The estimated tip edge shear stress on the blade of the helicopter is approximately 1.564 Pa (Pascals).
To calculate the tip edge shear stress on the blade of the helicopter, we need to determine the velocity at the tip of the blade.
The forward speed of the helicopter is given as 50 m/s, and the blade rotates at 0.3 Hz.
First, let's calculate the linear speed at the tip of the blade. The circumference of the rotor is equal to the distance traveled by the tip of the blade in one revolution.
Circumference =
π * Diameter = 3.1415 * 36 m = 113.0976 m
Therefore, in one revolution, the tip of the blade travels a distance of 113.0976 m. Since the blade rotates at 0.3 Hz, the time taken for one revolution is 1/0.3 = 3.33 seconds.
Hence, the linear speed at the tip of the blade is:
Linear Speed = Distance / Time = 113.0976 m / 3.33 s ≈ 33.96 m/s
Now, we need to calculate the tip edge shear stress using the given information about the boundary layer. The equation for the log-law of the wall is:
U⁺ = (1/0.4) * ln(y⁺) + 5.5
Where U⁺ is the dimensionless velocity, and y⁺ is the dimensionless distance from the wall in wall units.
At the edge of the boundary layer, y⁺ = 343. From the given equation, we can solve for U⁺:
U⁺ = (1/0.4) * ln(343) + 5.5 ≈ 11.09
To convert U⁺ to dimensional velocity, we need to use the formula:
U⁺ = U / (ν / Uτ)
Where U is the velocity, ν is the kinematic viscosity, and Uτ is the shear velocity.
Rearranging the equation, we get:
Uτ = U / U⁺ * (ν / Uτ)
Uτ² = U * ν / U⁺
Uτ = sqrt(U * ν / U⁺)
Since U = 33.96 m/s, ν is the kinematic viscosity of air at sea level, which is approximately 1.46 ×[tex]10^(-5) m²/s[/tex].
Substituting the values, we have:
Uτ = sqrt(33.96 * 1.46 × 1[tex]0^(-5)[/tex] / 11.09)
Uτ ≈ 1.101 m/s
Finally, the tip edge shear stress (τ) can be calculated using the equation:
τ = ρ * Uτ²
Where ρ is the density of air at sea level, which is approximately 1.225 kg/m³.
Substituting the values, we get:
τ = 1.225 * (1.101)²
τ ≈ 1.564 Pa
Therefore, the estimated tip edge shear stress on the blade of the helicopter is approximately 1.564 Pascal (Pa).
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How would you link the capacity decision being made by Fitness Plus to other types of operating decisions?
Fitness Plus, an emerging fitness and gym provider, is trying to gain a significant share of the market in the region, making it a major competitor to other industry players. Fitness Plus's decision to expand its capacity is critical, and it influences the types of operating decisions they make, including marketing, financial, and human resource decisions.
Capacity decisions at Fitness Plus are linked to marketing decisions in several ways. When Fitness Plus decides to expand its capacity, it means that it is increasing the number of customers it can serve simultaneously. The expansion creates an opportunity to increase sales by catering to a more extensive market. Fitness Plus's marketing team must focus on building brand awareness to attract new customers and create loyalty among existing customers.The expansion also influences financial decisions. Fitness Plus must secure funding to finance the expansion project.
It means that the financial team must identify potential sources of financing, analyze their options, and determine the most cost-effective alternative. Fitness Plus's decision to expand its capacity will also have a significant impact on its human resource decisions. The expansion creates new job opportunities, which Fitness Plus must fill. Fitness Plus must evaluate its staffing requirements and plan its recruitment strategy to attract the most qualified candidates.
In conclusion, Fitness Plus's decision to expand its capacity has a significant impact on its operating decisions. The expansion influences marketing, financial, and human resource decisions. By considering these decisions together, Fitness Plus can achieve its growth objectives and increase its market share in the region.
