In the design practice for a continuous fibre reinforced composite for aerospace application, Ti was chosen as the matrix and alumina (Sumitomo fibre) fibre as the reinforcing agent. One of the best fabrication routes is the Hot Isostatic Pressing (HIP) method. In this process, the powder and fibres are first mixed and then subjected to hot pressing.The process is carried out at temperatures of about 1300°C and high pressure, which results in a completely homogeneous structure with almost no voids.
Ti can then be infiltrated into the structure by HIP, resulting in a complete infiltration of the fibre matrix structure.The selected fabrication route is Hot Isostatic Pressing (HIP). This method was chosen because it results in a completely homogeneous structure with almost no voids. Ti can then be infiltrated into the structure by HIP, resulting in a complete infiltration of the fibre matrix structure.
The HIP method is a very cost-effective method that requires a relatively low investment in equipment. In addition, it is a very efficient method of producing composite materials with a high level of consistency and quality. The HIP process also produces very high quality composites, and the consistency of the final product is very good.The HIP method requires very high temperatures and high pressures, which can be an expensive process, but the benefits of producing high-quality composites with consistent properties outweigh the costs.
In summary, the Hot Isostatic Pressing (HIP) method is the best fabrication route for producing continuous fibre reinforced composite for aerospace application because it produces high-quality composites with consistent properties and it is a cost-effective method that requires a relatively low investment in equipment.
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The pipe network runs between two tanks (k for pipe entry of 0.2 and exit 1) over a distance of 25 m, with a total static lift of 10 m between the two tanks. This pipe network includes two long sweeping elbows, a Mitre bend and a gate valve (1/4 open). Calculate: iii) the total head loss [7] iv) the hydraulic output required to pump the mixture between the two tanks [3]
In this pipe network scenario between two tanks, the objective is to calculate the total head loss and the hydraulic output required to pump the mixture between the tanks. The network includes specific components such as long sweeping elbows.
iii) To calculate the total head loss in the pipe network, we need to consider the individual head losses caused by different components. The head loss due to pipe entry and exit can be determined using the provided values. The head loss due to pipe friction can be calculated using the pipe length, pipe diameter, and the Darcy-Weisbach equation. The head losses associated with the long sweeping elbows, Mitre bend, and the gate valve can be estimated using empirical data or fitting coefficients specific to each component. By summing up these individual head losses, we can determine the total head loss in the pipe network.
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a) Draw a fully labelled temperature/entropy diagram of the Brayton Cycle. (5 Marks) b) Using appropriate thermodynamic terms, explain the Brayton cycle
It is a method of compressing stress air, adding fuel to the compressed air, igniting the fuel-air mixture, and then expanding the air-fuel mixture to generate power.
a) The temperature-entropy (T-S) diagram for the Brayton cycle is shown below. In a gas turbine engine, the Brayton cycle is a thermodynamic cycle.
It is a method of compressing air, adding fuel to the compressed air, igniting the fuel-air mixture, and then expanding the air-fuel mixture to generate power. The following are the stages of the cycle: 1. Isentropic compression 2. Isobaric heat addition 3. Isentropic expansion 4. Isobaric heat rejectionIn a gas turbine engine, the Brayton cycle is used.
It is a cyclic operation that generates mechanical energy by operating on a closed loop. The loop consists of an inlet where air is taken in, a compressor where the air is compressed, a combustion chamber where fuel is mixed with the compressed air and burned to raise its temperature, a turbine where the high-temperature, high-pressure air is expanded and the power is extracted, and an outlet where the exhaust gas is released.
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Heat recovery steam boiler (HRSB) was designed to produce 4600 kg/h saturated steam at pressure 20 atm with exhaust gas flow mg = 34000 kg / h and temperatures Tgin = 540οC, Tgout = 260οC. During its operation with reduced load (mg = 22800 kg / h, Tgi = 510οC) the exhaust temperature of the exhaust gas Tgο = 271οC is measured. Can you comment on the possibility of deterioration of the boiler operation due to the formation of deposits?
The lower exhaust gas temperature observed during reduced load operation suggests a potential improvement in heat transfer efficiency, but a thorough assessment of the specific operating conditions and potential deposit formation is necessary to evaluate the overall impact on boiler performance.
The formation of deposits in a boiler can have negative effects on its operation. Deposits are usually formed by the condensation of impurities contained in the exhaust gas onto the heat transfer surfaces. These deposits can reduce heat transfer efficiency, increase pressure drop, and potentially lead to corrosion or blockage. In this case, the decrease in exhaust gas temperature (Tgο) from the designed operating conditions could suggest improved heat transfer due to reduced fouling or deposit formation. The lower exhaust gas temperature indicates that more heat is being transferred to the steam, resulting in a higher steam production temperature. However, it is important to consider other factors such as the composition of the exhaust gas and the properties of the deposits. Different impurities and operating conditions can lead to varying degrees of deposit formation. A comprehensive analysis, including a study of the exhaust gas composition, flue gas analysis, and inspection of the boiler surfaces, would be required to make a definitive conclusion about the possibility of boiler operation deterioration due to deposits.
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1. A 2.004 L rigid tank contains .04 kg of water as a liquid at 50°C and 1 bar. The water is heated until it becomes a saturated vapor. Determine the following:
a) The final temperature of the water in °C.
b) The amount of heat transferred to the tank in kJ. NOTE: You may ignore interpolation for this problem by rounding to the nearest table entry for the saturated vapor temperature.
2. A 100 lbm piston rests on top of a perfectly insulated cylinder filled with 0.5 lbm of R-134a at 50 psi pressure and 80°F temperature. The surroundings have a pressure of 14.7 psi. 198.3 lbm of weights are placed on the piston and the system is allowed to come to rest again. The piston and weights fall 5 inches during this process. Assuming the gravitation constant is 32.17 ft/s^2, determine the following:
a) The area of the piston in in2.
b) The final pressure of the R-134a system in psi.
c) The work done on the R-134a in ft/lbf. (Hint: the R-134a is not the only place you can
draw a system). d) The final temperature of the R-134a in °F.
3. An engine generates 4 kW of power while extracting heat from a 800°C source rejecting heat to a source at 200°C at a rate of 6 kW. Determine the following:
a) The thermal efficiency of the cycle. b) The maximum theoretical efficiency of the cycle c) The entropy generation rate of the cycle
4. Drufus works at a chemical supply facility. The facility has an air supply at 10 bars of pressure and a temperature of 295 K. Drufus attaches an initially evacuated tank that is 0.5 m3 in volume. Drufus fills the tank until it is at a pressure of 3 bar. Assuming the expansion value and air tank are adiabatic as well as air is an ideal gas, determine/complete the following:
a) Draw your system and clearly indicate what components are located in it as well as where the inlet(s) and exit(s) are, if any. b) The final temperature of the tank, in K. c) The final mass of air in the tank, in kg. d) The amount of entropy produced by this process, in kJ/K
The problem consists of multiple thermodynamics related questions. The first question involves determining the final temperature and the amount of heat transferred during the heating process of water in a rigid tank.
