To determine the stress at the given point (u = 20 mm, v = 35 mm) on the asymmetric profile, we need to calculate the normal stress along the UU and VV axes using the bending moment and second moments of area. The formula for calculating normal stress due to bending is: σ = (M * y) / I
Where σ is the stress, M is the bending moment, y is the perpendicular distance from the neutral axis, and I is the second moment of area. First, we need to find the distances (y_u and y_v) from the neutral axis along the UU and VV axes, respectively. We can use the given coordinates and the angle of the principal axes. y_u = u * cos(f) = 20 mm * cos(17°), y_v = v * sin(f) = 35 mm * sin(17°). Next, we can calculate the stresses along the UU and VV axes. σ_u = (M * y_u) / I_uu, σ_v = (M * y_v) / I_vv. Substituting the given values, we can calculate the stresses. σ_u = (2000 Nm * 20 mm * cos(17°)) / (1.48x10^-6 m^4),σ_v = (2000 Nm * 35 mm * sin(17°)) / (8.67x10^-7 m^4). Finally, convert the stresses to the appropriate units if needed. Please note that the calculations provided here assume linear elastic behavior and neglect other factors such as shear stress.
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If a logical address is 32 bits (4 bytes), what is the minimum header size at network layer of the TCP/IP protocol suite?
In a TCP/IP network, if the logical address is 32 bits (4 bytes), then the minimum header size at the network layer of the TCP/IP protocol suite is 20 bytes.
The Transmission Control Protocol/Internet Protocol (TCP/IP) is a suite of communication protocols used to interconnect network devices on the internet. It is divided into four layers: the application layer, the transport layer, the network layer, and the data link layer.The network layer is the third layer of the TCP/IP protocol suite. Its primary function is to provide logical addressing and routing services between different networks. The network layer header includes fields such as the source and destination IP addresses, the type of service, time to live (TTL), and protocol.
In a TCP/IP network, the minimum size of the network layer header is 20 bytes, regardless of the logical address size. This is because the network layer header is fixed in size and contains information such as the protocol, source and destination IP addresses, and other important fields that are necessary for the proper functioning of the network layer .Therefore, the minimum header size at the network layer of the TCP/IP protocol suite, when a logical address is 32 bits (4 bytes), is 20 bytes.
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Air with mass of 16 kg is heated at constant gage pressure of 9044 hPa from -24.0 °C to 184 °C. Determine the heat required in MJ, and the change of system volume (AV=V₂- V₁).
The heat required in MJ is 0.43488 MJ. The change of system volume is 0.0718 m³.
Given,
Mass of air, m = 16 kg
Initial temperature, T₁ = -24°C
Initial pressure, P₁ = 9044 hPa
Final temperature, T₂ = 184°C
We know that the heat required is given by
Q = mCp(T₂ - T₁)
where Cp is the specific heat capacity of air at constant pressure.
The change in volume, AV is given by
AV = V₂ - V₁
We know that for an ideal gas,
PV = mRT
where P is the pressure of the gas,
V is the volume of the gas,
m is the mass of the gas,
R is the universal gas constant and
T is the temperature of the gas.
We can write the above equation as
PV = nRT
where n is the number of moles of gas. We can write n in terms of mass as
n = m / MM
where MM is the molar mass of the gas.
For air,
MM = 28.97 g/mol
= 0.02897 kg/mol
Therefore,
n = m / 0.02897
The ideal gas law can be written as
PV = (m / MM)RT
or
PV = nRT
Also,
P / T = constant
Therefore,
P₁ / T₁ = P₂ / T₂
or
P₂ = (P₁ / T₁) x T₂
Therefore,
P₂ = (9044 / (273 - 24)) x (184 + 273)
= 123531.24 Pa
The volume of the gas can be found using the ideal gas law:
V₁ = (mRT₁) / P₁= (16 x 8.314 x (273 - 24)) / (9044 x 100)
V₁ = 0.1554 m³
V₂ = (mRT₂) / P₂= (16 x 8.314 x (184 + 273)) / (123531.24)
V₂ = 0.2272 m³
Therefore,
AV = V₂ - V₁
= 0.2272 - 0.1554
= 0.0718 m³
We know that
Cp = 1005 J/kg K
Therefore,
Q = mCp(T₂ - T₁)
= 16 x 1005 x (184 + 24)
= 434880 J
= 434.88 kJ
= 0.43488 MJ
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A. Considering a stoichiometric mixture of the fuel given below (C 3H 1O 2) and air, determine a. The stoichiometric reaction equation. b. The mole fraction of oxygen (O2). c. The air-fuel ratio.
The stoichiometric reaction equation for the fuel C3H10 and air is C3H10 + (13/2)O2 -> 3CO2 + 5H2O. The mole fraction of oxygen (O2) can be calculated by dividing the moles of O2 by the total moles of the mixture.
The air-fuel ratio is determined by dividing the moles of air (oxygen) by the moles of fuel, and in this case, it is 6.5:1.
a. The stoichiometric reaction equation for the fuel C3H10 is:
C3H10 + (13/2)O2 -> 3CO2 + 5H2O
b. To determine the mole fraction of oxygen (O2), we need to calculate the moles of oxygen relative to the total moles of the mixture. In the stoichiometric reaction equation, the coefficient of O2 is (13/2). Since the stoichiometric ratio is based on the balanced equation, the mole fraction of O2 can be calculated by dividing the moles of O2 by the total moles of the mixture.
c. The air-fuel ratio can be calculated by dividing the moles of air (oxygen) by the moles of fuel. In this case, the stoichiometric reaction equation indicates that 13/2 moles of O2 are required for 1 mole of C3H10. Therefore, the air-fuel ratio can be expressed as 13/2:1 or 6.5:1.
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a) Describe the following: i. Encoder ii. Decoder iii. RAM iv. ROM
b) Describe the operation of: i. Write and read ii. Basic binary decoder
a) i. Encoder: An encoder is an electronic device or circuit that is used to convert the data signal into a coded format that has a different format than the initial data signal.
ii. Decoder: A decoder is an electronic circuit that is used to convert a coded signal into a different format. It is the inverse of an encoder and is used to decode the coded data signal back to its original format.
iii. RAM: Random Access Memory (RAM) is a type of volatile memory that stores data temporarily. It is volatile because the data stored in RAM is lost when the computer is switched off or restarted. RAM is used by the computer's processor to store data that is required to run programs and applications.
iv. ROM: Read-Only Memory (ROM) is a type of non-volatile memory that stores data permanently. The data stored in ROM cannot be modified or changed by the user. ROM is used to store data that is required by the computer's operating system to boot up and start running.
b) i. Write and read: The write operation is used to store data in a memory location. The data is written to the memory location by applying a write signal to the memory chip. The read operation is used to retrieve data from a memory location. The data is retrieved by applying a read signal to the memory chip.
ii. Basic binary decoder: A basic binary decoder is a logic circuit that is used to decode a binary code into a more complex output code. The binary decoder takes a binary input code and produces a more complex output code that is based on the input code. The output code can be used to control other circuits or devices.
