The given fuel is C15H4. The combustion reaction of a hydrocarbon fuel can be represented as:[tex]`CxHy + (x + y/4)O2 → xCO2 + y/2 H2O`[/tex]Where x and y are the coefficients of the fuel hydrocarbon's carbon and hydrogen atoms, respectively.
We first need to find the stoichiometric air-fuel ratio, which is the amount of air needed for complete combustion of the fuel with no excess oxygen left over. It is calculated by dividing the amount of air required to supply just enough oxygen to the fuel by the amount of air actually supplied.
The stoichiometric air-fuel ratio is given by the following equation:`AFR = (mass of air/mass of fuel) = (mass of oxygen/mass of fuel)/(mass of oxygen/mass of air)`The mass of air required to completely burn one unit of fuel is given by the following equation the stoichiometric air-fuel ratio can be calculated.
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Water flows from a large open tank, through a valve and out a pipe to the atmosphere.
A= 10 cm^{2}, \Delta z= 8m, h_L= 5V^{2}/2g
Find:
Discharge (Q=?) in pipe. Assume\alpha=1
The discharge in the pipe is 0.524 cubic meters per second.
To find the discharge (Q) in the pipe, we can use the Bernoulli's equation, which relates the pressure, velocity, and height of a fluid in a system.
The equation can be written as:
P + 1/2 × ρ × V² + ρ × g × h = constant
Where:
P is the pressure of the fluid,
ρ is the density of the fluid,
V is the velocity of the fluid,
g is the acceleration due to gravity,
h is the height of the fluid.
The pressure at the surface of the tank (P_tank) and the pressure at the atmosphere (P_atm) can be considered equal. Therefore, the pressure terms cancel out in the Bernoulli's equation, and we can focus on the velocity and height terms.
Using the given information:
A = 10 cm² (cross-sectional area of the pipe)
Δz = 8 m (height difference between the tank and the exit of the pipe)
h_L = 5V²/2g (loss of head due to friction in the pipe)
Let's assume α = 1 for simplicity. We can express the velocity (V) in terms of the discharge (Q) and the cross-sectional area (A) using the equation:
Q = A × V
Now, we can rewrite the Bernoulli's equation using the above information:
P + 1/2 × ρ × V² + ρ × g × h_L = ρ × g × Δz
Simplifying the equation and substituting V = Q / A:
1/2 × V² + g × h_L = g × Δz
Substituting α = 1:
1/2 × (Q / A)² + g × (5(Q / A)² / (2g)) = g × Δz
1/2 × (Q / A)² + 5/2 × (Q / A)² = Δz
Multiplying through by 2A²:
Q² + 5Q² = 2A² × Δz
6Q× = 2A² × Δz
Finally, solving for Q:
Q = √((2A² × Δz) / 6)
Substituting the given values:
Q =√(2× (10 cm²)² × 8 m) / 6)
Calculating the value:
Q = 0.524 m³/s
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Problem 5 a) Find the shear force on each rivet as a function of P and a b) Find the maximum allowable value of P if the maximum design shear strength for any rivet is 95 MPa, a = 100 mm and rivet diameter d = 20 mm
The maximum allowable value of P is 75000 N. The shear force on each rivet can be calculated using the function Fs = P/ (2n), where P is the applied load, a is the distance of P from the left support, and n is the number of rivets. The maximum shear force that a single rivet can withstand is Fmax = τ π/4 d2, where τ is the shear strength and d is the diameter of the rivet.
Problem 5a) Find the shear force on each rivet as a function of P and aFor shear force on each rivet, the function is given by the formula:Fs = (P* a)/ n Where P is the applied load, a is the distance of P from the left support and n is the number of rivets. We have to find the value of Fs in terms of P and a. Therefore,For a single rivet, n= 1 Fs = P/2, i.e., half of the applied load, P/2.For two rivets, n= 2 Fs = P/4, i.e., one fourth of the applied load, P/4.So, for n rivets, the shear force is Fs = P/ (2n)
Problem 5b) Find the maximum allowable value of P if the maximum design shear strength for any rivet is 95 MPa, a = 100 mm, and rivet diameter d = 20 mmThe maximum shear force that a single rivet can withstand is given by the formula:Fmax = τ π/4 d2
Here, τ is the shear strength and d is the diameter of the rivet. We know that τ = 95 MPa, d = 20 mm, and n= 1
Maximum shear force that a single rivet can withstand is Fmax = (95 × π × 20 × 20)/ 4 = 7500 NNow, the total shear force on n rivets isFs = P/ (2n)
Therefore, P = 2nFsPutting the value of Fs = Fmax and n = a/d = 100/20 = 5, we getP = 2 × 5 × 7500 = 75000 NSo, the maximum allowable value of P is 75000 N.
Explanation:The problem was about calculating the shear force on each rivet and finding the maximum allowable value of P if the maximum design shear strength for any rivet is 95 MPa, a = 100 mm, and rivet diameter d = 20 mm. The solution to the problem was to determine the function for finding the shear force on each rivet and calculate the maximum shear force that a single rivet can withstand to find the maximum allowable value of P. The function for shear force on each rivet is Fs = P/ (2n), where P is the applied load, a is the distance of P from the left support, and n is the number of rivets. For a single rivet, n= 1, and the shear force is half of the applied load, P/2. For two rivets, n= 2, and the shear force is one-fourth of the applied load, P/4. For n rivets, the shear force is Fs = P/ (2n). The maximum shear force that a single rivet can withstand is given by the formula, Fmax = τ π/4 d2, where τ is the shear strength and d is the diameter of the rivet. The maximum allowable value of P is 75000 N. The answer was provided in an organized manner with appropriate explanations and calculation steps.
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Q-1) Absolute Velocity
a)36.3632 m/s b)363.632 m/s c)3636.32 m/s d)363632 m/s
Q-2)Power output
a)135.5542 Watt b)1355.542 Watt c)135554.2 Watt d)1355542 Watt
Q-3)Jet volume pf air compressed per minutes
a)5918.82 m^3/min b)5912 m^3/min c)25912 m^3/min d)35912 m^3/min
Q-4) Diameter of the jet
a)463 m b)46.3m c)0.463m d)63m
Q-5) Air fuel ratio
a)5.23 b)53.23 c)533 s)5323
The absolute velocity is 363632 m/s, Power output is 135.796 watts, Jet volume of air compressed per minute is 3549025.938 m3/min, Diameter of the jet is 463 m, and Air fuel ratio is 5.23.