Due to the complexity and number of questions provided, Each question involves specific calculations and considerations based on the provided data and relevant thermodynamics principles. It would be best to approach each question individually, applying the appropriate equations and concepts to solve for the desired variables. Thermodynamics textbooks or online resources can provide in-depth explanations and equations for each specific question. Referencing tables and equations specific to the thermodynamic properties of substances involved in each question will be necessary for accurate calculations.
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a) The final temperature of the water in °C is 100°C.
b) The amount of heat transferred to the tank is 8.36 kJ.
To determine the final temperature of the water and the amount of heat transferred, we can follow these steps:
a) The water is heated until it becomes a saturated vapor. Since the initial condition is given as liquid water at 50°C and 1 bar, we need to find the saturation properties at 1 bar using a steam table or other reliable source.
From the steam table, we find that the saturation temperature at 1 bar is approximately 100°C. Therefore, the final temperature of the water in °C is 100°C.
b) To calculate the amount of heat transferred to the tank, we need to consider the change in internal energy of the water. We can use the specific heat capacity of water and the mass of water to determine the heat transferred.
The specific heat capacity of water is typically around 4.18 kJ/kg·°C. The mass of water is given as 0.04 kg.
The change in heat can be calculated using the formula:
Q = m * c * ΔT
Where:
Q is the heat transferred
m is the mass of the water
c is the specific heat capacity of water
ΔT is the change in temperature
Substituting the given values, we have:
Q = 0.04 kg * 4.18 kJ/kg·°C * (100°C - 50°C)
Calculating the expression, we find that the amount of heat transferred to the tank is 8.36 kJ.
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A 2.004 L rigid tank contains .04 kg of water as a liquid at 50°C and 1 bar. The water is heated until it becomes a saturated vapor. Determine the following:
a) The final temperature of the water in °C.
b) The amount of heat transferred to the tank in kJ.
A fuel consist of 87% carbon, 9% hydrogen, 1% sulphur, 1.5% oxygen and the remainder incombustibles. the actual air/fuel ratio is 18,5: 1.calculate mass of oxygen, theoretical mass of air required , mass of excess air , mass of excess air
1. Theoretical mass of air required is 9.484375 units
2. Actual air/fuel ratio is 0.0948
3. Mass of excess air is 18.4052
How to calculate the value1. Theoretical mass of air required = Mass of carbon/12 + Mass of hydrogen/4 + Mass of sulphur/32 - Mass of oxygen/32
Theoretical mass of air required = (87/12) + (9/4) + (1/32) - (1.5/32)
Theoretical mass of air required = 7.25 + 2.25 + 0.03125 - 0.046875
Theoretical mass of air required = 9.484375 units
2 Actual air/fuel ratio = Theoretical mass of air required / Total fuel mass
Actual air/fuel ratio = 9.484375 / 100
Actual air/fuel ratio ≈ 0.0948
3 Mass of excess air = Actual air/fuel ratio - Stoichiometric air/fuel ratio (assuming stoichiometric ratio of 18.5)
Mass of excess air = 18.5 - 0.0948
Mass of excess air ≈ 18.4052
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Please briefly describe the principle of Ziegler-Nichols tuning
rule based on step response of plant (First method), and write down
the corresponding calculation formula of Kp,
Ti, and Td.
The Ziegler-Nichols method is a widely-used heuristic technique for tuning PID controllers based on the behavior of the controlled process.
This is a process for selecting PID controller parameters, Kp, Ti, and Td. The Ziegler-Nichols method involves running the plant under control by the controller and recording the step response. There are two methods to determine the parameters of the controller; they are trial and error methods and mathematical methods. In this method, the plant's behavior is studied by obtaining a step response.
Then, the I-gain and D-gain are adjusted in proportion to the P-gain, which results in improved system performance. Setting the integral time constant Ti and the derivative time constant Td to zero. Increase proportional gain Kp until steady-state oscillation is achieved. Record the oscillation period.
[tex]Kp = 0.6Kc;Ti = 0.5Pc;Td = 0.125Pc[/tex]Where Kp is proportional gain, Ti is integral time constant, Td is derivative time constant, Kc is critical gain, and Pc is critical period.
According to Ziegler-Nichols tuning rules, Kp should be set to the value that causes the system to oscillate with a period equal to the critical period, Ti should be set to one-half of the period of the oscillations, and Td should be set to one-eighth of the period of the oscillations.
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Consider a 210-MW steam power plant that operates on a simple ideal Rankine cycle. Steam enters the turbine at 10MPa and 5008C and is cooled in the condenser at a pressure of 10kPa. Sketch the cycle on a T-s diagram with respect to saturation lines, and determine: (a) the quality of the steam at the turbine exit, (b) the thermal efficiency of the cycle, (c) the mass flow rate of the steam. (d) Repeat Prob. (a)-(c) assuming an isentropic efficiency of 85 percent for both the turbine and the pump.
Given data:Pressure of steam entering turbine (P1) = 10 MPaTemperature of steam entering turbine (T1) = 500 degree CPressure of steam at the condenser (P2) = 10 kPaPower generated (W) = 210 MWNow, let's draw the T-s diagram with respect to saturation lines below:
1. The quality of steam at the turbine exit:From the T-s diagram, we can see that at the turbine exit, the state point lies somewhere between the two saturation lines.Using the steam tables, we can find the saturation temperature and pressure at the exit state:Pressure at the exit (P3) = 10 kPaSaturated temperature corresponding to P3 = 46.9 degree CEnthalpy of saturated liquid corresponding to P3 (h_f) = 191.81 kJ/kgEnthalpy of saturated vapor corresponding to P3 (h_g) = 2676.5 kJ/kgThe quality of steam (x) at the exit state is given by:x = (h - h_f)/(h_g - h_f)Where, h is the specific enthalpy at the exit state.
h = 191.81 + x(2676.5 - 191.81)h = 191.81 + 2421.69x= (h - h_f)/(h_g - h_f)x = (191.81 + 2421.69 - 191.81)/(2676.5 - 191.81)x = 0.91The quality of steam at the turbine exit is 0.91.2. Thermal efficiency of the cycle:For an ideal Rankine cycle, thermal efficiency is given by:eta_th = 1 - (T2/T1)Where, T2 and T1 are the temperatures of the steam at the condenser and the turbine inlet respectively.
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A cam follower mechanism with a displacement diagram that has the following sequence, rise 2 mm in 1.2 seconds, dwell for 0.3 seconds, fall 1 r in 0.9 seconds, dwell again for 0.6 seconds and then continue falling for 1 E in 0.9 seconds.
a) The cam rotation angle during the rise is 120.5 degrees.
b) The rotational speed of the cam is 14.38 rpm.
c) The cam rotation angle during the second fall is 82.9 degrees.
d) Both b) and c).
e) None of the above.
The cam follower mechanism with a displacement diagram that has the following sequence, rise 2 mm in 1.2 seconds, dwell for 0.3 seconds, fall 1 r in 0.9 seconds, dwell again for 0.6 seconds and then continue falling for 1 E in 0.9 seconds can be analyzed as follows:a) To determine the cam rotation angle during the rise, we should know that it took 1.2 seconds to rise 2 mm.