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You are instructed by the plant Operations Manager to install a pump to lift 30L/s of water at 22degC from a sump to a tank. The tank pressure is 200Kpag. The water level in the tank is 20m above the pump centerline and the pump is 4m above the water level in the sump. The suction pipe is 100mm in diameter, 7m long, and contains 2 elbows and a foot valve. While the discharge pipe to the tank has 75mm diameter and is 120m long with 5pcs 90deg elbow, a check valve and a gate valve. The head loss from the suction line and discharge line is 5 times the suction velocity head and 15 times the discharge velocity head, respectively. for a mechanical efficiency of 80%. Determine the required motor output power (kW).
By determining the required induction motor output power for the pump, we need to consider the total head required and the efficiency of the pump.
First, let's calculate the total head required for the pump:
1. Suction Side:
- Convert the flow rate to m³/s: 30 L/s = 0.03 m³/s.
- Calculate the suction velocity head (Hv_suction) using the diameter and velocity: Hv_suction = (V_suction)² / (2g), where V_suction = (0.03 m³/s) / (π * (0.1 m)² / 4).
- Calculate the total suction head (H_suction) by adding the elevation difference and head loss: H_suction = 4 m + Hv_suction + 5 * Hv_suction.
2. Discharge Side:
- Calculate the discharge velocity head (Hv_discharge) using the diameter and velocity: Hv_discharge = (V_discharge)² / (2g), where V_discharge = (0.03 m³/s) / (π * (0.075 m)² / 4).
- Calculate the total discharge head (H_discharge) by adding the elevation difference and head loss: H_discharge = 20 m + Hv_discharge + 15 * Hv_discharge.
3. Total Head Required: H_total = H_suction + H_discharge.
Next, we can calculate the pump power using the following formula:
Pump Power = (Q * H_total) / (ρ * η * g), where Q is the flow rate, ρ is the density of water, g is the acceleration due to gravity, and η is the mechanical efficiency.
Substituting the given values and solving for the pump power will give us the required motor output power in kilowatts (kW).
Please note that the density of water at 22°C can be considered approximately 1000 kg/m³.
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Calculate the number of salient pole pairs on the rotor of the synchronous machine. with rated power of 4000 hp, 200 rpm, 6.9 kV, 50 Hz. Submit your numerical answer below.
The number of salient pole pairs on the rotor of the synchronous machine is determined to be 374.
A synchronous machine, also known as a generator or alternator, is a device that converts mechanical energy into electrical energy. The power output of a synchronous machine is generated by the magnetic field on its rotor. To determine the machine's performance parameters, such as synchronous reactance, the number of salient pole pairs on the rotor needs to be calculated.
Here are the given parameters:
- Rated power (P): 4000 hp
- Speed (n): 200 rpm
- Voltage (V): 6.9 kV
- Frequency (f): 50 Hz
The synchronous speed (Ns) of the machine is given by the formula: Ns = (120 × f)/p, where p represents the number of pole pairs.
In this case, Ns = 6000/p.
The rotor speed (N) can be calculated using the slip (s) equation: N = n = (1 - slip)Ns.
The slip is determined by the formula: s = (Ns - n)/Ns.
By substituting the values, we find s = 0.967.
Therefore, N = n = (1 - s)Ns = (1 - 0.967) × (6000/p) = 195.6/p volts.
The induced voltage in each phase (E) is given by: E = V/Sqrt(3) = 6.9/Sqrt(3) kV = 3.99 kV.
The voltage per phase (Vph) is E/2 = 1.995 kV.
The flux per pole (Øp) can be determined using the equation: Øp = Vph/N = 1.995 × 10³/195.6/p = 10.19/p Webers.
The synchronous reactance (Xs) is calculated as: Xs = (Øp)/(3 × E/2) = (10.19/p)/(3 × 1.995 × 10³/2) = 1.61/(p × 10³) Ω.
The impedance (Zs) is given by jXs = j1.61/p kΩ.
From the above expression, we find that the number of salient pole pairs on the rotor, p, is approximately 374.91. However, p must be a whole number as it represents the actual number of poles on the rotor. Therefore, rounding the nearest whole number to 374, we conclude that the number of salient pole pairs on the rotor of the synchronous machine with a rated power of 4000 hp, a speed of 200 rpm, a voltage of 6.9 kV, and a frequency of 50 Hz is 374.
In summary, the number of salient pole pairs on the rotor of the synchronous machine is determined to be 374.
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(6) Explain the spatial enumeration representation scheme of the CAD models. Use a table to compare its advantages and disadvantages against B-Rep representation scheme. (6 marks)
The spatial enumeration representation scheme is one of the most used schemes of representing CAD models. This type of representation scheme is used to model.
Solids and surfaces and represents geometry by the use of ordered or unordered sets of volumes or surfaces. The tables below show a comparison of the advantages and disadvantages of the spatial enumeration representation scheme of CAD models against the B-Rep representation scheme.
The Spatial Enumeration Representation Scheme is a method that is easy to learn and use. It has a fast computation time and low memory requirements. It is not suitable for modelling complex geometries and may not always be accurate.
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Periodic Review System: Target/Max Level Inventory A hardware company stocks nuts and bolts and orders them from a local supplier once every 2 weeks (10 working days). Lead time is 2 days. The company has determined that the average demand for 2-inch bolts of 150 per week (5 working days), and it wants to keep a safety stock of 3 days' supply on hand. An order is to be placed this week, stock on hand is 130 bolts. Compute . • The target inentory level • The number of 2 - bolts that should be ordered this time
To compute the target inventory level and the number of 2-inch bolts that should be ordered, we'll consider the average demand, lead time, and desired safety stock.
- Average demand for 2-inch bolts: 150 per week (5 working days)
- Lead time: 2 days
- Safety stock: 3 days' supply
- Stock on hand: 130 bolts
1. Compute the target inventory level:
Target Inventory Level = Average Demand * (Lead Time + Safety Stock)
Target Inventory Level = 150/5 * (2 + 3)
Target Inventory Level = 30 * 5
Target Inventory Level = 150 units
2. Compute the number of 2-inch bolts that should be ordered this time:
Number of Bolts to be Ordered = Target Inventory Level - Stock on Hand
Number of Bolts to be Ordered = 150 - 130
Number of Bolts to be Ordered = 20 units
Therefore, the target inventory level is 150 units and the number of 2-inch bolts that should be ordered this time is 20 units.
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Design a circuit which counts seconds, minutes and hours and displays them on the 7-segement display in 24 hour format. The clock frequency available is 36 KHz. Assume that Binary to BCD converter and BCD to 7-Segement display is already available for the design.