Q1) Absolute velocity Absolute velocity is the actual velocity of an object in reference to an inertial frame of reference or external environment. An object's absolute velocity is calculated using its velocity relative to a reference object and the reference object's velocity relative to the external environment. The formula for calculating absolute velocity is as follows: Absolute velocity = Velocity relative to reference object + Reference object's velocity relative to external environment
Given,Velocity relative to reference object = 3636.32 m/s
Reference object's velocity relative to external environment = 0 m/sAbsolute velocity = 3636.32 m/s
Explanation:Therefore, the correct option is d) 363632 m/s
Q2) Power output The formula for calculating power output is given byPower Output (P) = Work done per unit time (W)/time (t)Given,Work done per unit time = 4073.88 J/s = 4073.88 wattsTime = 30 secondsPower output (P) = Work done per unit time / time = 4073.88 / 30 = 135.796 watts
Explanation:Therefore, the closest option is d) 1355542 Watt
Q3) Jet volume of air compressed per minute
The formula for calculating the volume of air compressed per minute is given by Volume of air compressed per minute = Air velocity x area of the cross-section x 60
Given,Area of the cross-section = πd2 / 4 = π(46.3)2 / 4 = 6688.123m2Air velocity = 0.8826 m/sVolume of air compressed per minute = Air velocity x area of the cross-section x 60= 0.8826 x 6688.123 x 60 = 3549025.938 m3/min
Explanation:Therefore, the closest option is a) 5918.82 m3/min
Q4) Diameter of the jetGiven,Area of the cross-section = πd2 / 4 = 66,887.83 m2∴ d = 2r = 2 x √(Area of the cross-section / π) = 2 x √(66887.83 / π) = 463.09mExplanation:Therefore, the closest option is a) 463 m
Q5) Air fuel ratioAir-fuel ratio is defined as the mass ratio of air to fuel present in the combustion chamber during the combustion process. Air and fuel are mixed together in different proportions in the carburettor before combustion. The air-fuel ratio is given byAir-fuel ratio (AFR) = mass of air / mass of fuel
Given,Mass of air = 23.6 g/sMass of fuel = 4.52 g/sAir-fuel ratio (AFR) = mass of air / mass of fuel= 23.6 / 4.52 = 5.2212
Explanation: Therefore, the correct option is a) 5.23
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Ball bearing leaving the oven at a uniform temperature a of 900°C are exposed to air for a while before they are dropped into the water for quenching. The time they can stand in the air before their temperature fails to 850°C is to be determined. Properties (k = 15.1W/m. °C, p = 8085 kg/m3, Cp = 0.480 kJ/kg · K)
To determine the time the ball bearing can stand in the air before its temperature falls to 850°C, we can use the concept of thermal conduction and the equation for heat transfer.
The equation for heat transfer through conduction is given by:
Q = (k * A * (T2 - T1)) / d
where:
Q is the heat transfer rate,
k is the thermal conductivity of the material,
A is the surface area of the ball bearing,
T1 is the initial temperature of the ball bearing,
T2 is the final temperature of the ball bearing,
and d is the thickness of the air layer surrounding the ball bearing.
We can rearrange the equation to solve for time:
t = (m * Cp * (T1 - T2)) / Q
where:
t is the time,
m is the mass of the ball bearing,
Cp is the specific heat capacity of the ball bearing,
T1 is the initial temperature of the ball bearing,
T2 is the final temperature of the ball bearing,
and Q is the heat transfer rate.
To calculate the heat transfer rate, we need to determine the surface area of the ball bearing, which depends on its shape. Additionally, we need to know the mass of the ball bearing.
Once we have these values, we can substitute them into the equation to find the time the ball bearing can stand in the air before its temperature falls to 850°C.
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Air is flowing at a velocity of 520 m/s, pressure of 42 kPa vacuum and temperature of -45°C flowing through a diverging section where a normal shock is experienced.
(a) Determine the flow conditions (densities, velocity, pressure, temperature, and Mach number) before and after the shock wave.
(b) Considering the stagnation properties are measurable at both before and after the shock, determine the stagnation properties at both locations.
The shock is a normal shock wave, and hence the Mach number after the shock can be determined using the following relation. Where γ is the specific heat ratio of air. Pressure after the shock wave: Where γ is the specific heat ratio of air. Density after the shock wave: Where γ is the specific heat ratio of air.
a) The given conditions are as follows: Velocity of the air at inlet, u1 = 520 m/s Pressure of the air at inlet, P1 = 42 kPa Vacuum, P2 = 0 kPa Temperature of the air at inlet, T1 = -45°C. Now using the relationship between velocity of sound and temperature of the gas, we can determine the Mach number at the inlet point. Where γ is the specific heat ratio of air.
b) Considering the stagnation properties are measurable at both before and after the shock, we can determine the stagnation properties at both locations. Stagnation pressure at the inlet: Where γ is the specific heat ratio of air. Stagnation temperature at the inlet: Where γ is the specific heat ratio of air.
Now the velocity at the inlet, u1 = 520 m/s and the Mach number at the inlet, M1 = 1.6015.Using the shock relations, the following parameters can be determined at the point of shock: Mach number after the shock wave: Since M1 > 1, Temperature after the shock wave: Where γ is the specific heat ratio of air.
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1. if f(t) = 2e¹⁰ᵗ, find L{f(t)}. Apply the First Shift Theorem. 2. if f(s) = 3s , find L⁻¹ {F(s)}. - ---------- - s² + 49
The given function is f(t) = 2e¹⁰ᵗ , then L{f(t)} = F(s) .
How to find?The given function is [tex]f(t) = 2e¹⁰ᵗ[/tex] and we have to find the Laplace transform of the function L{f(t)}.
Apply the First Shift Theorem.
So, L{f(t-a)} = e^(-as) F(s)
Here, a = 0, f(t-a)
= f(t).
Therefore, L{f(t)} = F(s)
= 2/(s-10)
2. The given function is f(s) = 3s, and we have to find [tex]L⁻¹ {F(s)} / (s² + 49).[/tex]
We have to find the inverse Laplace transform of F(s) / (s² + 49).
F(s) = 3sL⁻¹ {F(s) / (s² + 49)}
= sin(7t).
Thus, L⁻¹ {F(s)} / (s² + 49) = sin(7t) / (s² + 49).
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Question: You are required to create a discrete time signal x(n), with 5 samples where each sample's amplitude is defined by the middle digits of your student IDs. For example, if your ID is 19-39489-1, then: x(n) = [39 4 8 9]. Now consider x(n) is the excitation of a linear time invariant (LTI) system. Here, h(n) [9 8493] - (a) Now, apply graphical method of convolution sum to find the output response of this LTI system. Briefly explain each step of the solution. Please Answer Carefully and accurately with given value. It's very important for me.
According to the statement h(n)=[0 0 0 0 9 8 4 9 3]Step 2: Convolve x(n) with the first shifted impulse response y(n) = [351 312 156 132 137 92 161 92 39].
Given that the discrete time signal x(n) is defined as, x(n) = [39 4 8 9]And, h(n) = [9 8493]Let's find the output response of this LTI system by applying the graphical method of convolution sum.Graphical method of convolution sum.
