The energy equation for an adiabatic turbine may be written as:
P1 + (v1^2/2g) + h1 = P2 + (v2^2/2g) + h2
Where:P1 = inlet pressurev1 = inlet velocity
h1 = inlet specific enthalpy
P2 = outlet pressure
v2 = outlet velocity
h2 = outlet specific enthalpy
g = acceleration due to gravity, 9.81 m/s^2
From the provided data, inlet conditions are:P1 = 85 m/s and h1 = 3450 + 8 = 3458 kJ/kgOutlet conditions are:P2 = 190 m/s and h2 = 2630 + 6 = 2636 kJ/kgHere, N = 8 and M = 6.
Power generated per 1 kg of the steam it the process is adiabatic:Adiabatic turbine is a type of turbine that is completely insulated, such that no heat is transferred to or from the surroundings, allowing for the conversion of work into heat and vice versa. Thus, there will be no heat loss or gain to the surroundings during the adiabatic process.Power generated per 1 kg of steam during an adiabatic process is given by:P = m * ((h1 - h2) - (v2^2 - v1^2)/(2*g))where m = 1kgUsing the formula:P = m * ((h1 - h2) - (v2^2 - v1^2)/(2*g))= 1 * ((3458 - 2636) - (190^2 - 85^2)/(2*9.81))= 1067.8 kJ/kg
Therefore, the power generated per 1 kg of steam it the process is adiabatic is 1067.8 kJ/kg.
The energy equation for an adiabatic turbine may be written as:P1 + (v1^2/2g) + h1 = P2 + (v2^2/2g) + h2Power generated per 1 kg of steam during an adiabatic process is given by:P = m * ((h1 - h2) - (v2^2 - v1^2)/(2*g))= 1 * ((3458 - 2636) - (190^2 - 85^2)/(2*9.81))= 1067.8 kJ/kgThe power generated per 1 kg of steam it the process is adiabatic is 1067.8 kJ/kg.
The power generated per 1 kg of the steam it the process is adiabatic is 1067.8 kJ/kg, while considering the heat lost to surroundings as 10kJ per 1kg of steam flowing through the turbine.
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The average flow speed in a constant-diameter section of the pipeline is 2.5 m/s. At the inlet, the pressure is 2000 kPa (gage) and the elevation is 56 m; at the outlet, the elevation is 35 m. Calculate the pressure at the outlet (kPa, gage) if the head loss = 2 m. The specific weight of the flowing fluid is 10000N/m³. Patm = 100 kPa.
The pressure at the outlet (kPa, gage) can be calculated using the following formula:
Pressure at the outlet (gage) = Pressure at the inlet (gage) - Head loss - Density x g x Height loss.
The specific weight (γ) of the flowing fluid is given as 10000N/m³.The height difference between the inlet and outlet is 56 m - 35 m = 21 m.
The head loss is given as 2 m.Given that the average flow speed in a constant-diameter section of the pipeline is 2.5 m/s.Given that Patm = 100 kPa.At the inlet, the pressure is 2000 kPa (gage).
Using Bernoulli's equation, we can find the pressure at the outlet, which is given as:P = pressure at outlet (gage), ρ = specific weight of the fluid, h = head loss, g = acceleration due to gravity, and z = elevation of outlet - elevation of inlet.
Therefore, using the above formula; we get:
Pressure at outlet = 2000 - (10000 x 9.81 x 2) - (10000 x 9.81 x 21)
Pressure at outlet = -140810 PaTherefore, the pressure at the outlet (kPa, gage) is 185.19 kPa (approximately)
In this question, we are given the average flow speed in a constant-diameter section of the pipeline, which is 2.5 m/s. The pressure and elevation are given at the inlet and outlet. We are supposed to find the pressure at the outlet (kPa, gage) if the head loss = 2 m.
The specific weight of the flowing fluid is 10000N/m³, and
Patm = 100 kPa.
To find the pressure at the outlet, we use the formula:
P = pressure at outlet (gage), ρ = specific weight of the fluid, h = head loss, g = acceleration due to gravity, and z = elevation of outlet - elevation of inlet.
The specific weight (γ) of the flowing fluid is given as 10000N/m³.
The height difference between the inlet and outlet is 56 m - 35 m = 21 m.
The head loss is given as 2 m
.Using the above formula; we get:
Pressure at outlet = 2000 - (10000 x 9.81 x 2) - (10000 x 9.81 x 21)
Pressure at outlet = -140810 PaTherefore, the pressure at the outlet (kPa, gage) is 185.19 kPa (approximately).
The pressure at the outlet (kPa, gage) is found to be 185.19 kPa (approximately) if the head loss = 2 m. The specific weight of the flowing fluid is 10000N/m³, and Patm = 100 kPa.
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Q.7. For each of the following baseband signals: i) m(t) = 2 cos(1000t) + cos(2000); ii) m(t) = cos(10000) cos(10,000+): a) Sketch the spectrum of the given m(t). b) Sketch the spectrum of the amplitude modulated waveform s(t) = m(t) cos(10,000t). c) Repeat (b) for the DSB-SC signal s(t). d) Identify all frequencies of each component in (a), (b), and (c). e) For each S(f), determine the total power Pr, single sideband power Pss, power efficiency 7, modulation index u, and modulation percentage.
a) For this spectrum, the frequencies of the two signals are:
f1= 1000 Hz, and f2 = 2000 Hz
b) The frequencies of the signals in this case are:
fc= 10,000 Hz, f1=9,000 Hz, and f2= 12,000 Hz
c) The frequencies of the signals in this case are:
fc= 10,000 Hz, f1= 1000 Hz, and f2 = 2000 Hz
d) For the DSB-SC wave the frequencies are:
f1= 1000 Hz and f2 = 2000 Hz
e) Modulation Percentage= 100%
(a) Sketch the spectrum of the given m(t)For the first signal,
m(t) = 2 cos(1000t) + cos(2000),
the spectrum can be represented as follows:
Sketch of spectrum of the given m(t)
For this spectrum, the frequencies of the two signals are:
f1= 1000 Hz, and f2 = 2000 Hz
(b) Sketch the spectrum of the amplitude modulated waveform
s(t) = m(t) cos(10,000t)
Sketch of spectrum of the amplitude modulated waveform
s(t) = m(t) cos(10,000t)
The frequencies of the signals in this case are:
fc= 10,000 Hz,
f1= 10,000 - 1000 = 9,000 Hz, and
f2 = 10,000 + 2000 = 12,000 Hz
(c) Repeat (b) for the DSB-SC signal s(t)Sketch of spectrum of the DSB-SC signal s(t)
The frequencies of the signals in this case are:
fc= 10,000 Hz,
f1= 1000 Hz, and
f2 = 2000 Hz
(d) Identify all frequencies of each component in (a), (b), and (c)
Given that the frequencies of the components are:
f1= 1000 Hz,
f2 = 2000 Hz,
fc = 10,000 Hz.
