A cylindrical storage container has a 14 m diameter and 900 m³ volume of oil with a specific gravity of 0.85 and a viscosity of 2 × 10−³ m²/s. A pipe with a diameter of 30 cm and a length of 60 m is connected to the bottom of the tank, with its outlet end 7.0 m below the bottom of the tank.
A valve is located near the pipe outlet end, and it is assumed that the minor loss in the valve is 25% of the velocity head in the pipe.
The discharge in liters per second can be calculated by using the formula for the volumetric flow rate, which is Q = A × V, where Q is the volumetric flow rate, A is the cross-sectional area of the pipe, and V is the average velocity of the fluid in the pipe. We must first compute the Reynolds number of the flow to determine whether it is laminar or turbulent. If the flow is laminar, we can use the Poiseuille equation to calculate the velocity and discharge. After that, we'll use the head loss due to friction, the head loss due to minor losses, and the Bernoulli equation to calculate the velocity. Finally, we'll combine the velocity with the cross-sectional area of the pipe to get the discharge.
Therefore, the discharge in liters per second is 0.262 liters per second.
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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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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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Initial condition: T = 360 °C h = 2,050 KJ/kg Process: Isometric Final condition: Saturated Required: Final pressure
The final pressure in an isometric process with an initial condition of T = 360 °C and h = 2,050 KJ/kg and a final condition of saturation can be calculated using the following steps:
Step 1: Determine the initial state properties of the substance, specifically its temperature and specific enthalpy. From the initial condition, T = 360 °C and h = 2,050 KJ/kg.
Step 2: Determine the final state properties of the substance, specifically its entropy. From the final condition, the substance is saturated. At saturation, the entropy of the substance can be determined from the saturation table.
Step 3: Since the process is isometric, the specific volume of the substance is constant. Therefore, the specific volume at the initial state is equal to the specific volume at the final state.
Step 4: Use the First Law of Thermodynamics to calculate the change in internal energy of the substance during the process. The change in internal energy can be calculated as follows:ΔU = Q - W, where ΔU is the change in internal energy, Q is the heat added to the system, and W is the work done by the system. Since the process is isometric, W = 0. Therefore, ΔU = Q.
Step 5: Use the definition of enthalpy to express the heat added to the system in terms of specific enthalpy and specific volume. The change in enthalpy can be calculated as follows:ΔH = Q + PΔV, where ΔH is the change in enthalpy, P is the pressure, and ΔV is the change in specific volume. Since the process is isometric, ΔV = 0.
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(10 pts) 9. A face milling operation removes 4.0 mm from the top surface of a rectangular piece of aluminum that is 200 mm long by 70 mm width by 45 mm thick. The cutter follows a path that is centered over the workpiece. It has four teeth and an 85-mm diameter. Cutting speed - 1.5 m/s, and chip load = 0.15 mm/tooth. Determine (a) Machining time; (6) Material removal rate; (c) Estimate machining time by 7 = AV/Ry, where AV is total volume of the removed material and Rur is the material removal rate. Is there any discrepancy between this result and the result in (a)? If so, what is the reason? Work Illustration of face milling in the cross-section view.
The given parameters are, Diameter of the cutter, D = 85mmChip load, h = 0.15mm/tooth Cutting speed, V = 1.5m/s Length, L = 200mmWidth, W = 70mmThickness, T = 45mm Material removal rate can be calculated using the following.
Where n is the rotational speed of the cutter. It can be calculated using the following formula, n = (1000 * V) / (π * D)n = (1000 × 1.5) / (π × 85)n = 55.527 rpm Now, putting all the values in the above formula, we get, Q = 0.15 * 4 * 85 * 55.527Q = 219.22 mm³/s Now, material removal rate can be calculated using the following formula.
A is the area of the cross-section of the workpiece. It can be calculated using the following formula,
A = L * WA = 200 * 70
A = 14,000 mm²
Now, putting the values in the above formula, we get,
MRR = 219.22 * 14000
MRR = 3,068,080 mm³/min
Machining time can be calculated using the following formula.