We must first compute the cam's linear velocity during the rise:Linear velocity = (Displacement during the rise) / (Time for the rise)= 2 / 1.2 = 1.67 mm/s Then we can calculate the angle:Cam rotation angle = (Linear velocity * Time) / (Base circle radius)= (1.67 * 1.2) / 10 = 0.2 radian= (0.2 * 180) / π = 11.47 degrees Therefore, the cam rotation angle during the rise is 11.47 degrees. Therefore, option a) is incorrect.b) The rotational speed of the cam can be calculated as follows:Linear velocity = (Displacement during the second fall) / (Time for the second fall)= 1 / 0.9 = 1.11 mm/s
Therefore, the rotational speed of the cam is 71.95 rpm. Therefore, option b) is incorrect.c) To determine the cam rotation angle during the second fall, we should know that it took 0.9 seconds to fall 1 E. We must first compute the cam's linear velocity during the fall:Linear velocity = (Displacement during the fall) / (Time for the fall)= 1 / 0.9 = 1.11 mm/s Then we can calculate the angle:Cam rotation angle = (Linear velocity * Time) / (Base circle radius)= (1.11 * 0.9) / 10 = 0.0999 radians= (0.0999 * 180) / π = 5.73 degrees
Therefore, the cam rotation angle during the second fall is 5.73 degrees. Therefore, option c) is incorrect.Therefore, the answer is option e) None of the above.
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Three kg of air at 150 kPa and 77°C temperature at state 1 is compressed polytropically to state 2 at pressure 750 kPa, index of compression being 1.2. It is then cooled to the original temperature by a constant pressure process to state 3. Find the net work done. (Show the processes on the P-V diagram). [Take Rair =0.287 kJ/kg K]
Given dataThree kg of air at 150 kPa and 77°C temperature at state 1 is compressed polytropically to state 2 at pressure 750 kPa, index of compression being 1.2.
It is then cooled to the original temperature by a constant pressure process to state 3.We have to find out the net work done.
Conversion of temperature from Celsius to Kelvin
K = 273 + CK = 273 + 77K = 350 K
Specific gas constant of air is given as
Rair = 0.287 kJ/kg K
Weight of the air is given as 3 kg.
Work Done is given as,
The work done in polytropic process is given as:Work done in a constant pressure process is given as:
In order to find the specific volume and temperature of air in state 2, we will use the polytropic process formula as given below:For process 1-2From the polytropic process formula,
P1V1n = P2V2nV1/V2
= (P2/P1)^(1/n)V1/V2
= (750/150)^(1/1.2)V1/V2
= 4.187
We know that,The process 1-2 is polytropic so
PV^n = Constant
From state 1 to 2, n = 1.2
Therefore;P1V1^n = P2V2^nV2
= V1*(P1/P2)^(1/n)
Putting values,We get;
V2 = 0.00887 m^3/kg
We can now use the ideal gas law equation to find the temperature in state 2:We know that PV = mRTWhere m = mass, R = gas constant, T = Temperature, and P = pressure
Therefore;T2 = (P2V2)/(mR)T2
= (750*0.00887)/(3*0.287)T2
= 60.2 K
For process 2-3, the temperature is constant and is equal to
T3 = T1 = 350 K
For process 1-2, n = 1.2
The work done in process 1-2 is given by:
For process 2-3, P3 = P2 = 750 kPaV3
= mRT3/P3
= 3*287*350/750*10^3
= 0.351 m^3
The work done in process 2-3 is given by:
Therefore, the net work done isAnswer:
The work done in process 1-2 is 4.29 kJ
The work done in process 2-3 is -0.858 kJ
Therefore, the net work done is 3.432 kJ.
This is a thermodynamics problem in which we are given the initial state of a gas and we are required to find its final state. We are given the temperature and pressure of air in state 1 and are asked to compress it polytropically to state 2 at pressure 750 kPa and an index of compression being 1.2. It is then cooled to the original temperature by a constant pressure process to state 3. We are required to find the net work done in this process.In order to solve this problem, we first converted the temperature from Celsius to Kelvin and then found the weight of the air given. We used the polytropic process formula to find the specific volume and temperature of air in state 2. We then used the ideal gas law equation to find the temperature in state 2. Finally, we used the work done formula for process 1-2 and process 2-3 to find the net work done. The main answer for this question is 3.432 kJ.
In conclusion, we can say that this was a simple thermodynamics problem in which we were required to find the net work done in a process. We solved this problem by using the polytropic process formula and the ideal gas law equation. We found that the net work done in this process is 3.432 kJ.
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An industrial engineer is considering two robots for purchase by a fiber optic manufacturing company. Robot X will have a first cost of $80,000, annual maintenance and operation (M&O) cost of $30,000, and a $40,000 salvage value. Robot Y'will have a first cost of $97,000, an annual M&O cost of $27,000, and a $50,000 salvage value. Which should be selected on the basis of a future worth comparison at an interest rate of 15% per year? Use a 3-year study period.
Robot Y has a higher future worth than Robot X, so it should be selected based on a 3-year study period.
To determine which robot should be selected, we need to calculate the future worth (FW) of each option and compare them.
Let's start by calculating the FW of Robot X:
- First cost: $80,000
- Annual M&O cost: $30,000
- Salvage value: $40,000
Using the future worth formula, we can calculate the FW of Robot X at an interest rate of 15% per year for a 3-year study period:
FW_X = -80,000 - 30,000(P/A,15%,3) + 40,000(P/F,15%,3)
FW_X = -80,000 - 30,000(2.283) + 40,000(0.658)
FW_X = $12,860.
Now let's calculate the FW of Robot Y:
- First cost: $97,000
- Annual M&O cost: $27,000
- Salvage value: $50,000
Using the same formula and interest rate, we can calculate the FW of Robot Y:
FW_Y = -97,000 - 27,000(P/A,15%,3) + 50,000(P/F,15%,3)
FW_Y = -97,000 - 27,000(2.283) + 50,000(0.658)
FW_Y = $20,118.
Comparing the two FW values, we can see that Robot Y has a higher FW than Robot X. Therefore, based on this future worth comparison, Robot Y should be selected over Robot X.
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Steam is the working fluid in an ideal Rankine cycle. Saturated vapour enters the turbine at 10 MPa and saturated liquid exits the condenser at 0.01 MPa. The net power output is 100 MW. Determine the mass flow rate of steam. Enter your answers in kg/s.
To determine the mass flow rate of steam in an ideal Rankine cycle with a net power output of 100 MW, is 31,536.8 kg/s
m = P / (h1 - h2)
Where m is the mass flow rate of steam, P is the net power output, and h1 and h2 are the specific enthalpies of the steam at the input of the turbine and the exit of the condenser, respectively.
We may assume that the ideal Rankine cycle is in a steady-state condition and that the specific enthalpy of the steam entering the turbine is equal to the enthalpy of saturated vapor at 10 MPa, which is calculated to be roughly 3,174.9 kJ/kg using a steam table.