The 24-hour clock has two digits for hours, two digits for minutes, and two digits for seconds. Binary Coded Decimal (BCD) is a technique for representing decimal numbers using four digits in which each decimal digit is represented by a 4-bit binary number.
A 7-segment display is used to display the digits from 0 to 9.
Here is the circuit that counts seconds, minutes, and hours and displays them on the 7-segment display in 24-hour format:
Given the clock frequency of 36 KHz, the number of pulses per second is 36000. The seconds counter requires 6 digits, or 24 bits, to count up to 59. The minutes counter requires 6 digits, or 24 bits, to count up to 59. The hours counter requires 5 digits, or 20 bits, to count up to 23.The clock signal is fed into a frequency divider that produces a 1 Hz signal. The 1 Hz signal is then fed into a seconds counter, minutes counter, and hours counter. The counters are reset to zero when they reach their maximum value.
When the seconds counter reaches 59, it generates a carry signal that increments the minutes counter. Similarly, when the minutes counter reaches 59, it generates a carry signal that increments the hours counter.
The outputs of the seconds, minutes, and hours counters are then converted to BCD format using a binary to BCD converter. Finally, the BCD digits are fed into a BCD to 7-segment display decoder to produce the display on the 7-segment display.Here's a block diagram of the circuit: Block diagram of the circuit
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(4) Why is belt drive usually arranged in high-speed class and chain drive generally at the low-speed in the design of the transmission system? (6scores) (5) What are the failure modes of gear transmission? (4 scores) (6) What are the design criteria for gear transmission? (5 scores)
(4) Belts possess high flexibility and elasticity which allow for smooth power transfer over long distances.
(5) Fatigue failures, wear failures, tooth fractures, and skipping teeth.
(6) Load capacity, material selection, transmission ratios, lubrication, and sound level.
Explanation:
In the design of a transmission system, the belt drive is usually arranged in the high-speed class while the chain drive is arranged in the low-speed class. This is because belts possess high flexibility and elasticity which allow for smooth power transfer over long distances.
Additionally, they have a low noise level, are long-lasting, and do not require frequent lubrication. Due to these features, belts are suitable for high-speed machinery.
On the other hand, chain drives are ideal for low-speed, high-torque applications. While they can transmit more power than belt drives, they tend to be noisier, less flexible, and require more lubrication. Hence, chain drives are best suited for low-speed applications.
The failure modes of gear transmission can be categorized into fatigue failures, wear failures, tooth fractures, and skipping teeth. Fatigue failures occur when a component experiences fluctuating loads, leading to cracking, bending, or fracture of the material. Wear failures happen when two parts rub against each other, resulting in material loss and decreased fit. Tooth fractures occur when high stress levels cause a tooth to break off. Skipping teeth, on the other hand, are caused by poor gear engagement, leading to the teeth skipping over one another, causing further wear and damage.
The design criteria for gear transmission include load capacity, material selection, transmission ratios, lubrication, and sound level. The load capacity refers to the ability to handle the transmitted load adequately. Material selection should consider factors such as sufficient strength, good machinability, good wear resistance, and corrosion resistance. The design must fulfill the transmission requirements such as speed and torque requirements. Lubrication is also critical as it helps reduce friction and wear. Finally, the noise level produced during gear transmission should be minimized.
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A 230 V D.C. shunt motor takes 32 A at full load. Find the back e.m.f. full load if the resistance of motor armature and shunt field winding are 0.2 and 115 1 respectively
The back e.m.f. of the motor at full load is -3468.2 V.
Given: Voltage of DC motor, V = 230 V Current taken by DC motor at full load, I = 32 A
Resistance of motor armature, Ra = 0.2 ΩResistance of shunt field winding, Rs = 115.1 Ω
Formula Used: Back e.m.f. of DC motor, E = V - I (Ra + Rs) Where, V = Voltage of DC motor I = Current taken by DC motor at full load Ra = Resistance of motor armature Rs = Resistance of shunt field winding
Calculation: The back e.m.f. of the motor is given by the equation
E = V - I (Ra + Rs)
Substituting the given values we get,
E = 230 - 32 (0.2 + 115.1)
E = 230 - 3698.2
E = -3468.2 V (negative sign shows that the motor acts as a generator)
Therefore, the back e.m.f. of the motor at full load is -3468.2 V.
Shunt motors are constant speed motors. These motors are also known as self-regulating motors. The motor is connected in parallel with the armature circuit through a switch called the shunt. A shunt motor will maintain a nearly constant speed over a wide range of loads. In this motor, the field winding is connected in parallel with the armature. This means that the voltage across the field is always constant. Therefore, the magnetic field produced by the field winding remains constant.
As we know, the back EMF of a motor is the voltage induced in the armature winding due to rotation of the motor. The magnitude of the back EMF is proportional to the speed of the motor. At no load condition, when there is no load on the motor, the speed of the motor is maximum. So, the back EMF of the motor at no load is also maximum. As the load increases, the speed of the motor decreases. As the speed of the motor decreases, the magnitude of the back EMF also decreases. At full load condition, the speed of the motor is minimum. So, the back EMF of the motor at full load is also minimum.
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For the common faults in rotating machinery shown below (i)-(v), list the expected frequencies at which a vibration spectrum is likely to show peaks in terms of its RPM.
Unbalance
Misalignment
Bent shaft
Mechanical looseness
Oil Whirl in journal bearing
Rotating machinery is used in almost every industry for their respective purposes. When rotating machinery has faults, they generate vibrations, which can cause damage and, in extreme cases, the entire machine can fail.
The expected frequencies for the peaks in terms of the RPM for the common faults in rotating machinery are discussed below: I. Unbalance: Unbalance occurs when the mass distribution of a rotating object is not even.
It can be caused by the accumulation of dirt or corrosion, unbalanced bearing support, or excessive of components. A peak frequency of 1x RPM (rotation per minute) is expected for the unbalance fault in a vibration spectrum. Misalignment: Misalignment occurs when the shaft centerlines of the machines are not properly aligned.
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Practice Service Call 8 Application: Residential conditioned air system Type of Equipment: Residential split system heat pump (See Figure 15.45.) Complaint: System heats when set to cool. Symptoms: 1. System heats adequately. 2. With thermostat fan switch on, the fan operates properly. 3. Outdoor fan motor is operating. 4. Compressor is operating. 5. System charge is correct. 6. R to O on thermostat is closed. 7. 24 volts are being supplied to reversing valve solenoid.
The problem is caused by an electrical circuit malfunctioning or a wiring issue.
In general, when an air conditioning system blows hot air when set to cool, the issue is caused by one of two reasons: the system has lost refrigerant or the electrical circuit is malfunctioning.