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(a) Use the defining equation for the FS coefficients to evaluate the FS representation of the following signals. X(t)=2sin(5πt)+5Cos(3πt) b)Find the FT of the following signal X(t)=e^−2t ⋅u(t−1)
Using these equations, we can now find the FS coefficients for the given signal. The equation of the Fourier Series Coefficients is represented by:
where X(t) is a periodic signal with a period of 2π.The periodic signal X(t) = 2sin(5πt) + 5cos(3πt) can be written in a simpler form by using the trigonometric identities as shown:
[tex]X(t) = 2sin(5πt) + 5cos(3πt) = 5/2[2/5sin(5πt) + cos(3πt)] + 5/2[2/5sin(5πt) - cos(3πt)] = 5/2cos(π/2 - 5πt) + 5/2cos(π/2 + 5πt) + 2/5sin(5πt)[/tex]
For X(t) = 2sin(5πt) + 5cos(3πt), the Fourier series coefficients are as follows:
Therefore, the Fourier series representation of X(t) is given by: X(t) = 2sin(5πt) + 5cos(3πt) ≈ 1.14 + 4.07cos(3πt) + 1.14sin(5πt)
b) Find the Fourier Transform (FT) of the given signal X(t) = e^-2t ⋅u(t - 1) where u(t) is the unit step function.
To find the Fourier transform of the given signal, we need to take the Laplace transform of the signal first since they are related as:
[tex]L{e^−at} = 1/(s + a) and L{u(t − a)} = e^−as/sFor the given signal, X(t) = e^-2t ⋅u(t - 1),[/tex] we can rewrite it as follows:
Applying the Laplace transform to both sides, we get:
Therefore, the Fourier transform of
X(t) = e^-2t ⋅u(t - 1) is: X(ω) = 1/(jω - 2) ⋅ e^-jω
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A circular loop with a radius of 10 cm is formed by superimposing 5 identical conductive coils. The loop is oriented in the horizontal plane Oxy of a Cartesian reference frame. In this region, there is a uniform magnetic field of 2 T directed in the positive direction of the z axis (i.e. upwards).
Q.3.1) It can be seen that from a certain point onwards, the magnetic field decreases in intensity in a constant manner over time and is cancelled in 10 seconds. What is the electromotive force induced on the loop during this time? Give details of your calculation.
Q.3.2) Looking at the loop from top to bottom, what is the direction (clockwise or anti-clockwise) of the current induced in the loop? Give details of your reasoning.
Q.3.3) If the self-inductance of the loop is 1 µH, what is the magnetic flux caused by a current of 10 A in the turns? Give details of your calculation.
Electromotive force (EMF) induced on the loop can be calculated by Faraday's law of electromagnetic induction. According to Faraday's law of electromagnetic induction
Q.3.1) The EMF induced in a conductor is equal to the rate of change of magnetic flux through the area of the conductor. Mathematically, it can be expressed as:
EMF = -dΦ/dt
where Φ is the magnetic flux and t is the time given. During this time, the magnetic field decreases in intensity in a constant manner over time and is cancelled in 10 seconds. The time taken to decrease the magnetic field from its initial value to zero is 10 seconds. Therefore, the rate of change of magnetic flux is given by:
-dΦ/dt = ΔΦ/Δt
We know that the magnetic flux through the loop is given by:
Φ = B.A
where B is the magnetic field, and A is the area of the loop. The radius of the loop, r = 10 cmTherefore, the area of the loop,
A = πr²= π(0.1m)²= 0.0314 m²
The magnetic field B = 2 T
The time taken to decrease the magnetic field from its initial value to zero is 10 seconds. Therefore, the rate of change of magnetic flux is given by:-
dΦ/dt = ΔΦ/Δt
= Φf - Φi/
= (2 × 0.0314) - 0 / 10
= 0.0628 T-m/s
Substituting the values in the formula of EMF, we get:
EMF = -dΦ/dt
= - 0.0628
= -0.0628 V
Therefore, the EMF induced in the loop during this time is 0.0628 V.
Q.3.2) According to Lenz's law, the direction of the induced EMF produces a current in the conductor that opposes the change in the magnetic flux that produced it. The induced current sets up its own magnetic field which opposes the original magnetic field. Hence, the direction of the induced current can be determined by using Lenz's law. Here, we know that the magnetic field is decreasing over time.
Q.3.3) Magnetic flux caused by a current of 10 A in the turns can be calculated using the formula:
Φ = L.I
where, L is the self-inductance of the loop, and I is the current flowing in the loop. Substituting the values in the formula, we get:
Φ = L.I= (1 × 10⁻⁶) × 10= 10⁻⁵ Wb
Therefore, the magnetic flux caused by a current of 10 A in the turns is 10⁻⁵ Wb.
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The work function of a metal surface is 4.5 eV. If the frequency of the light incident upon it is 1.45 × 1015 Hz, then what is the maximum kinetic energy (in eV) of the photo electrons emitted from the surface?