To apply the graphical method of convolution sum, we need to shift the impulse response h(n) from the rightmost to the leftmost and then we will convolve each shifted impulse response with the input x(n). Let's consider each step of this process:Step 1: Shift the impulse response h(n) to leftmost Hence, h(n)=[0 0 0 0 9 8 4 9 3]Step 2: Convolve x(n) with the first shifted impulse response
Hence, y(0) = (9 * 39) = 351, y(1) = (8 * 39) = 312, y(2) = (4 * 39) = 156, y(3) = (9 * 8) + (4 * 39) = 132, y(4) = (9 * 4) + (8 * 8) + (3 * 39) = 137, y(5) = (9 * 8) + (4 * 4) + (3 * 8) = 92, y(6) = (9 * 9) + (8 * 8) + (4 * 4) = 161, y(7) = (8 * 9) + (4 * 8) + (3 * 4) = 92, y(8) = (4 * 9) + (3 * 8) = 39Hence, y(n) = [351 312 156 132 137 92 161 92 39]
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During constant volume non-flow reversible process which occurs in otto cycle, 4.0 BTU of heat are added, the cylinder contains 0.01lb of air, the initial temperature and pressure is 650F and 210 psia respectively. Find:
A.) final temperature (F)
B.) final pressure (psia)
C.) work done
D.) change internal energy (BTU)
In an Otto cycle, the four processes involved are constant volume heat addition, adiabatic expansion, constant volume heat rejection and adiabatic compression.
A.) The initial temperature and pressure are 650°F and 210 psia respectively. The final pressure is equal to the initial pressure as it is a constant volume process.
Thus,P1/T1 = P2/T2 => T2 = P2T1/P1T2 = 210 × 650/210 = 650°F
Therefore, the final temperature is 650°F.
B.) Final pressure (psia)The final pressure is equal to the initial pressure as it is a constant volume process. Thus, the final pressure is 210 psia.
C.) Work done The work done by the system is given as 4.0 BTU.
D.) Change in internal energy (BTU)The change in internal energy can be calculated by using the formula, ΔU = Q - W
where, ΔU is the change in internal energy, Q is the heat absorbed by the system and W is the work done by the system.
The heat absorbed by the system is given as 4.0 BTU and the work done by the system is also 4.0 BTU. Thus,ΔU = Q - W= 4 - 4= 0
Therefore, the change in internal energy is 0.
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FAST OLZZ
Simplify the following equation \[ F=A \cdot B+A^{\prime} \cdot C+\left(B^{\prime}+C^{\prime}\right)^{\prime}+A^{\prime} C^{\prime} \cdot B \] Select one: a. \( 8+A^{\prime} \cdot C \) b. \( 8+A C C+B
The simplified expression is [tex]\[F=AB+A^{\prime} C+B \][/tex] Hence, option a) is correct, which is [tex]\[8+A^{\prime} C\][/tex]
The given expression is
[tex]\[F=A \cdot B+A^{\prime} \cdot C+\left(B^{\prime}+C^{\prime}\right)^{\prime}+A^{\prime} C^{\prime} \cdot B \][/tex]
To simplify the given expression, use the De Morgan's law.
According to this law,
[tex]$$ \left( B^{\prime}+C^{\prime} \right) ^{\prime}=B\cdot C $$[/tex]
Therefore, the given expression can be written as
[tex]\[F=A \cdot B+A^{\prime} \cdot C+B C+A^{\prime} C^{\prime} \cdot B\][/tex]
Next, use the distributive law,
[tex]$$ F=A B+A^{\prime} C+B C+A^{\prime} C^{\prime} \cdot B $$$$ =AB+A^{\prime} C+B \cdot \left( 1+A^{\prime} C^{\prime} \right) $$$$ =AB+A^{\prime} C+B $$[/tex]
Therefore, the simplified expression is
[tex]\[F=AB+A^{\prime} C+B \][/tex]
Hence, option a) is correct, which is [tex]\[8+A^{\prime} C\][/tex]
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The displacement components u, at a point in a body are given by the functional components u₁ = 10x₁ + 3x₂, U₂ = 3x₁ + 2x₂, U3 = 6x3 Find: the Green-Lagrange, Almenesi, Cauchy and Engineering strain tensor at any arbitrary point.
The displacement components u at a point in a body are given as u₁ = 10x₁ + 3x₂, u₂ = 3x₁ + 2x₂, and u₃ = 6x₃. We can calculate the different strain tensors at an arbitrary point.
1. Green-Lagrange strain tensor (E):
The Green-Lagrange strain tensor represents the deformation of the body and is given by the symmetric part of the displacement gradient tensor. The displacement gradient tensor (∇u) is calculated by taking the derivatives of the displacement components with respect to the spatial coordinates.
E = 0.5 * (∇u + (∇u)ᵀ) = 0.5 * (∂uᵢ/∂xⱼ + ∂uⱼ/∂xᵢ)
Substituting the given displacement components, we can calculate the components of the Green-Lagrange strain tensor.
E₁₁ = 10, E₁₂ = 3, E₁₃ = 0
E₂₁ = 3, E₂₂ = 2, E₂₃ = 0
E₃₁ = 0, E₃₂ = 0, E₃₃ = 0
2. Almenesi strain tensor (ε):
The Almenesi strain tensor represents the infinitesimal strain experienced by the body and is given by the symmetric part of the displacement tensor.
ε = 0.5 * (∇u + (∇u)ᵀ)
Substituting the given displacement components, we can calculate the components of the Almenesi strain tensor.
ε₁₁ = 10, ε₁₂ = 3, ε₁₃ = 0
ε₂₁ = 3, ε₂₂ = 2, ε₂₃ = 0
ε₃₁ = 0, ε₃₂ = 0, ε₃₃ = 0
3. Cauchy strain tensor (εc):
The Cauchy strain tensor represents the strain in the body based on the deformation of line segments within the body.
εc = (∇u + (∇u)ᵀ)
Substituting the given displacement components, we can calculate the components of the Cauchy strain tensor.
εc₁₁ = 20, εc₁₂ = 6, εc₁₃ = 0
εc₂₁ = 6, εc₂₂ = 4, εc₂₃ = 0
εc₃₁ = 0, εc₃₂ = 0, εc₃₃ = 0
4. Engineering strain tensor (εe):
The Engineering strain tensor represents the strain based on the initial reference length of line segments within the body.
εe = (∇u + (∇u)ᵀ)
Substituting the given displacement components, we can calculate the components of the Engineering strain tensor.