The following are the most likely reasons:
1. The thermostat isn't working properly.
2. The reversing valve is malfunctioning.
3. The defrost thermostat is malfunctioning.
4. The reversing valve's solenoid is malfunctioning.
5. There's a wiring issue.
6. The unit's compressor isn't functioning correctly.
7. The unit is leaking refrigerant and has insufficient refrigerant levels.
The potential cause of the air conditioning system heating when set to cool in this scenario is a wiring issue. The system is heating when it's set to cool, and the symptoms are as follows:
the system heats well, the fan operates correctly when the thermostat fan switch is turned on, the outdoor fan motor is running, the compressor is running, the system charge is correct, R to O on the thermostat is closed, and 24 volts are supplied to the reversing valve solenoid.
Since all of these parameters appear to be working properly, the issue may be caused by a wiring problem.
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A mass of 2.6 kg of saturated water liquid at 300 kPa is heated until it completely vaporized. Calculate the total amount latent heat released during the process.
The total amount of latent heat released during the process is 5865.6 kJ.
Given : Mass of saturated water, m = 2.6 kgPressure, P1 = 300 kPaLatent heat of vaporisation of water, Lv = 2256 kJ/kgSince the water is heated until it is completely vaporised, the process is isobaric (constant pressure) and isothermal (constant temperature).
During the process of vaporisation of water, the temperature remains constant. Hence the temperature at which the water starts vaporising will be the same as the temperature at which it completely vaporises.
From Steam Tables, at 300 kPa, the saturation temperature of water (i.e. the temperature at which water starts vaporising) is 127.6°C.So, initial temperature of water, T1 = 127.6°CLatent heat released during the process = Latent heat of vaporisation of water × mass of saturated water Latent heat released during the process = Lv × m= 2256 kJ/kg × 2.6 kg= 5865.6 kJ
Therefore, the total amount of latent heat released during the process is 5865.6 kJ.
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3. This question requires you to demonstrate your knowledge and understanding of rocket engines. An end-burning composite solid rocket motor is designed to produce an optimal thrust of 5 kN at an altitude of 5 km where the ambient pressure is 54 kPa. The propellant properties are given in Table Q3. a) Draw a process diagram for this engine showing your stage notation. [2 marks] b) Assuming an ideal rocket, determine: (i) the nozzle exit velocity; [8 marks] (ii) the nozzle exit diameter; [4 marks] (iii) the throat velocity; [3 marks] (iv) the propellant grain diameter; [3 marks] [3 marks] (v) the burning time if the propellant charge length is 1.4 m; the thrust if the altitude of the rocket increases to 7 km where the ambient pressure is 41 kPa. (vi) [2 marks] Propellant grain density 1,820 kg/m³ Vielle's law burning rate indices; n = 0.4 and k = 3.6 Combustion temperature 2,800 K Combustion pressure 12 MPa Exhaust gas constants Y 1.2 R 290 J/(kg-K) Table Q3: Rocket engine performance data.
The propellant properties, such as Vielle's law burning rate indices, combustion temperature, combustion pressure, and exhaust gas constants, are provided in Table Q3. By solving these calculations and utilizing the given data, we can determine the required parameters for the rocket engine and gain insights into its performance characteristics.
a) The process diagram for the engine can be represented using stage notation, which indicates the different stages or components of the rocket engine, such as the combustion chamber, nozzle, and propellant grain. The diagram should illustrate the flow of gases and the expansion of exhaust gases through the nozzle.
b) To determine the required parameters, we assume the rocket engine operates ideally.
(i) The nozzle exit velocity can be calculated using the ideal rocket equation, which relates the exhaust velocity to the specific impulse and the gravitational constant.
(ii) The nozzle exit diameter can be determined using the area ratio between the throat and the exit.
(iii) The throat velocity can be calculated using the specific impulse and the exhaust gas constants.
(iv) The propellant grain diameter is not directly provided in the question, so additional information or assumptions are needed to determine this parameter.
(v) The burning time can be calculated using the propellant charge length and the burning rate of the propellant. The thrust at a different altitude can be estimated by adjusting for the change in ambient pressure.
(vi) Additional information is not provided in the question to calculate the propellant grain density.
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Two materials are being considered for an application in which electrical conductivity is important
Material Working strength MN/m² Electrical conductance % A 500 50
B 1000 40
The weighting factor on strength is 3 and 10 for conductance. Which material is pre- ferred based on the weighted property index? Does the choice change if the weighting factor on strength is 6 and the weighting factor on conductance is 4?
Material B is still preferred based on the weighted property index even when the weighting factor on strength is 6 and the weighting factor on conductance is 4.
To determine which material is preferred based on the weighted property index, we use the formula:
Weighted property index = (Weighting factor 1 * Property 1) + (Weighting factor 2 * Property 2)
where, Weighting factor 1 and 2 are the weightings assigned to the first and second property, and Property 1 and 2 are the values of the first and second properties for the materials.
Using the above formula, the weighted property index for materials A and B are calculated below:For Material A, the weighted property index = (3*500) + (10*50) = 1500 + 500 = 2000
For Material B, the weighted property index = (3*1000) + (10*40) = 3000 + 400 = 3400
Therefore, Material B is preferred based on the weighted property index.
Now, let's consider the case where the weighting factor on strength is 6 and the weighting factor on conductance is 4.
Weight of Strength = 6
Weight of Conductance = 4For Material A, the weighted property index = (6*500) + (4*50) = 3000 + 200 = 3200
For Material B, the weighted property index = (6*1000) + (4*40) = 6000 + 160 = 6160
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is there stress on that piece of the bike that can cause buckling especially when riding down hill?
Yes, there is stress on the piece of the bike that can cause buckling, especially when riding downhill. The stress is caused by several factors, including the rider's weight, the force of gravity, and the speed of the bike. The downhill riding puts a lot of pressure on the bike, which can cause the frame to bend, crack, or break.
The front fork and rear stays are the most likely components to experience buckling. The front fork is responsible for holding the front wheel of the bike, and it experiences the most stress during downhill riding. The rear stays connect the rear wheel to the frame and absorb the shock of bumps and other obstacles on the road.
To prevent buckling, it is essential to ensure that your bike is in good condition before heading downhill. Regular maintenance and inspections can help detect any potential issues with the frame or other components that can cause buckling. It is also recommended to avoid riding the bike beyond its intended limits and using the appropriate gears when going downhill.
Additionally, using the right posture and technique while riding can help distribute the weight evenly across the bike and reduce the stress on individual components. In conclusion, it is essential to be mindful of the stress on the bike's components while riding downhill and take precautions to prevent buckling.
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A concrete-coated steel gas pipeline is to be laid between two offshore platforms in 100 m water depth where the maximum environmental conditions include waves of 20 m wave height and 14 s period. The pipeline outside diameter is 46 cm, and the clay bottom slope is 1 on 100. Determine the submerged unit weight of the pipe. Assume linear wave theory is valid and that the bottom current is negligible.