For the Amplitude Modulated wave the frequencies are:
f1= 9000 Hz and f2 = 12000 Hz
For the DSB-SC wave the frequencies are:
f1= 1000 Hz and f2 = 2000 Hz
(e) For each S(f), determine the total power Pr, single sideband power Pss, power efficiency 7, modulation index u, and modulation percentage.
Using the formula for total power,
PT=0.5 * (Ac + Am)^2/ R
For the first signal,
Ac = Am = 1 V,
and
R = 1 Ω, then PT = 1 W
For the amplitude modulated signal:
Total power Pr = PT = 2 W
Single sideband power Pss = 0.5 W
Power efficiency η = Pss/PT = 0.25
Modulation Index, μ = Ac/Am = 1
Modulation Percentage = μ*100 = 100%
For the DSB-SC signal, Pss = PT/2 = 1 WPt = 2 W
Power efficiency η = Pss/PT = 0.5
Modulation Index, μ = Ac/Am = 1
Modulation Percentage = μ*100 = 100%
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Briefly describe 3 sources or reasons for needing nonlinear simulation. Provide an example of each. Why do these simulations take longer to run than linear simulation?
Nonlinear simulations are necessary when dealing with large deformations or displacements, nonlinear material properties, or complex contact interactions.
Large deformations or displacements change the geometry significantly during deformation, invalidating the assumption of small displacements in linear analyses. For example, analyzing the large bending of a cantilever beam under a heavy load would require nonlinear simulation. Nonlinear material properties refer to materials that do not obey Hooke's Law, such as rubber, which stretches non-linearly with load. Complex contact interactions, such as multiple bodies in contact, may also require nonlinear analysis, for example, the engagement and disengagement of gear teeth in a gearbox. Nonlinear simulations take longer to run because they often require iterative solution methods, which necessitate repeated calculation until the solution converges to a set limit, thereby consuming more computational resources and time.
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A Direct Numerical Simulation of a high Reynolds number is performed with a wall time of 29 seconds required to compute each time step. The prohibitive computational cost forces the user to redesign the grid so that LES can be employed instead. Given that 80% of the computing time in a DNS is typically used to resolve the smallest 20% of scales of motion, calculate the wall time required to compute each time step on a wall designed LES grid.
To calculate the wall time required to compute each time step on a wall-designed LES grid, we need to consider the reduction in computational cost achieved by switching from DNS (Direct Numerical Simulation) to LES (Large Eddy Simulation).
The given information states that in DNS, the wall time required to compute each time step is 29 seconds. Additionally, it mentions that 80% of the computing time in DNS is typically used to resolve the smallest 20% of scales of motion.
In LES, the computational cost is significantly reduced by modeling the smallest scales of motion instead of resolving them explicitly. The LES grid is designed to capture the larger-scale turbulent motions while modeling the smaller, unresolved scales.
Let's calculate the reduction in computational cost achieved by LES compared to DNS:
Computational Cost Reduction = 80% (portion of computing time used in DNS to resolve smallest scales) / 20% (portion of scales modeled in LES) = 4
This means that the computational cost of LES is four times lower than DNS for each time step.
To calculate the wall time required for each time step on the LES grid:
Wall Time (LES) = Wall Time (DNS) / Computational Cost Reduction
Wall Time (LES) = 29 seconds / 4
Wall Time (LES) = 7.25 seconds
Therefore, the wall time required to compute each time step on a wall-designed LES grid is approximately 7.25 seconds.
By switching from DNS to LES, the computational cost is reduced by a factor of four, resulting in a significantly shorter wall time required for each time step. This reduction in computational cost allows for a more feasible and practical simulation of high Reynolds number flows.
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Lyapunov Stability. For the following system: *1 =-2 +23 12 = -21 +2 (a) Find all equilibrium points. (b) Evaluate the stability of each equilibrium point using Lyapunov's indirect method.
Answer:(a) Equilibrium points: (x,y) = (2,2), (0,0)
Answer (b) Stability of equilibrium points :Equilibrium point at (2,2): unstable
Equilibrium point at (0,0): stable
Given system is:1 = -2 + 2x3 - x2 2 = -2x1 + 2y
Solution (a) To find all the equilibrium points, we need to solve for x and y, such that dx/dt and dy/dt becomes zero. In other words, we need to find (x, y) such that f(x,y) = 0, where f(x,y) = [dx/dt, dy/dt]
From the given system, we can say, dx/dt = -2 + 2x3 - x2
dy/dt = -2x1 + 2y
We need to solve for dx/dt = 0 and dy/dt = 0 => x2 - 2x3 = 2=> x2/2 - x3 = 1... equation (1)
And, -2x1 + 2y = 0 => x1 = y
We can substitute x1 with y, to get 2y - 2y = 0 => 0 = 0... equation (2)
From equation (1), we have: x2/2 = x3 + 1 => x2 = 2(x3 + 1) => x2 = 2x3 + 2
We can substitute x2 and x1 with the above relations, in the original system :dx/dt = -2 + 2x(2x3 + 2) - (2x3 + 1) => dx/dt = -4x3 - 2dy/dt = -2y + 2y = 0
So, equilibrium points are at: (x,y) = (2,2), (0,0)
(b) Lyapunov's Indirect method tells us to check the nature of eigenvalues of the jacobian matrix at the equilibrium point. The stability is dependent on the nature of the eigenvalues.
Jacobian Matrix is:J(x,y) = [[df/dx, df/dy], [dg/dx, dg/dy]]
where f(x,y) and g(x,y) are the two equations of the system.
Here, f(x,y) = dx/dt and g(x,y) = dy/dt
So, we have: J(x,y) = [[-2x2 + 6, 2], [-2, 2]]
(i) Equilibrium point at (2,2):J(2,2) = [[2, 2], [-2, 2]]
Characteristics equation: |J - λI| = (2-λ)(2-λ) - 2(-2) => λ2 - 4λ + 6 = 0 => λ = 2 ± i√2
Since both eigenvalues have non-zero real part, the equilibrium point is unstable
(ii) Equilibrium point at (0,0):J(0,0) = [[-2, 2], [-2, 2]]
Characteristics equation: |J - λI| = (-2-λ)(2-λ) - 2(-2) => λ2 + 2λ = 0 => λ = -2, 0
Since both eigenvalues have negative or zero real part, the equilibrium point is stable.
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A jet of water issues out from a fire hydrant nozzle fitted at a height of 3 m from the ground at an angle of 45° with the horizontal. If the jet under a particular flow condition strikes the ground at a horizontal distance of 15 m from the nozzle, Find the jet velocity, in m/s. Determine the maximum height the jet can reach above the nozzle, in meters. How far horizontally is the location of the maximum height from the nozzle, in meters?
Given data, Height of the nozzle from the ground (h) = 3 m Angle made by the jet with the horizontal (θ) = 45°Horizontal distance of the jet from the nozzle (d) = 15 m We need to find.
Velocity of the jet Maximum height the jet can reach above the nozzle Distance of the maximum height from the nozzle Using the principle of conservation of energy, we can calculate the velocity of the jet as follows, Initial energy = Kinetic energy + Potential energy0 = 1/2 mv² + mg h Where, v = Velocity of the jet h = Height of the nozzle above the ground.
