The temperatures will depend on the amount of solar radiation that is received, which will vary throughout the day. As a result, the temperatures may not be the same every day.
A Flat Plate Solar Collector (FPSC) is a device that absorbs solar radiation and converts it into thermal energy, which can be used for heating water or other fluids. It consists of a flat, rectangular plate that is tilted at an angle to the ground to maximize the amount of solar radiation it receives.
In this problem, we are given a FPSC with 2 m surface area and 30° tilted from the ground, and we are asked to create a table including hours of the day and fill it with collector inlet& outlet, tank initial & final temperatures. The mass flow rate is 0.02 kg/s.
To solve this problem, we need to calculate the temperatures at the inlet and outlet of the collector, as well as the initial and final temperatures of the tank. We can use the following equations to do this:
q = m*c*(T2 - T1)
q = m*c*(T4 - T3)
where q is the heat transferred, m is the mass flow rate, c is the specific heat capacity of the fluid, T1 is the inlet temperature of the collector, T2 is the outlet temperature of the collector, T3 is the initial temperature of the tank, and T4 is the final temperature of the tank.
We can assume that the specific heat capacity of the fluid is constant and equal to 4.18 kJ/kg.K, which is the specific heat capacity of water.
We can also assume that the collector and tank are well-insulated, so there is no heat loss to the surroundings.
Using these equations, we can calculate the temperatures at different times of the day and fill them in the table.
The table should have columns for the time of day, collector inlet temperature, collector outlet temperature, tank initial temperature, and tank final temperature. Here is an example of what the table might look like:
Time of day | Inlet temp. | Outlet temp. | Initial temp. | Final temp.
------------|-------------|--------------|----------------|-------------
8:00 am | 20°C | 40°C | 25°C | 35°C
10:00 am | 25°C | 45°C | 30°C | 40°C
12:00 pm | 30°C | 50°C | 35°C | 45°C
2:00 pm | 35°C | 55°C | 40°C | 50°C
4:00 pm | 30°C | 50°C | 35°C | 45°C
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What is the net entropy change per second of a 1 m^2 solar
panel absorbing 1000 W/m^2 of sunlight (T = 5800 K) and radiating "waste" heat into
the environment at a temperature of T = 70 C into an environment at 25 C?
The net entropy change per second of a 1 m² solar panel absorbing 1000 W/m² of sunlight (T = 5800 K) and radiating "waste" heat into the environment at a temperature of T = 70°C into an environment at 25°C is 2.67 J/Ks.
What is entropy change?The entropy change of a thermodynamic system is the difference between its final and initial entropy values. The entropy of a system increases as its disorderliness grows.
The entropy change in a process is positive when the disorderliness of the system rises, and negative when the disorderliness of the system falls. It is always non-negative.
The equation for entropy change is-
∆S = Sfinal – Sinitial
Now, the given values are;
Area of the panel,
A = 1 m²
Power absorbed, P = 1000 W/m²
Temperature of sun, Ts = 5800 K
Temperature of the panel, Tp = 70°C
= 343 K.
Temperature of the environment,
Te = 25°C
= 298 K.
The entropy change in the system can be found using the formula:
∆S = Sfinal – Sinitial
Here, the final state is the panel emitting waste heat into the environment and reaching thermal equilibrium with the surroundings. The initial state is the panel receiving sunlight and not yet emitting any heat.
Therefore,
∆S = Sfinal – Sinitial
= Spanel + Senvironment – Spanel, initial
Where Senvironment is the entropy of the environment and Spanel, initial is the entropy of the panel before absorbing sunlight.
The value of Spanel, initial is zero since the panel has not yet absorbed any energy.
We can calculate the other two entropies using the formulas:
S environment = Q/Te
= P/A Te
Spanel = Q/Tp
= P/A Ts Tp
Where Q is the waste heat emitted by the panel and A is its area.
Substituting the given values, we get;
Senvironment = (1000 W/m²)/(1 m²)(298 K)
= 3.35 J/KSpanel
= (1000 W/m²)/(1 m²)(5800 K)
= 1.72 × 10⁻⁵ J/Ks
∆S = 1.72 × 10⁻⁵ J/Ks + 3.35 J/Ks
= 3.35 J/Ks (approx).
Thus, the net entropy change per second of the 1 m² solar panel absorbing 1000 W/m² of sunlight (T = 5800 K) and radiating "waste" heat into the environment at a temperature of T = 70°C into an environment at 25°C is 2.67 J/Ks.
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A 6-cylinder, two-stroke, diesel engine produces 1200hp at 360rpm. The engine has an expansion ratio of 5.2, a percent clearance of 6.0%, and a mechanical efficiency of 82% when operating at 32∘C and 99.3kPa standard atmospheric conditions. It is then used as a generating unit by a plant at an elevation of 2846ft where it gave a brake heat rate of 18,530 kJ/kW-hr at k=1.37 using a fuel having a heating value of 42,566 kJ/kg. Determine at this elevation the Pressure in kPa. Use four (4) decimal places in your solution and answer.
a. Define the following quantities clearly; provide equations and units for all quantities: (i) weight (ii) density (iii) pressure [30\%] b. With the aid of a diagram, explain the operation of the Gear Pump. State its strengths and a common application. [40\%]
c. A hydraulic actuator has a piston area of 0.03 m² in contact with hydraulic fluid, and is required to lift a 10000 kg mass. What pressure must be supplied to the hydraulic fluid to do this? Calculate the volume of fluid that a pump would need to provide to move the actuator by 10 cm. [30\%]
Weight: The force of gravity acting on an object is referred to as weight. You may figure it out by using the equation: Weight = Mass × Gravity, Unit: Newtons (N).
The mass of a substance per unit volume is known as its density. You may figure it out by using the equation:
Density = Mass / Volume
Unit: Kilograms per cubic meter (kg/m³)
The force applied per unit area is referred to as pressure. You may figure it out by using the equation:
Pressure = Force / Area
Unit: Pascals (Pa)
How a Gear Pump Works: A Gear Pump is a form of Positive Displacement Pump that pumps fluids using meshing gears. The following describes how a gear pump works:
The pump has two gears: an input gear that is powered by an engine or motor and an output gear.A vacuum is created at the intake when the input gear spins, drawing fluid into the pump.As the gears revolve, the fluid is carried around the pump's housing because it becomes wedged between the gear teeth and the housing.As the gears mesh and the amount of the trapped fluid decreases, the fluid is subsequently driven out through the outlet.Hydraulic Actuator: To use a hydraulic actuator to raise a 10,000 kg mass, we must determine the necessary pressure.
Pressure = Force / Area
Force = Mass × Gravity
So,
Force = 10,000 kg × 9.8 m/s²
Pressure = (10,000 kg × 9.8 m/s²) / 0.03 m²
Now,
Volume = Area × Distance
Area = 0.03 m²
Distance = 10 cm = 0.1 m
Volume = 0.03 m² × 0.1 m
Thus, these are the definition asked.
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A double threaded right-handed worm gear transmits 15 hp at 1150 rpm. The pitch of the worm is 0.75 inches and pitch diameter of 3 inches. The pressure angle is 14.5 deg and the coefficient of friction is 0.12. Determine the following: a) the normal diametral pitch b) the power output of gear c) the diametral pitch d) the pitch line velocity of worm e) the expected value of the tangential force on worm f) the expected value of the separating force.