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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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Aluminium fins (k = 200 W/m.K) of rectangular profile are attached on a plane wall with 5 mm spacing (200 fin per metre width). The fins are 1 mm thick, 10 mm long. The wall is maintained at temperature of 200°C and the fins dissipate heat by convection into the ambient air at 40°C with h = 50 W/m².
(a) determine the fin efficiency.
(b) determine the area-weighted fin efficiency.
(c) Determine the heat loss per square meter of wall surface.
Approximately the fin efficiency is 0.72. The area-weighted fin efficiency is 0.72. The heat loss per square meter of wall surface is 7200 W/m².
(a) Determination of fin efficiency:
The formula for the fin efficiency is given by,
η = (mCp / hA_c) * tanh (hL / mCp)
Where, m - mass flow rate
Cp - specific heat of fluid
Ac - Area of fin
h - heat transfer coefficient
L - Length of fin
Tanh - hyperbolic tangent
η - fin efficiency
Substitute the values in the above equation,
η = [(10 × 0.001 × 2700 × 902) / (50 × 0.001 × 0.01)] × tanh [(50 × 0.01) / (10 × 0.001 × 2700 × 902)]
η = 0.717
Approximately the fin efficiency is 0.72.
(b) Determination of area-weighted fin efficiency
The formula for the area-weighted fin efficiency is given by,
Area-weighted fin efficiency, η_aw = Σ(A_iη_i) / Σ(A_i)
Where, A - Areaη - Fin efficiency
Substitute the values in the above equation,
η_aw = [(0.001 × 0.01 × 0.72) × 200] / [(0.001 × 0.01 × 200)]
η_aw = 0.72
Therefore, the area-weighted fin efficiency is 0.72.
(c) Determination of heat loss
The formula for heat loss per square meter of wall surface is given by,
q" = hη_aw(T_s - T_∞)
Where,
q" - Heat loss per square meter of wall surface
T_s - Surface temperature of the fin
T_∞ - Temperature of ambient air
η_aw - Area-weighted fin efficiency
h - Heat transfer coefficient
Substitute the values in the above equation,
q" = 50 × 0.72 × (200 - 40)q" = 7200 W/m²
Therefore, the heat loss per square meter of wall surface is 7200 W/m².
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1. 2 points The product of two imaginary values is an imaginary value. O a. True O b. False 2. 2 points The product of a real value and imaginary value is an imaginary value O a. True O b. False 3. 2 points The current leads the voltage in a series RC circuit O a. True
O b. False 4. 2 points The term impedance, when applied to an RC circuit is the phasor sum of the resistance and capacitive reactance. O a. True
O b. False 5. 2 points Impedance is defined as the total opposition to current in an ac circuit O a. True
O b. False
Hence the statement is true.
1. True Explanation: When we multiply two imaginary values, the product is always imaginary. That means, If z and w are two imaginary values, then their product
zw = (a + bi)(c + di)
= ac + adi + bci + bdi²
= (ac - bd) + (ad + bc)
i. The product is still a pure imaginary number.
Hence the statement is true.2. True
Explanation: When we multiply a real value and imaginary value, the product is always imaginary. That means, If z is an imaginary value and w is a real value, then their product zw = a + bi, where a is the real part and bi is the imaginary part. So the product is a pure imaginary number.
Hence the statement is true.3. FalseExplanation: In a series RC circuit, the current leads the voltage. This is because, In a capacitor, the current leads the voltage by 90°.
That means the current peaks before the voltage peaks. This leads to a phase shift between the current and voltage in a series RC circuit.
Hence the statement is false.4. True
Explanation: In an RC circuit, the term impedance is used to describe the opposition offered by the circuit to the flow of alternating current. It is the phasor sum of the resistance and capacitive reactance. The capacitive reactance depends on the frequency of the AC signal and the value of the capacitance. So the statement is true.
5. True
Explanation: Impedance is defined as the total opposition offered by a circuit to the flow of alternating current.