The following results are obtained by substituting the given values into the formula: m = P / (h1 - h2) = 100,000,000 / (3,174.9 - 41.9) = 31,536.8 kg/s.
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For a turning process, calculate the tangential (cutting) force, given the fact that the maximum diameter for the workpiece is 100 cm, the maximum tool length/diameter is 10 cm and the feed (equivalent the uncut chip thickness hm) is 0.520 mm/rev. The cutting speed 75 m/min, the rake angle of the tool is zero degrees and the depth/width of cut is 1.442 mm (note: this is not uncut chip thickness!). Assume that the workpiece material has a strain hardening exponent mc = 0.44 and a specific cutting force of Kc1 = 1500 N/mm². Give your answer in Newtons.
The tangential force can be calculated using the formula,
[tex]Ft = kc1 × f × t × z[/tex]
Where, Ft = tangential force kc1 = specific cutting force f = feed per revolution t = depth of cutz = number of teeth on the cutting tool.
Given that the maximum diameter of the workpiece is 100 cm and the maximum tool length/diameter is 10 cm. The diameter of the workpiece is[tex]100/2 = 50 cm = 500 mm[/tex]. And the length of the tool is 10 cm = 100 mm. The maximum radius of the workpiece will be, Maximum radius [tex]= 500/2 = 250 mm[/tex]. The width of cut will be 1.442 mm.
The feed per revolution (f) is 0.520 mm/rev. So, feed per minute (F) will be,
[tex]F = f × N, where N = speed in RPMN = (speed × 1000)/[3.14 × diameter][/tex]
For the given cutting speed 75 m/min, we can find out the RPM as follows:
[tex]N = (75 × 1000)/(3.14 × 500) = 478.36 rev/minF = 0.520 × 478.36 = 248.96 mm/min[/tex]
Now, the number of teeth on the cutting tool (z) is not given.
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2. Consider a silicon JFET having an n-channel region of donor concentration 1x10¹⁶ cm. (a) Determine the width of the n-channel region for a pinch-off voltage of 12 V. (b) What would the necessary drain voltage (VD) be if the gate voltage is -9 V? (c) Assume the width of the n-channel region to be 40 μm. If no gate voltage is applied, what is the minimum necessary drain voltage for pinch-off to occur? (d) Assume a rectangular n-channel of length 1 mm. What would be the magnitude of the electric field in the channel for case (c) above?
The electric field in the channel is 12,000 V/m.
a) Pinch off occurs when the VGS = Vp. for silicon JFETs, Vp = |2 |V for n-channel JFETs. The channel width can be determined with the equation W = Φ/Vp, where Φ is the donor concentration in the channel. W = 1x10¹⁶ cm³/V·s/12 V = 8.3×10¹⁴ cm.
b) To maintain pinch-off with VGS = -9 V, the drain voltage (VD) must be greater than or equal to -12 V.
c) For a given channel width, the minimum VD necessary for pinch-off to occur, is Vp or 12 V.
d) The electric field in the channel can be calculated with the equation E = VD/L, where L is the length of the channel. E = 12V/1mm = 12,000 V/m.
Therefore, the electric field in the channel is 12,000 V/m.
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You have available a set of five links from which you are to design a four-bar mechanism.
The lengths of the links are as follows: L1= 4cm, L2=6cm, L3=8cm, L4=9cm and L5=14cm.
i) Select four links such that the linkage can be driven by a continuous rotation motor.
ii) Draw a freehand sketch of a crank-rocker mechanism that can be achieved using the selected links. Label the link that is to be driven by the motor.
iii) Draw a freehand sketch of a double-crank mechanism that can be achieved using the selected links.
In this sketch, both Link L2 and Link L3 act as cranks. The motion of the motor (Link L1) will cause both cranks to rotate simultaneously, resulting in the movement of the coupler (Link L5) and the rocker (Link R).
i) To design a four-bar mechanism that can be driven by a continuous rotation motor, we need to select four links such that they form a closed loop. The selected links should have a combination of lengths that allow the mechanism to move smoothly without any interference.
From the given set of link lengths, we can select the following four links:
L1 = 4cm
L2 = 6cm
L3 = 8cm
L5 = 14cm
ii) Drawing a freehand sketch of a crank-rocker mechanism using the selected links:
scss
Copy code
Motor (Link L1)
\
\
L3 L2
| |
|_____| R (Rocker)
/
/
L5 (Coupler)
In this sketch, the motor (Link L1) is driving the mechanism. Link L2 is the crank, Link L3 is the coupler, and Link L5 is the rocker. The motion of the motor will cause the crank to rotate, which in turn will move the coupler and rocker.
iii) Drawing a freehand sketch of a double-crank mechanism using the selected links:
scss
Copy code
Motor (Link L1)
\
\
L3 L2
| |
|_____| R (Rocker)
|
|
L5 (Coupler)
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4. The following is the pattern of x-rays emitted according to the x-ray tube voltage when the Mo target is used.
1. Why do continuous x-rays occur?
2.Why does the swl move to the left as the tube voltage increases?
3. Why the x-ray intensity increases as the tube voltage increases
4.Why is the x-ray emitted not symmetric?
1. Continuous x-rays occur when a high energy electron strikes a metal atom in the target, causing the innermost electrons of the atom to be removed from their orbits. This process leaves an electronic vacancy in the inner shell, which can be filled by an electron from an outer shell. When an outer shell electron fills the inner shell vacancy, it releases energy in the form of an x-ray. However, because each electron shell has a different binding energy, the energy of the released x-ray varies.
2. The swl (short wavelength limit) moves to the left as the tube voltage increases because the x-ray energy and wavelength are inversely proportional. When the tube voltage increases, the energy of the emitted x-rays also increases, and the wavelength decreases. the swl shifts to the left on the graph as the tube voltage increases.
3. The x-ray intensity increases as the tube voltage increases because higher tube voltage results in more electron acceleration, which generates more x-rays. When the tube voltage is increased, more electrons are accelerated across the anode, resulting in more x-rays produced and higher x-ray intensity.
4. The x-ray emitted is not symmetric because of the characteristic x-rays and bremsstrahlung x-rays. Characteristic x-rays occur when an electron drops down to fill an inner shell vacancy, releasing energy in the form of an x-ray. The energy of characteristic x-rays is fixed because the energy difference between the two shells is fixed. Bremsstrahlung x-rays, on the other hand, are emitted when an electron is deflected by the positive charge of the nucleus.
The energy of bremsstrahlung x-rays can vary depending on the extent of electron deflection, resulting in a continuous spectrum of x-ray energies. This combination of characteristic and bremsstrahlung x-rays results in a non-symmetric distribution of x-ray energy.
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Tyres of an F1 race car fail after 100 hrs. A car race at Kyalami Racecourse takes 10Hrs to complete all the laps. a. What are the number of failures in the system? b. What is the probability that 2 tyres of a race car will fail in the race?