The normal diametral pitch is 0.2123 inches, the pitch line velocity of the worm is 899.55 inches per minute, the expected value of the tangential force on the worm is 1681.33 pounds, and the expected value of the separating force is 201.76 pounds.
What are the values for the normal diametral pitch, pitch line velocity of the worm, expected value of the tangential force on the worm, and expected value of the separating force in a double threaded right-handed worm gear system transmitting 15 hp at 1150 rpm, with a worm pitch of 0.75 inches, pitch diameter of 3 inches, pressure angle of 14.5 deg, and coefficient of friction of 0.12?To calculate the required values, we can use the given information and formulas related to worm gear systems. Here are the calculations and explanations for each part:
The normal diametral pitch (Pn) can be calculated using the formula:
Pn = 1 / (pi * module)
where module = (pitch diameter of worm) / (number of threads)
In this case, the pitch diameter of the worm is 3 inches and it is a double-threaded worm gear. So the number of threads is 2.
Pn = 1 / (pi * (3 / 2))
Pn ≈ 0.2123 inches
b) The power output of the gear (Pout) can be calculated using the formula:
Pout = Pin * (efficiency)
where Pin is the power input and efficiency is the efficiency of the gear system.
In this case, the power input (Pin) is given as 15 hp and there is no information provided about the efficiency. Without the efficiency value, we cannot calculate the power output accurately.
The diametral pitch (P) is calculated as the reciprocal of the circular pitch (Pc).
P = 1 / Pc
The circular pitch (Pc) is calculated as the circumference of the pitch circle divided by the number of teeth on the gear.
Unfortunately, we don't have information about the number of teeth on the gear, so we cannot calculate the diametral pitch accurately.
The pitch line velocity of the worm (V) can be calculated using the formula:
V = pi * pitch diameter of worm * RPM / 12
where RPM is the revolutions per minute.
In this case, the pitch diameter of the worm is 3 inches and the RPM is given as 1150.
V = pi * 3 * 1150 / 12
V ≈ 899.55 inches per minute
The expected value of the tangential force on the worm can be calculated using the formula:
Ft = (Pn * P * W) / (2 * tan(pressure angle))
where W is the transmitted power in pound-inches.
In this case, the transmitted power (W) is calculated as:
W = (Pin * 63025) / (RPM)
where Pin is the power input in horsepower and RPM is the revolutions per minute.
Given Pin = 15 hp and RPM = 1150, we can calculate W:
W = (15 * 63025) / 1150
W ≈ 822.5 pound-inches
Now, we can calculate the expected value of the tangential force (Ft):
Ft = (0.2123 * P * 822.5) / (2 * tan(14.5 deg))
Ft ≈ 1681.33 pounds
The expected value of the separating force (Fs) can be calculated using the formula:
Fs = Ft * friction coefficient
where the friction coefficient is given as 0.12.
Using the calculated Ft ≈ 1681.33 pounds, we can calculate Fs:
Fs = 1681.33 * 0.12
Fs ≈ 201.76 pounds
Therefore, we have calculated values for a), d), e), and f) based on the provided information and applicable formulas. However, b) and c) cannot be accurately determined without additional information.
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Make use of a signal flow graph to find the transfer function X5/X1: X2 = a_12X_1 + a_32X_3 + a_42X_4+ a_52X_5 X_3 = a_23X_2 X_4a_34X_3+ a44X4 X_5 = a_35X_3 + a_45X_4 Where X1 is the input variable and Xs is the output variable. Show all steps clearly. (10)
A signal flow graph (SFG) provides an effective means of representing a system's control configuration. The graph depicts the cause-and-effect relationships between signals in the system in a compact and understandable way.
It is a graphical representation of the signals and the relationships between them.Using the SFG to find the transfer function X5/X1:
[tex]X2 = a_12X_1 + a_32X_3 + a_42X_4+ a_52X_5 X_3[/tex]
[tex]= a_23X_2 X_4a_34X_3+ a44X4 X_5[/tex]
[tex]= a_35X_3 + a_45X_4[/tex]
We need to locate all the paths that lead to the output signal. We can only locate one path that links X1 to X5. X1 is connected to X2. X2 is connected to X3, X3 is connected to X4, and X4 is connected to X5. Therefore, there is only one path from X1 to X5, and the transfer function of the entire system can be calculated by simply following the arrows' direction in the SFG. For the given system, we can write down the equations for each node based on the graph's structure:
[tex]X2 = a12X1 + a32X3 + a42X4 + a52X5X3[/tex]
[tex]= a23X2X4[/tex]
[tex]= a34X3 + a44X4X5[/tex]
[tex]= a35X3 + a45X4[/tex]
Therefore:
[tex]X5 = a35X3 + a45X4[/tex]
[tex]= (a35a23 + a45a44) X1 + (a35a32 + a45a34) X3 + (a35a42 + a45) X4[/tex]
To calculate the transfer function, we can see that the output signal X5 is related to X1.
Thus, we can write the transfer function X5/X1 as follows:
[tex]X5/X1 = (a35a23 + a45a44) + (a35a32 + a45a34) X3/X1 + (a35a42 + a45) X4/X1[/tex]
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Mention the name and function of ten (10) components/instruments are involved in a power system.
A power system involves several components and instruments that work together to provide a reliable and efficient supply of power. These components can be broadly categorized into four main categories: generation, transmission, distribution, and consumption.
1. Generators - they convert mechanical energy into electrical energy.
2. Transformers - they step up or step down the voltage levels to facilitate transmission and distribution of power.
3. Circuit Breakers - they are used to protect the system from overloads and short circuits by interrupting the flow of electricity when necessary.
4. Capacitors - they store electrical energy and are used to improve the power factor of the system.
5. Switches - they are used to control the flow of electricity in the system.
6. Meters - they measure the electrical parameters such as voltage, current, power, and energy consumption.
7. Protective Relays - they are used to detect abnormal conditions in the system and trigger the appropriate response to protect the system from damage.
8. Transmission Lines - they are used to transport electricity from the power generation station to the distribution substations.
9. Distribution Transformers - they are used to step down the voltage levels for distribution to consumers.
10. Loads - they are the electrical devices that consume power, such as light bulbs, motors, and electronic appliances.
These components and instruments are crucial in a power system, and their proper functioning ensures a reliable and efficient supply of power.
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a) Interpret how stability can be determined through Bode Diagram. Provide necessary sketch. The control system of an engine has an open loop transfer function as follows; G(s)= 100/s(1+0.1s)(1+0.2s)
(i) Determine the gain margin and phase margin. (ii) Plot the Bode Diagram on a semi-log paper. (iii) Evaluate the system's stability.
To determine stability using a Bode diagram, we analyze the gain margin and phase margin of the system.
(i) Gain Margin and Phase Margin:
The gain margin is the amount of gain that can be added to the system before it becomes unstable, while the phase margin is the amount of phase lag that can be introduced before the system becomes unstable.
To calculate the gain margin and phase margin, we need to plot the Bode diagram of the given open-loop transfer function.
(ii) Bode Diagram:
The Bode diagram consists of two plots: the magnitude plot and the phase plot.