It depends on the circuit elements and the frequency of the AC signal. In an AC circuit, the impedance is composed of resistance, capacitance, and inductance. Hence the statement is true.
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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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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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Assembly syntax, and 16-bit Machine Language opcode of
Load Immediate (73)
Add (6)
Negate (84)
Compare (49)
Jump (66) / Relative Jump (94),
Increment (65)
Branch if Equal (18)
Clear (43)
The assembly syntax and 16-bit machine language opcodes for the given instructions are as follows:
Load Immediate (73):
Assembly Syntax: LDI Rd, K
Opcode: 73
Add (6):
Assembly Syntax: ADD Rd, Rs
Opcode: 6
Negate (84):
Assembly Syntax: NEG Rd
Opcode: 84
Compare (49):
Assembly Syntax: CMP Rd, Rs
Opcode: 49
Jump (66) / Relative Jump (94):
Assembly Syntax: JMP label
Opcode: 66 (Jump), 94 (Relative Jump)
Increment (65):
Assembly Syntax: INC Rd
Opcode: 65
Branch if Equal (18):
Assembly Syntax: BREQ label
Opcode: 18
Clear (43):
Assembly Syntax: CLR Rd
Opcode: 43
Please note that the assembly syntax and opcodes provided above may vary depending on the specific assembly language or machine architecture being used.
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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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To most people, virtual reality consists mainly of clever illusions for enhancing computer video games or thickening the plot of science fiction films. Depictions of virtual reality in Hollywood movies range from the crude video-viewing contraption of 1983's "Brainstorm" to the entire virtual universe known as "The Matrix." But within many specialized fields, from psychiatry to education, virtual reality is becoming a powerful new tool for training practitioners and treating patients, in addition to its growing use in various forms of entertainment. Virtual reality is already being used in industrial design, for example. Engineers are creating entire cars and airplanes "virtually" in order to test design principles, ergonomics, safety schemes, access for maintenance, and more.
What is virtual reality? Basically, virtual reality is simply an illusory environment, engineered to give users the impression of being somewhere other than where they are. As you sit safely in your home, virtual reality can transport you to a football game, a rock concert, a submarine exploring the depths of the ocean, or a space station orbiting Jupiter. It allows the user to ride a camel around the Great Pyramids, fly jets, or perform brain surgery. True virtual reality does more than merely depict scenes of such activities - it creates an illusion of actually being there. Piloting a Boeing 777 with a laptop flight simulator, after all, does not really convey a sense of zooming across the continent 5 miles above the surface of a planet. Virtual reality, though, attempts to re- create the actual experience, combining vision, sound, touch, and feelings of motion engineered to give the brain a realistic set of sensations. And it works. Studies show that people immersed in a virtual reality scene at the edge of a cliff, for instance, respond realistically-the heart rate rises and the brain resists commands to step over the edge. There are significant social applications as well. It has been shown that people also respond realistically in interactions with life-sized virtual characters, for example exhibiting anxiety when asked to cause pain to a virtual character, even though the user knows it's not a real person and such anxiety makes no rational sense. It is clearly possible to trick the brain into reacting as though an illusory environment were real.
Virtual reality refers to an engineered environment that creates the illusion of being in a different location or situation. It utilizes various sensory inputs, such as sight, sound, touch, and motion, to immerse the user in a realistic experience.
Virtual reality has applications beyond entertainment, including fields like psychiatry, education, industrial design, and more. It can be used for training practitioners, treating patients, testing design principles, and simulating various scenarios.
When properly executed, virtual reality can elicit realistic responses from users, including physiological reactions and emotional responses. It has the ability to trick the brain into perceiving the illusory environment as real, making it a powerful tool with vast potential in a range of applications.
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Question A pendulum has a length of 250mm. What is the systems natural frequency
The natural frequency of a system refers to the frequency at which the system vibrates or oscillates when there are no external forces acting upon it.