Acceleration due to gravity We can express the initial energy in terms of the kinetic energy only since the jet leaves the nozzle horizontally, so there is no initial potential energy. Thus, we get. Initial energy = Kinetic energy1/2 m0² = 1/2 mv²v = √(2gh) Putting the given values in the above equation.
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Metal sheets are to be flanged on a pneumatically operated bending tool. After clamping the component by means of a single acting cylinder (A), it is bent over by a double acting cylinder (B), and subsequently finish bent by another double acting cylinder (C). The operation is to be initiated by a push-button. The circuit is designed such that one working cycle is completed each time the start signal is given.
In this setup, metal sheets are flanged using a pneumatically operated bending tool.
The process involves clamping the component using a single-acting cylinder (A), followed by bending over using a double-acting cylinder (B), and finally finish bending using another double-acting cylinder (C). A push-button initiates the operation, and each cycle completes when the start signal is given. The single-acting cylinder (A) is responsible for clamping the metal sheet in place, providing stability during the bending process. The double-acting cylinder (B) is then activated to bend the metal sheet over, shaping it according to the desired angle or curvature. Finally, the second double-acting cylinder (C) performs the finish bending to achieve the desired form. This circuit design ensures that each working cycle starts when the push-button is pressed, allowing for efficient and controlled flanging of metal sheets.
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Problem 3. A specimen of a 4340 steel alloy having a plane strain fracture toughness of 45 MPam is exposed to a stress of 1000 MPa. Will this specimen experience fracture if it is known that the largest surface crack is 0.75 mm long? Why or why not? Assume that the parameter Y has a value of 1.0. Solution
To determine whether the specimen will experience fracture, we can use the fracture mechanics concept and the stress intensity factor (K) equation.Please provide the calculation for the stress intensity factor (K) so that we can determine whether the specimen will experience fracture or not.
Plane strain fracture toughness (K_IC): 45 MPam
Applied stress (σ): 1000 MPa
Largest surface crack length (a): 0.75 mm
Parameter (Y): 1.0
The stress intensity factor (K) can be calculated using the equation:
K = Y * σ * √(π * a)
Substituting the given values into the equation:
K = 1.0 * 1000 MPa * √(π * 0.75 mm)
Now, we need to compare the calculated value of K with the plane strain fracture toughness (K_IC) to determine whether fracture will occur. If K is greater than or equal to K_IC, fracture will occur. If K is less than K_IC, fracture will not occur.
If the calculated value of K is greater than or equal to 45 MPam, then the specimen will experience fracture. If the calculated value of K is less than 45 MPam, the specimen will not experience fracture.
To determine the result, we need to perform the calculation for the stress intensity factor (K) and compare it with the given plane strain fracture toughness (K_IC). Unfortunately, the specific calculation of K is missing from the information provided. Please provide the calculation for the stress intensity factor (K) so that we can determine whether the specimen will experience fracture or not.
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MFL1601 ASSESSMENT 3 QUESTION 1 [10 MARKSI Figure 21 shows a 10 m diameter spherical balloon filled with air that is at a temperature of 30 °C and absolute pressure of 108 kPa. Determine the weight of the air contained in the balloon. Take the sphere volume as V = nr. Figure Q1: Schematic of spherical balloon filled with air
Figure 21 shows a 10m diameter spherical balloon filled with air that is at a temperature of 30°C and absolute pressure of 108 kPa. The task is to determine the weight of the air contained in the balloon. The sphere volume is taken as V = nr.
The weight of the air contained in the balloon can be calculated by using the formula:
W = mg
Where W = weight of the air in the balloon, m = mass of the air in the balloon and g = acceleration due to gravity.
The mass of the air in the balloon can be calculated using the ideal gas law formula:
PV = nRT
Where P = absolute pressure, V = volume, n = number of moles of air, R = gas constant, and T = absolute temperature.
To get n, divide the mass by the molecular mass of air, M.
n = m/M
Rearranging the ideal gas law formula to solve for m, we have:
m = (PV)/(RT) * M
Substituting the given values, we have:
V = (4/3) * pi * (5)^3 = 524.0 m³
P = 108 kPa
T = 30 + 273.15 = 303.15 K
R = 8.314 J/mol.K
M = 28.97 g/mol
m = (108000 Pa * 524.0 m³)/(8.314 J/mol.K * 303.15 K) * 28.97 g/mol
m = 555.12 kg
To find the weight of the air contained in the balloon, we multiply the mass by the acceleration due to gravity.
g = 9.81 m/s²
W = mg
W = 555.12 kg * 9.81 m/s²
W = 5442.02 N
Therefore, the weight of the air contained in the balloon is 5442.02 N.
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Design a four-bar mechanism such that the length of the fixed link ( r1) is 50 mm, the length of the rocker (r₄) is 40 mm, the rocking angle (β) is 60°, and the time ratio (λ) is 1.2
The length of the fixed link (r1) is 50 mm, the length of the coupler (r2) is 20 mm, the length of the output link (r3) is 60 mm, and the length of the rocker (r4) is 40 mm.
A four-bar mechanism can be designed based on certain specifications and requirements. Given specifications include the length of the fixed link ( r1) is 50 mm, the length of the rocker (r₄) is 40 mm, the rocking angle (β) is 60°, and the time ratio (λ) is 1.2.
Following is the step-by-step solution for designing a four-bar mechanism:
Step 1: Draw a rough sketch of the four-bar mechanism with given measurements
Step 2: Determine the length of the coupler (r2) using cosine law
cos(α )=(r2^2+r1^2-r4^2)/(2*r1*r2)
cos(α) = (r2² + r1² - r4²)/(2*r1*r2)
cos(60°) = (r2² + 50² - 40²)/(2*50*r2) 0.5
= (r2² + 2500 - 1600)/(100*r2)r2² + 900
= 50r2 r2² - 50r2 + 900
= 0 (r2 - 30)(r2 - 20)
= 0
Hence, r2 = 20 mm or 30 mm.
Step 3: Calculate the angle between the coupler and rocker (γ) using sin law
sin(γ )=(r4*sinβ)/r2
sin(γ) = (r4*sinβ)/r2
sin(γ) = (40*sin60°)/20
sin(γ) = 0.866
Hence, γ = sin⁻¹(0.866)
= 60.24°
Step 4: Calculate the length of the output link (r3) using cosine law
cos(α )=(r3^2+r2^2-r4^2)/(2*r2*r3)
cos(α) = (r3² + r2² - r4²)/(2*r2*r3)
cos(α) = (r3² + 20² - 40²)/(2*20*r3)
cos(α) = (r3² - 1200)/(40r3)
cos(α)*40r3 = r3² - 1200 40r3
= r3² - 1200 r3² - 40r3 - 1200 = 0
(r3 - 60)(r3 + 20) = 0
r3 = 60 mm or -20 mm.
Since length can not be negative so, the value of r3 = 60 mm.