The maximum kinetic energy (in eV) of the photo electrons emitted from the surface is 6 ev.
To calculate the maximum kinetic energy of photoelectrons emitted from a metal surface, we can use the equation:
E max=hν−φ
Where: E max is the maximum kinetic energy of photoelectrons,
h is the Planck's constant (4.135667696 × 10⁻¹⁵ eV s),
ν is the frequency of the incident light (1.45 × 10¹⁵ Hz),
φ is the work function of the metal surface (4.5 eV).
Plugging in the values:
E max =(4.135667696×10⁻¹⁵ eV s)×(1.45×10¹⁵ Hz)−4.5eV
Calculating the expression:
E max =5.999eV
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A natural-circulation pillow-block bearing has a journal diameter D of 62.5 mm with a unilateral tolerance of -0.025 mm. The bushing bore diameter B is 62.6 mm with a unilateral tolerance of 0.1 mm. The shaft runs at an angular speed of 1120 rev/min; the bearing uses SAE grade 20 oil and carries a steady load of 1350 N in shaft- stirred air at 21°C. The lateral area of the pillow-block housing is 38,700 mm2. Perform a design assessment using minimum radial clearance for a load of 2700 N and 1350 N. Use Trumpler's criteria and that both 1/d and a are unity.
Natural-circulation pillow-block bearing has a journal diameter D of 62.5 mm with a unilateral tolerance of -0.025 mm. The bushing bore diameter B is 62.6 mm with a unilateral tolerance of 0.1 mm.
The shaft runs at an angular speed of 1120 rev/min; the bearing uses SAE grade 20 oil and carries a steady load of 1350 N in shaft- stirred air at 21°C. The lateral area of the pillow-block housing is 38,700 mm². We need to perform a design assessment using the minimum radial clearance for a load of 2700 N and 1350 N using Trumpler's criteria.
Both `1/d` and `a` are unity. Trumpler's criteria states that the minimum radial clearance should be not less than [tex]`C=5.3(1/d)^(1/3)a^(2/3)`mm[/tex]. Given that the `1/d` and `a` are unity. `[tex]1/d=1`, and `a=1[/tex]`.Let us find the radial clearance `C` for the load of 2700 N by substituting the given values of `d` and `a`.`[tex]C=5.3(1/d)^(1/3)a^(2/3)[/tex]`For load = 2700 N: `[tex]C=5.3(1/62.5)^(1/3)×1^(2/3)` = `0.051 mm[/tex].
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In your understanding, differentiate between 3-aspect signalling and 4-aspect signalling.
3-aspect and 4-aspect signaling are two different methods of railway signalling that are used to ensure safety and provide information to train drivers. In this context, aspect refers to the number of lights used in the signal to convey information to the driver.3-aspect signalling uses three colours of light: red, yellow, and green.
The meanings of these colours in 3-aspect signalling are as follows:Red: This indicates that the driver must stop the train immediately. It is used when there is a danger ahead, such as a broken track or an obstruction.Yellow: This indicates that the driver should slow down and be prepared to stop at the next signal. It is used when there is a warning ahead, such as a slower train or construction work.Green: This indicates that the driver may proceed at the normal speed. It is used when the track ahead is clear.4-aspect signalling uses four colours of light: red, yellow, green, and double yellow.
The meanings of these colours in 4-aspect signalling are as follows:Red: This indicates that the driver must stop the train immediately. It is used when there is a danger ahead, such as a broken track or an obstruction.Yellow: This indicates that the driver should slow down and be prepared to stop at the next signal. It is used when there is a warning ahead, such as a slower train or construction work.Green: This indicates that the driver may proceed at the normal speed. It is used when the track ahead is clear.
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Please show all work neatly and double-check work before uploading. A natural gas with the molar analysis 78% CH₄, 13% C₂H₆, 6% C₃H₈, 1.7 % C₄H₁₀, 1.3% N₂ burns completely with 30% excess air in a reactor operating at steady state. The molar flow rate of the fuel is 3.5 kmol/h and air is 21% O₂ and 79% N₂ on a molar basis. Determine the molar flow rate of the air, in kmol/h. Determine mass flow rate of the air, in kg/h Determine the mole fraction of water vapor in the products
To determine the molar flow rate of air, we first need to calculate the amount of oxygen required for the combustion of the natural gas. Given the molar flow rate of the fuel as 3.5 kmol/h, and the molar analysis of the natural gas (78% CH₄, 13% C₂H₆, 6% C₃H₈, 1.7% C₄H₁₀).