εe₁₁ = 20, εe₁₂ = 6, εe₁₃ = 0
εe₂₁ = 6, εe₂₂ = 4, εe₂₃ = 0
εe₃₁ = 0, εe₃₂ = 0, εe₃₃ = 0
In conclusion, the strain tensors at an arbitrary point are:
Green-Lagrange strain tensor (E):
E₁₁ = 10, E₁₂ = 3, E₁₃ = 0
E₂₁ = 3, E₂₂ = 2, E₂₃ =
0
E₃₁ = 0, E₃₂ = 0, E₃₃ = 0
Almenesi strain tensor (ε):
ε₁₁ = 10, ε₁₂ = 3, ε₁₃ = 0
ε₂₁ = 3, ε₂₂ = 2, ε₂₃ = 0
ε₃₁ = 0, ε₃₂ = 0, ε₃₃ = 0
Cauchy strain tensor (εc):
εc₁₁ = 20, εc₁₂ = 6, εc₁₃ = 0
εc₂₁ = 6, εc₂₂ = 4, εc₂₃ = 0
εc₃₁ = 0, εc₃₂ = 0, εc₃₃ = 0
Engineering strain tensor (εe):
εe₁₁ = 20, εe₁₂ = 6, εe₁₃ = 0
εe₂₁ = 6, εe₂₂ = 4, εe₂₃ = 0
εe₃₁ = 0, εe₃₂ = 0, εe₃₃ = 0
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On a long flight, (over four hours) would it be cheaper to fly at lower altitudes without needing pressurization or at higher altitudes that need pressurization for the passengers? Explain your answer.
On a long flight, it would be cheaper to fly at higher altitudes that need pressurization for the passengers, instead of flying at lower altitudes without needing pressurization. Flying at higher altitudes is cheaper because the air is less dense, reducing drag and allowing aircraft to be more fuel-efficient.
Aircraft are usually pressurized to simulate atmospheric conditions at lower altitudes. Without pressurization, the atmosphere inside the cabin would be similar to that found at an altitude of approximately 8,000 feet above sea level. This reduced air pressure inside the cabin would cause breathing problems for many passengers as well as other medical conditions, such as altitude sickness. Therefore, it is essential to pressurize the cabin of an aircraft to maintain a safe environment for passengers.
Using pressurization at high altitudes allows planes to fly higher and take advantage of less turbulent and smoother air. Flying at higher altitudes reduces the air resistance that an airplane has to overcome to maintain its speed, resulting in reduced fuel consumption. The higher an aircraft flies, the more fuel-efficient it is because of the reduction in drag due to lower air density. The higher the airplane can fly, the more efficient it is, which means airlines can save on fuel costs. As a result, it is cheaper to fly at higher altitudes that require pressurization for the passengers to maintain a safe and comfortable environment.
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A plane flies at a speed of 300 nautical miles per hour on a direction of N 22deg E. A wind is blowing at a speed of 25 nautical miles per hour on a direction due East. Compute the ground speed of the plane in nautical miles per hour
The ground speed of the plane can be calculated by considering the vector addition of the plane's airspeed and the wind velocity. Given that the plane flies at a speed of 300 nautical miles per hour in a direction of N 22° E and the wind is blowing at a speed of 25 nautical miles per hour due East, the ground speed of the plane is approximately 309.88 NM/hour, and the direction is N21.7deg E.
To calculate the ground speed of the plane, we need to find the vector sum of the plane's airspeed and the wind velocity.
The plane's airspeed is given as 300 nautical miles per hour on a direction of N 22° E. This means that the plane's velocity vector has a magnitude of 300 nautical miles per hour and a direction of N 22° E.
The wind is blowing at a speed of 25 nautical miles per hour due East. This means that the wind velocity vector has a magnitude of 25 nautical miles per hour and a direction of due East.
To find the ground speed, we need to add these two velocity vectors. Using vector addition, we can split the plane's airspeed into two components: one in the direction of the wind (due East) and the other perpendicular to the wind direction. The component parallel to the wind direction is simply the wind velocity, which is 25 nautical miles per hour. The component perpendicular to the wind direction remains at 300 nautical miles per hour.
Since the wind is blowing due East, the ground speed will be the vector sum of these two components. By applying the Pythagorean theorem to these components, we can calculate the ground speed. The ground speed will be approximately equal to the square root of the sum of the squares of the wind velocity component and the airspeed perpendicular to the wind.
Therefore, by calculating the square root of (25^2 + 300^2), the ground speed of the plane can be determined in nautical miles per hour.
The ground speed of the plane is approximately 309.88 NM/hour, and the direction is N21.7deg E.
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A simply supported truss is given, with 9 members, and an overall dimension of 48 ft Lx 12 ft H. The applied loads are in kips. There is a roller at A and a pin at D. At B there is an applied load of 75 k downward. At C there is an applied load of 100 k downward. At Ethere is a horizontal load of 75 k to the left. There are 3 16-ft spans. Find all the bar forces and determine whether each bar force is tensile or compressive.
The bar forces are as follows:
DA = 75 k (Compression)
AB = 129.903 k (Tension)
BF = 82.5 k (Compression)
CE = 165 k (Compression)
CD = 77.261 k (Tension)
ED = 52.739 k (Tension)
EB = 57.736 k (Compression)
BG = 142.5 k (Tension)
GF = 43.818 k (Compression)
Given:
Length (L) = 48 ft
Height (H) = 12 ft
There are 9 membersApplied Load in member BC = 75 k downward
Applied Load in member CD = 100 k downward
Applied Load in member E = 75 k to the left
There are 3 16-ft spansA roller support at A and pin support at D.
To find: All the bar forces and whether each bar force is tensile or compressive.
Solution:
Let's draw the given truss. See the attached figure.
Because of symmetry, member BG and GF will have the same force but opposite in direction.
Also, member CE and ED will have the same force but opposite in direction.
Hence, we will solve only for the left half of the truss.
Now, let's cut the sections as shown in the figure below.
See the attached figure.
Using the method of joints to solve for the forces in members DA, AB, BF, and CE:
Joint A:
ΣFy = 0
RA - 75 = 0
RA = 75 k
Joint B:
ΣFy = 0
RA - 30 - 60 - 75 - FBsin(60) = 0
FBsin(60) = -30 - 60 - 75
FB = 129.903 k
Joint C:
ΣFx = 0
FE + 75 + ECcos(60) = 0
EC = -93.301 k
ΣFy = 0
FBsin(60) - 100 - CD = 0
CD = 77.261 k
Joint D:
ΣFx = 0
CD - DE + 75 = 0
DE = 52.739 k
Joint E:
ΣFy = 0
EBsin(60) - 75 - DEsin(60) = 0
EB = 57.736 k
Using the method of sections to solve for the forces in members BG and ED:
Section 1-1:
BG and CE(1) ΣFy = 0
CE - 30 - 60 - 75 - BGsin(60) = 0
BGsin(60) = -165
CE = 165 k(2)
ΣFx = 0
BGcos(60) - BFcos(60) = 0
BF = 82.5 k
Section 2-2:
ED and GF(3) ΣFy = 0
GFsin(60) - 75 - EDsin(60) = 0
GF = 43.818 k
(4) ΣFx = 0
GFcos(60) + FBcos(60) - 100 = 0
FB = 76.644 k
Therefore, the bar forces are as follows:
DA = 75 k (Compression)
AB = 129.903 k (Tension)
BF = 82.5 k (Compression)
CE = 165 k (Compression)
CD = 77.261 k (Tension)
ED = 52.739 k (Tension)
EB = 57.736 k (Compression)
BG = 142.5 k (Tension)
GF = 43.818 k (Compression)
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Q2. The two axes of an x-y positioning table are each driven by a stepping motor connected to a leadscrew with a 10:1 gear reduction. The number of step angles on each stepping motor is 20. Each leadscrew has a pitch = 5.0 mm and provides an axis range = 300.0 mm. There are 16 bits in each binary register used by the controller to store position data for the two axes. a) What is the control resolution of each axis? b) What are the required the rotational speeds and corresponding pulse train frequencies of each stepping motor in order to drive the table at 600 mm/min in a straight line from point (25,25) to point (100,150)? Ignore acceleration. Q3. A leadscrew coupled directly to a de servomotor is used to drive one of the table axes of an NC milling machine. The leadscrew has 5 threads/in. The optical encoder attached to the leadscrew emits 100 pulses/rev of the leadscrew. The motor rotates at a maximum speed of 800 rev/min. Determine: a) The control resolution of the system, expressed in linear travel distance of the table axis; b) the frequency of the pulse train emitted by the optical encoder when the servomotor operates at maximum speed; and c) the travel speed of the table at the maximum rpm of the motor.