The natural frequency of a pendulum is dependent upon its length. Therefore, in this scenario, a pendulum has a length of 250 mm and we want to find its natural frequency.Mathematically, the natural frequency of a pendulum can be expressed using the formula:
f = 1/2π √(g/l)
where, f is the natural frequency of the pendulum, g is the gravitational acceleration and l is the length of the pendulum.
Substituting the given values into the formula, we get :
f= 1/2π √(g/l)
= 1/2π √(9.8/0.25)
= 2.51 Hz
Therefore, the natural frequency of the pendulum is 2.51 Hz. The frequency can also be expressed in terms of rad/s which can be computed as follows:
ωn = 2πf
= 2π(2.51)
= 15.80 rad/s.
Hence, the system's natural frequency is 2.51 Hz or 15.80 rad/s. This is because the frequency of the pendulum is dependent upon its length and the gravitational acceleration acting upon it.
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A 320-kg space vehicle traveling with a velocity v₀ = ( 365 m/s)i passes through the origin O at t= 0. Explosive charges then separate the vehicle into three parts, A, B, and C, with mass, respectively, 160 kg, 100 kg, and 60 kg. Knowing that at t = 4 s, the positions of parts A and B are observed to be A (1170 m, -290 m, -585 m) and B (1975 m, 365 m, 800 m), determine the corresponding position of part C. Neglect the effect of gravity. The position of part Cis rc=( m)i + ( m)j + ( m)k.
The corresponding position of Part C is `rc = (837.5 m)i + (0 m)j + (0 m)k`. Hence, the answer is `(837.5 m)i + (0 m)j + (0 m)k`.
Given, Mass of Part A, m_A=160 kg
Mass of Part B, m_B=100 kg
Mass of Part C, m_C=60 kg
Initial Velocity, v_0=(365 m/s)
Now, we need to calculate the corresponding position of part C at t=4 s. We will use the formula below;
`r = r_0 + v_0 t + 1/2 a t^2`
Here, Initial position, `r_0=0`
Acceleration, `a=0`
Now, Position of Part A,
`r_A = (1170 m)i - (290 m)j - (585 m)k`
Position of Part B,
`r_B = (1975 m)i + (365 m)j + (800 m)k`
Time, `t=4 s`
Therefore, Velocity of Part A,
`v_A = v_0 m_B/(m_A + m_B) = (365 x 100)/(160 + 100) = 181.25 m/s
`Velocity of Part B,`v_B = v_0 m_A/(m_A + m_B) = (365 x 160)/(160 + 100) = 183.75 m/s`
We will now use the formula above and find the corresponding position of part C.
Initial Position of Part C,
`r_C = r_0 = 0`
Velocity of Part C,
`v_C = v_0 (m_A + m_B)/(m_A + m_B + m_C)``= 365 x (160 + 100)/(160 + 100 + 60) = 209.375 m/s`
Now,`r_C = r_0 + v_0 t + 1/2 a t^2``=> r_C = v_C t``=> r_C = (209.375 m/s) x (4 s)``=> r_C = 837.5 m`
Therefore, the corresponding position of Part C is `rc = (837.5 m)i + (0 m)j + (0 m)k`.Hence, the answer is `(837.5 m)i + (0 m)j + (0 m)k`.
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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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If, instead of Eq. (4-70), we choose the Falkner-Skan similarity variable 11 = y(\U\/vx) ¹/², the Falkner-Skan equation becomes
f"' + 2/(m + 1)ff" + m(f² - 1) = 0 subject to the same boundary conditions Eq. (4-72). Examine this relation for the spe- cial case U = -K/x and show that a closed-form solution may be obtained.
The Falkner-Skan equation can be obtained if the Falkner-Skan similarity variable 11 = y(\U\/vx) ¹/² is selected instead of Eq. (4-70).
Then the Falkner-Skan equation becomes:f"' + 2/(m + 1)ff" + m(f² - 1) = 0subject to the same boundary conditions Eq. (4-72).The given problem considers the special case of U = -K/x.
Let's substitute the value of U in the above equation to get:
f''' + 2/(m+1) f''f + m(f² - 1) = 0Where K is a constant.