Therefore, the length of the fixed link (r1) is 50 mm, the length of the coupler (r2) is 20 mm, the length of the output link (r3) is 60 mm, and the length of the rocker (r4) is 40 mm.
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New Product Development Process
You have been tasked by a manufacturing company that manufactures all sorts of luggage and travel equipment, to develop a new product that helps travelers and will enable the company make profits on the long run.
You need to develop a product for travelers going through the seven stages of new product development. You can use the below table as a reference:
(2%) Idea Generation:
What is the idea, concept, or theme that you came up with as a team and what is the reason behind your choice?
How does your idea help the customers and ensure that the company will make profits on the long run?
You will need to conduct some research on the current needs of travelers and the behaviors of travelers to support your
Our idea: Smart luggage tracking system to prevent lost luggage, benefiting customers with real-time tracking and ensuring long-term profitability by meeting travel industry demands.
As a team, we came up with the idea of developing a smart luggage tracking system. The reason behind this choice is to address the common problem faced by travelers of lost or mishandled luggage.
Our smart luggage tracking system will help customers by providing real-time location tracking of their luggage through a mobile application. It will also have additional features such as weight monitoring, security alerts, and personalized travel recommendations.
This idea helps customers by giving them peace of mind and saving them from the inconvenience and stress of lost luggage. It ensures long-term profitability for the company by tapping into the growing travel industry and meeting the increasing demand for smart and innovative travel solutions. Our research on the current needs of travelers and their behaviors indicates a strong market potential for such a product, with a high willingness to pay for enhanced luggage tracking and security features.
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A 2000 CC four stroke CI engine intakes air at a rate (mass flow rate) of 0.0401 Kg/s, pressure of 1.5 bar and temperature of 42°C. Calculate the volumetric efficiency of that engine.
The volumetric efficiency of the engine, we need to determine the actual air mass flow rate (m_actual) and compare it to the theoretical air mass flow rate (m_theoretical) that would be achieved if the engine operated under ideal conditions.
Engine displacement (Vd) = 2000 cc = 0.002 m³
Mass flow rate (m) = 0.0401 kg/s
Pressure (P) = 1.5 bar = 150 kPa
Temperature (T) = 42°C = 42 + 273.15 = 315.15 K
calculate the density of air(ρ) using ideal gas law:
ρ = P / (R * T)
R is the specific gas constant for air (287.1 J/(kg·K)).
ρ = 150 kPa/(287.1J/(kg·K)*315.15 K)≈ 1.591kg/m³
Next, we can calculate the theoretical air mass flow rate using the equation:
m_theoretical = ρ * Vd * N / (60 * A)
where N is engine speed in revolutions per minute, A is cross-sectional area of the intake port.
Since the engine speed (N) is not provided in the question, we cannot calculate the theoretical mass flow rate and therefore cannot directly determine the volumetric efficiency.
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The data from a series of flow experiments is given to you for analysis. Air is flowing at a velocity of
2.53 m/s and a temperature of 275K over an isothermal plate at 325K. If the transition from laminar to
turbulent flow is determined to happen at the end of the plate, please illuminate the following:
A. What is the length of the plate?
B. What are the hydrodynamic and thermal boundary layer thicknesses at the end of the plate?
C. What is the heat rate per plate width for the entire plate?
For parts D & E, the plate length you determined in part A above is increased by 42%. At the end of
the extended plate what would be the
D. Reynolds number?
E. Hydrodynamic and thermal boundary laver thicknesses?
Using the concepts of boundary layer theory and the Reynolds number. The boundary layer is a thin layer of fluid near the surface of an object where the flow velocity and temperature gradients are significant. The Reynolds number (Re) is a dimensionless parameter that helps determine whether the flow is laminar or turbulent. The transition from laminar to turbulent flow typically occurs at a critical Reynolds number.
A. Length of the plate:
To determine the length of the plate, we need to find the location where the flow transitions from laminar to turbulent.
Given:
Air velocity (V) = 2.53 m/s
Temperature of air (T) = 275 K
Temperature of the plate (T_pl) = 325 K
Assuming the flow is fully developed and steady-state:
Re = (ρ * V * L) / μ
Where:
ρ = Density of air
μ = Dynamic viscosity of air
L = Length of the plate
Assuming standard atmospheric conditions, ρ is approximately 1.225 kg/m³, and the μ is approximately 1.79 × 10^(-5) kg/(m·s).
Substituting:
5 × 10^5 = (1.225 * 2.53 * L) / (1.79 × 10^(-5))
L = (5 × 10^5 * 1.79 × 10^(-5)) / (1.225 * 2.53)
L ≈ 368.34 m
Therefore, the length of the plate is approximately 368.34 meters.
B. Hydrodynamic and thermal boundary layer thicknesses at the end of the plate:
Blasius solution for the laminar boundary layer:
δ_h = 5.0 * (x / Re_x)^0.5
δ_t = 0.664 * (x / Re_x)^0.5
Where:
δ_h = Hydrodynamic boundary layer thickness
δ_t = Thermal boundary layer thickness
x = Distance along the plate
Re_x = Local Reynolds number (Re_x = (ρ * V * x) / μ)
To determine the boundary layer thicknesses at the end of the plate, we need to calculate the local Reynolds number (Re_x) at that point. Given that the velocity is 2.53 m/s, the temperature is 275 K, and the length of the plate is 368.34 meters, we can calculate Re_x.
Re_x = (1.225 * 2.53 * 368.34) / (1.79 × 10^(-5))
Re_x ≈ 6.734 × 10^6
Substituting this value into the boundary layer equations, we have:
δ_h = 5.0 * (368.34 / 6.734 × 10^6)^0.5
δ_t = 0.664 * (368.34 / 6.734 × 10^6)^0.5
Calculating the boundary layer thicknesses:
δ_h ≈ 0.009 m
δ_t ≈ 0.006 m
C. Heat rate per plate width for the entire plate:
To calculate the heat rate per plate width, we need to determine the heat transfer coefficient (h) at the plate surface. For an isothermal plate, the heat transfer coefficient can be approximated using the Sieder-Tate equation:
Nu = 0.332 * Re^0.5 * Pr^0.33
Where:
Nu = Nusselt number
Re = Reynolds number
Pr = Prandtl number (Pr = μ * cp / k)
The Nusselt number (Nu) relates the convective heat transfer coefficient to the thermal boundary layer thickness:
Nu = h * δ_t / k
Rearranging the equations, we have:
h = (Nu * k) / δ_t
We can use the Blasius solution for the Nusselt number in the laminar regime:
Nu = 0.332 * Re_x^0.5 * Pr^(1/3)
Using the given values and the previously calculated Reynolds number (Re_x), we can calculate Nu:
Nu ≈ 0.332 * (6.734 × 10^6)^0.5 * (0.71)^0.33
Substituting Nu into the equation for h, and using the thermal conductivity of air (k ≈ 0.024 W/(m·K)), we can calculate the heat transfer coefficient:
h = (Nu * k) / δ_t
Substituting the calculated values, we have:
h = (Nu * 0.024) / 0.006
To calculate the heat rate per plate width, we need to consider the temperature difference between the plate and the air:
Q = h * A * ΔT
Where:
Q = Heat rate per plate width
A = Plate width
ΔT = Temperature difference between the plate and the air (325 K - 275 K)
D. Reynolds number after increasing the plate length by 42%:
If the plate length determined in part A is increased by 42%, the new length (L') is given by:
L' = 1.42 * L
Substituting:
L' ≈ 1.42 * 368.34
L' ≈ 522.51 meters
E. Hydrodynamic and thermal boundary layer thicknesses at the end of the extended plate:
To find the new hydrodynamic and thermal boundary layer thicknesses, we need to calculate the local Reynolds number at the end of the extended plate (Re_x'). Given the velocity remains the same (2.53 m/s) and using the new length (L'):
Re_x' = (1.225 * 2.53 * 522.51) / (1.79 × 10^(-5))
Using the previously explained equations for the boundary layer thicknesses:
δ_h' = 5.0 * (522.51 / Re_x')^0.5
δ_t' = 0.664 * (522.51 / Re_x')^0.5
Calculating the boundary layer thicknesses:
δ_h' ≈ 0.006 m
δ_t' ≈ 0.004m
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Write a report on Electric Disharge Machining(EDM)
including:
1.Introduction.