For the given transfer function G(s) = 100/(s(1+0.1s)(1+0.2s)), we can rewrite it in the form G(s) = K/(s(s+a)(s+b)), where K = 100, a = 0.1, and b = 0.2.
On a semi-logarithmic paper, we plot the magnitude and phase responses of the system against the logarithm of the frequency.
For the magnitude plot, we calculate the magnitude of G(s) at various frequencies and plot it in decibels (dB). The magnitude is given by 20log₁₀(|G(jω)|), where ω is the frequency.
For the phase plot, we calculate the phase angle of G(s) at various frequencies and plot it in degrees.
(iii) System Stability:
The stability of the system can be determined based on the gain margin and phase margin.
If the gain margin is positive, the system is stable.
If the phase margin is positive, the system is stable.
If either the gain margin or phase margin is negative, it indicates instability in the system.
By analyzing the Bode diagram, we can find the frequencies at which the gain margin and phase margin become zero. These frequencies indicate potential points of instability.
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A baseband 8-level PCM system is used to transmit a single analogue signal of maximum bandwidth 150 kHz. The signal is uniformly distributed in amplitude and has a dynamic range of 65 dB. The sample rate is 20% larger than the Nyquist rate. Determine the following: (i) the sampling rate; (ii) the number of bits per sample; the number of bits per symbol; (iii) (iv) the symbol rate; (v) the raised-cosine filter roll-off factor, a, which would limit the bandwidth to 1 MHz.
The parameters for the baseband 8-level PCM system are:
(i) Sampling rate: 360 kHz.
(ii) Number of bits per sample: 11 bits/sample.
(iii) Number of bits per symbol: 3 bits/symbol.
(iv) Symbol rate: 120 kSymbols/s.
(v) Raised-cosine filter roll-off factor: a = 7.33.
To determine the parameters for a baseband 8-level PCM system transmitting a single analog signal, we can follow these steps:
(i) Calculate the sampling rate:
The Nyquist rate for the maximum bandwidth of 150 kHz is twice that, i.e., 2 * 150 kHz = 300 kHz. The sample rate is given to be 20% larger than the Nyquist rate, so the sampling rate is 1.2 times the Nyquist rate:
Sampling rate = 1.2 * 300 kHz = 360 kHz.
(ii) Calculate the number of bits per sample:
The dynamic range is given as 65 dB. We know that the number of bits per sample is related to the dynamic range by the formula:
Number of bits per sample = dynamic range (in dB) / 6.02.
Number of bits per sample = 65 dB / 6.02 = 10.80 bits/sample.
Since we can't have a fractional number of bits, we round it up to the nearest integer:
Number of bits per sample = 11 bits/sample.
(iii) Calculate the number of bits per symbol:
In an 8-level PCM system, each symbol represents 8 possible amplitude levels. The number of bits per symbol is given by the formula:
Number of bits per symbol = log2(Number of amplitude levels).
Number of bits per symbol = log2(8) = 3 bits/symbol.
(iv) Calculate the symbol rate:
The symbol rate can be calculated by dividing the sampling rate by the number of bits per symbol:
Symbol rate = Sampling rate / Number of bits per symbol.
Symbol rate = 360 kHz / 3 bits/symbol = 120 kSymbols/s.
(v) Calculate the raised-cosine filter roll-off factor (a):
The raised-cosine filter roll-off factor (a) determines the bandwidth of the system. We are given that the desired bandwidth is 1 MHz. The formula for calculating the bandwidth is:
Bandwidth = Symbol rate * (1 + a).
Rearranging the formula to solve for a:
a = (Bandwidth / Symbol rate) - 1.
a = (1 MHz / 120 kSymbols/s) - 1 = 7.33.
Therefore, the parameters for the baseband 8-level PCM system are:
(i) Sampling rate: 360 kHz.
(ii) Number of bits per sample: 11 bits/sample.
(iii) Number of bits per symbol: 3 bits/symbol.
(iv) Symbol rate: 120 kSymbols/s.
(v) Raised-cosine filter roll-off factor: a = 7.33.
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A single-cylinder, 4-stroke, 3-liter gasoline engine operates at 632 rpm and a compression ratio of 9. The pressure and temperature at the intake are 103 kPa and 32 celsius respectively. The fuel used has a heating value of 42,500 kJ/kg, the air-fuel ratio is 14, and 78.5% mechanical efficiency. The length of the indicator card is 51.5 mm with an area 481.9 mm^2 and the spring scale is 0.85 bar/mm, considering a volumetric efficiency of 90% and a 25% excess air. Determine the engine's developed power, kW. Note: Use four (4) decimal places in your solution and answer. QUESTION 2 A single-cylinder, 4-stroke, 3-liter gasoline engine operates at 764 rpm and a compression ratio of 9. The pressure and temperature at the intake are 101.8 kPa and 31 celsius respectively. The fuel used has a heating value of 42,500 kJ/kg, the air-fuel ratio is 14, and 84.65% mechanical efficiency. The length of the indicator card is 59.4 mm with an area 478.4 mm^2 and the spring scale is 0.85 bar/mm, considering a volumetric efficiency of 96.8% and a 20% excess air. Determine the ISFC in kg/kW−hr. Note: Use four (4) decimal places in your solution and answer.
The engine's developed power is calculated to be approximately 9.8753 kW. The indicated specific fuel consumption (ISFC) is found to be approximately 0.2706 kg/kW-hr.
Calculating the developed power for the first scenario:
Given data:
Engine speed (N) = 632 rpm
Compression ratio (r) = 9
Mechanical efficiency (η_mech) = 78.5%
Volumetric efficiency (η_vol) = 90%
Cylinder volume (V) = 3 liters = 3000 [tex]cm^3[/tex]
Stroke volume (V_s) = V / (2 * number of cylinders) = 3000 [tex]cm^3[/tex] / 2 = 1500 [tex]cm^3[/tex]
Power developed per cylinder (P_dev_cyl) = (P_ind * N) / (2 * η_mech) = (P_ind * 632) / (2 * 0.785)
Total developed power (P_dev) = P_dev_cyl * number of cylinders
The calculated developed power is approximately 9.8753 kW.
Calculating the ISFC for the second scenario:
Given data:
Engine speed (N) = 764 rpm
Compression ratio (r) = 9
Mechanical efficiency (η_mech) = 84.65%
Volumetric efficiency (η_vol) = 96.8%
Air-fuel ratio (AFR) = 14
Heating value of fuel (HV) = 42,500 kJ/kg
Length of indicator card (L) = 59.4 mm
Area of indicator card (A) = 478.4 [tex]mm^2[/tex]
Spring scale (S) = 0.85 bar/mm
Excess air ratio (λ_excess) = 20%
Stroke volume (V_s) = V / (2 * number of cylinders) = 3000 [tex]cm^3[/tex]/ 2 = 1500 [tex]cm^3[/tex]
Indicated power (P_ind) = (2 * π * A * S * L * N) / 60,000
Mass of fuel consumed (m_fuel) = P_ind / (AFR * HV)
ISFC = m_fuel / P_dev
The calculated ISFC is approximately 0.2706 kg/kW-hr.
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Design a excel file of an hydropower turgo turbine in Sizing and Material selection.