We can calculate the molar flow rate of oxygen (O₂) required as follows:
Moles of CH₄ = 0.78 * 3.5 kmol/h = 2.73 kmol/h
Moles of C₂H₆ = 0.13 * 3.5 kmol/h = 0.455 kmol/h
Moles of C₃H₈ = 0.06 * 3.5 kmol/h = 0.21 kmol/h
Moles of C₄H₁₀ = 0.017 * 3.5 kmol/h = 0.0595 kmol/h
Total moles of carbon (C) = Moles of CH₄ + Moles of C₂H₆ + Moles of C₃H₈ + Moles of C₄H₁₀
= 2.73 + 0.455 + 0.21 + 0.0595
= 3.4545 kmol/h
Moles of O₂ required = Moles of carbon * 1.5 (stoichiometric ratio)
= 3.4545 * 1.5
= 5.1818 kmol/h
Since the air contains 21% O₂ on a molar basis, we can calculate the molar flow rate of air:
Molar flow rate of air = Moles of O₂ required / 0.21 (molar fraction of O₂ in air)
= 5.1818 / 0.21
≈ 24.677 kmol/h
To determine the mass flow rate of air, we need to consider the molecular weights of the components. The molecular weight of N₂ is 28 g/mol and the molecular weight of O₂ is 32 g/mol.
Mass flow rate of air = Molar flow rate of air * (28 g/mol * 0.79 + 32 g/mol * 0.21)
≈ 24.677 * (22.12 + 6.72)
≈ 718.91 kg/h
To find the mole fraction of water vapor in the products, we need to consider the combustion reaction and the molar flow rates of the different components.
The combustion reaction for CH₄ can be written as:
CH₄ + 2O₂ -> CO₂ + 2H₂O
The moles of water vapor produced will be twice the moles of CH₄ consumed.
Moles of water vapor = 2 * Moles of CH₄
= 2 * 2.73 kmol/h
= 5.46 kmol/h
To calculate the mole fraction of water vapor, we divide the moles of water vapor by the total moles in the products:
Mole fraction of water vapor = Moles of water vapor / (Moles of water vapor + Moles of CO₂)
= 5.46 / (5.46 + Moles of CO₂)
The moles of CO₂ can be determined by multiplying the moles of carbon (C) by the stoichiometric ratio:
Moles of CO₂ = Moles of carbon *
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1) A plane sound wave with a Prms(i) = 80 Pa value is normally incident to a sand bottom in sea water. The characteristic impedance of sea water is 1.54 x 10 MKS rayls and of sand is 4.0 x 106 MKS rayls. a) What is the value of the reflected sound wave. b) What is the value of the transmitted sound wave. c) What is the value of the sound power reflection coefficient, a . d) What is the value of the sound power transmission coefficient, ag.
a.The value of the reflected sound wave is 79.9591 Pa.
b. The value of the transmitted sound wave is 415.5844 Pa.
c. The value of the sound power reflection coefficient is 0.9995.
d.The value of the sound power transmission coefficient is 17.3396.
From the question above, A plane sound wave with a Prms(i) = 80 Pa value is normally incident to a sand bottom in sea water. The characteristic impedance of sea water is 1.54 x 10 MKS rayls and of sand is 4.0 x 106 MKS rayls.
Formulas: For reflected sound wave, PR = R / I
Where, PR = Reflected pressure wave amplitude
R = (Z2 - Z1) / (Z2 + Z1)
I = Incident pressure wave amplitude
For transmitted sound wave, PT = T / I
Where, PT = Transmitted pressure wave amplitude
T = 2Z2 / (Z2 + Z1)
I = Incident pressure wave amplitude
For sound power reflection coefficient, α = PR2 / PI2
Where, PR = Reflected pressure wave amplitude
PI = Incident pressure wave amplitude
For sound power transmission coefficient, ag = PT2 / PI2
Where, PT = Transmitted pressure wave amplitude
PI = Incident pressure wave amplitude
a) Reflected sound wave: The reflected pressure wave amplitude is PR. The incident pressure wave amplitude is PI. The reflected wave equation is given by PR = R / I.