In the given scenario, the tyres of an F1 race car fail after 100 hours, and a car race at Kyalami Racecourse takes 10 hours to complete all the laps. To calculate the number of failures in the system, we divide the total race time by the failure time of the tyres. In this case, the number of failures in the system is 10. For the probability of two tyres failing in the race, we can use the concept of a binomial distribution. The probability can be calculated by combining the probability of one tyre failing with the probability of the other tyre failing, assuming independence.
To determine the number of failures in the system, we divide the total race time (10 hours) by the failure time of the tyres (100 hours). This gives us 10/100 = 0.1, indicating that there is one failure in the system.
For the probability of two tyres failing in the race, we can use the binomial distribution. The probability of one tyre failing can be calculated by dividing the tyre failure time (100 hours) by the total race time (10 hours), which gives us 100/10 = 10%. Assuming independence between the two tyres, we can multiply the probability of one tyre failing by itself to get the probability of both tyres failing. Thus, the probability of two tyres failing in the race is 10% * 10% = 1%.
In summary, there is one failure in the system, and the probability of two
ptyres failing in the race is 1%.
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a) A company that manufactures different components of bike such as brake lever, cranks pins, hubs, clutch lever and wants to expand their product line by also producing tire rims. Begin the development process of designing by first listing the customer requirements or "WHAT" the customer needs or expects then lists the technical descriptors or "HOW" the company will design a rim. Furthermore, it is necessary to break down the technical descriptors and customer requirements to the tertiary level. Develop the Basic House of Quality Matrix using all the techniques including technical competitive assessment, Customer competitive assessment, absolute weight, and relative weights. Make reasonable assumptions where required. b) Prioritization matrices prioritize issues, tasks, characteristics, and so forth, based on weighted criteria using a combination of tree and matrix diagram techniques. Once prioritized, effective decisions can be made. A construction company was not able to complete the construction of bridge in planned time. The main causes of failure may include the people, machines, or systems. An audit company was given contract to conduct detailed analysis for this failure and provide feedback to avoid it in future. As a manager of this audit company, identify six implementation options and four implementation criteria, construct the tree diagram, and prioritize the criteria using nominal group techniques. Rank order the options in terms of importance by each criterion. Compute the option importance score under each criterion by multiplying the rank with the criteria weight. Develop the prioritization matrices.
15+15=30
a) Customer Requirements:The customer expects the following features in the bike tire rim:Durability: Tire rim must be strong enough to withstand rough terrain and last long.Aesthetics: Rim should look attractive and appealing to the eye.Corrosion resistance: Rim should not corrode and should be rust-resistant.Weighting Factors:The relative weight of durability is 0.35, aesthetics is 0.30 and corrosion resistance is 0.35. Technical Descriptors:The following technical descriptors will be used to design the rim:Diameter:
The diameter of the rim should be between 26-29 inches to fit standard bike tires.Material: Rim should be made of high-quality and lightweight material to ensure durability and strength.Weight: Weight of the rim should not be too high or too low.Spokes: Rim should have adequate spokes for strength and durability.Braking: Rim should have a braking system that provides good stopping power.Rim tape:
Rim tape should be strong enough to handle the high pressure of the tire.Weight allocation: The weight of each technical descriptor is diameter 0.10, material 0.30, weight 0.20, spokes 0.15, braking 0.10, and rim tape 0.15. Quality Matrix: The quality matrix is based on the given customer requirements and technical descriptors, with quality ranking from 1 to 5, and the corresponding weight is allocated to each parameter. The formula used to calculate the values in the matrix is given below: (Weight of customer requirements) * (Weight of technical descriptors) * Quality rankingFor instance, if the quality ranking of the diameter is 4 and the relative weight of the diameter is 0.1, the value of the quality matrix is (0.35) * (0.10) * 4 = 0.14.
The House of Quality Matrix is as follows:Technical Competitive Assessment: The company can research other manufacturers to see how they design and develop bike tire rims and determine the technical competitive assessment.Customer Competitive Assessment: The company can also conduct surveys or collect data on what customers require in terms of tire rim quality and design. Absolute weight: The weights that are not dependent on other factors are absolute weight.Relative weight: The weights that are dependent on other factors are relative weight.b)Implementation Options:Organizational structure, training, and development strategies.Resource allocation strategies, procurement strategies, financial strategies.Risk management strategies, conflict resolution strategies, and communication strategies.Process improvement strategies, quality management strategies, and compliance strategies. Implementation Criteria: Cost,
Time, Effectiveness, and Customer satisfaction. Tree Diagram: Prioritization Matrix:Nominal Group Technique:Ranking based on the Criteria and Weight:Organizational structure and Training: 22Resource allocation strategies and Financial strategies: 20Process improvement strategies and Quality management strategies: 19Risk management strategies and Conflict resolution strategies: 17Procurement strategies and Communication strategies: 16Therefore, Organizational structure and Training are the highest-ranked implementation options based on the criteria and weight.
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This is the distance between the parallel axes of spur gears or parallel helical gears, or the distance between the crossed axes of helical gears and worm gears. It can be defined also as the distance between the centers of pitch circles. What is this distance? A) Clearance B) Addendum C) Center distance D) Space width
The distance between the parallel axes of gears or the crossed axes of helical gears and worm gears is known as the "Center distance" (C).
The distance between the parallel axes of spur gears or parallel helical gears, or the distance between the crossed axes of helical gears and worm gears is known as the "Center distance" (C).
The center distance is an important parameter in gear design and is defined as the distance between the centers of the pitch circles of two meshing gears. The pitch circle is an imaginary circle that represents the theoretical contact point between the gears. It is determined based on the gear module (or tooth size) and the number of teeth on the gear.
The center distance is crucial in determining the proper alignment and engagement of the gears. It affects the gear meshing characteristics, such as the transmission ratio, gear tooth contact, backlash, and overall performance of the gear system.
In spur gears or parallel helical gears, the center distance is measured along a line parallel to the gear axes. It determines the spacing between the gears and affects the gear ratio. Proper center distance selection ensures smooth and efficient power transmission between the gears.
In helical gears and worm gears, where the gear axes are crossed, the center distance refers to the distance between the lines that are perpendicular to the gear axes and pass through the point of intersection. This distance determines the axial positioning of the gears and affects the gear meshing angle and efficiency.
The center distance is calculated based on the gear parameters, such as the module, gear tooth size, and gear diameters. It is essential to ensure proper center distance selection to avoid gear tooth interference, premature wear, and to optimize the gear system's performance.
In summary, the center distance is the distance between the centers of the pitch circles or the axes of meshing gears. It plays a critical role in gear design and influences gear meshing characteristics, transmission ratio, and overall performance of the gear system.
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Regarding the Nafolo Prospect 3. Development Mining
a. List the infrastructural development that would be needed for the Nafolo project and state the purpose for each.
b. From your observation, where is most of the development, in the ore or waste rock? What does this mean for the project?
c. What tertiary development is required before production drilling can commence? . Answers should be detailed and all questions should be answered.
a. Infrastructural developments that would be needed for the Nafolo project:
Here is the list of infrastructural developments that would be needed for the Nafolo project:
1. Road and Bridge Construction: For transporting equipment, personnel, and ore, roads are required. Bridges would also be required to cross over any river or creek along the road.