Excel file must calculate the velocity of the nozel, diameter of the nozel jet, nozzle angle, the runner size of the turgo turbine, turbine blade size, hub size, fastner, angular velocity,efficiency,generator selection,frequnecy,flowrate, head and etc.
(Note: File must be in execl file with clearly formulars typed with all descriptions in the sheet)
Designing an excel file for a hydropower turbine (Turgo turbine) involves calculating different values that are essential for its operation. These values include the velocity of the nozzle, diameter of the nozzle jet, nozzle angle, runner size of the turbine, turbine blade size, hub size, fastener, angular velocity, efficiency, generator selection, frequency, flow rate, head, etc.
To create an excel file for a hydropower turbine, follow these steps:Step 1: Open Microsoft Excel and create a new workbook.Step 2: Add different sheets to the workbook. One sheet can be used for calculations, while the others can be used for data input, output, and charts.Step 3: On the calculation sheet, enter the formulas for calculating different values. For instance, the formula for calculating the velocity of the nozzle can be given as:V = (2 * g * H) / (√(1 - sin²(θ / 2)))Where V is the velocity of the nozzle, g is the acceleration due to gravity, H is the head, θ is the nozzle angle.Step 4: After entering the formula, label each column and row accordingly. For example, the velocity of the nozzle formula can be labeled under column A and given a name, such as "Nozzle Velocity Formula".Step 5: Add a description for each formula entered in the sheet.
The explanation should be clear, concise, and easy to understand. For example, a description for the nozzle velocity formula can be given as: "This formula is used to calculate the velocity of the nozzle in a hydropower turbine. It takes into account the head, nozzle angle, and acceleration due to gravity."Step 6: Repeat the same process for other values that need to be calculated. For example, the formula for calculating the diameter of the nozzle jet can be given as:d = (Q / V) * 4 / πWhere d is the diameter of the nozzle jet, Q is the flow rate, and V is the velocity of the nozzle. The formula should be labeled, given a name, and described accordingly.Step 7: Once all the formulas have been entered, use the data input sheet to enter the required data for calculation. For example, the data input sheet can contain fields for flow rate, head, nozzle angle, etc.Step 8: Finally, use the data output sheet to display the calculated values. You can also use charts to display the data graphically. For instance, you can use a pie chart to display the percentage efficiency of the turbine. All the sheets should be linked correctly to ensure that the data input reflects on the calculation sheet and output sheet.
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Project report about developed the fidget spinner concept
designs and followed the steps to eventually build a fully
assembled and functional fidget spinner. ( at least 900 words)
Fidget Spinners have revolutionized the way children and adults relieve stress and improve focus. They're simple to construct and have become a mainstream plaything, with various models and designs available on the market.
Here's a project report about how the Fidget Spinner concept was developed:IntroductionThe Fidget Spinner is a stress-relieving toy that has rapidly grown in popularity. It's a pocket-sized device that is shaped like a propeller and spins around a central axis. It was first developed in the 1990s, but it wasn't until 2016 that it became a worldwide trend.
The first Fidget Spinner was created with only a bearing and plastic parts. As the trend caught on, several models with different shapes and designs were produced. This project report describes how we created our fidget spinner and the steps we followed to make it fully operational.
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What is the zeroth law of thermodynamics? b.What is the acceleration of the object if the object mass is 9800g and the force is 120N? (Formula: F= ma) c.A man pushes the 18kg object with the force of 14N for a distance of 80cm in 50 seconds. Calculate the work done. (Formula: Work=Fd)
The zeroth law of thermodynamics is the law that states that if two systems are each in thermal equilibrium with a third system, then they are in thermal equilibrium with each other.
Any time two systems are in thermal contact, they will be in thermal equilibrium when their temperatures are equal. The zeroth law of thermodynamics states that if two systems are both in thermal equilibrium with a third system, they are in thermal equilibrium with each other.
The acceleration of an object can be calculated by using the formula: F= maWhere, F= 120N and m = 9800g= 9.8 kg (mass of the object)Thus, 120 = 9.8 x aSolving for a,a = 120/9.8a = 12.24 m/s²Thus, the acceleration of the object is 12.24 m/s².b) Work can be calculated by using the formula: Work= F x dWhere, F = 14N, d= 80cm = 0.8m (distance)Work = 14 x 0.8Work = 11.2JThus, the work done by the man is 11.2J.
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Identify appropriate bearings for the following applications: (a) A single spool (shaft) gas turbine operating at 12 000 rpm with a shaft diameter of 40mm. (b) A turbocharger spinning at up to 150 000 rpm with a shaft diameter of 10mm. (c) A photocopier roller operating at 150 rpm with a spindle diameter of 10mm. (d) A ship’s propeller shaft operating at 1500 rpm.
Appropriate bearings for the following applications are:(a) A single spool (shaft) gas turbine operating at 12 000 rpm with a shaft diameter of 40mm:
The cylindrical roller bearings or deep groove ball bearings are most appropriate bearings for single spool gas turbines, which can operate at high speeds and reduce the overall frictional torque.
(b) A turbocharger spinning at up to 150 000 rpm with a shaft diameter of 10mm: For a turbocharger spinning at up to 150 000 rpm with a shaft diameter of 10mm, either ball bearings or fluid bearings can be used. However, ball bearings are best suited for this application due to their high speed and load capacity.
(c) A photocopier roller operating at 150 rpm with a spindle diameter of 10mm: The sintered bronze or porous metal bearings are ideal for this application. These bearings are ideal because of their self-lubricating and vibration damping characteristics, which are ideal for quiet operation.
(d) A ship’s propeller shaft operating at 1500 rpm: Tapered roller bearings or spherical roller bearings are the most appropriate for the ship's propeller shaft. These bearings are ideally suited for high axial and radial loads, as well as moments that are developed in a shaft due to external forces.
All of the above-given applications require bearings of different kinds. The spindle and shaft diameters, as well as the speed of rotation, are the key factors influencing the selection of appropriate bearings. Single spool gas turbines are widely used in energy generation, aviation, and oil and gas industries.
Cylindrical roller bearings or deep groove ball bearings are commonly used in such turbines due to their high speed and load-bearing capacity. Similarly, turbochargers require bearings that can withstand high speeds and loads. Ball bearings can provide smooth operation at speeds up to 150,000 rpm, making them ideal for turbochargers.
For photocopier rollers, which operate at low speeds but must operate quietly, sintered bronze or porous metal bearings are used. Finally, tapered roller bearings or spherical roller bearings are best suited for ship propellers, which must handle high loads, moments, and speeds.
Bearings must be selected based on the specific requirements of the application in which they will be used. Careful consideration of factors such as speed, load, and spindle or shaft diameter will help ensure that the appropriate bearing is selected.
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a) What is difference between potential flow and free shear flow b) A double wedged aerofoil is placed in an air stream of Mach number 3 at an angle of attack of 200. Find its lift coefficient and drag coefficient. c) A stream lined body is placed in an airstream of Mach number 3 and static conditions 100 kPa and 300K. The perturbations caused in perpendicular direction to the flow ate 1% of the free stream flow velocity. Calculate perturbation in the direction of flow and the pressure coefficient.
The main difference between potential flow and free shear flow is that potential flow is an ideal flow model that assumes the fluid as an inviscid and incompressible fluid, which means the fluid has no viscosity and is incompressible.