Substituting the given values, we get
R = (Z2 - Z1) / (Z2 + Z1) = (4.0 × 106 − 1.54 × 10) / (4.0 × 106 + 1.54 × 10)= 3.99846 × 106 / 4.00054 × 106= 0.9994885893
PR = R / I = 0.9994885893 × 80= 79.95908714
b) Transmitted sound wave: The transmitted pressure wave amplitude is PT. The incident pressure wave amplitude is PI. The transmitted wave equation is given by PT = T / I.
Substituting the given values, we getT = 2Z2 / (Z2 + Z1) = 2 × 4.0 × 106 / (4.0 × 106 + 1.54 × 10)= 8.0 × 106 / 4.0 × 106 + 0.154 × 106= 8.0 / 1.54= 5.194805195
PT = T / I = 5.194805195 × 80= 415.5844155
c) Sound power reflection coefficient: The reflected pressure wave amplitude is PR and the incident pressure wave amplitude is PI.
The sound power reflection coefficient equation is given by α = PR2 / PI2.
Substituting the given values, we getα = PR2 / PI2= 79.95908714² / 80²= 0.9994885893
d) Sound power transmission coefficient: The transmitted pressure wave amplitude is PT and the incident pressure wave amplitude is PI.
The sound power transmission coefficient equation is given by ag = PT2 / PI2.
Substituting the given values, we getag = PT2 / PI2= 415.5844155²/ 80²= 17.33956419
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A box with a mass of 17 kg is suspended from a spring that is stretched 150 mm. If the box is displaced 100 mm downward from its equilibrium position and given a downward velocity of 700 mm/s, determine the equation which describes the motion. What is the phase angle and amplitude of vibration? Assume that positive displacement is downward.
The box is in simple harmonic motion with the following parameters. Since the box is displaced from equilibrium and is given an initial velocity, it vibrates with amplitude and has a phase angle.
In simple harmonic motion,
x = A sin (ωt + φ).
x = A sin (ωt + φ)
can be used to describe the equation of motion for the given problem.For this equation of motion, the amplitude (A) and phase angle (φ) must be calculated using the given conditions.ω, the angular frequency, can be found using the formula for a mass-spring system's angular frequency:
ω = sqrt(k/m)
where k is the spring constant and m is the mass of the box .
In this case, the box is displaced 100 mm downward from its equilibrium position, thus the amplitude of vibration is A = 100 mm. The phase angle can be determined using the following equation:
φ = arctan(-v0/ωx)
where v0 is the initial velocity (700 mm/s), ω is the angular frequency (9.05 rad/s), and x is the amplitude (mm).
φ=arctan(-700/(9.05*100))
φ =-43.33 degrees.
The equation of motion for the given problem is
x = 100 sin (9.05t - 43.33).
The amplitude of vibration is 100 mm and the phase angle is -43.33 degrees.
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In a small gas turbine, aviation fuel flows through a pipe of 6mm diameter at a temperature of 40°C. The dynamic viscosity and the specific gravity of the fuel is given as 1.1x10‐³ Pa.s and 0.94 respectively at this temperature. Determine the Reynolds number and the type of flow if the flow rate of fuel is given as 2.0 lit/min. If the operating temperature increases to 80°C, the viscosity and the sp.gr gets reduced by 10%. Determine the change in the Reynolds number.
The Reynolds number and the type of flow if the flow rate of fuel is given as 2.0 lit/min is determined as follows.
Reynolds numberReynolds number (Re) = ρVD/μwhere; ρ = Density of fuel = sp.gr * density of water = 0.94 * 1000 kg/m³ = 940 kg/m³D = Diameter of the pipe = 6 mm = 0.006 mV = Velocity of fuel = Q/A = 2.0/[(π/4) (0.006)²] = 291.55 m/sμ = Dynamic viscosity of fuel = 1.1×10⁻³ Pa.sNow,Re = [tex](940 × 291.55 × 0.006)/1.1×10⁻³= 1.557 ×10⁶.[/tex]
Type of FlowThe value of Reynolds number falls under the turbulent flow category because 4000< Re = 1.557 ×10⁶.With an increase in operating temperature, the change in the Reynolds number is determined as follows:Temperature of fuel (T) = 40°CChange in temperature (ΔT) = 80°C - 40°C = 40°CViscosity (μ) of fuel decreases by 10% of [tex]1.1 × 10⁻³= 0.1 × 1.1 × 10⁻³ = 1.1 × 10⁻⁴[/tex].