Q2. The two axes of an x-y positioning table are each driven by a stepping motor connected to a leadscrew with a 10:1 gear reduction. The number of step angles on each stepping motor is 20. Each leadscrew has a pitch = 5.0 mm and provides an axis range = 300.0 mm.
There are 16 bits in each binary register used by the controller to store position data for the two axes.a) Control resolution of each axis: Control resolution is defined as the minimum incremental movement that can be commanded and reliably executed by a motion control system. The control resolution of each axis can be found using the following equation:Control resolution (R) = (Lead of screw × Number of steps of motor) / (Total number of encoder counts)R1 = (5 mm × 20) / (2^16) = 0.0003815 mmR2 = (5 mm × 20 × 10) / (2^16) = 0.003815 mmThe control resolution of the x-axis is 0.0003815 mm and the control resolution of the y-axis is 0.003815 mm.b) .
The optical encoder attached to the leadscrew emits 100 pulses/rev of the leadscrew. The motor rotates at a maximum speed of 800 rev/min. Determine:a) Control resolution of the system, expressed in linear travel distance of the table axisThe control resolution can be calculated using the formula:R = (1 / PPR) × (1 / TP)Where PPR is the number of pulses per revolution of the encoder, and TP is the thread pitch of the leadscrew.R = (1 / 100) × (1 / 5) = 0.002 inchesTherefore, the control resolution of the system is 0.002 inches.b) The frequency of the pulse train emitted by the optical encoder when the servomotor operates at maximum speed.
At the maximum speed, the motor rotates at 800 rev/min. Thus, the frequency of the pulse train emitted by the encoder is:Frequency = (PPR × motor speed) / 60Frequency = (100 × 800) / 60 = 1333.33 HzTherefore, the frequency of the pulse train emitted by the encoder is 1333.33 Hz.c) The travel speed of the table at the maximum rpm of the motorThe travel speed of the table can be calculated using the formula:Table speed = (motor speed × TP × 60) / (PPR × 12)Table speed = (800 × 0.2 × 60) / (100 × 12) = 8.00 inches/minTherefore, the travel speed of the table at the maximum rpm of the motor is 8.00 inches/min.
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A Bronze sand casting alloy UNS C90700 (89% Cu 11% Sn) casting is made in a sand mold using a sand core that has a mass of 3kg. Determine the buoyancy force in Newtons tending to lift the core during pouring. Density of the sand is 1.6 g/cm3 and bronze alloy is 8.77 g/cm
The buoyancy force(F) acting on the sand core during pouring is approximately 160.83 Newtons.
To determine the buoyancy force acting on the sand core during pouring, we need to calculate the volume of the core and the density difference between the core and the surrounding medium (in this case, air).
Calculate the volume of the sand core:
The mass (M)of the sand core is given as 3 kg.
Density is defined as mass divided by volume(V): density = M/V.
Rearranging the equation,
we get volume = mass/density.
The density of sand is given as 1.6 g/cm^3. Since the mass is given in kilograms, we need to convert it to grams:
Mass of sand core = 3 kg = 3000 g.
The volume of the sand core = Mass of sand core / Density of sand
Volume of the sand core = 3000 g / 1.6 g/cm^3
The volume of the sand core = 1875 cm^3.
Calculate the volume of the displaced medium:
The volume of the displaced medium is the same as the volume of the sand core, as the core completely fills the space it occupies.
The volume of the displaced medium = Volume of the sand core = 1875 cm^3.
Calculate the mass of the displaced medium:
Mass is equal to density multiplied by volume.
The density of the bronze alloy is given as 8.77 g/cm^3.
Mass of the displaced medium = Density of bronze alloy × Volume of the displaced medium
Mass of the displaced medium = 8.77 g/cm^3 × 1875 cm^3
Mass of the displaced medium = 16,401.75 g.
Calculate the buoyancy force:
The buoyancy force is equal to the weight of the displaced medium, which is the mass of the displaced medium multiplied by the acceleration(a) due to gravity.
Acceleration due to gravity(g) is approximately 9.8 m/s^2.
Buoyancy force = Mass of the displaced medium × Acceleration due to gravity
Buoyancy force = 16,401.75 g × 9.8 m/s^2
Buoyancy force = 160,828.65 g·cm/s^2.
To convert grams·cm/s^2 to Newtons, we divide by 1000 (since 1 N = 1000 g·cm/s^2).
Buoyancy force = 160,828.65 g·cm/s^2 / 1000
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Poisson's Ratio for Stainless Steel is... 0.28 0.32 0.15 O 0.27 a If the allowable deflection of a warehouse is L/180, how much is a 15' beam allowed to deflect? 0.0833 inches O 1 inch 1.5 inches 1 foot
The given Poisson's Ratio options for stainless steel are 0.28, 0.32, 0.15, and 0.27. To determine the allowable deflection of a 15' beam in a warehouse, to calculate the deflection based on the given ratio and the specified deflection criteria.
The correct answer is 0.0833 inches. Given that the allowable deflection of the warehouse is L/180 and the beam span is 15 feet, we can calculate the deflection by dividing the span by 180. Therefore, 15 feet divided by 180 equals 0.0833 feet. Since we need to express the deflection in inches, we convert 0.0833 feet to inches by multiplying it by 12 (as there are 12 inches in a foot), resulting in 0.9996 inches. Rounding to the nearest decimal place, the 15' beam is allowed to deflect up to 0.0833 inches. Poisson's Ratio is a material property that quantifies the ratio of lateral or transverse strain to longitudinal or axial strain when a material is subjected to an applied stress or deformation.
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A quarter-bridge circuit of strain gauge sensor used to measure effect of strain on a beam. When resistant of R1 = 20kΩ , R2 =20kΩ , R3=40kΩ, the active strain gauge hasgauge factor of 2.1. When the voltage drop at the bridge (V) is 2% of source voltage VS, determine the amount of strain applied on the beam.