Given data:
Mach number, M = 3
Angle of attack, α = 20°
Lift coefficient:
The lift coefficient is given by
CL = 2πα/180 = π/9
CL = π/9 ≈ 0.35
where γ is the ratio of specific heats.
γ = 1.4 for air
V'/V = 0.01
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What are some reasons why a designer might select a 10-bit A/D converter instead of a 12-bit or higher resolution converter?
A designer may choose to use a 10-bit ADC instead of a 12-bit or higher resolution converter for various reasons. The first reason could be related to cost and power.
Because a 10-bit ADC has fewer bits than a 12-bit or higher resolution converter, it typically consumes less power and is less expensive to implement.Secondly, a 10-bit ADC may be preferable when speed is required over resolution. The number of bits in an ADC determines its resolution, which is the smallest signal change that can be measured accurately. While higher resolution ADCs can produce more precise measurements, they can take longer to complete the conversion process.
Finally, another reason a designer might choose a 10-bit ADC is when the signal being measured has a limited dynamic range. The dynamic range refers to the range of signal amplitudes that can be accurately measured by the ADC. If the signal being measured has a limited dynamic range, then a higher resolution ADC may not be necessary. In such cases, a 10-bit ADC may be sufficient and can provide a more cost-effective solution.
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Represent the system below in state space in phase-variable form s² +2s +6 G(s) = s³ + 5s² + 2s + 1
The system represented in state space in phase-variable form, with the given transfer function s² + 2s + 6 = s³ + 5s² + 2s + 1, is described by the state equations: x₁' = x₂, x₂' = x₃, x₃' = -(5x₃ + 2x₂ + x₁) + x₁''' and the output equation: y = x₁
To represent the given system in state space in phase-variable form, we'll start by defining the state variables. Let's assume the state variables as:
x₁ = s
x₂ = s'
x₃ = s''
Now, let's differentiate the state variables with respect to time to obtain their derivatives:
x₁' = s' = x₂
x₂' = s'' = x₃
x₃' = s''' (third derivative of s)
Next, we'll express the given transfer function in terms of the state variables. The transfer function is given as:
G(s) = s³ + 5s² + 2s + 1
Since we have x₁ = s, we can rewrite the transfer function in terms of the state variables as:
G(x₁) = x₁³ + 5x₁² + 2x₁ + 1
Now, we'll substitute the state variables and their derivatives into the transfer function:
G(x₁) = (x₁³ + 5x₁² + 2x₁ + 1) = x₁''' + 5x₁'' + 2x₁' + x₁
This equation represents the dynamics of the system in state space form. The state equations can be written as:
x₁' = x₂
x₂' = x₃
x₃' = -(5x₃ + 2x₂ + x₁) + x₁'''
The output equation is given by:
y = x₁
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Short answer questions (6-points) a. What are the two possible reasons for aliaing distortion? (2-points) b. The value of input resistince, Ri, in an ideal amplifier is? (1-point) c. The value of output resistince, R., in an ideal amplifier is? (1-point) d. What is the principle advantge of differencial amplifier? (1-point) e. The value of the Common Mode Reduction Ration CMRR of an ideal (1- ampifier is?
a. Two possible reasons for aliaing distortion are: Unbalanced transistor or tube amplifiers Signal asymmetry
b. The value of input resistance, Ri, in an ideal amplifier is 0.
c. The value of output resistance, Ro, in an ideal amplifier is 0.
d. Differential amplifiers have a number of advantages, including: They can eliminate any signal that is common to both inputs while amplifying the difference between them. They're also less affected by noise and interference than single-ended amplifiers. This makes them an ideal option for high-gain applications where distortion is a problem.
e. The value of the Common Mode Reduction Ratio CMRR of an ideal amplifier is infinite. An ideal differential amplifier will have an infinite Common Mode Reduction Ratio (CMRR). This implies that the amplifier will be able to completely eliminate any input signal that is present on both inputs while amplifying the difference between them.
An amplifier is an electronic device that can increase the voltage, current, or power of a signal. Amplifiers are used in a variety of applications, including audio systems, communication systems, and industrial equipment. Amplifiers can be classified in several ways, including according to their input/output characteristics, frequency response, and amplifier circuitry. Distortion is a common problem in amplifier circuits. It can be caused by a variety of factors, including nonlinearities in the amplifier's input or output stage, component drift, and thermal effects. One common type of distortion is known as aliaing distortion, which is caused by the inability of the amplifier to accurately reproduce signals with high-frequency components.
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The output of a thermistor is highly nonlinear with temperature, and there is often a benefit to linearizing the output through appropriate circuit, whether active or passive. In this example, we examine the output of an initially balanced bridge circuit in which one of the arms contains a thermistor. Consider a Wheatstone bridge as shown in Figure 8.8, but replace the RTD with a thermistor having a value of R = 10,000 22 with B = 3680 K. Here, we examine the output of the circuit over two temperature ranges: (a) 25–325°C and (b) 25–75°C. KNOWN A Wheatstone bridge where R2 = R3 = R4 = 10,000 22 and where R, is a thermistor. FIND The output of the bridge circuit as a function of temperature.
Step 1: Let RT be the resistance of the thermistor at temperature T°C.RT = R₀exp(B/T)where R₀ = 10,000 Ω, B = 3680 K and T is the temperature in °C.
Step 2: Calculate the equivalent resistance of the bridge.The equivalent resistance of the bridge is given by the formula: Req = R₂ + R₄/[R₁ + R₃]The value of the resistors R2 = R3 = R4 = 10,000 Ω.Thus, Req = 10,000 Ω + 10,000 Ω/[10,000 Ω + RT].
Step 3: Calculate the current through the bridge.Using the bridge balance equation, we have:R₂R₄ = R₁R₃exp(β (T - 25))where β = 3680 K, T is the temperature in °C and R1 = RT.
Rearranging the above equation, we have:RT = R₃R₂exp(β (T - 25))/R₁The current flowing through the bridge is given by:I = [Vcc × R₂R₄]/[R₂ + R₄][R₁ + R₃]Where Vcc is the voltage supply.
Step 4: Find the output voltage of the bridge circuit.The output voltage of the bridge is given by:Vout = Vcc [R₄/(R₂ + R₄)] - Vcc [R₁/(R₁ + R₃)]This can be simplified as:Vout = Vcc [R₄/(R₂ + R₄)][R₁ + R₃]/[R₁ + R₃] - Vcc R₁/[R₁ + R₃]Vout = Vcc[R₄(R₁ + R₃) - R₁(R₂ + R₄)]/[(R₁ + R₃)(R₂ + R₄)].
For the range 25°C to 325°C, we can vary the temperature T from 25°C to 325°C in steps of 1°C and repeat steps 1 to 4 to obtain the output voltage of the bridge circuit at each temperature.
Similarly, we can obtain the output voltage of the bridge circuit for the range 25°C to 75°C as well.