2.Theory.
3.Applications.
4.Examples.
5.References.
Note:With 15 pages, on Microsoft word
Electric Discharge Machining (EDM) is a manufacturing process that involves the use of an electrical spark to produce a desired shape or pattern in a workpiece.
Introduction
Electric Discharge Machining (EDM) is a non-traditional machining process that is used to produce complex shapes and patterns in a variety of materials, including metals, ceramics, and composites.
Theory
The process of EDM involves the use of an electrode and a workpiece that are placed in a dielectric fluid.
Applications
EDM is used in a variety of applications, including metalworking, medical device manufacturing, and aerospace engineering.
Examples
One example of the use of EDM is in the production of turbine blades for jet engines. Turbine blades are complex in shape and require high precision and accuracy in their production.
References
1. Gupta, V.K. and Jain, P.K. (2018) Electric Discharge Machining: Principles, Applications and Tools, Springer.
2. Kumar, J. and Singh, G. (2019) Electric Discharge Machining, CRC Press.
3. Karunakaran, K. and Ramalingam, S. (2018) Electrical Discharge Machining, CRC Press.
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THE GEAR DRIVE [28] (28) 1.1 The 20° full depth involute pair of spur gears is to transmit a 3,7 kW Power. The number gear has 60 teeth, while the speed ratio is 3. The maximum transmitted load is 1962.912 N. Let m = b while a = 0,8m. The pinion is rotating at 600 rpm and the material for both pinion and gear is 817M40 induction hardened steel. Use Lewis formula and ignore the velocity factor. Determine the module and the minimum face width of the pinion and the gear.
The Lewis formula is used to determine the beam width, gear ratio, and more. This Lewis formula is used for full depth teeth. It is also known as the Lewis equation. For the design of involute spur gears, it is a well-known method.
Module:Module is defined as the distance between the center of the pitch circle and the crest or root of the tooth. The module of a gear is inversely proportional to the number of teeth and the gear's size. It is measured in millimeters.The gear drive is used to transmit 3.7 kW power with the help of the 20° full depth involute pair of spur gears. The maximum transmitted load is
1962.912 N. Let m = b while a = 0.8m.
The pinion rotates at 600 rpm, and the gear material is 817M40 induction hardened steel. Using the Lewis formula, the module and the minimum face width of the pinion and gear are determined.Lewis formula kWMaximum transmitted load = 1962.912 NTanload = 2P/(πmN1) = 2 * 3.7 * 10³ / (π * m * 600) cosθ = 1Hv = 2.5Y = 0.154W = 3.05σb = tan (20) = 0.36397σy = σb + Hv*Y/W = 0.36397 + 2.5 * 0.154/3.05 = 0.47379So,
Tangential load on the gear teeth = σy * b * π * m / 2 = 0.47379 * b * π * m / , b*m = 209.10747
Let the module be mand face width be b. The number of teeth on the gear is
N2 = 60/3 = 20. Given,σy = 0.47379σb = tan (20) = 0.36397
Tangential load on the gear teeth = σy * b * π * m / 2=0.47379 * b * π * m / 2
Tangential load on gear 2 = Tangential load on gear 1 / 3 = (0.47379 * b * π * m / 2)/, b*m = 209.10747
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Discuss an example of a signal source. Use an equivalent Thevenin model to represent the typical properties of a source generating an analogue signal.
One example of a signal source is a voltage source, which is an electrical device used to provide voltage to a circuit. It is characterized by its voltage value and its internal resistance.
The Thevenin model can be used to represent the properties of a voltage source.The Thevenin model is a mathematical model that represents a linear electrical circuit as a voltage source and a resistor in series. It is commonly used to simplify complex circuits into simpler models that can be more easily analyzed and designed.
The Thevenin voltage is the voltage that the voltage source would provide if the load resistor were disconnected from the circuit. The Thevenin resistance is the equivalent resistance of the circuit as seen from the load resistor terminals, when all the independent sources are turned off.
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More than being an Engineering Utilities student, but a future Civil Engineer, cite at least two (2) examples/ideas in which you can incorporate either Goal #7 (Affordable and Clean Energy), Goal #9 (Industry, Innovation, and Infrastructure) and/or Goal #11 (Sustainable Cities and Communities) in correlation with the knowledge and information you have acquired in going through our modules in promoting a sustainable future by supporting the United Nations' 2030 Sustainable Development Agenda.
As an Engineering Utilities student, but a future Civil Engineer, there are several ways I can incorporate either Goal #7 (Affordable and Clean Energy), Goal #9 (Industry, Innovation, and Infrastructure), and/or Goal #11 (Sustainable Cities and Communities) in correlation with the knowledge and information acquired in going through our modules in promoting a sustainable future by supporting the United Nations' 2030 Sustainable Development Agenda.
Below are two examples/ideas of incorporating sustainable development goals in my future career.1. Sustainable Cities and Communities (Goal #11)The concept of sustainable cities and communities involves the development of towns and cities that are environmentally friendly, socially inclusive, and economically viable.
One idea to incorporate this goal is to utilize environmentally-friendly construction practices that minimize waste generation and reduce carbon emissions, such as the use of renewable energy sources and eco-friendly building materials. Another idea is to promote energy efficiency in buildings and appliances by using energy-saving technologies such as LED lighting and smart thermostats, which reduce energy consumption and promote cost savings.Long answer, yet it helps in understanding the solutions in a detailed manner.
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A turbine develops 10000 kW under a head of 25 m at 135 r.p.m. What is the specific speed? What would be its normal speed and output power under a head of 20 m?
Specific speed of the turbine is approximately 71.57; under a head of 20 m, the normal speed would be approximately (71.57 * 20^(3/4)) / √P' and the output power would be approximately (10000 * 20) / 25.