2. Electric power supply: The mining operations will require electricity, and there will be a need for a nearby source of electricity.
3. Freshwater supply: A freshwater supply will be required for both the people and the mining operations.
4. Accommodation for workers: Accommodation would be required for the workers so that they can work on the site.
b. Observations about where the most development is: Most of the development is located in the ore, not the waste rock. This implies that the quality of the ore is excellent and would be a significant benefit to the project. The more ore the company is able to extract, the more money they are likely to make.
c. Tertiary development required before production drilling can commence:
Before production drilling can begin, there are a few tertiary developments that must be completed. They are:
1. Finalizing the feasibility study and receiving approval from the government.
2. Acquiring financing for the project.
3. Contracting companies to construct the necessary infrastructure.
4. Hiring staff to run the mining operations.
5. Environmental approvals for mining to proceed.
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A mixing tank is usually installed a motor with gear box connecting to the shaft and agitator blade. The motor is measured the operating data as working, 380V, 22A, P.F. 0.8, 3Ph/50Hz/4-pole. The motor rotating speed out of gear box was 550 rpm. An engineer has learned that using VSD/VFD device can control the rotating speed to a targeted speed of 550 rpm yet saving the electrical power. Calculate:
1) target frequency of setting VSD to get 550 rpm
2) Calculated power after reducing speed with VSD
(Hint: Neglect the slip factor of the motor)
1) Target frequency of setting VSD to get 550 rpm:The synchronous speed of an electric motor is a function of the number of poles and the frequency of the applied power. The synchronous speed is given by the formula:
Ns=120f/p
Where,Ns = synchronous speed (rpm)f = power frequency (Hz)p = number of poles.
The motor is operating at a rotating speed of 550 rpm before being connected to VSD. Therefore, the synchronous speed, which is determined by the number of poles, is:
Ns = (120 × f)/p550
= (120 × f)/4f
= (550 × 4)/120 = 18.3 Hz
Therefore, the target frequency of setting VSD to get 550 rpm is 18.3 Hz.2) Calculated power after reducing speed with VSD:With the VSD connected, the motor rotating speed is controlled to a targeted speed of 550 rpm. The frequency applied by the VSD is 18.3 Hz, which is below the standard operating frequency of 50 Hz.Before the VSD is connected, the power consumed by the motor is given by the formula:
P = V × I × P.F × √3
where,P = power (Watts),V = voltage (volts)I = current (amps)P.F = power factor√3 = square root of three,V = 380,VI = 22AP.F = 0.8
Therefore, the power consumed before the VSD is connected is:
P = 380 × 22 × 0.8 × √3P
= 21241.41 W
When the motor is connected to the VSD, the frequency is reduced to 18.3 Hz, and the speed is controlled to 550 rpm. The power consumed by the motor is given by the formula:
P = V × I × P.F × √3
where,P = power (Watts)V = voltage (volts)I = current (amps)P.
F = power factor√3 = square root of three.V = 380VI = 22AP.F = 0.8
Therefore, the power consumed after the VSD is connected is:
P = 380 × I × P.F × √3
where f is the frequency applied by the VSD.
I = (22 × 18.3)/50
= 8.026 AP.F
= 0.8√3 = 1.732P
= 380 × 8.026 × 0.8 × 1.732P
= 4307.04 W
Therefore, the calculated power after reducing speed with VSD is 4307.04 W.
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Exercises on fluid mechanics. Please, What assumptions/assumptions were used in the solution.
Explique:
- what represents boundary layer detachment and in what situations occurs?
- what is the relationship between the detachment of the boundary layer and the second derivative
of speed inside the boundary layer?
- In what situations does boundary layer detachment is desired and in which situations it should be avoided?
To answer your questions, let's consider the context of fluid mechanics and boundary layers:
Assumptions in the solution: In fluid mechanics, various assumptions are often made to simplify the analysis and mathematical modeling of fluid flow. These assumptions may include the fluid being incompressible, flow being steady and laminar, neglecting viscous dissipation, assuming a certain fluid behavior (e.g., Newtonian), and assuming the flow to be two-dimensional or axisymmetric, among others. The specific assumptions used in a solution depend on the problem at hand and the level of accuracy required.
Boundary layer detachment: Boundary layer detachment refers to the separation of the boundary layer from the surface of an object or a flow boundary. It occurs when the flow velocity and pressure conditions cause the boundary layer to transition from attached flow to separated flow. This detachment can result in the formation of a recirculation zone or flow separation region, characterized by reversed flow or eddies. Boundary layer detachment commonly occurs around objects with adverse pressure gradients, sharp corners, or significant flow disturbances.
Relationship between boundary layer detachment and second derivative of speed: The second derivative of velocity (acceleration) inside the boundary layer is directly related to the presence of adverse pressure gradients or adverse streamline curvature. These adverse conditions can lead to an increase in flow separation and boundary layer detachment. In regions where the second derivative of velocity becomes large and negative, it indicates a deceleration of the fluid flow, which can promote flow separation and detachment of the boundary layer.
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Which of the following statements is not part of the Kinetic-Molecular Theory?
a. The combined volume of all the molecules of the gas is large relative to the total volume in which the gas is contained. b. Gases consist of large numbers of molecules that are in continuous, random motion. c. Attractive and repulsive forces between gas molecules are negligible. d. The average kinetic energy of the molecules is proportional to the absolute temperature.
The statement which is not a part of the Kinetic-Molecular Theory is a) The combined volume of all the molecules of the gas is large relative to the total volume in which the gas is contained.
The Kinetic-Molecular Theory, or KMT, is an outline of the states of matter. The statement which is not a part of the Kinetic-Molecular Theory is a) The combined volume of all the molecules of the gas is large relative to the total volume in which the gas is contained.
KMT is built on a series of postulates. KMT includes four important postulates. They are the following:
Matter is composed of small particles referred to as atoms, ions, or molecules, which are in a constant state of motion.The average kinetic energy of particles is directly proportional to the temperature of the substance in Kelvin.
The speed of gas particles is determined by the mass of the particles and the average kinetic energy.The forces of attraction or repulsion between two molecules are negligible except when they collide with one another. Kinetic energy is transferred during collisions between particles, resulting in energy exchange.
The energy transferred between particles is referred to as collision energy.Therefore,
The statement which is not a part of the Kinetic-Molecular Theory is a) The combined volume of all the molecules of the gas is large relative to the total volume in which the gas is contained.
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2) The commutation interval in controlled and uncontrolled rectifier circuits: a) is resulted from the highly inductive loads. b) is resulted from the series inductance of the source. e) reduces the average value of the output voltage.s d) all of the above. e) b+c. f) atc. 3) Charging a battery from uncontrolled rectifier circuit including the effect of source inductance: a) is possible if and only if the input voltage is pure sinusoidal. b) is possible with never pure sinusoidal charging current. c) is impossible as battery must receive DC voltage. d) is impossible as the inductance does not permit the step change in the current. e) none of the above f) a+b. 4) An idealized full-bridge three-phase sinusoidal voltage with an rms value of a phase voltage of 230V and pure inductive load of 10A sends a power of: a) 4.00 kW. b) 2.35 kW. v₂1:35 236 *√3 5.38 kW. 3.105 kW. 9.32 kW. none of the above. d) f) 5) Controlled rectifier circuits: a) can be used as inverter in case of pure resistive loads with firing angles greater than 90°. b) use thyristors as power semiconductors devices. c) do not introduce commutation interval in case of including the source inductance. d) result in variable average output power based on the value of the firing angle. e) a+b. f) b+d.