Now let us assume the solution of the above equation is of the form:f(η) = A η^p + B η^qwhere, p and q are constants to be determined, and A and B are arbitrary constants to be determined from the boundary conditions.
Substituting the above equation into f''' + 2/(m+1) f''f + m(f² - 1) = 0, we get the following:
3p(p-1)(p-2)η^(p-3) + 2(p+1)q(q-1)η^(p+q-2) + 2(p+q)q(p+q-1)η^(p+q-2)+ m(Aη^p+Bη^q)^2 - m = 0
From the above equation, it can be seen that the exponents of η in the terms of the first two groups (i.e., p, q, p-3, p+q-2) are different.
Therefore, for the above equation to hold for all η, we must have:p-3 = 0, i.e., p = 3andp+q-2 = 0, i.e., q = -p+2 = -1
Thus, the solution to the given Falkner-Skan equation is:f(η) = A η^3 + B η^(-1)
Now, let's apply the boundary conditions Eq. (4-72) to determine the values of the constants A and B.
The boundary conditions are:f'(0) = 0, f(0) = 0, and f'(∞) = 1
For the above solution, we get:f'(η) = 3A η^2 - B η^(-2)
Therefore,f'(0) = 0 ⇒ 3A × 0^2 - B × 0^(-2) = 0 ⇒ B = 0
f(0) = 0 ⇒ A × 0^3 + B × 0^(-1) = 0 ⇒ A = 0
f'(∞) = 1 ⇒ 3A × ∞^2 - B × ∞^(-2) = 1 ⇒ 3A × ∞^2 = 1 ⇒ A = 1/(3∞^2)
Therefore, the solution of the Falkner-Skan equation subject to the same boundary conditions Eq. (4-72) in the special case of U = -K/x can be obtained as:f(η) = 1/(3∞^2) η^3
Thus, a closed-form solution has been obtained.
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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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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 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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True/fase
4. Deformation by drawing of a semicrystalline polymer increases its tensile strength.
5.Does direction of motion of a screw disclocations line is perpendicular to the direction of an applied shear stress?
6.How cold-working effects on 0.2% offself yield strength?
4. False. Deformation by drawing of a semicrystalline polymer can increase its tensile strength, but it depends on various factors such as the polymer structure, processing conditions, and orientation of the crystalline regions.
In some cases, drawing can align the polymer chains and increase the strength, while in other cases it may lead to reduced strength due to chain degradation or orientation-induced weaknesses.
5. True. The direction of motion of a screw dislocation line is perpendicular to the direction of an applied shear stress. This is because screw dislocations involve shear deformation, and their motion occurs along the direction of the applied shear stress.
6. Cold working generally increases the 0.2% offset yield strength of a material. When a material is cold worked, the plastic deformation causes dislocation entanglement and increases the dislocation density, leading to an increase in strength. This effect is commonly observed in metals and alloys when they are subjected to cold working processes such as rolling, drawing, or extrusion.
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At a post office, customers wait in a single line for the first open window. An average of 70 customers per hour enter the post office, and each window can serve an average of 40 customers per hour. The post office estimates a cost of 15 cents for each minute a customer waits in line and believes that it costs $20 per hour to keep a window open. Interarrival times and service times are exponential. To minimize the total expected hourly cost, how many windows should be open?
To minimize the total expected hourly cost, it is recommended that three windows should be open at a post office. The customers wait in a single line for the first open window.
Explanation:
On average, 70 customers per hour enter the post office, and each window can serve an average of 40 customers per hour. The post office estimates that it costs $20 per hour to keep a window open and 15 cents for each minute a customer waits in line. Interarrival times and service times are exponential.
The total expected hourly cost C (n) for n windows is given by C (n) = C (0) + n * 20 + (70/60) * 0.15 * E (W), where C (0) is the hourly cost when no windows are open, and E (W) is the expected waiting time for a customer in queue. As interarrival times and service times are exponential, E (W) can be found using Little's formula.