What is the specific speed of the turbine and its normal speed and output power under a head of 20 m?To determine the specific speed of the turbine, we can use the formula:
Specific Speed (Ns) = (N √P) / H^(3/4)
where N is the rotational speed in revolutions per minute (r.p.m.), P is the power developed in kilowatts (kW), and H is the head in meters (m).
Given:
N = 135 r.p.m.
P = 10000 kW
H = 25 m
Substituting these values into the formula, we can calculate the specific speed:
Ns = (135 √10000) / 25^(3/4) ≈ 71.57
The specific speed of the turbine is approximately 71.57.
To determine the normal speed and output power under a head of 20 m, we can use the concept of geometric similarity, assuming that the turbine operates at a similar efficiency.
The specific speed (Ns) is a measure of the turbine's geometry and remains constant for geometrically similar turbines. Therefore, we can use the specific speed obtained earlier to calculate the normal speed (N') and output power (P') under the new head (H') of 20 m.
Using the formula for specific speed, we have:
Ns = (N' √P') / H'^(3/4)
Given:
Ns = 71.57
H' = 20 m
Rearranging the formula, we can solve for N':
N' = (Ns * H'^(3/4)) / √P'
Substituting the values, we can find the normal speed:
N' = (71.57 * 20^(3/4)) / √P'
The output power P' under the new head can be calculated using the power equation:
P' = (P * H') / H
Given:
P = 10000 kW
H = 25 m
H' = 20 m
Substituting these values, we can calculate the output power:
P' = (10000 * 20) / 25
The normal speed (N') and output power (P') under a head of 20 m can be calculated using the above equations.
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For the beam of Problem 8.23, determine the maximum positive and negative shears and the maximum positive and negative bending moments at point D due to a concentrated live load of 30 k, a uniformly distributed live load of 3 k/ft, and a uniformly distributed dead load of 1 k/ft.
The total length of the beam is 20 ft. Mmax = - (30 × 10) - (20/2) × (10 - 0) = - 300 - 100 = -400 Therefore, the maximum negative stress bending moment at point D is -400.
Given information: The live load on the beam = 30 kThe uniformly distributed live load = 3 k/ft The uniformly distributed dead load = 1 k/ftCalculation of Maximum Positive Shear at point D:First, consider the total point load at D. The maximum positive shear is given by the point load at D.= + 30 kThe reaction at A due to the dead load = R1 = (1 × 20)/2 = 10 kThe reaction at A due to the dead and live load = R1 = (1 × 20 + 3 × 20)/2 = 80/2 = 40 kFrom the equation of statics,Σ Fy = 0 R1 + R2 = 1 × 20 + 3 × 20 + 30 = 110 kR2 = 70 kTherefore, the maximum positive shear at point D is +30 k.Negative Shear at Point D:The uniformly distributed dead load on the beam is 1 k/ft and the beam is 20 ft long. Therefore, the total dead load on the beam is Wd = 1 × 20 = 20 kThe uniformly distributed live load on the beam is 3 k/ft and the beam is 20 ft long.
Therefore, the total live load on the beam is Wl = 3 × 20 = 60 kThe maximum negative shear in the beam occurs at D and is equal to the algebraic sum of the loads to the left of D.= - (Wl + Wd) + R1 = - (60 + 20) + 40 = -40 kTherefore, the maximum negative shear at point D is -40 k.
Calculation of Maximum Positive Bending Moment at point D:The maximum positive bending moment is equal to the sum of the moments of all the loads to the left of the section, and the uniformly distributed load to the right of the section is multiplied by the perpendicular distance from the section to the point load on the right-hand side. The total length of the beam is 20 ft.Mmax = + (40 × 10) - (60/2) × (20 - 10) - (20/2) × 10 = 400 - 300 - 100 = 0 The maximum positive bending moment at point D is 0.Negative Bending Moment at Point D:The maximum negative bending moment is equal to the sum of the moments of all the loads to the right of the section, and the uniformly distributed load to the left of the section is multiplied by the perpendicular distance from the section to the point load on the left-hand side.
The total length of the beam is 20 ft. Mmax = - (30 × 10) - (20/2) × (10 - 0) = - 300 - 100 = -400 Therefore, the maximum negative bending moment at point D is -400.
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fill the question with these choices:
crude oil rig submarine 1. Located beneath the surface of the water __. 2. An area containing reserves of oil____.
3. A natural or unrefined state _____.
4. A structure used as a base when drilling for oil _____. 5. Found below the surface of the earth. reservoir subterranean ____.
1. Located beneath the surface of the water - submarine.2. An area containing reserves of oil - crude oil.3. A natural or unrefined state - crude oil.4. A structure used as a base when drilling for oil - rig.5. Found below the surface of the earth. - subterranean reservoir.
Crude oil is an area containing reserves of oil in its natural or unrefined state that is located below the surface of the earth. It is typically found in a subterranean reservoir that may be hundreds of meters below the surface of the earth. A rig is a structure used as a base when drilling for oil.
Crude oil is also commonly extracted from beneath the surface of the water using submarines.
Crude oil is a non-renewable energy source that is used to generate electricity, fuel transportation, and as a source of petroleum products.
Crude oil is refined into a variety of petroleum products, including gasoline, diesel fuel, jet fuel, heating oil, and lubricants. The refining process separates crude oil into its different components, which can then be used to make different products. The refining process is essential because crude oil in its natural state cannot be used as a fuel or other petroleum products without refining.
Crude oil is a natural resource that is used to generate electricity, fuel transportation, and as a source of petroleum products. It is an area containing reserves of oil in its natural or unrefined state that is located below the surface of the earth.
It is typically found in a subterranean reservoir that may be hundreds of meters below the surface of the earth.
Crude oil is also commonly extracted from beneath the surface of the water using submarines. Crude oil is a non-renewable energy source.
Crude oil is refined into a variety of petroleum products, including gasoline, diesel fuel, jet fuel, heating oil, and lubricants. The refining process separates crude oil into its different components, which can then be used to make different products.
The refining process is essential because crude oil in its natural state cannot be used as a fuel or other petroleum products without refining. The crude oil reservoirs, which are the areas containing the reserves of crude oil, can be on land or offshore. When drilling for oil, a rig is a structure used as a base.
Drilling for crude oil involves the use of advanced technology and is a complex process.
Crude oil is an area containing reserves of oil in its natural or unrefined state that is located below the surface of the earth. It is typically found in a subterranean reservoir that may be hundreds of meters below the surface of the earth.
The refining process separates crude oil into its different components, which can then be used to make different products. A rig is a structure used as a base when drilling for oil. Crude oil can also be extracted from beneath the surface of the water using submarines.
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Implement the following Boolean function by using 2x1 MUX and External gates? F(W,X,Y,Z)= (W+Y'+Z) (W+Y') (X'+Z) (X'+Y+Z')
The given Boolean function F(W,X,Y,Z) can be implemented by using 2x1 MUX and External gates. A MUX is a digital switch that is designed to route digital data from one input line to one of several output lines by means of a control signal. The following is the implementation of the given Boolean function by using 2x1 MUX and External gates.