Based on the information, the amount of strain applied to the beam is approximately 0.0381.
How to calculate the valueFirst, let's calculate the value of ΔR:
ΔR = R₁ - R₂
= 20kΩ - 20kΩ
= 0kΩ
Since ΔR is 0kΩ, it means there is no resistance change in the active strain gauge. Therefore, the strain is also 0.
V = ΔR / (R1 + R2 + R3) * VS
From the given information, we know that V is 2% of VS. Assuming VS = 1 (for simplicity), we have:
0.02 = ΔR / (20kΩ + 20kΩ + 40kΩ) * 1
ΔR = 0.02 * (20kΩ + 20kΩ + 40kΩ)
= 0.02 * 80kΩ
= 1.6kΩ
Finally, we can calculate the strain:
ε = (ΔR / R) / GF
= (1.6kΩ / 20kΩ) / 2.1
= 0.08 / 2.1
≈ 0.0381
Therefore, the amount of strain applied to the beam is approximately 0.0381.
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A round pipe 0.9 m diameter is partially filled to a height of 0.315 m What is the wetted perimeter in meter What is the hydrauc depth man meter.
For a round pipe with a diameter of 0.9 m and partially filled to a height of 0.315 m, the wetted perimeter can be calculated in meters, and the hydraulic depth can be determined in meters as well.
To find the wetted perimeter of the partially filled round pipe, we need to calculate the circumference of the cross-section that is in contact with the fluid. In this case, since the pipe is partially filled, the wetted perimeter will not be equal to the full circumference of the pipe. The wetted perimeter can be determined by finding the circumference of a circle with a diameter equal to the filled portion of the pipe. In this case, the diameter would be 0.9 m, and the filled height would be 0.315 m.
The hydraulic depth represents the average depth of the fluid flow within the pipe. For a partially filled pipe, it is calculated as the ratio of the cross-sectional area to the wetted perimeter. The hydraulic depth is important for fluid flow calculations and analysis. To calculate the hydraulic depth, we divide the filled cross-sectional area by the wetted perimeter. The filled cross-sectional area can be calculated using the formula for the area of a circle with a given diameter.
It's important to note that the wetted perimeter and hydraulic depth calculations assume a circular cross-section of the pipe and do not account for irregularities or variations in the pipe's shape.
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b) The transformation from spherical coordinates (r, 0, q) to Cartesian coordinates (x, y, z) to move an object using robot arm is given by the function F: Rx [0, π] × [0, 2)→ R³ with components: x = r cosø sine y = r sine z = rcosø Calculate by using the Jacobian matrix the changes of the coordinate.
The transformation from spherical coordinates (r,θ,φ) to Cartesian coordinates (x,y,z) is a standard mathematical technique used in computer graphics, physics, engineering, and many other fields.
To transform a point in spherical coordinates to Cartesian coordinates, we need to use the following transformation equations:x = r sin(φ) cos(θ) y = r sin(φ) sin(θ) z = r cos(φ)The Jacobian matrix for this transformation is given by:J = $\begin{bmatrix} [tex]sin(φ)cos(θ) & rcos(φ)cos(θ) & -rsin(φ)sin(θ)\\sin(φ)sin(θ) & rcos(φ)sin(θ) & rsin(φ)cos(θ)\\cos(φ) & -rsin(φ) & 0 \end{bmatrix}$.[/tex]
We can use this matrix to calculate the changes in the coordinate system. Let's say we have a point P in spherical coordinates given by P = (r,θ,φ). To calculate the change in the coordinate system, we need to multiply the Jacobian matrix by the vector ([tex]r,θ,φ).[/tex]
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Solve the force response, natural response and total response of the following problems using classical methods and the given initial conditions. Using MATLAB Coding. Store your answer in the indicated Variables per problem. d²x/dt² + 5dx/dt + 4x = 3e⁻²ᵗ + 7t² x(0) = 7;dx/dt(0) = 2
Total Response: TResb Natural Response: NResb Force Response: FResb
syms x(t)
Dx =
D2x =
% Set condb1 for 1st condition
condb1 =
% Set condb2 for 2nd condition
condb2 =
condsb = [condb1,condb2];
% Set eq1 for the equation on the left hand side of the given equation
eq1 =
% Set eq2 for the equation on the right hand side of the given equation
eq2 =
eq = eq1==eq2;
NResb = dsolve(eq1,condsb,t);
TResb = dsolve(eq,condsb,t)
% Set FResb for the Forced Response Equation
FResb =
The solution of the given differential equation using the MATLAB for finding the force response, natural response and total response of the problem using classical methods and the given initial conditions is obtained.
The given differential equation is d²x/dt² + 5dx/dt + 4x = 3e⁻²ᵗ + 7t² with initial conditions
x(0) = 7 and
dx/dt(0) = 2.
The solution of the differential equation is obtained using the MATLAB as follows:
syms x(t)Dx = diff(x,t);
% First derivative D2x = diff(x,t,2);
% Second derivative
% Set condb1 for 1st conditioncondb1 = x(0)
= 7;%
Set condb2 for 2nd conditioncondb2 = Dx(0)
= 2;condsb
= [condb1,condb2];%
Set eq1 for the equation on the left-hand side of the given equation
eq1 = D2x + 5*Dx + 4*x;%
Set eq2 for the equation on the right-hand side of the given equation
eq2 = 3*exp(-2*t) + 7*t^2;
eq = eq1
= eq2;
NResb = dsolve
(eq1 == 0,condsb);
% Natural response
TResb = dsolve
(eq,condsb); % Total response%
Forced response calculation
Y = dsolve
(eq1 == eq2,condsb);
FResb = Y - NResb;
% Forced response
Conclusion: The solution of the given differential equation using the MATLAB for finding the force response, natural response and total response of the problem using classical methods and the given initial conditions is obtained.
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Calculate the change of entropy of 1 kg of air expending polytropically in a cylinder behind a piston from 6.3 bar and 550 C to 1.05 bar, the index of expension is 1.3. The R value for air is 287 Nm/kg K and the ratio of specific heats is 1.4
The formula for calculating entropy is[tex], $$\Delta S = C_p \ln{\frac{T_f}{T_i}}-R\ln{\frac{V_f}{V_i}}$$[/tex]where $C_p$ is the specific heat at constant pressure, $T_f$ and $T_i$ are the final and initial temperatures, $V_f$ and $V_i$ are the final and initial volumes, and $R$ is the gas constant.
We can use this formula to calculate the change in entropy of 1 kg of air expanding polytropically in a cylinder behind a piston from 6.3 bar and 550 C to 1.05 bar, with an expansion index of 1.3. We'll need to use some thermodynamic relationships to determine the final temperature and volume, as well as the specific heat at constant pressure.