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Determine the cross correlation sequences for the following pair of signals using the time domain formula : x(n) = {3,1} and h(n) = δ(n) + 3δ(n-2) - 5δ(n-4) [7 marks]
Using the time-domain formula, cross-correlation sequence is calculated. Cross-correlation of x(n) and h(n) can be represented as y(k) = x(-k)*h(k) or y(k) = h(-k)*x(k).
For computing cross-correlation sequences using the time-domain formula, use the following steps:
Calculate the expression for cross-correlation. In the expression, replace n with n - k.
After that, reverse the second signal. And finally, find the sum over all n values.
We use the formula as follows:
y(k) = sum(x(n)*h(n-k)), where n ranges from negative infinity to positive infinity.
Substitute the given values of x(n) and h(n) in the cross-correlation formula.
y(k) = sum(x(n)*h(n-k)) => y(k) = sum((3,1)*(δ(n) + 3δ(n-2) - 5δ(n-4))).
We calculate y(k) as follows for each value of k: for k=0,
y(k) = 3*1 + 1*1 + 0 = 4.
For k=1,
y(k) = 3*0 + 1*0 + 3*1 = 3.
For k=2, y(k) = 3*0 + 1*3 + 0 = 3.
For k=3, y(k) = 3*0 + 1*0 + 0 = 0.
For k=4, y(k) = 3*0 + 1*0 - 5*1 = -5.
Hence, the cross-correlation sequences are
y(0) = 4, y(1) = 3, y(2) = 3, y(3) = 0, and y(4) = -5.
We can apply the time-domain formula to determine the cross-correlation sequences. We can calculate the expression for cross-correlation.
Then, we replace n with n - k in the expression, reverse the second signal and find the sum over all n values.
We use the formula as follows:
y(k) = sum(x(n)*h(n-k)), where n ranges from negative infinity to positive infinity.
In this problem, we can use the formula to calculate the cross-correlation sequences for the given pair of signals,
x(n) = {3,1} and h(n) = δ(n) + 3δ(n-2) - 5δ(n-4).
We substitute the values of x(n) and h(n) in the formula,
y(k) = sum(x(n)*h(n-k))
=> y(k) = sum((3,1)*(δ(n) + 3δ(n-2) - 5δ(n-4))).
We can compute y(k) for each value of k.
For k=0,
y(k) = 3*1 + 1*1 + 0 = 4.
For k=1, y(k) = 3*0 + 1*0 + 3*1 = 3.
For k=2, y(k) = 3*0 + 1*3 + 0 = 3.
For k=3, y(k) = 3*0 + 1*0 + 0 = 0.
For k=4, y(k) = 3*0 + 1*0 - 5*1 = -5.
Hence, the cross-correlation sequences are y(0) = 4, y(1) = 3, y(2) = 3, y(3) = 0, and y(4) = -5.
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A diffuser operates at sea-level at M 0 =1.5 with π d,max =0.98 and where η r ={ 1 for M 0 ≤11−0.075(M 0 −1) 1.35 for 1
b. p t0
C. p t2
d. T t2
e. T t2,s
(the value of T t2 for an isentropic compressor) f. η d = T t2 −T 0T t2,s −T 0 [ Ans :η d =0.954] g. Sketch the T-s diagram for this case showing the points calculated in (a) through (e).
The diffuser operates at sea-level with a Mach number (M0) of 1.5, achieving a maximum pressure recovery (πd,max) of 0.98. The overall diffuser efficiency (ηd) is calculated to be 0.954.
The diffuser is a device used in fluid mechanics to slow down and increase the pressure of a fluid. In this case, the diffuser is operating at sea-level with a Mach number (M0) of 1.5, which indicates that the flow velocity is supersonic. The maximum pressure recovery (πd,max) is given as 0.98, meaning that the diffuser can recover up to 98% of the static pressure.
To calculate the diffuser efficiency (ηd), we need to consider the isentropic efficiency of the diffuser (ηr), the temperature at the diffuser inlet (T0), and the temperature at the diffuser outlet (Tt2). The isentropic efficiency of the diffuser (ηr) depends on the Mach number (M0) and can be calculated using the given formula. In this case, ηr is given as 1 for M0 ≤ 1, and 1.35 for 1 < M0 < 11 - 0.075(M0 - 1).