Diameter of the pipeline (d) = 46 cm = 0.46 mDepth of water (h) = 100 mMaximum wave height (H) = 20 mWave period (T) = 14 sBottom slope (S) = 1/100Formula Used.
Submerged weight = (pi * d² / 4) * (1 - ρ/γ)Where, pi = 3.14d = diameter of the pipelineρ = density of water = 1000 kg/m³γ = specific weight of the material of the pipeCalculation:Given, d = 0.46 mρ = 1000 kg/m³γ = ?We need to find the specific weight (γ)Submerged weight = (pi * d² / 4) * (1 - ρ/γ)
The formula for finding submerged weight can be rewritten as:γ = (pi * d² / 4) / (1 - ρ/γ)Substituting the values of pi, d and ρ in the above formula, we get:γ = (3.14 * 0.46² / 4) / (1 - 1000/γ)Simplifying the above equation, we get:γ = 9325.56 N/m³Thus, the submerged unit weight of the pipe is 9325.56 N/m³. Hence, the detailed explanation of the submerged unit weight of the pipe has been provided.
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A 2 DOF system has mode shapes given by Φ₁ = {1}
{-2}
and
Φ₂ =
{1}
{3}
A force vector F = {1}
{p}
sin(Ωt) is acting on the system. Find the value of P if the system steady state response is purely in mode 1.
A 2 DOF (Degree of Freedom) system has mode shapes given by Φ₁ = {1} {-2} and Φ₂ = {1} {3}. A force vector F = {1} {p} sin(Ωt) is acting on the system, where P is the value of the steady-state response in mode
1.The system response can be given by the equation,
M = M₀ + M₁ sin(Ωt + φ₁) + M₂ sin(2Ωt + φ₂)
Here,Ω = 1 (the driving frequency)
φ₁ is the phase angle of the first modeφ₂ is the phase angle of the second modeM₀ is the static deflection
M₁ is the amplitude of the first mode
M₂ is the amplitude of the second mode
So, the response of the system can be given by:
M = M₁ sin(Ωt + φ₁)
Now, substituting the values,
M = Φ₁ F = {1} {-2} {1} {p} sin(Ωt) = {1-2p sin(Ωt)}
In order for the steady-state response to be purely in mode 1, M₂ = 0
So, the equation for the response becomes,
M = M₁ sin(Ωt + φ₁) ⇒ {1-2p sin(Ωt)} = M₁ sin(Ωt + φ₁)
Comparing both sides, we get,
M₁ sin(Ωt + φ₁) = 1 and -2p sin(Ωt) = 0sin(Ωt) ≠ 0, as Ω = 1, so -2p = 0P = 0
Therefore, the value of P if the system steady-state response is purely in mode 1 is 0
In this problem, we are given a 2 DOF (Degree of Freedom) system having mode shapes Φ₁ and Φ₂.
The mode shapes of a system are the deflected shapes that result from the system vibrating in free vibration. In the absence of any external forcing, these deflected shapes are called natural modes or eigenmodes. The system is also subjected to a force vector F = {1} {p} sin(Ωt).
We have to find the value of P such that the system's steady-state response is purely in mode 1. Steady-state response refers to the long-term behavior of the system after all the transient vibrations have decayed. The steady-state response is important as it helps us predict the system's behavior over an extended period and gives us information about the system's durability and reliability.
In order to find the value of P, we first find the system's response. The response of the system can be given by the equation,
M = M₀ + M₁ sin(Ωt + φ₁) + M₂ sin(2Ωt + φ₂)
where M₀, M₁, and M₂ are constants, and φ₁ and φ₂ are the phase angles of the two modes.
In this case, we are given that Ω = 1 (the driving frequency), and we assume that the system is underdamped. Since we want the steady-state response to be purely in mode 1, we set M₂ = 0.
Hence, the equation for the response becomes,
M = M₁ sin(Ωt + φ₁)
We substitute the values of Φ₁ and F in the above equation to get,{1-2p sin(Ωt)} = M₁ sin(Ωt + φ₁)
Comparing both sides, we get,
M₁ sin(Ωt + φ₁) = 1 and -2p sin(Ωt) = 0sin(Ωt) ≠ 0, as Ω = 1, so -2p = 0P = 0
Therefore, the value of P if the system steady-state response is purely in mode 1 is 0.
The value of P such that the system steady-state response is purely in mode 1 is 0.
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c) In a solar module, what is the function of blocking and bypassing diodes? d) Draw the schematic diagram for a grid-connected solar PV system designed to supply both AC and DC loads
Blocking and bypassing diodes play crucial roles in a solar module. The blocking diode prevents reverse current flow, ensuring that electricity generated by the module does not flow back into the solar cells during periods of low or no sunlight. On the other hand, bypass diodes offer an alternative path for the current to bypass shaded or faulty cells, optimizing the overall efficiency of the module.
The function of blocking and bypassing diodes in a solar module is essential for maintaining its performance and protecting the cells from potential damage. Let's take a closer look at each diode's role:
1. Blocking Diode: The blocking diode, also known as an anti-reverse diode, is typically placed in series between the solar module and the charge controller or battery bank. Its primary purpose is to prevent reverse current flow. During periods when the solar module is not generating electricity, such as at night or when shaded, the blocking diode acts as a one-way valve, ensuring that the current does not flow back into the solar cells. This helps to prevent power losses and potential damage to the cells.
2. Bypass Diodes: Solar modules are typically made up of several interconnected solar cells. When a single cell or a portion of the module becomes shaded or fails to generate electricity efficiently, it can create a "hotspot." A hotspot occurs when the shaded or faulty cell acts as a resistance, potentially causing overheating and reducing the overall output of the module. Bypass diodes provide an alternate pathway for the current to flow around the shaded or faulty cells, minimizing the impact of the hotspot and allowing the module to continue generating power effectively.
By incorporating bypass diodes, solar modules can mitigate the negative effects of shading or individual cell failure, ensuring optimal performance even in partially shaded conditions. These diodes divert the current around the shaded or faulty cells, allowing the unshaded cells to continue generating electricity. This helps to maximize the overall energy output of the solar module and improve its reliability.
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If the production of a material increases by r% every year, show
that the doubling time is given by 70/r.
When the production of a material increases at the rate of r% every year, the doubling time is given by 70/r. Assume that the initial production rate is P₀ at the start of the year, and after t years, it will be P.
After the first year, the production rate will be
P₁ = P₀ + (r/100)P₀
P₁ = (1 + r/100)P₀.
In general, the production rate after t years is given by the formula
P = (1 + r/100)ᵗP₀.
when the production of a material is doubled, the following equation is satisfied:
2P₀ = (1 + r/100)ᵗP₀
Applying the logarithm to both sides of the equation, we obtain:
log 2 = tlog(1 + r/100)
Dividing both sides by log(1 + r/100), we get:
t = log 2 / log(1 + r/100)
This expression shows the number of years required for the production of a material to double at a constant rate of r% per year. Using the logarithm property, we can rewrite the above equation as:
t = 70/ln(1 + r/100)
In the above expression, ln is the natural logarithm.