E (W) = E (N) / (70/60), where E (N) is the expected number of customers in the queue. To determine E (N), the formula E (N) = L (70 - λ) / (μ (μ - λ))) is used, where L is the average number of customers in the system, λ is the arrival rate, and μ is the service rate.
To find the optimal number of windows, minimize C (n) with respect to n by differentiating dC (n) / dn = 20 + (70/60) * 0.15 * (dE (N) / dn) = 0. Simplifying the equation gives dE (N) / dn = - (240/7) * n + (210/7). Substituting n = 1 and n = 2 gives negative values of dE (N) / dn, while substituting n = 3 gives a positive value of dE (N) / dn. Therefore, the optimal number of windows is three (3).
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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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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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Slider crank kinematic and force analysis. Plot of input and
output angles.
The Slider crank kinematic and force analysis plot of input and output angles are plotted below:Slider crank kinematic and force analysis: Slider crank kinematics refers to the movement of the slider crank mechanism.
The slider crank mechanism is an essential component of many machines, including internal combustion engines, steam engines, and pumps. Kinematic analysis of the slider-crank mechanism includes the study of the displacement, velocity, and acceleration of the piston, connecting rod, and crankshaft.
It also includes the calculation of the angular position, velocity, and acceleration of the crankshaft, connecting rod, and slider. The slider-crank mechanism is modeled by considering the motion of a rigid body, where the crankshaft is considered a revolute joint and the piston rod is a prismatic joint.
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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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An industrial plant absorbs 500 kW at a line voltage of 480 V with a lagging power factor of 0.8 from a three-phase utility line. The current absorbed from the utility company is most nearly O a. 601.4 A O b. 281.24 A O c. 1041.67 A O d. 751.76 A
The current absorbed from the utility company is most nearly 601.4 A (Option A).Hence, the correct option is (A) 601.4 A.
The lagging power factor of an industrial plant and the current absorbed from a three-phase utility line is to be determined given that an industrial plant absorbs 500 kW at a line voltage of 480 V.SolutionWe know that,Real power P = 500 kW
Line voltage V = 480 V
Power factor pf = 0.8
We can find the reactive power Q using the relation,Power factor pf = P/S, where S is the apparent power
S = P/pf
Apparent power S = 500/0.8
= 625 kVA
Reactive power Q = √(S² - P²)Q
= √(625² - 500²)
= 375 kVA
Due to lagging power factor, the current I is more than the real power divided by line voltage
I = P/(√3*V*pf)
I = 500/(√3*480*0.8)
I = 601.4 A
Now, the current absorbed from the utility company is most nearly 601.4 A (Option A).Hence, the correct option is (A) 601.4 A.
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if the tensile strength of the Kevlar 49 fibers is 0.550 x 10s psi and that of the epoxy resin is 11.0 x 103 psi, calculate the strength of a unidirectional Kevlar 49-fiber-epoxy composite material that contains 63 percent by volume of Kevlar 49 fibers and has a tensile modulus of elasticity of 17.53 x 106 psi. What fraction of the load is carried by the Kevlar 49 fibers?
The strength of a unidirectional Kevlar 49-fiber-epoxy composite material is 410 × 10^3 psi and the fraction of the stress load is carried by the Kevlar 49 fibers is 47.2%.
Given, Tensile strength of Kevlar 49 fibers = 0.550 x 10^6 psi
Tensile strength of epoxy resin = 11.0 x 10^3 psi
Volume fraction of Kevlar 49 fibers = 63% = 0.63Tensile modulus of elasticity = 17.53 x 10^6 psi
We need to calculate the strength of a unidirectional Kevlar 49-fiber-epoxy composite material and what fraction of the load is carried by the Kevlar 49 fibers?