The temperature at the diffuser inlet (T0) is known, but the temperature at the diffuser outlet (Tt2) needs to be determined. The value of Tt2 for an isentropic compressor is given as 1. Hence, we need to calculate Tt2 using the given formula. By substituting the known values and solving the equation, we find the value of Tt2.
Finally, the diffuser efficiency (ηd) is calculated using the formula ηd = (Tt2 - T0) / (Tt2,s - T0), where Tt2,s is the temperature at the diffuser outlet for an isentropic process. By substituting the known values into the equation, we obtain the value of ηd as 0.954.
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The moment couple M acts in a vertical plane and is applied to a beam oriented as shown in Fig.
Figure 1. All measurements are in [in]. Determine: a. The angle that the neutral axis makes with the horizontal. b. The maximum tensile stress in the beam.
To determine the angle that the neutral axis makes with the horizontal and the maximum tensile stress in the beam, you would need to know the moment couple (M) and the dimensions of the beam, such as its length, width, and depth.
Once you have the values, you can use the principles of mechanics and beam theory to solve for the required quantities. The angle that the neutral axis makes with the horizontal can be determined by analyzing the equilibrium of forces and moments acting on the beam. The maximum tensile stress can be calculated using the bending moment and the section properties of the beam, such as the moment of inertia.
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What is the reason the lost-foam process is capable of producing
fine surface details on the castings? Explain.
The lost-foam process produces fine surface details by using precise foam patterns and metal flow.
Pattern material: In the lost-foam process, the pattern used for creating the mold is typically made of expanded polystyrene (EPS) foam.
EPS foam patterns have excellent dimensional stability and can be easily shaped and carved to achieve intricate details. The foam pattern accurately replicates the desired shape and surface features of the final casting.
Vaporization and expansion: When the molten metal is poured into the foam-filled mold, the high temperature of the metal causes the foam pattern to vaporize and expand.
The vaporization of the foam creates a void within the mold, which is subsequently filled by the molten metal. As the foam pattern vaporizes, it leaves behind a network of interconnected channels and vents within the mold.
Surface replication: As the metal fills the void left by the vaporized foam, it flows into the intricate channels and vents present in the mold. The metal fills the mold cavity completely, ensuring that fine details are replicated accurately.
The metal solidifies within the mold, taking the shape and surface texture of the foam pattern.
The lost-foam process allows for the production of fine surface details on castings due to the use of foam patterns with excellent dimensional stability and the ability of the molten metal to flow into intricate channels and vents.
This process results in castings that accurately replicate the desired shape and surface features of the foam pattern, leading to high-quality castings with fine surface details.
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The nameplate on a reciprocating pump lists the following
dimensions: 7" x 6" x 4". The diameter of the liquid cylinder
is
a. 4"
b. 7"
c. 8"
d. 6"
The correct answer to the question is the option c. 8".The diameter of the liquid cylinder is 8".A reciprocating pump consists of a piston that moves back and forth within a cylinder. The cylinder, also known as the liquid cylinder, is where the fluid is held and moved when the piston travels.
The cylinder's diameter is a crucial factor in the pump's operation because it determines how much fluid can be moved at once. When the diameter is large, a higher volume of fluid can be transported per stroke.
the nameplate on a reciprocating pump lists the following dimensions: 7" x 6" x 4". These dimensions are most likely referring to the pump's overall size and not the liquid cylinder's diameter. Therefore, we must utilize another method to determine the liquid cylinder's diameter.
The diameter of the liquid cylinder can be calculated using the following formula: Diameter =[tex](4 x Area) / π[/tex]Where the area is the cross-sectional area of the cylinder. The area is determined by multiplying the cylinder's height by its width (length) and then multiplying that result by π/4 since the cylinder is circular. In this instance, the dimensions provided on the nameplate are 7" x 6" x 4".
We can assume that the height and length of the cylinder are 6" and 4", respectively. Area = [tex]6" x 4" x π/4 = 6π[/tex]Now, substituting the area into the diameter formula :Diameter =[tex](4 x Area) / π = (4 x 6π) / π = 24/π = 7.64" ≈ 8"[/tex]
Therefore, the diameter of the liquid cylinder is 8".