By substituting ln(2) = 0.693 into the equation, we can obtain:
t = 0.693 / ln(1 + r/100)
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What is spectrum (spectra) plot? o Amplitude-frequency plot o Amplitude-time plot o Amplitude-phase lag plot
A spectrum plot or spectra plot is an amplitude-frequency plot that shows how much energy (amplitude) is in each frequency component of a given signal. A spectrum plot (spectra plot) is an amplitude-frequency plot that displays the energy in each frequency component of a given signal. This plot is used to represent a signal in the frequency domain.
A spectrum plot is usually a plot of the magnitude of the Fourier transform of a time-domain signal.
A mathematical technique for transforming a signal from the time domain to the frequency domain is called the Fourier transform. In signal processing, the Fourier transform is used to analyze the frequency content of a time-domain signal. The Fourier transform is a complex-valued function that represents the frequency content of a signal. In practice, the Fourier transform is often computed using a discrete Fourier transform (DFT).
The amplitude is a measure of the strength of a signal. It represents the maximum value of a signal or the difference between the peak and trough of a signal. The amplitude is usually measured in volts or decibels (dB). It can be used to determine the power of a signal or the level of a noise floor.
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Why does polysilicon etch prefer chlorine instead of fluorine as the main etchant?
Polysilicon etches prefer chlorine instead of fluorine as the main etchant because of the nature of silicon. Polysilicon is a type of silicon used in the semiconductor industry to make microprocessors, solar cells, and other electronics. It is made up of a large number of silicon atoms arranged in a repeating pattern, hence the name polysilicon.
Polysilicon etches are chemicals that are used to remove or etch away certain parts of the polysilicon layer during the manufacturing process. Chlorine and fluorine are two of the most common etchants used for this purpose, but they differ in their effectiveness.
Chlorine is a better etchant for polysilicon than fluorine because chlorine reacts more readily with silicon than fluorine does. Chlorine has a larger atomic radius and a higher electronegativity than fluorine, which means that it is better able to form bonds with the silicon atoms in polysilicon.
Furthermore, chlorine is more reactive and more volatile than fluorine, which makes it easier to handle in the manufacturing process. It also has a lower etch rate than fluorine, which means that it etches more uniformly and is less likely to cause damage to the underlying substrate.
In summary, polysilicon etches prefer chlorine over fluorine as the main etchant because it is more reactive, has a larger atomic radius and a higher electronegativity, and is more volatile and easier to handle in the manufacturing process.
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A steam generator with economizer and air heater has an overall draft loss of 21.78 cmWG. If the stack gases are at 117°C and if the atmosphere is at 101.3 kPa and 26°C, what theoretical height of stack in meters is needed when no draft fans are used? Assume that the gas constant for the flue gases is the same as that for air.
580 m
560 m
570 m
550 m
The theoretical height of the stack in meters needed when no draft fans are used is 575 m (approx). The correct option is option(c).
Given that the overall draft loss of the steam generator with economizer and air heater is 21.78 cmWG. The stack gases are at 117°C and the atmosphere is at 101.3 kPa and 26°C.
The theoretical height of the stack in meters when no draft fans are used is to be calculated. Assuming that the gas constant for the flue gases is the same as that for air, we have:
We know that:
Total draft loss = Hf + Hc + Hi + H o
Hf = Frictional losses in the fuel bed
Hc = Frictional losses in the fuel passages
Hi = Loss of draft in the chimney caused by the change of temperature of the flue gases
H o = Loss of draft in the chimney due to the wind pressure
Let's assume that there is no wind pressure, then the total draft loss =
Hf + Hc + Hi
Putting the values in the above equation:
21.78 = Hf + Hc + Hi
We know that the loss of draft Hi due to a change in temperature is given by:
Hi = Ht (t1 - t2)/t2
Ht = Total height of the chimney from fuel bed to atmosphere
= Hf + Hc + Hch + Hah1
= Temperature of flue gases leaving the chimney in K = (117 + 273) K
= 390 K
h2 = Temperature of the atmospheric air in K = (26 + 273) K
= 299 KK
= Gas constant
= R/M = 0.287/29 kg/mol
= 0.00989 kg/mol
Hch = Height of the chimney from the point of exit of flue gases to the top of the chimney
Hah = Height of the air heater above the point of exit of the flue gases
Let's assume Hah = 0
We know that,
Hc = l ρV²/2g
where
l = Length of flue passages
ρ = Density of flue gases
V = Velocity of flue gases
g = Acceleration due to gravity
Substituting the given values, we get
Hc = 0.7 ρV² .......... (1)
We also know that,
Hf = l ρV²/2g
where l = Length of the fuel bed
ρ = Density of fuel
V = Velocity of fuel
g = Acceleration due to gravity
Substituting the given values, we get
Hf = 1.2 ρV² .......... (2)
Now, combining equation (1) and (2), we get:
21.78 = Hf + Hc + Hi1.2 ρV² + 0.7 ρV² + Ht (t1 - t2)/t2 = 21.78
Let's assume that V = 10 m/s
We know that, ρ = p/RT
where
p = Pressure of flue gases in Pa
R = Gas constant of the flue gases
T = Temperature of flue gases in K
Substituting the given values, we get
ρ = 101.3 × 10³/ (0.287 × 390) = 8.44 kg/m³
Substituting the given values in the equation
21.78 = 1.2 ρV² + 0.7 ρV² + Ht (t1 - t2)/t2, we get:
Ht = 574.68 m
The theoretical height of the stack in meters needed when no draft fans are used is 575 m (approx). Therefore, the correct option is 570 m.
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If you need to heat 10 liters of water from 0°C to 100 °C using kitchen natural gas system. I kg of liquefied Pressurized gas (LPG) has a useful energy value of 20.7 MJ/kg, (the ideal energy value is 34.8 MJ/kg). The energy required to heat 1 g of water from 0°C to 100 °C = 100 calories 1 kcal = 4186 J, 1 kWh = 3.16 * 10 Joule, 1000 g of water = 1 liter of water. If the cost of 1 kg natural gas (LPG) = 0.5 Jordanian Dinars, what will be the cost of heating 10 liters of water from 0°C to 100 °C in JD?
The cost of heating 10 liters of water from 0°C to 100°C using the kitchen natural gas system would be approximately 49 Jordanian Dinars (JD).
To calculate the cost of heating 10 liters of water from 0°C to 100°C using the kitchen natural gas system, we need to determine the energy required and then calculate the cost based on the cost of 1 kg of natural gas (LPG).