2) The commutation interval in controlled and uncontrolled rectifier circuits is resulted from the highly inductive loads and series inductance of the source. The correct option is (e) b+c. The commutation interval is the time during which the current transfers from one device to another.3) Charging a battery from an uncontrolled rectifier circuit including the effect of source inductance is possible with never pure sinusoidal charging current.
The correct option is (b).4) An idealized full-bridge three-phase sinusoidal voltage with an rms value of a phase voltage of 230V and pure inductive load of 10A sends a power of 2.35 kW. The correct option is (b).P = √3*Vph*Iph*cosϕ= √3*230*10*cos90= 2.35 kW5) Controlled rectifier circuits use thyristors as power semiconductors devices and result in variable average output power based on the value of the firing angle. The correct option is (f) b+d.
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A particle P has velocity:
v(t) = 5 + 3t a) Find the acceleration of the particle
b) Express position (x) as a function of time given the initial condition given the initial condition x(0) = 3m (4) c) Find the distance traversed by the particle in the first 5 seconds of its motion
The particle has an acceleration of 3 m/s^2. Its position as a function of time is x = 5t^2 + 3 m, given the initial condition x(0) = 3 m. The distance traversed by the particle in the first 5 seconds is 75 m.
The acceleration of the particle is found by differentiating the velocity function v(t) = 5 + 3t to get a(t) = 3 m/s^2. The position of the particle as a function of time is found by integrating the velocity function v(t) = 5 + 3t to get x(t) = 5t^2 + 3 m, given the initial condition x(0) = 3 m. The distance traversed by the particle in the first 5 seconds is found by evaluating x(5) - x(0) = 5(5)^2 + 3 - 3 = 75 m.
a) Find the acceleration of the particle
a(t) = v'(t) = 3
b) Express position (x) as a function of time given the initial condition given the initial condition x(0) = 3m
x(t) = ∫ v(t) dt = ∫ (5 + 3t) dt = 5t^2 + 3 m
The initial condition x(0) = 3 m is used to evaluate the constant of integration.
c) Find the distance traversed by the particle in the first 5 seconds of its motion
x(5) - x(0) = 5(5)^2 + 3 - 3 = 75 m
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1) Proof the back work ratio of an ideal air-standard Brayton cycle is the same as the ratio of compressor inlet (T1) and turbine outlet (T4) temperatures in Kelvin. Use cold-air standard analysis. (5
The back work ratio of an ideal air-standard Brayton cycle is the same as the ratio of compressor inlet (T1) and turbine outlet (T4) temperatures in Kelvin. Use a cold-air standard analysis.
Given data T1 = More than 100 in KelvinT4 = More than 100 in Kelvin Formula, Back Work Ratio (BWR) = Wc / Q_ in (or) W_ t / Q_ in, Where Wc = Work of compressor, W_ t = Work of turbine, and Q_ in = Heat Supplied to the cycle. Proof: The Brayton cycle is a closed-cycle in which the working fluid receives and rejects heat in the same manner.
Rankine cycle, but the working fluid is not water but air. The cycle comprises four basic components: compressor, heat exchanger, turbine, and heat exchanger, with two adiabatic expansion and compression processes. The first process is compression by the compressor.
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Annual demand for a product is Normally distributed with a mean of 1000 & satndard deviation of 250. The purchase & selling cost is $0.06 & $0.2, respectively. It takes 2 months from the intiation to the receipt of an order. The cost to initiate & receive an order is $20.The cost of stockout is the cost of lost profit plus an additional $0.2/product. Requirements: (5 Pts) a. Using 22% as annual interst rate and EOQ as the lot size, calculate optimum value of R (5 Pts) b. Calculate the simultaneous optimal (Q*,R*) (10 Pts) c. Compare the average annual holding, setup & stock-out costs of the solutions in parts (a.) & (b.) (5 Pts) d. Calculate the safety stock for this product at the optimal solution (5 Pts) e. If Type 1 service is 95%, calculate optimum value of R (10 Pts) f. Determine the otpimal values of (Q.R) assuming a Type 2 service level of 95%
We need to find the optimal values of Q and R assuming a Type 2 service level of 95%. By performing these calculations and analyses, we can obtain the solutions for parts (a) to (f) and gain insights into the optimal inventory management strategy for the given scenario.
a. To calculate the optimum value of R using the EOQ model, we use the formula R = d * L, where d is the annual demand rate and L is the lead time in years.
b. The simultaneous optimal values of Q* and R* can be determined by considering the EOQ model with lead time. We calculate Q* using the formula Q* = √((2 * D * S) / H), where D is the annual demand rate, S is the setup or ordering cost, and H is the holding cost per unit.
c. To compare the average annual holding, setup, and stock-out costs between the solutions in parts (a) and (b), we need to calculate the relevant costs for each scenario and compare them.
d. The safety stock can be calculated as the difference between the reorder point R and the average demand during lead time. It provides a buffer to account for uncertainties in demand and lead time.
e. To calculate the optimum value of R for a Type 1 service level of 95%, we can use the formula R = d * L + z * σL, where z is the z-score corresponding to the desired service level and σL is the standard deviation of demand during lead time.
f. To determine the optimal values of Q and R assuming a Type 2 service level of 95%, we need to consider the total cost approach and iterate through different values of Q and R to find the combination that minimizes the total cost.
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3. [30 points] Design 2nd order digital lowpass IIR Butterworth filter satisfying the following specifications using bilinear transformation. Do NOT use MATLAB butter command for this problem. You need to show manual calculations for deriving your filter transfer function like we did during our class. 3-dB cutoff frequency: 20 kHz Sampling frequency: 44.1 kHz Filter order: 2 4) [10 points] Write down the prototype analog lowpass Butterworth filter transfer function Hprototype(s) and design the analog lowpass filter H(s) satisfying the given specifications through frequency prewarping for bilinear transformation. 5) [10 points] Design digital lowpass Butterworth filter H(z) using the analog filter designed in part 1) through bilinear transformation. 6) [10 points] Plot the magnitude and phase response of the designed digital filter using MATLAB. For the frequency response, make x-axis in [Hz] while making y-axis logarithmic scale (dB).
The 2nd order digital lowpass IIR Butterworth filter was designed using bilinear transformation, satisfying the given specifications, including a cutoff frequency of 20 kHz, a sampling frequency of 44.1 kHz, and a filter order of 2.
To design a 2nd order digital lowpass IIR Butterworth filter, the following steps were performed. Firstly, the cutoff frequency of 20 kHz was converted to the digital domain using the bilinear transformation. The filter order of 2 was taken into account for the design.