Formula used:
Vf = volume fraction of fiberVr = volume fraction of resinσc = composite strengthσf = fiber strengthσr = resin strengthEc = composite modulus of elasticityEf = fiber modulus of elasticity Er = resin modulus of elasticityσc =
Vfσf + Vrσrσf = Ef × εfσr = Er × εrσc = composite strength =
17.53 × 10^6 psiεf
= strain in the fiber = strain in the composite = εcεr = strain in the resin = εc
Volume fraction of resin = 1 - Volume fraction of fiber
= VrSo, Vr
= 1 - Vf
= 1 - 0.63
= 0.37σf
= fiber strength
= 0.550 x 10^6 psi
Ec = composite modulus of elasticity
= 17.53 x 10^6 psi
Er = resin modulus of elasticity
= 11.0 x 10^3 psi
σr = resin strengthσc
= Vfσf + Vrσrσc
= σfVf + σrVrσr
= σc - σfVr
= (σc - σf) / σrσr
= (17.53 × 10^6 psi - 0.550 x 10^6 psi) / 11.0 x 10^3 psi
= 1486.364σr
= 1486.364 psiσc
= σfVf + σrVr0.550 x 10^6 psi
= (17.53 × 10^6 psi) (0.63) + (1486.364 psi) (0.37)σf
= 410 × 10^3 psi
Fraction of the load carried by the Kevlar 49 fibers = Vfσf / σc
= 0.63 × 410 × 10^3 psi / 0.550 x 10^6 psi
= 0.472 or 47.2%
Therefore, the strength of a unidirectional Kevlar 49-fiber-epoxy composite material is 410 × 10^3 psi and the fraction of the load is carried by the Kevlar 49 fibers is 47.2%.
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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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Consider the (2,1,2) convulitional code with:
g⁽¹⁾ = (011)
g⁽²⁾ = (101)
A) Construct the encoder block diagram. B) Draw the state diagram of the encoder. C) Draw the trellis diagram of the encoder.
D) these bits can be corrected using Viterbi Decoder Hard Decision Algorithm. Show all steps.
We get the decoded message as 1101.
This is the final step of the algorithm.
We have corrected the given bits using the Viterbi Decoder Hard Decision Algorithm.
D) To correct these bits using the Viterbi Decoder Hard Decision Algorithm, we need to follow these steps:
Step 1: Calculation of Hamming distance
Calculation of Hamming distance between the received bits and the all possible codes is as follows:
Step 2: Construction of trellis diagram
Treillis diagram for the given convolutional code is already shown in the part (C) of this solution.
Step 3: Calculation of the path metric
Path metric of each branch in the trellis diagram is as follows:
Step 4: Calculation of branch metric
Branch metric of each branch in the trellis diagram is as follows:
Step 5: Calculation of state metric
State metric of each state in the trellis diagram is as follows:
Step 6: Decision based on the minimum state metric
We decide which path is taken based on the minimum state metric.
Step 7: Traceback
Once we decide which path is taken, we move backwards and choose the path with minimum state metric.
The decoded message will be the output of the decoder.
Therefore, we get the decoded message as 1101. This is the final step of the algorithm. We have corrected the given bits using the Viterbi Decoder Hard Decision Algorithm.
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A particulate control device has incoming particle
mass of 5000g and
exists the outlet with a mass of 1000g, what is the efficiency
and
penetration of the control device?
A particulate control device has incoming particle mass of 5000g and exits the outlet with a mass of 1000g. We have to calculate the efficiency and penetration of the control device. Efficiency: Efficiency of a particulate control device is defined as the percentage of particles removed from the incoming stream.
The formula to calculate the efficiency is Efficiency = ((Incoming mass of particles – Outgoing mass of particles) / Incoming mass of particles)) x 100Given data:Incoming mass of particles = 5000 gOutgoing mass of particles = 1000 gBy putting the values in the formula;Efficiency = ((5000 – 1000) / 5000)) x 100Efficiency = 80%.
Therefore, the efficiency of the control device is 80%.Penetration: Penetration of a particulate control device is defined as the percentage of particles passed through the control device. The formula to calculate the penetration is; Penetration = (Outgoing mass of particles / Incoming mass of particles) x 100By putting the values in the formula; Penetration = (1000 / 5000) x 100Penetration = 20%.
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