Given:
Energy required to heat 1 g of water from 0°C to 100°C = 4186 J
Energy value of 1 kg of LPG = 20.7 MJ = 20.7 * 10^6 J
Cost of 1 kg of natural gas (LPG) = 0.5 JD
1: Calculate the total energy required to heat 10 liters of water:
10 liters of water = 10 * 1000 g = 10,000 g
Energy required = Energy per gram * Mass of water = 4186 J/g * 10,000 g = 41,860,000 J
2: Convert the total energy to kilojoules (kJ):
Energy required in kJ = 41,860,000 J / 1000 = 41,860 kJ
3: Calculate the amount of LPG required in kilograms:
Amount of LPG required = Energy required in kJ / Energy value of 1 kg of LPG
Amount of LPG required = 41,860 kJ / 20.7 * 10^6 J/kg
4: Calculate the cost of the required LPG:
Cost of LPG = Amount of LPG required * Cost of 1 kg of LPG
Cost of LPG = (41,860 kJ / 20.7 * 10^6 J/kg) * 0.5 JD
5: Simplify the expression and calculate the cost in JD:
Cost of heating 10 liters of water = (41,860 * 0.5) / 20.7
Cost of heating 10 liters of water = 1,015.5 / 20.7
Cost of heating 10 liters of water ≈ 49 JD (rounded to two decimal places)
Therefore, the approximate cost of heating 10 liters of water from 0°C to 100°C using the kitchen natural gas system would be 49 Jordanian Dinars (JD).
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A process has the following transfer functions: Process g= 1 / (20s+1)(10s+1) Control valve: gv = -0.25 / 0.5s+1 PI controller: gc = Kc (1 + 1/Tis) i. Design a feedback PI controller for the above process, use Cohen-Coon tuning rule (Table 15.3). Please remember to submit all programs that you use to design the controller. For example, the program to find the approximate model and/or spreadsheet to calculate the controller parameters ii. In Simulink, develop a block diagram with the designed PI controller. Then generate the output response, for a step change of magnitude 3 at time 4s in the setpoint. Before time 4, the setpoint and the output stays at the value of 1.
i. The feedback PI controller for the process G is [tex]$$Gc = K_c\left(1+\frac{1}{T_i s}\right) = 1.5416 \left(1+\frac{2.3884}{s}\right) = \frac{1.5416s+3.6726}{s}$$[/tex]
ii. It can be observed that the output eventually stabilizes at the new setpoint.
i. Designing a feedback PI controller using Cohen-Coon tuning rule:
Cohen-Coon tuning rule is used to tune PI controllers. This method has an approximate model of the system and is not suitable for tuning PID controllers.
Cohen-Coon tuning rule:
[tex]$$\begin{aligned} &K_c = \frac{1}{K_p}\left[ {\frac{28}{13} + \frac{{3{{\tau }}_p }}{{{{{\left( {3{{\tau }}_p +{{\tau }}_d } \right)}}}}}} \right] \\ &\frac{1}{{{T_i}}} = \frac{1}{\theta }\left[ {\frac{4}{13} + \frac{{{{\tau }}_d }}{{3{{\tau }}_p +{{\tau }}_d }}} \right] \\ \end{aligned}$$[/tex]
Given: Process G = 1/(20s+1)(10s+1)
Control Valve: gv = -0.25 / 0.5s+1
We need to find out the feedback PI controller for the process G.
Approximate the model and determine the process parameters. Using the given transfer functions, we can determine the time constant and the time delay.
[tex]$$G = \frac{1}{(20s + 1)(10s + 1)}$$$$G = \frac{1}{200s^2 + 30s + 1}$$$$\tau_p \\= \frac{1}{\omega_p} \\= \frac{1}{\sqrt{200}} \\= 0.0707$$\\$$\omega_d = 0.1$$[/tex]
Therefore, from the Cohen-Coon tuning rule, we can determine the values of Kc and Ti.
[tex]$$K_c = 1.5416$$\\$$Ti = 0.4183$$[/tex]
Hence the feedback PI controller for the process G is [tex]$$Gc = K_c\left(1+\frac{1}{T_i s}\right) = 1.5416 \left(1+\frac{2.3884}{s}\right) = \frac{1.5416s+3.6726}{s}$$[/tex]
ii. Developing a Simulink block diagram with the designed PI controller
Here is the Simulink block diagram with the designed PI controller. As mentioned above, the setpoint and the output stays at the value of 1 before time 4, and after 4s there is a step change of magnitude 3 in the setpoint.
The block diagram is designed such that it simulates the response for the next 20 seconds. The controller output is shown in red and the process variable in blue.
The final output response is shown below. The output response, after 4 seconds of time and a setpoint change of 3, is similar to the response of a standard PI controller.
It can be observed that the output eventually stabilizes at the new setpoint.
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The spacecraft has 4 solar panels. Each panel has the dimension of 2m x 1m x 20mm with a density of 7830 kg/m3 and is connected to the body by aluminum rods of a length of 0.4 m and a diameter of 20mm. Determine the natural frequency of vibration of each panel about the axis of the connecting rod. Use G = 26GPa. Im= m (w2 + h2)/12 =
The spacecraft has four solar panels, and each of them has a dimension of 2m x 1m x 20mm. These panels have a density of 7830 kg/m³. The solar panels are connected to the body by aluminum rods that have a length of 0.4m and a diameter of 20mm.
We are required to find the natural frequency of vibration of each panel about the axis of the connecting rod. We use
[tex]G = 26 GPa and Im = m(w² + h²)/12[/tex]
to solve this problem. The first step is to calculate the mass of each solar panel. Mass of each
s[tex ]olar panel = density x volume = 7830 x 2 x 1 x 0.02 = 313.2 kg.[/tex]
The next step is to calculate the moment of inertia of the solar panel.
[tex]Im = m(w² + h²)/12 = 313.2(2² + 1²)/12 = 9.224 kgm².[/tex]
Now we can find the natural frequency of vibration of each panel about the axis of the connecting rod.The formula for the natural frequency of vibration is:f = (1/2π) √(k/m)where k is the spring constant, and m is the mass of the solar panel.To find the spring constant, we use the formula:k = (G x A)/Lwhere A is the cross-sectional area of the rod, and L is the length of the rod.
[tex]k = (26 x 10⁹ x π x 0.02²)/0.4 = 83616.7 N/m[/tex]
Now we can find the natural frequency of vibration:
[tex]f = (1/2π) √(k/m) = (1/2π) √(83616.7/313.2) = 5.246 Hz[/tex]
Therefore, the natural frequency of vibration of each panel about the axis of the connecting rod is 5.246 Hz.