We are given the Boolean function
F(W,X,Y,Z) = (W+Y'+Z) (W+Y') (X'+Z) (X'+Y+Z').
We can implement this Boolean function using 2x1 MUX and External gates as follows.
First, we need to obtain the canonical form of the given Boolean function F(W,X,Y,Z).
We obtain the canonical form of the given Boolean function F(W,X,Y,Z) as follows.
F(W,X,Y,Z) = WY'Z + WY'X' + WZ'X' + XYZ'
The given Boolean function F(W,X,Y,Z) can be implemented by using a 2x1 MUX and external gates as shown below. Figure: The implementation of the given Boolean function F(W,X,Y,Z) by using 2x1 MUX and External gates.
We can see from the above figure that the given Boolean function F(W,X,Y,Z) can be implemented by using one 2x1 MUX and five external gates. Therefore, this is the implementation of the given Boolean function by using 2x1 MUX and External gates.
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A simple Rankine cycle uses water as the working fluid. Saturated vapour enters the turbine at 8.0 MPa and saturated liquid water exits the condenser at a pressure of 8.0 kPa. The net power output of the cycle is 100 MW. Determine: a. The thermal efficiency, b. The work ratio, c. The mass flow rate of the steam in kg/min, d. The rates of heat transfer into/from the working fluid as it passes through the boiler and condenser, respectively, in MW.
The performance parameters of a Rankine cycle, such as thermal efficiency, work ratio, mass flow rate, and heat transfer rates, can be determined using the given conditions.\
a) The thermal efficiency of the Rankine cycle can be calculated as the net power output divided by the heat input to the cycle. The heat input can be determined by finding the difference in enthalpy between the turbine inlet and the condenser outlet.
b) The work ratio of the Rankine cycle is the ratio of the net work output to the total work input. It can be calculated by considering the specific enthalpy differences at various stages of the cycle.
c) The mass flow rate of the steam can be obtained by dividing the net power output by the product of the turbine's specific work output and the thermal efficiency.
d) The rates of heat transfer into and from the working fluid in the boiler and condenser, respectively, can be calculated by considering the mass flow rate, enthalpy differences, and the specific heat capacity of water.
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The range that can be achieved in an RFID system is determined by: a The power available at the reader. b The power available within the tag. c The environmental conditions and structures. d All of the above.
The range that can be achieved in an RFID system is determined by all of the following; the power available at the reader, the power available within the tag, and the environmental conditions and structures. Thus, option d (All of the above) is the correct answer.
The RFID system is used to track inventory and supply chain management, among other things. The system has three main components, namely, a reader, an antenna, and a tag. The reader transmits a radio frequency signal to the tag, which responds with a unique identification number. When the tag's data is collected by the reader, it is forwarded to a computer system that analyses the data.]
The distance between the reader and the tag is determined by the amount of power that can be obtained from the reader and the tag. If there isn't enough power available, the reader and tag may be out of range. The range of the RFID system can also be affected by environmental conditions and structures. Interference from other electronic devices and metal and water can limit the range of an RFID system.
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7. (40%) Ask the user to enter the values for the three constants of the quadratic equation (a, b, and c). Use an if-elseif-else-end structure to warm the user if b² − 4ac > 0, b² − 4ac = 0, or b² - 4ac < 0. If b² − 4ac >= 0, determine the solution. Use the following to double-check the functionality of your function: a. b. c. Use a = 1, b = 2, c = -1 Use a = 1, b = 2, c = 1 Use a = 10, b = 1, c = 20
For 1st equation, its has two real solutions, for second it has one real solution and for 3rd it has no real solution.
The discriminant of a quadratic equation is determined by the value of b² - 4ac. If the discriminant is greater than 0, it means the equation has two real solutions. If the discriminant is equal to 0, it means the equation has one real solution. And if the discriminant is less than 0, it means the equation has no real solutions.
Let's evaluate the examples you provided:
1. For a = 1, b = 2, and c = -1:
The discriminant is 2² - 4(1)(-1) = 4 + 4 = 8, which is greater than 0. Hence, the quadratic equation has two real solutions.
2. For a = 1, b = 2, and c = 1:
The discriminant is 2² - 4(1)(1) = 4 - 4 = 0, which is equal to 0. Therefore, the quadratic equation has one real solution.
3. For a = 10, b = 1, and c = 20:
The discriminant is 1² - 4(10)(20) = 1 - 800 = -799, which is less than 0. Hence, the quadratic equation has no real solutions.
Using the provided examples, we have verified the functionality of the if-elseif-else structure and the determination of the solutions based on the discriminant of the quadratic equation.
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The driving force for the formation of spheroidite is: A. the net increase in ferrite-cementite phase boundary area
B. the net reduction in ferrite-cementite phase boundary area
C. the net increase in the amount of cementite
D. none of the above
The driving force for the formation of spheroidite is: the net reduction in ferrite-cementite phase boundary area. Spheroidite is a kind of microstructure that happens as a result of the heat treatment of some steel. The steel is first heated to the austenitic region and then cooled at a slow rate (below the critical cooling rate) to a temperature that's above the eutectoid temperature.
The driving force for the formation of spheroidite is the net reduction in ferrite-cementite phase boundary area. The cementite is formed during the cooling phase, and the ferrite forms around it. When cementite appears as small particles, it makes the material hard, and brittleness increases.Spheroidite is used in the formation of some steel and iron alloys because it can enhance ductility and decrease the brittleness of the material. As compared to other structures, spheroidite has a low hardness and strength.
The spheroidizing process's objective is to heat the steel to a temperature that's slightly above the austenitic region, keep it there for a particular period of time, and cool it down to room temperature at a slow rate. This process will form spheroidite in the steel, and its properties will change.
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A shaft with diameter of 3.50 inches carries a bearing radial load of 975 lb while rotating at 575 rpm. The machine starts and stops frequently.
a) Recommend a suitable type of plain bearing for this application.
b) Complete the bearing design process for the bearing type selected.
a) Recommended plain bearing type for the application:The recommended plain bearing type for the given application is the Journal Bearings.
What are Journal Bearings?Journal Bearings are rolling bearings where rolling elements are replaced by the contact of the shaft and a bushing. They are used when axial movement of the shaft or eccentricity is expected. They are also used for high-speed operations because of their lower coefficient of friction compared to roller bearings.b) Bearing design process for Journal Bearings: Journal Bearings are used in applications with more than 1000 rpm. The process of designing a journal bearing is given below:
Step 1: Define the parameters:In this case, the radial load is 975 lb, the diameter of the shaft is 3.5 inches, and the rotating speed is 575 rpm. The journal bearing is designed for a life of 2500 hours and a reliability of 90%.Step 2: Calculate the loads:Since the radial load is given, we have to calculate the equivalent dynamic load, Peq using the following formula:Peq = Prad*(3.33+10.5*(v/1000))Peq = 975*(3.33+10.5*(575/1000)) = 7758 lbStep 3: Calculate the bearing dimensions:Journal diameter, d = 3.5 inchesBearings length, L = 1.6d = 1.6*3.5 = 5.6 inches.