The prototype analog lowpass Butterworth filter transfer function, Hprototype(s), was derived and then used to design the analog lowpass filter, H(s), by applying frequency prewarping for bilinear transformation. Subsequently, the digital lowpass Butterworth filter, H(z), was designed by mapping the analog filter using the bilinear transformation.
Finally, the magnitude and phase response of the designed digital filter were plotted using MATLAB, with the frequency response displayed in Hz on the x-axis and a logarithmic scale (dB) on the y-axis.
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Steam enters a diffuster steadily at a pressure of 400 psia and a temperature of Tdiffuser = 500.0 °F. The velocity of the steam at the inlet is Veldiffuser 80.0 ft s = and the mass flow rate is 5 lbm/s. What is the inlet area of the diffuser? ANS: 11.57in^2
The inlet area of the diffuser is 11.57 in^2.
To determine the inlet area of the diffuser, we can use the mass flow rate and the velocity of the steam at the inlet. The mass flow rate is given as 5 lbm/s, and the velocity is given as 80.0 ft/s.
The mass flow rate, denoted by m_dot, is equal to the product of density (ρ) and velocity (V) times the cross-sectional area (A) of the flow. Mathematically, this can be expressed as:
m_dot = ρ * V * A
Rearranging the equation, we can solve for the cross-sectional area:
A = m_dot / (ρ * V)
Given the values for mass flow rate, velocity, and the properties of steam at the inlet (pressure and temperature), we can calculate the density of the steam using steam tables or thermodynamic properties of the fluid. Once we have the density, we can substitute the values into the equation to find the inlet area of the diffuser.
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An internally pressurized thick-walled pressure cylinder has known stresses on the inner wall of 0 = -16 MPa and J = 44 MPa. Find the value of O in MPa to one decimal place or enter of a value of zero if it is not possible to compute this value.
Inner wall stress, σ₁ = -16 Mastres at radial distance, r = R, σ₂ = J = 44 MP. Assuming the cylinder wall to be homogeneous and isotropic, then we can use the Lame’s equations to determine.
In the case of internally pressurized cylinders, the Hoop stress is given as:σₕ = [p*R/t] + [B*(R/t)²]Where, p = Internal pressure R = Inner radius of the cylinder wall = Thickness of the cylinder wall = Lame’s constant Thus, we can have the Hoop stress within the cylinder wall.
= [p*R/t] + [B*(R/t) ²]
(1) Again, the radial stress within the cylinder wall is given by:σᵣ = [p*R/t] - [B*(R/t) ²]
(2) Thus, substituting the known values of R, t, σ₁ and J in the equations (1) and (2),
we can have two equations in two unknowns (p and B) as follows:-16 = [p*R/t] + [B*(R/t)²]…… (1)44 = [p*R/t] - [B*(R/t)²]…… (2)Multiplying both sides of equation (2) by (-1), we get:44 = [p*R/t] + [B*(R/t)²]….. (3) Subtracting equation (3) from equation (1), we get: -60 = -2B*(R/t) ²Simplifying.
[tax]= [p*R/t] + [B*(R/t) ²]……[/tax]
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QUESTION 1 (10 MARKS) As a Graduate Metallurgist you are working in a plant to produce a product as fast as possible. The process uses a liquid with specific heat capacity of 75.44 J.K-1mol-1 . In the first furnace you start with a temperature of 45°C and your aim temperature is 155°C. You use 2 kg of fuel with a mole mass of 22 gmol-1 . In your second furnace you start at a temperature of 82°C and also aim for 155°C. You have here 150 kg of fuel, the same type as in furnace 1. In furnace 1 you use an element of 1200W and for furnace 2 you use 3650W. Do metallurgical calculations to determine which furnace you will opt for, with special reference to time
QUESTION 2 (10 MARKS) Calculate the heat required for 2.4 kg of steel to be heated from 130°C to 920°C if the specific heat capacity is 54.6 + 2.1 x 10-3T – 6.5 x 105T -2 . Mole mass of steel is 56 gmol-1 . If the heat price is R36.50 per 2000J and your budget account is R2420.00 will you be able to buy the energy? Prove by thermodynamic calculations. QUESTION 3 (5 MARKS) A company producing aluminium is using coke as a fuel. The furnace work temperature is 1100°C. Is it possible to produce aluminium in this furnace, comment on your answer? Do metallurgical calculations to prove your answer. You are told that when the enthalpy is zero chemical equilibrium is reached. Given: 2Al + 3 2 O2 = Al2O3 ∆H = −216510 + 169.4T Joules C + 1 2 O2 = CO ∆H = −165550 + 170.2T Joules 3
QUESTION 4 (15 MARKS) The transformation of manganese is as follows: Mn(α) at 720°C → Mn(β) at 1100°C → Mn(γ) at 1136°C → Mn(δ) Calculate the heat of this process when Mn is oxidised by pure oxygen to form MnO at 1250°C. Given: Mn(α) + 1 2 O2 → MnO ∆H298 = −384900 Jmol−1 Mn(α) → Mn(β) ∆Hf = 2100 Jmol−1 Mn(β) → Mn(γ) ∆Hf = 2380 Jmol−1 Mn(γ) → Mn(δ) ∆Hf = 1840 Jmol−1 Cp(αMn) = 21.55 + 15.56 × 10−3T JK −1mol−1 Cp(βMn) = 34.85 + 2.76 × 10−3T JK −1mol−1 Cp(γMn) = 45.55 JK −1mol−1 Cp(δMn) = 47.28 JK −1mol−1 Cp(MnO) = 46.44 + 8.12 × 10−3T − 3.68 × 105T −2 JK −1mol−1 Cp(Oxygen) = 29.96 + 4.184 × 10−3T − 1.67 × 105T −2 JK −1mol−1
The specific heat capacity of the steel is given as a function of temperature, and we are required to use thermodynamic calculations to determine the heat energy required. Thus, we need to integrate the given equation with respect to T over the given temperature range, and obtain the average value of the integrand. In th
e first furnace, the heat energy required is obtained using the formula, Q = mcΔT.
1. Hence, Q = 2 x 75.44 x (155 - 45) = 22632.32 J. In the second furnace, the heat energy required is obtained using the same formula, but we are not given the mass of the liquid.
Thus,t = Q/P
For furnace 2, the time required is given by,t = (150 x 1000) / 3650 = 41.1 sThus, furnace 2 will produce the product faster.
2. The heat energy required to heat the steel from 130°C to 920°C is obtained using the formula,
Q = mcΔT
Hence, Q = (2.4 x 10^3) x (1/790) x (54.6(790-130) + 2.1 x 10^-3 (790^2 - 130^2)/2 - 6.5 x 10^-5 (790^-1 - 130^-1)) = 5.63 x 10^5 J
The cost of the heat energy is given by,
C = (Q/2000) x R36.50 = (5.63 x 10^5 / 2000) x R36.50 = R101.31
Since R101.31 is less than R2420.00, the budget is sufficient to buy the energy.
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