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A 5-stage Marx impulse generator has individual and load condensers of 0.27μF and 0.04μF. The circuit's wavefront and wavetail resistances were observed to be 30.223Ω and 2004.817Ω. i. Find the front and the tail times of the waveform generated by the generator. ii. Calculate the time taken to reach the impulse generator's peak value (t max
). iii. Determine the nominal energy storage if the charging voltage is 170kV.
i) The front time is 0.178 microseconds and the tail time is 17.717 microseconds.
ii) The time taken to reach the peak value is 17.895 microseconds.
iii) The nominal energy storage is: 0.693 J
What is the Nominal Energy Storage?The given parameters are:
Individual capacitor (C₁) = 0.27 μF
Load capacitor (C₂) = 0.04 μF
Wave front resistance (R₁) = 30.223 Ω
Wave tail resistance (R₂) = 2004.817 Ω
Charging voltage (V) = 170 kV
i) The front time (t_f) and tail time (t_t) can be calculated using the following formulas:
t_f = 2.2 × R₁ × C₁
t_t = 2.2 × R₂ × C₂
Plugging in the relevant values, we have:
t_f = 2.2 × 30.223 Ω × 0.27 μF
t_t = 2.2 × 2004.817 Ω × 0.04 μF
Calculating the values:
t_f = 0.178 microseconds
t_t = 17.717 microseconds
ii) The time taken to reach the peak value (t_max) can be calculated as:
t_max = t_f + t_t
Plugging in the values we calculated earlier:
t_max = 0.178 microseconds + 17.717 microseconds
Calculating the value:
t_max = 17.895 microseconds
iii) The nominal energy storage (E) can be calculated using the formula:
E = 0.5 × (C₁ + C₂) × V²
Plugging in the given values:
E = 0.5 × (0.27 μF + 0.04 μF) × (170 kV)²
Converting the units to the appropriate form:
E = 0.5 × (0.27 × 10⁻⁶ F + 0.04 × 10⁻⁶ F) × (170 × 10³ V)²
Calculating the value:
E ≈ 0.693 Joules
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Methane gas at 120 atm and −18°C is stored in a 20−m³ tank. Determine the mass of methane contained in the tank, in kg, using the
(a) ideal gas equation of state. (b) van der Waals equation. (c) Benedict-Webb-Rubin equation.
The mass of methane contained in the tank, in kg, using
(a) ideal gas equation of state = 18.38 kg
(b) van der Waals equation = 18.23 kg
(c) Benedict-Webb-Rubin equation = 18.21 kg.
(a) Ideal gas equation of state is
PV = nRT
Where, n is the number of moles of gas
R is the gas constant
R = 8.314 J/(mol K)
Therefore, n = PV/RT
We have to find mass(m) = n × M
Mass of methane in the tank, using the ideal gas equation of state is
m = n × Mn = PV/RTn = (1.2159 × 10⁷ Pa × 20 m³) / (8.314 J/(mol K) × 255 K)n = 1145.45 molm = n × Mm = 1145.45 mol × 0.016043 kg/molm = 18.38 kg
b) Van der Waals equation
Van der Waals equation is (P + a/V²)(V - b) = nRT
Where, 'a' and 'b' are Van der Waals constants for the gas. For methane, the values of 'a' and 'b' are 2.25 atm L²/mol² and 0.0428 L/mol respectively.
Therefore, we can write it as(P + 2.25 aP²/RT²)(V - b) = nRT
At given conditions, we have
P = 120 atm = 121.59 × 10⁴ Pa
T = 255 K
V = 20 m³
n = (P + 2.25 aP²/RT²)(V - b)/RTn = (121.59 × 10⁴ Pa + 2.25 × (121.59 × 10⁴ Pa)²/(8.314 J/(mol K) × 255 K) × (20 m³ - 0.0428 L/mol))/(8.314 J/(mol K) × 255 K)n = 1138.15 molm = n × Mm = 1138.15 mol × 0.016043 kg/molm = 18.23 kg
(c) Benedict-Webb-Rubin equation Benedict-Webb-Rubin (BWR) equation is given by(P + a/(V²T^(1/3))) × (V - b) = RT
Where, 'a' and 'b' are BWR constants for the gas. For methane, the values of 'a' and 'b' are 2.2538 L² kPa/(mol² K^(5/2)) and 0.0387 L/mol respectively.
Therefore, we can write it as(P + 2.2538 aP²/(V²T^(1/3)))(V - b) = RT
At given conditions, we haveP = 120 atm = 121.59 × 10⁴ PaT = 255 KV = 20 m³n = (P + 2.2538 aP²/(V²T^(1/3)))(V - b)/RTn = (121.59 × 10⁴ Pa + 2.2538 × (121.59 × 10⁴ Pa)²/(20 m³)² × (255 K)^(1/3) × (20 m³ - 0.0387 L/mol))/(8.314 J/(mol K) × 255 K)n = 1135.84 molm = n × Mm = 1135.84 mol × 0.016043 kg/molm = 18.21 kg
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Consider the stoichiometric overall reaction for methane in air with the global reaction rate considering only complete products is:
RR = -8.3 X 105 exp [-15098/T] [CH₂1-0.3 [0₂]¹.³ gmol/cm³.s
a) If the reactant mixture is suddenly brought to a temperature of 2000 K and 1 atmospheric pressure, what is the initial rate of reaction?
b) If the temperature held constant at 2000 K and the volume is constant, what is the rate of reaction when a 50% of the original fuel has been converted to products? The reaction rate unit is gmol/cm³ s.
c) Also calculate the time required to convert the 50% of the original fuel into products of (b) case above. (Hint: the reaction rate could be assumed as an average of above two cases).
(Note: R. = 8.314 J/(gmol.K) is the universal gas constant)
The global reaction rate, considering only complete products is given by:RR = -8.3 × 105 exp[-15098/T][CH41-0.3[O21.3]]gmol/cm³swhere, RR = reaction rate; T = temperature; CH4 = methane; O2 = oxygen.The activation energy, E = 15098 J/molThe gas constant, R = 8.314 J/mol KT = 2000 KThe pressure, P = 1 atmThe initial concentration of methane and oxygen = 1 atm.
The reaction rate equation can be rewritten by substituting the given values as follows:RR = -8.3 × 105 exp[-15098/2000][1.0 1-0.3[1.0]1.3]]RR = -8.3 × 105 exp(-25.25)RR = -8.3 × 105 × 2.68 × 10-11RR = 2.224 gmol/cm³sThe initial rate of reaction is 2.224 gmol/cm³s.b) When 50% of the original fuel has been converted to products, the remaining 50% fuel concentration = 0.5 atm The product concentration = 0.5 atm
Therefore, the reaction rate at 50% conversion,R1 = R02/2. The rate of reaction when 50% of the original fuel has been converted to products is R1 = 2.224/2 = 1.112 gmol/cm³s. Thus, the rate of reaction when 50% of the original fuel has been converted to products is 1.112 gmol/cm³s.c) To calculate the time required to convert the 50% of the original fuel into products of (b) case above substituting the given values, the time required to convert 50% of the original fuel into products is given by:t = ln(1 - 0.5) /(-1.668) = 0.2087 s (approx).
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