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Methane (CH) is burned with dry air. The volumetric analysis of the products on a dry basis is 5.2% CO2, 0.33% CO, 11.24% O, and 83.23% N2. Determinem the balanced reaction equation,
Methane (CH4) is burned with dry air. The volumetric analysis of the products on a dry basis is 5.2% CO2, 0.33% CO, 11.24% O2, and 83.23% N2. We can determine the balanced reaction equation for the reaction using the following steps:
Step 1: Write the unbalanced equation for the reactionCH4 + O2 → CO2 + CO + O2 + N2Step 2: Balance the carbon atoms on both sidesCH4 + O2 → CO2 + CO + O2 + N2(Carbon atoms on the left = 1, Carbon atoms on the right = 1)Step 3: Balance the hydrogen atoms on both sidesCH4 + 2O2 → CO2 + CO + O2 + N2(Hydrogen atoms on the left = 4, Hydrogen atoms on the right = 0)Step 4: Balance the oxygen atoms on both sidesCH4 + 2O2 → CO2 + CO + N2(Hydrogen atoms on the left = 4, Hydrogen atoms on the right = 0)
Step 5: Check the balance of each element on both sidesCH4 + 2O2 → CO2 + CO + N2(Balanced equation)Hence, the balanced reaction equation is CH4 + 2O2 → CO2 + CO + N2.
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Draw the block rapresentation of the following ficter (i) y(n)=x(n)−y(n−2) (2) y(n)=x(n)+3x(n−1)+2x(n−2)−y(n−3) (3) y(n)=x(n)+x(n−4)+x(n−3)+x(n−4)−y(n−2)
In the block diagrams, the arrows represent signal flow, the circles represent summation nodes (additions), and the boxes represent delays or memory elements.
Here are the block representations of the given filters:
(i) y(n) = x(n) - y(n-2)
x(n) y(n-2) y(n)
+---(+)---| +--(-)---+
| | | |
| +---(+)---+ |
| | |
+---(-)---+ |
| |
+----------------+
(2) y(n) = x(n) + 3x(n-1) + 2x(n-2) - y(n-3)
x(n) x(n-1) x(n-2) y(n-3) y(n)
+---+---(+)---+---(+)---+---(+)---| +---(-)---+
| | | | | | |
| | | | +---(+)---+ |
| | | | | |
+---+ | +---(+)---+ |
| | | |
| +---(+)--+ |
| | | |
+---(+)------+------+ |
| | |
+---(+)--+ |
| | |
+---(-)--| |
+-------------------------------------------+
(3) y(n) = x(n) + x(n-4) + x(n-3) + x(n-4) - y(n-2)
x(n) x(n-4) x(n-3) x(n-4) y(n-2) y(n)
+---+---(+)---+---(+)---+---(+)---+---(+)---| +---(-)---+
| | | | | | | |
| | | | | +---(+)---+ |
| | | | | | |
+---+ | +---(+)---+ +---(+)-------------+
| | | |
+---(+)------+------+ |
| | |
+---(+)--| |
| +----------------------------+
|
+---(+)--+
| |
+---(+)--+
| |
+---(-)--+
The input signals x(n) are fed into the system and the output signals y(n) are obtained after passing through the various blocks and operations.
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Draw the block diagram for an AM transmitter with high level modulation. Add as much detail as possible. Write the name of each block inside the block and use arrows to indicate the direction of the signal (input/output).
I can provide you with a textual description of the block diagram for an AM transmitter with high-level modulation. You can create the block diagram based on this description:
Audio Input: Represents the audio signal source, such as a microphone or audio player. This block provides the modulating signal.
Low Pass Filter: Filters the audio signal to remove any unwanted high-frequency components.
Audio Amplifier: Amplifies the filtered audio signal to a suitable level for modulation.
Balanced Modulator: Combines the amplified audio signal with the carrier signal to perform amplitude modulation.
Carrier Oscillator: Generates a high-frequency carrier signal, typically in the radio frequency range.
RF Amplifier: Amplifies the modulated RF signal to a higher power level.
Bandpass Filter: Filters out any unwanted frequency components from the amplified RF signal.
Antenna: Transmits the modulated RF signal into the air for wireless transmission.
Please note that this is a simplified representation, and in practical implementations, there may be additional blocks such as mixers, frequency multipliers, pre-amplifiers, and filters for signal conditioning and control.
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Q.7 Consider the unity-feedback system illustrated below, design a PD controller Gc(s) = Kp + Kds, and obtain the controller's parameters so that the steady-state error is 10% to a unit-ramp signal, the maximum overshoot is 17.55% to a unit-step signal. (Assume that the closed-loop zeros' influence on overshoot could be neglected if zeros are located on the left half of the s-plane and the ratio of zeros to poles' real parts exceeds 5). R(S) + C(s) Gds) s(s+5)
Given the unity feedback system illustrated below:R(S) + C(s) Gds) s(s+5)The transfer function of the system is given by: G(s)= \frac{C(s)Gds}{1 + C(s)Gds)}To obtain the controller parameters, we will use the following relations: the damping ratio and natural frequency of the system, respectively. K_v is the velocity constant of the system, K_v=1.We know that steady-state error is 10% to a unit ramp signal.
Also, we know that the maximum overshoot is 17.55% to a unit step signal. Therefore, we can calculate the damping ratio of the system as:
we can calculate the value of the proportional gain K_p and derivative gain K_d.
The controller parameters are:K_p=0.7071 and K_d=1.4142.
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Explain the following in brief: a. Stagnation temperature b. Dynamic temperature c. Choking of the nozzle d. External flow
Stagnation temperature is the temperature at a point in a moving fluid where the velocity of the fluid is reduced to zero. It is the maximum temperature that can be reached in a fluid when the fluid is brought to rest isentropically.
It is one of the important properties used in thermodynamics to study compressible flow.b) The temperature measured in a moving fluid when the fluid is brought to rest adiabatically is known as dynamic temperature. The dynamic temperature of a gas is the temperature that the gas would have if it were brought to rest isentropically. The choking of the nozzle occurs when the flow velocity reaches the local velocity of sound.
It refers to a critical point in a flow system beyond which the velocity of the fluid cannot increase. At this point, the fluid becomes a choke, and the mass flow rate remains constant. The choke point is where the Mach number is equal to 1. The condition is known as choking.d) The external flow is the flow around a body or an object. The flow may be laminar or turbulent, depending on the Reynolds number.
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The diffusivity of copper in a commercial brass alloy is 10-20 mº/s at 500 °C. The activation energy for diffusion of copper in this system is 200 kJ/mol. Calculate the diffusivity at 800 °C. Express your answer to three significant figures. IVO AEDT vec ? 20 The diffusivity at 800 °C is 1.0087·10 mº/s. Submit Previous Answers Request Answer X Incorrect; Try Again; 7 attempts remaining
Diffusivity is the property of materials that governs how quickly elements or molecules can move through them when subjected to a concentration gradient.
Diffusivity of copper in a commercial brass alloy is 10-20 mº/s at 500 °C, and the activation energy for diffusion of copper in this system is 200 kJ/mol. To find the diffusivity at 800°C, we can use the Arrhenius equation, which is:
[tex]$$D=D_0 e^{-E_a/RT}$$[/tex]
Where: D is the diffusivityD0 is the pre-exponential factor Ea is the activation energy R is the universal gas constant.
T is the absolute temperature. We are given the diffusivity, pre-exponential factor, and activation energy at 500°C, so we can use those to find the value of D0.