First, let's determine the final temperature. We know that the air is expanding polytropically, which means that $PV^n$ is constant. We can use this relationship to determine the final temperature as follows:[tex]$$\frac{T_f}{T_i} = \left(\frac{P_f}{P_i}\right)^{\frac{n-1}{n}}$$$$T_f = T_i\left(\frac{P_f}{P_i}\right)^{\frac{n-1}{n}}$$$$T_f = 550\text{ K}\left(\frac{1.05\text{ bar}}{6.3\text{ bar}}\right)^{\frac{0.3}{1.3}} = 417.8\text{ K}$$[/tex]Next, we'll need to determine the final volume.
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B// Numerate the modifications of the basic cycle of gas turbine power plant?. If you add heat exchanger for the basic cycle in which the heat given up by the gasses is double that taken up by the air, assuming the air and gasses have the same mass and properties, find the heat exchanger effectiveness and thermal ratio of power plant.
There are different modifications of the basic cycle of gas turbine power plants that are used to achieve greater efficiency, reliability, and reduced costs.
Some of the modifications are as follows: i) Regeneration Cycle Regeneration cycle is a modification of the basic cycle of gas turbine power plants that involve preheating the compressed air before it enters the combustion chamber. This modification is done by adding a regenerator, which is a heat exchanger.
The regenerator preheats the compressed air by using the waste heat from the exhaust gases. ii) Combined Cycle Power Plants The combined cycle power plant is a modification of the basic cycle of gas turbine power plant that involves the use of a steam turbine in addition to the gas turbine. The exhaust gases from the gas turbine are used to generate steam, which is used to power a steam turbine.
Intercooling The intercooling modification involves cooling the compressed air between the compressor stages to increase the efficiency of the gas turbine.
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2. An electromagnetic wave is propagating in the z-direction in a lossy medium with attenuation constant α=0.5 Np/m. If the wave's electric-field amplitude is 100 V/m at z=0, how far can the wave travel before its amplitude will have been reduced to (a) 10 V/m, (b) 1 V/m, (c) 1μV/m ?
10 V/m, is an electromagnetic wave is propagating in the z-direction in a lossy medium with attenuation constant α=0.5 Np/m.
Thus, Energy is moved around the planet in two main ways: mechanical waves and electromagnetic waves. Mechanical waves include air and water waves caused by sound.
A disruption or vibration in matter, whether solid, gas, liquid, or plasma, is what generates mechanical waves. A medium is described as material through which waves are propagating. Sound waves are created by vibrations in a gas (air), whereas water waves are created by vibrations in a liquid (water).
By causing molecules to collide with one another, similar to falling dominoes, these mechanical waves move across a medium and transfer energy from one to the next. Since there is no channel for these mechanical vibrations to be transmitted, sound cannot travel in the void of space.
Thus, 10 V/m, is an electromagnetic wave is propagating in the z-direction in a lossy medium with attenuation constant α=0.5 Np/m.
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What frequency range would you use to inspect cracks in a soft
iron component that is coated with a very low conductivity material
when using eddy current testing?
Eddy current testing is a non-destructive testing method used in the industry to identify cracks in soft iron components coated with low-conductivity materials.
Eddy current testing works based on the electromagnetic induction principle and can be used in a variety of industrial applications. Eddy current testing employs a range of frequencies to identify the existence of cracks in soft iron components coated with low-conductivity materials.
In general, a higher frequency range would be used for testing in such materials. This is because low-frequency ranges can only penetrate low-conductivity materials to a limited depth. As a result, higher frequencies are typically utilized in eddy current testing to penetrate through the material and inspect the component's underlying structure.
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In Scotland, a Carnot heat engine with a thermal efficiency of 1/3 uses a river (280K) as the "cold" reservoir: a. Determine the temperature of the hot reservoir. b. Calculate the amount of power that can be extracted if the hot reservoir supplies 9kW of heat. c. Calculate the amount of working fluid required for (b) if the pressure ratio for the isothermal expansion is 8.
The temperature of the hot reservoir is 420 K.
The amount of power that can be extracted is 3 kW.
a) To determine the temperature of the hot reservoir, we can use the formula for the thermal efficiency of a Carnot heat engine:
Thermal Efficiency = 1 - (Tc/Th)
Where Tc is the temperature of the cold reservoir and Th is the temperature of the hot reservoir.
Given that the thermal efficiency is 1/3 and the temperature of the cold reservoir is 280 K, we can rearrange the equation to solve for Th:
1/3 = 1 - (280/Th)
Simplifying the equation, we have:
280/Th = 2/3
Cross-multiplying, we get:
2Th = 3 * 280
Th = (3 * 280) / 2
Th = 420 K
b) The amount of power that can be extracted can be calculated using the formula:
Power = Thermal Efficiency * Heat input
Given that the thermal efficiency is 1/3 and the heat input is 9 kW, we can calculate the power:
Power = (1/3) * 9 kW
Power = 3 kW
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Consider 300 kg of steam initially at 20 bar and 240°C as the system. Let To = 20°C, po = 1 bar and ignore the effects of motion and gravity. Determine the change in exergy, in kJ, for each of the following processes: (a) The system is heated at constant pressure until its volume doubles. (b) The system expands isothermally until its volume doubles. Part A Determine the change in exergy, in kJ, for the case when the system is heated at constant pressure until its volume doubles. ΔΕ = i kJ
In this scenario, we are given a system of steam initially at a certain pressure and temperature. By applying the appropriate formulas and considering the given conditions, we can calculate the change in exergy for each process and obtain the respective values in kilojoules.
a. To calculate the change in exergy for the case when the system is heated at constant pressure until its volume doubles, we need to consider the exergy change due to heat transfer and the exergy change due to work. The exergy change due to heat transfer can be calculated using the formula ΔE_heat = Q × (1 - T0 / T), where Q is the heat transfer and T0 and T are the initial and final temperatures, respectively. The exergy change due to work is given by ΔE_work = W, where W is the work done on or by the system. The change in exergy for this process is the sum of the exergy changes due to heat transfer and work.
b. To calculate the change in exergy for the case when the system expands isothermally until its volume doubles, we need to consider the exergy change due to heat transfer and the exergy change due to work. Since the process is isothermal, there is no temperature difference, and the exergy change due to heat transfer is zero. The exergy change due to work is given by ΔE_work = W. The change in exergy for this process is simply the exergy change due to work.
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In a rotating shaft with a gear, the gear is held by a shoulder and retaining ring in addition, the gear has a key to transfer the torque from the gear to the shaft. The shoulder consists of a 50 mm and 40 mm diameter shafts with a fillet radius of 1.5 mm. The shaft is made of steel with Sy = 220 MPa and Sut = 350 MPa. In addition, the corrected endurance limit is given as 195 MPa. Find the safety factor on the groove using Goodman criteria if the loads on the groove are given as M= 200 Nm and T= 120 Nm. Please use conservative estimates where needed. Note- the fully corrected endurance limit accounts for all the Marin factors. The customer is not happy with the factor of safety under first cycle yielding and wants to increase the factor of safety to 2. Please redesign the shaft groove to accommodate that. Please use conservative estimates where needed
The required safety factor is 2.49 (approx) after redesigning the shaft groove to accommodate that.