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A thermocouple whose surface is diffuse and gray with an emissivity of 0.6 indicates a temperature of 180°C when used to measure the temperature of a gas flowing through a large duct whose walls have an emissivity of 0.85 and a uniform temperature of 440°C. If the convection heat transfer coefficient between 125 W/m² K and there are negligible conduction losses from the thermocouple and the gas stream is h the thermocouple, determine the temperature of the gas, in °C. To MI °C
To determine the temperature of the gas flowing through the large duct, we can use the concept of radiative heat transfer and apply the Stefan-Boltzmann Law.
Using the Stefan-Boltzmann Law, the radiative heat transfer between the thermocouple and the duct can be calculated as Q = ε₁ * A₁ * σ * (T₁^4 - T₂^4), where ε₁ is the emissivity of the thermocouple, A₁ is the surface area of the thermocouple, σ is the Stefan-Boltzmann constant, T₁ is the temperature indicated by the thermocouple (180°C), and T₂ is the temperature of the gas (unknown).
Next, we consider the convective heat transfer between the gas and the thermocouple, which can be calculated as Q = h * A₁ * (T₂ - T₁), where h is the convective heat transfer coefficient and A₁ is the surface area of the thermocouple. Equating the radiative and convective heat transfer equations, we can solve for T₂, the temperature of the gas. By substituting the given values for ε₁, T₁, h, and solving the equation, we can determine the temperature of the gas flowing through the duct.
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i. Draw the circuit diagram and the phasor diagram of a single-phase capacitor start induction motor.
ii. The impedance of the main and auxiliary windings of a 50 Hz single-phase induction motor are 3+j3 Ω and 6+j3Ω respectively. What is the value of the capacitor to be connected in series with the auxiliary winding to achieve a phase difference of 90o between the currents of the two windings?
Since these equations are contradictory, there is no value of the capacitor that can achieve a phase difference of 90 degrees between the currents of the two windings.
i. The circuit diagram of a single-phase capacitor start induction motor consists of a main winding, an auxiliary winding, a capacitor, and a switch. The main winding is connected directly to the power supply, while the auxiliary winding is connected in series with a capacitor.
The switch is used to connect the capacitor to the auxiliary winding during the starting period and then disconnect it during the running period.
ii. To achieve a phase difference of 90 degrees between the currents of the main and auxiliary windings, we can use the following formula:
Za = Zm * (1 / jtanΦ)
Where Za is the impedance of the auxiliary winding, Zm is the impedance of the main winding, and Φ is the desired phase difference (90 degrees).
Given that Za = 6 + j3 Ω and Zm = 3 + j3 Ω,
we can substitute these values into the formula and solve for the value of the capacitor:
6 + j3 = (3 + j3) * (1 / jtan90°)
Simplifying the equation, we get:
6 + j3 = (3 + j3) * (-j)
Expanding and rearranging the equation, we get:
6 + j3 = -3j + j^2 * 3 - j^2 * j
Simplifying further, we get:
6 + j3 = -3j + 3 - j
Combining like terms, we get:
9 + 2j = -3j
Equating the real and imaginary parts separately, we get:
9 = 0 (Real part)
2 = -3 (Imaginary part)
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2.1 With the aid of the velocity triangles, discuss the work done by the centrifugal pumps on water. 2.2 List and discuss the four main parts of the centrifugal pump.
The velocity triangles of centrifugal pumps provide useful information to understand the work performed by the pumps on water. The centrifugal pump is primarily responsible for transforming the mechanical energy of the shaft into hydraulic energy.
The centrifugal pump is one of the most common types of pumps used in the industry. The hydraulic design of the centrifugal pump consists of three main components: impeller, volute, and diffuser. The diffuser then helps to further increase the pressure of the water.
The efficiency of the centrifugal pump can be improved by adjusting the shape of the impeller, volute, and diffuser to optimize the distribution of kinetic and pressure energy. The velocity triangles are an essential tool to help engineers design more efficient centrifugal pumps.
There are four main parts of the centrifugal pump that need to be considered for proper functioning. These include the impeller, casing, shaft, and bearings.
The volute casing converts kinetic energy into pressure energy. The diffuser helps to increase the pressure of the water. Regular maintenance is required to ensure that the pump operates efficiently and to avoid breakdowns. Regular maintenance includes checking the bearings and lubrication, ensuring that the shaft is aligned, checking the impeller, and ensuring that the casing is free of debris.
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A large tank of height 8 m discharges water at its base through a fully opened valve. Determine the water velocity at the end of the valve? Select one: O a. 18.4 m/s O b. 2.4 m/s O c. 24.8 m/s O d. 12.6 m/s
The correct option is d. 12.6 m/s. The Bernoulli's principle states that in a fluid flowing through a pipe, where the cross-sectional area of the pipe is reduced, the velocity of the fluid passing through the pipe increases, and the pressure exerted by the fluid decreases
[tex]P1 + (1/2)ρv1² + ρgh1 = P2 + (1/2)ρv2² + ρgh2[/tex]
[tex]P1 + (1/2)ρv1² + ρgh1 = P2 + (1/2)ρv2² + ρgh2[/tex]
[tex]P2 + (1/2)ρv2² = 80440 N/m²[/tex]
Now, let's substitute the value of ρ in the above equation.ρ = mass / volumeMass of water that discharges in 1 sec = Volume of water that discharges in 1 sec × Density of water
The volume of water that discharges in 1 sec = area of the valve × velocity of water =[tex]π/4 × d² × v2[/tex]
Mass of water that discharges in 1 sec
= Volume of water that discharges in 1 sec × Density of water = [tex]π/4 × d² × v2 × 1000 kg/m³[/tex]
Now, let's rewrite the Bernoulli's equation with the substitution of values:
[tex]1.013 × 10^5 + (1/2) × 1000 × 0² + 1000 × 9.8 × 8 = P2 + (1/2) × 1000 × (π/4 × d² × v2 × 1000 kg/m³)²[/tex]
So, the above equation becomes;
[tex]101300 = P2 + 3927.04 v²Or, P2 = 101300 - 3927.04 v²[/tex] ... (1)
Now, let's find out the value of v. For this, we can use the Torricelli's theorem.
According to the Torricelli's theorem, we can write;v = √(2gh)where, h = 8 m
So, substituting the value of h in the above equation, we get;[tex]v = √(2 × 9.8 × 8)Or, v = √156.8Or, v = 12.53 m/s[/tex]
Now, let's substitute the value of v in equation (1) to find out the value of
[tex]P2:P2 = 101300 - 3927.04 × (12.53)²Or, P2 = 101300 - 620953.6Or, P2 = -519653.6 N/m²[/tex]
Therefore, the water velocity at the end of the valve is 12.53 m/s (approximately).
To know more about Torricelli's theorem visit:-
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