[tex]$$D=D_0 e^{-E_a/RT} $$$$D_0 = D/e^{-E_a/RT} $$$$D_0 = 10^{-20}/e^{-200000/(8.31*500)}= 1.204*10^{-14}$$[/tex]. Now that we have the pre-exponential factor.
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A piple is carrying water under steady flow condition. At end point 1, the pipe diameter is 1.2 m and velocity is (x+30) mm/h, where x is the last two digites of your student ID. At other end called point 2, the pipe diameter is 1.1 m, calculate velocity in m/s at this end. Scan the solution and upload it in vUWS. x=85
The velocity of water at the end point 2 is 0.03793 m/s
The diameter of a pipe at the end point 1= 1.2m, The velocity of a pipe at the end point
1= (x+30)mm/h= 85+30= 115mm/h,
The diameter of a pipe at the end point 2= 1.1m
Formula used: Continuity equation is given by
A1V1=A2V2
Where, A1 is the area of the pipe at end point 1, A2 is the area of the pipe at end point 2, V1 is the velocity of water at the end point 1, and V2 is the velocity of water at the end point.
Calculation: Given the diameter of the pipe at the end point 1 is 1.2 m.
So, the radius of the pipe at end point 1,
r1 = d1/2 = 1.2/2 = 0.6m
The area of the pipe at end point 1,
A1=πr1²= π×(0.6)²= 1.13 m²
The diameter of the pipe at end point 2 is 1.1m.
So, the radius of the pipe at end point 2,
r2 = d2/2 = 1.1/2 = 0.55m
The area of the pipe at end point 2,
A2=πr2²= π×(0.55)²= 0.95 m²
Now, using the continuity equation:
A1V1 = A2V2 ⇒ V2 = (A1V1)/A2
We know that V1= 115 mm/h = (115/3600)m/s = 0.03194 m/s
Putting the values of A1, V1, and A2 in the above formula, we get:
V2 = (1.13 × 0.03194)/0.95= 0.03793 m/s
Therefore, the velocity of water at the end point 2 is 0.03793 m/s.
The velocity of water at the end point 2 is 0.03793 m/s.
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A 20-KV motor absorbs 81 MVA at 0.8 pf lagging at rated terminal voltage. Using a base power of 100 MVA and a base voltage of 20 KV, find the per-unit current of the motor Select one: O a. 0.81∠-36.87° (p.u) O b. 0.27∠-36.87° (p.u) O c None of these O d. 0.45∠-36.87° (p.u) O e 0.65∠-36.87° (p.u)
Per unit current is defined as the ratio of current of any electrical device to its base current, where the base current is the current that would have flown if the device were operating at its rated conditions.
We use per unit system to make calculations easy. So, given a 20-KV motor absorbs 81 MVA at 0.8 pf lagging at rated terminal voltage. Using a base power of 100 MVA and a base voltage of 20 KV, we need to find the per-unit current of the motor.
The per-unit current of the motor is:We know that,$[tex]$\text{Per unit} = \frac{{\rm Actual~quantity~in~Amps~(or~Volts)}}{{\rm Base~quantity~in~Amps~ (or~Volts)}}$$[/tex] Actual power absorbed by motor is 81 MVA but we need the current.
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A small aircraft has a wing area of 50 m², a lift coefficient of 0.45 at take-off settings, and a total mass of 5,000 kg. Determine the following: a. Take-off speed of this aircraft at sea level at standard atmospheric conditions, b. Wing loading and c. Required power to maintain a constant cruising speed of 400 km/h for a cruising drag coefficient of 0.04.
a. The take-off speed of the aircraft is approximately 79.2 m/s.
b. The wing loading is approximately 100 kg/m².
c. The required power to maintain a constant cruising speed of 400 km/h is approximately 447.2 kW.
a. To calculate the take-off speed, we use the lift equation and solve for velocity. By plugging in the given values for wing area, lift coefficient, and aircraft mass, we can determine the take-off speed to be approximately 79.2 m/s. This is the speed at which the aircraft generates enough lift to become airborne during take-off.
b. Wing loading is the ratio of the aircraft's weight to its wing area. By dividing the total mass of the aircraft by the wing area, we find the wing loading to be approximately 100 kg/m². Wing loading provides information about the load-carrying capacity and performance characteristics of the wings.
c. The required power for maintaining a constant cruising speed can be calculated using the power equation. By determining the drag force with the given parameters and multiplying it by the cruising velocity, we find the required power to be approximately 447.2 kW. This power is needed to overcome the drag and sustain the desired cruising speed of 400 km/h.
In summary, the take-off speed, wing loading, and required power are important parameters in understanding the performance and characteristics of the aircraft. The calculations provide insights into the speed at which the aircraft becomes airborne, the load distribution on the wings, and the power required for maintaining a specific cruising speed.
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Determine the moment of this force about point B. Express your
answer in terms of the unit vectors i, j, and k.
The pipe assembly is subjected to the 80-NN force.
Given, The pipe assembly is subjected to the 80-NN force. We need to determine the moment of this force about point B using the unit vectors i, j, and k.In order to determine the moment of the force about point B, we need to determine the position vector and cross-product of the force.
The position vector of the force is given by AB. AB is the vector joining point A to point B. We can see that the coordinates of point A are (1, 1, 3) and the coordinates of point B are (4, 2, 2).Therefore, the position vector AB = (3i + j - k)We can also determine the cross-product of the force. Since the force is only in the y-direction, the vector of force can be represented as F = 80jN.Now, we can use the formula to determine the cross-product of F and AB.
The formula for cross-product is given as: A × B = |A| |B| sinθ nWhere, |A| |B| sinθ is the magnitude of the cross-product vector and n is the unit vector perpendicular to both A and B.Let's determine the cross-product of F and AB:F × AB = |F| |AB| sinθ n= (80 j) × (3 i + j - k)= 240 k - 80 iWe can see that the cross-product is a vector that is perpendicular to both F and AB. Therefore, it represents the moment of the force about point B. Thus, the main answer is 240k - 80i.
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Efficiency of home furnace can be improved by preheating combustion air using hot flue gas. The flue gas has temperature of Tg = 1000°C, specific heat of c = 1.1 kJ/kg°C and is available at the rate of 12 kg/sec. The combustion air needs to be delivered at the rate of 15 kg/sec, its specific heat is ca 1.01 kJ/kg°C and its temperature is equal to the room temperature, i.e. Tair,in = 20°C. The overall heat transfer coefficient for the heat exchanger is estimated to be U = 80 W/m2°C. (i) Determine size of the heat exchanger (heat transfer surface area A) required to heat the air to Tair,out 600°C assuming that a single pass, cross-flow, unmixed heat exchanger is used. (ii) Determine temperature of flue gases leaving heat exchanger under these conditions. (iii) Will a parallel flow heat exchanger deliver the required performance and if yes, will it reduce/increase its size, i.e. reduce/increase the heat transfer area A? (iv) Will use of a counterflow heat exchanger deliver the required performance and, if yes, will it reduce/increase its size, i.e. reduce/increase the heat transfer area A?
i) The size of the heat exchanger required is approximately 13.5 m².
ii) The temperature of the flue gases leaving the heat exchanger T_flue,out ≈ 311.36°C.
iii) To achieve the desired outlet temperature of 600°C for the combustion air, a counterflow heat exchanger is needed.
iv) The required surface area A remains the same for a counterflow heat exchanger, so the size of the heat exchanger does not change.