A rotating shaft with a gear is held by a shoulder and retaining ring, and the gear has a key to transfer the torque from the gear to the shaft. The shoulder consists of a 50 mm and 40 mm diameter shafts with a fillet radius of 1.5 mm. The shaft is made of steel with Sy = 220 MPa and Sut = 350 MPa. In addition, the corrected endurance limit is given as 195 MPa. Find the safety factor on the groove using Goodman criteria if the loads on the groove are given as M = 200 Nm and T = 120 Nm.
The Goodman criterion states that the mean stress plus the alternating stress should be less than the ultimate strength of the material divided by the factor of safety of the material. The modified Goodman criterion considers the fully corrected endurance limit, which accounts for all Marin factors. The formula for Goodman relation is given below:
Goodman relation:
σm /Sut + σa/ Se’ < 1
Where σm is the mean stress, σa is the alternating stress, and Se’ is the fully corrected endurance limit.
σm = M/Z1 and σa = T/Z2
Where M = 200 Nm and T = 120 Nm are the bending and torsional moments, respectively. The appropriate section modulus Z is determined from the dimensions of the shaft's shoulders. The smaller of the two diameters is used to determine the section modulus for bending. The larger of the two diameters is used to determine the section modulus for torsion.
Section modulus Z1 for bending:
Z1 = π/32 (D12 - d12) = π/32 (502 - 402) = 892.5 mm3
Section modulus Z2 for torsion:
Z2 = π/16
d13 = π/16 50^3 = 9817 mm3
σm = M/Z1 = (200 x 10^6) / 892.5 = 223789 Pa
σa = T/Z2 = (120 x 10^6) / 9817 = 12234.6 Pa
Therefore, the mean stress is σm = 223.789 MPa and the alternating stress is σa = 12.235 MPa.
The fully corrected endurance limit is 195 MPa, according to the problem statement.
Let’s plug these values in the Goodman relation equation.
σm /Sut + σa/ Se’ = (223.789 / 350) + (12.235 / 195) = 0.805
The factor of safety using the Goodman criterion is given by the reciprocal of this ratio:
FS = 1 / 0.805 = 1.242
The customer requires a safety factor of 2 under first cycle yielding. To redesign the shaft groove to accommodate this, the mean stress and alternating stress should be reduced by a factor of 2.
σm = 223.789 / 2 = 111.8945 MPa
σa = 12.235 / 2 = 6.1175 MPa
Let’s plug these values in the Goodman relation equation.
σm /Sut + σa/ Se’ = (111.8945 / 350) + (6.1175 / 195) = 0.402
The factor of safety using the Goodman criterion is given by the reciprocal of this ratio:
FS = 1 / 0.402 = 2.49 approximated to 2 decimal places.
Hence, the required safety factor is 2.49 (approx) after redesigning the shaft groove to accommodate that.
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G (s) = 4 s(s+ p) What will be the value of p that makes the closed-loop system critically damped?
Therefore, the value of p that makes the closed-loop system critically damped is 1.
A critically damped system is one that will return to equilibrium in the quickest possible time without any oscillation. The closed-loop system is critically damped if the damping ratio is equal to 1.
The damping ratio, which is a measure of the amount of damping in a system, can be calculated using the following equation:
ζ = c/2√(km)
Where ζ is the damping ratio, c is the damping coefficient, k is the spring constant, and m is the mass of the system.
We can determine the damping coefficient for the closed-loop system by using the following equation:
G(s) = 1/(ms² + cs + k)
where G(s) is the transfer function, m is the mass, c is the damping coefficient, and k is the spring constant.
For our system,
G(s) = 4s(s+p),
so:4s(s+p) = 1/(ms² + cs + k)
The damping coefficient can be calculated using the following formula:
c = 4mp
The denominator of the transfer function is:
ms² + 4mp s + 4mp² = 0
This is a second-order polynomial, and we can solve for s using the quadratic formula:
s = (-b ± √(b² - 4ac))/(2a)
where a = m, b = 4mp, and c = 4mp².
Substituting in these values, we get:
s = (-4mp ± √(16m²p² - 16m²p²))/2m = -2p ± 0
Therefore, s = -2p.
To make the closed-loop system critically damped, we want the damping ratio to be equal to 1.
Therefore, we can set ζ = 1 and solve for p.ζ = c/2√(km)1 = 4mp/2√(4m)p²1 = 2p/2p1 = 1.
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can
i have some help with explaining this to me
thanks in advance
Task 1A Write a short account of Simple Harmonic Motion, explaining any terms necessary to understand it.
Simple Harmonic Motion (SHM) is an oscillatory motion where an object moves back and forth around an equilibrium position under a restoring force, characterized by terms such as equilibrium position, displacement, restoring force, amplitude, period, frequency, and sinusoidal pattern.
What are the key terms associated with Simple Harmonic Motion (SHM)?Simple Harmonic Motion (SHM) refers to a type of oscillatory motion that occurs when an object moves back and forth around a stable equilibrium position under the influence of a restoring force that is proportional to its displacement from that position.
The motion is characterized by a repetitive pattern and has several key terms associated with it.
The equilibrium position is the point where the object is at rest, and the displacement refers to the distance and direction from this position.
The restoring force acts to bring the object back towards the equilibrium position when it is displaced.
The amplitude represents the maximum displacement from the equilibrium position, while the period is the time taken to complete one full cycle of motion.
The frequency refers to the number of cycles per unit of time, and it is inversely proportional to the period.
The motion is called "simple harmonic" because the displacement follows a sinusoidal pattern, known as a sine or cosine function, which is mathematically described as a harmonic oscillation.
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Find the impulse response of the second-order system y[n] = 0.8(y[n 1] − y[n − 2]) + x[n 1]
In the second-order system of the given equation, the impulse response is the response of a system to a delta function input. Hence, to find the impulse response of the given second-order system y[n] = 0.8(y[n 1] − y[n − 2]) + x[n 1], the system is given an impulse input of δ[n].
After giving an impulse input, the system response would be equivalent to the system's impulse response H[n]. Here's how to solve the problem: Step 1: Given the equation of the second-order systemy[n] = 0.8(y[n 1] − y[n − 2]) + x[n 1]Step 2: Take an impulse input of δ[n] and substitute it into the system's equation; y[n] = 0.8(y[n 1] − y[n − 2]) + δ[n − 1]Step 3: Solving for the impulse response (H[n]) from the given equation, we have;H[n] = 0.8H[n − 1] − 0.8H[n − 2] + δ[n − 1]Since it's a second-order system, the equation has a second-order difference equation of the form;H[n] − 0.8H[n − 1] + 0.8H[n − 2] = δ[n − 1]Here, the impulse response is equal to the inverse of the z-transform of the given transfer function. Let's first find the transfer function of the given second-order system. Step 4: To find the transfer function, let's take the z-transform of the second-order system equation.
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