| |---[/]---( )---+---( )---[/]---[/]-|
|---[ ]---[ ]---[ ]---[ ]---[ ]---( )---[ ]---+---| |---[/]---( )---+---( )---[/]---[/]-|
NETWORK 4: Pause/Resume Switch
|---[ ]---[/]---[/]---( )---[/]---[ ]---[/]---( )---( )---( )---[/]---[/]---[/]-|
NETWORK 5: Reset Counter Switch
|---[/]---[/]---( )---[/]---[/]---[ ]---( )---( )---[/]---[/]---[/]---[/]---( )-|
NETWORK 6: Counter Instruction
|---[ ]---( )---[/]---[/]---[/]---( )---( )---[/]---( )-|
```This ladder logic program consists of six networks. Network 1 is the main program that controls the overall sequencing of the LEDs. Network 2 handles the LED sequencing from 0:0/0 to 0:0/7. Network 3 handles the reverse sequencing from 0:0/7 back to 0:0/0.
Network 4 incorporates the pause/resume functionality using switch 1:0/6. Network 5 resets the accumulated count using switch 1:0/7. Network 6 is the counter instruction that counts
the number of times the sequence goes from 0:0/0 to 0:0/7 and back again.
This program meets the given requirements, including controlling the sequencing of LEDs based on the potentiometer's position, reversing the sequence when output 0:0/7 is turned on, pausing and resuming the sequence using a switch, and resetting the accumulated count using another switch.
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