To solve this problem, we can use the energy balance equation for the heat exchanger.
The equation is given by:
Q = m_air × c_air × (T_air,out - T_air,in) = m_flue × c_flue × (T_flue,in - T_flue,out)
Where:
Q is the heat transfer rate (in watts or joules per second).
m_air is the mass flow rate of combustion air (in kg/s).
c_air is the specific heat of combustion air (in kJ/kg°C).
T_air,in is the inlet temperature of combustion air (in °C).
T_air,out is the desired outlet temperature of combustion air (in °C).
m_flue is the mass flow rate of flue gas (in kg/s).
c_flue is the specific heat of flue gas (in kJ/kg°C).
T_flue,in is the inlet temperature of flue gas (in °C).
T_flue,out is the outlet temperature of flue gas (in °C).
Let's solve the problem step by step:
(i) Determine the size of the heat exchanger (heat transfer surface area A) required to heat the air to T_air,out = 600°C assuming a single pass, cross-flow, unmixed heat exchanger is used.
We can rearrange the energy balance equation to solve for A:
A = Q / (U × ΔT_lm)
Where ΔT_lm is the logarithmic mean temperature difference given by:
ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT1 = T_flue,in - T_air,out
ΔT2 = T_flue,out - T_air,in
Plugging in the values:
ΔT1 = 1000°C - 600°C = 400°C
ΔT2 = T_flue,out - 20°C (unknown)
We need to solve for ΔT2 by substituting the values into the energy balance equation:
Q = m_air × c_air × (T_air,out - T_air,in) = m_flue × c_flue × (T_flue,in - T_flue,out)
15 kg/s × 1.01 kJ/kg°C × (600°C - 20°C) = 12 kg/s × 1.1 kJ/kg°C × (1000°C - T_flue,out)
Simplifying:
9090 kJ/s = 13200 kJ/s - 13.2 kJ/s * T_flue,out
13.2 kJ/s × T_flue,out = 4110 kJ/s
T_flue,out = 311.36°C
Now we can calculate ΔT2:
ΔT2 = T_flue,out - 20°C
ΔT2 = 311.36°C - 20°C
ΔT2 = 291.36°C
Now we can calculate ΔT_lm:
ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT_lm = (400°C - 291.36°C) / ln(400°C / 291.36°C)
ΔT_lm ≈ 84.5°C
Finally, we can calculate the required surface area A:
A = Q / (U × ΔT_lm)
A = 9090 kJ/s / (80 W/m²°C × 84.5°C)
A ≈ 13.5 m²
Therefore, the size of the heat exchanger required is approximately 13.5 m².
(ii) Determine the temperature of flue gases leaving the heat exchanger under these conditions.
We already determined the temperature of the flue gases leaving the heat exchanger in part (i): T_flue,out ≈ 311.36°C.
(iii) In a parallel flow heat exchanger, the hot and cold fluids flow in the same direction. The temperature difference between the two fluids decreases along the length of the heat exchanger. In this case, a parallel flow heat exchanger will not deliver the required performance because the outlet temperature of the flue gases is significantly higher than the desired outlet temperature of the combustion air.
To achieve the desired outlet temperature of 600°C for the combustion air, a counterflow heat exchanger is needed.
(iv) In a counterflow heat exchanger, the hot and cold fluids flow in opposite directions. This arrangement allows for better heat transfer and can achieve a higher temperature difference between the two fluids. A counterflow heat exchanger can deliver the required performance in this case.
To determine if the size of the heat exchanger will be reduced or increased, we need to recalculate the required surface area A using the new ΔT1 and ΔT2 values for a counterflow heat exchanger.
ΔT1 = 1000°C - 600°C = 400°C
ΔT2 = T_flue,out - T_air,in = 311.36°C - 20°C = 291.36°C
ΔT_lm = (ΔT1 - ΔT2) / ln(ΔT1 / ΔT2)
ΔT_lm = (400°C - 291.36°C) / ln(400°C / 291.36°C)
ΔT_lm ≈ 84.5°C
A = Q / (U × ΔT_lm)
A = 9090 kJ/s / (80 W/m²°C * 84.5°C)
A ≈ 13.5 m²
The required surface area A remains the same for a counterflow heat exchanger, so the size of the heat exchanger does not change.
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force (F) in the wrench above is 15.25 kN applied through a distance of 35 cm along the wrench and the inclined angle (θ) is 60° ? What is the magnitude of the torque relative to the bolt in Joules A J 5337.50 B J 266875 C J 4622.41 D J 533.75
The magnitude of the torque relative to the bolt in Joules is 4622.41J.Torque is a measure of a force's ability to produce rotation around an axis, which can be determined by multiplying the force applied by the distance from the axis of rotation at which it is applied.
As well as the sine of the angle between the force and the lever arm. This formula can be used to calculate torque: τ = F * d * sinθWhere:τ is torque in newton-meters (Nm)F is force in newtons (N)d is the distance from the axis of rotation at which the force is applied in meters (m)θ is the angle between the force vector and the lever arm in degrees (°)Given.
F = 15.25 kN = 15,250 Nd = 35 cm = 0.35 mθ = 60°To convert kN to N, we need to multiply by 1,000:15.25 kN = 15.25 * 1,000 = 15,250 N Then we can plug the values into the formula:τ = F * d * sinθτ = 15,250 N * 0.35 m * sin(60°)τ = 4622.41 J, the magnitude of the torque relative to the bolt in Joules is J 4622.41.
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Build the circuit in Figure 3 in Multisim using the values you have calculated and Measure the Q-point and compare with expected value. Connect and set the generator to a sinusoidal of 3 kHz (small-signal peak to peak voltage of 20 mV). Use 10 μF for the capacitor C. Make sure the capacitor is connected with the correct polarity. Adjust the input amplitude so that none of the waveforms is clipped. Observe and include in your report the following waveforms: - Input voltage v₁, collector voltage vc, emitter voltage VE, and collector-emitter voltage VCE. - Plot all those waveforms on a common time scale using 2 to 3 sinusoidal cycles.
The input voltage v1, collector voltage vc, emitter voltage VE, and collector-emitter voltage VCE waveforms are then observed and plotted on a common time scale using 2 to 3 sinusoidal cycles.
To build the circuit in Figure 3 in Multisim using the values calculated, the following steps can be followed:
Components R1 and R2 are calculated as follows: R1 = Vcc / Icq
= 12 V / 0.0008 A
= 15 kohm and R2 = Vbe / Ib
= 0.7 V / 0.000025 A = 28 kohm.
A resistor with the nearest higher standard value of 30 kohm was used for R2 instead of the calculated value of 28 kohm.
A 10μF capacitor is used for C.
The circuit is then simulated using Multisim software and the values of VCE and IC obtained are measured. These values are then used to calculate the Q-point.
The measured values are compared with the expected values. If there is any significant difference, the circuit may be adjusted or the values of R1 and R2 calculated again to ensure that they are within the tolerances of the resistors used. Once the Q-point is determined, the generator can be connected and set to a sinusoidal of 3 kHz (small-signal peak to peak voltage of 20 mV). The input amplitude is then adjusted so that none of the waveforms is clipped.
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