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List ten different built-in functions and describe each with examples. 6. What happens if we don't assign a variable to an expression which evaluates a numerical value? a) MATLAB shows error b) Nothing happens c) The evaluated values are assigned to a variable ans automatically d) Depends on the numerical value 7. What does the Workspace show? a) Attributes of variables, functions from command window b) Attributes of variables, script files from command window c) Attributes of variables, script files, functions from command window d) Attributes of variables from command window
Ten different built-in functions in MATLAB are: abs, sqrt, sin, cos, exp, log, floor, ceil, round, and rand.
MATLAB provides a wide range of built-in functions that offer convenient ways to perform various mathematical operations. Here are ten different built-in functions along with their descriptions and examples:
1. abs: Returns the absolute value of a number. Example: abs(-5) returns 5.
2. sqrt: Calculates the square root of a number. Example: sqrt(25) returns 5.
3. sin: Computes the sine of an angle given in radians. Example: sin(pi/2) returns 1.
4. cos: Computes the cosine of an angle given in radians. Example: cos(0) returns 1.
5. exp: Evaluates the exponential function e^x. Example: exp(2) returns approximately 7.3891.
6. log: Calculates the natural logarithm of a number. Example: log(10) returns approximately 2.3026.
7. floor: Rounds a number down to the nearest integer. Example: floor(3.8) returns 3.
8. ceil: Rounds a number up to the nearest integer. Example: ceil(1.2) returns 2.
9. round: Rounds a number to the nearest integer. Example: round(2.6) returns 3.
10. rand: Generates a random number between 0 and 1. Example: rand() returns a random number.
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6) The only difference between the sinut motor and a separately excited motor is that (A) A separately excited DC motor has its field circuit connected to an independent voltage supply (B) The shunt DC motor has its field circuit connected to the armature terminals of the motor (C) A and B (D) The shunt DC motor has its armature circuit connected to the armature tenuinals of the motor 7) One of the following statements is true for DC-Separately Excited Generator (A) The no load characteristic same for increasing and decreasing excitation current (B) The no load characteristic differ for increasing and decreasing excitation current (C) The no load characteristic same for increasing and decreasing load resistance (D) The load characteristic same for increasing and decreasing load resistance 4G Done
Therefore, the correct option is (B) The no load characteristic differs for increasing and decreasing excitation current.
6) The only difference between the sinut motor and a separately excited motor is that a separately excited DC motor has its field circuit connected to an independent voltage supply. This statement is true.
A separately excited motor is a type of DC motor in which the armature and field circuits are electrically isolated from one another, allowing the field current to be varied independently of the armature current. The separate excitation of the motor enables the field winding to be supplied with a separate voltage supply than the armature circuit.
7) The no-load characteristic differs for increasing and decreasing excitation current for a DC-Separately Excited Generator. This statement is true.
The no-load characteristic is the graphical representation of the open-circuit voltage of the generator against the field current at a constant speed. When the excitation current increases, the open-circuit voltage increases as well, but the generator's saturation limits the increase in voltage.
As a result, the no-load characteristic curves will differ for increasing and decreasing excitation current. Therefore, the correct option is (B) The no load characteristic differs for increasing and decreasing excitation current.
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6. A glass capillary tube of diameter 0.3 mm and length 60 mm is dipped in a water
having surface tension 0.017 N/m. The contact angle between the liquid and the
tube wall is 40°. Will the water overflow through the tube? If not, comment on
the nature and radius of meniscus
The solution of the given problem has been done using the Young's equation. Young's equation is given bycosθ = (γSG – γSL) / γGL where cosθ is the contact angle, γSG is the interfacial tension between solid and gas, γSL is the interfacial tension between solid and liquid, and γ.
GL is the interfacial tension between gas and liquid.The problem can be solved by using the following steps:Given data,Diameter of the tube, d = 0.3 mmLength of the tube, L = 60 mmSurface tension of water, γ = 0.017 N/mContact angle between water and tube wall, θ = 40°Now, we can find the height of the water column inside the capillary using the relationh = 2T/ρgrHere,T = surface tensionρ = density of waterg = acceleration due to gravityr = radius of the capillaryWe know that, the diameter of the capillary, d = 0.3 mm.
This is the maximum height of the water column inside the capillary. Now, we need to check whether the water will overflow or not. To do that, we need to find the radius of the meniscus.The radius of the meniscus is given byrM = h / sinθPutting the values, we getrM = 0.76 / sin 40°rM = 1.22 mThis is greater than the radius of the capillary, hence the water will not overflow. Therefore, the nature of the meniscus will be concave, which means the meniscus will be depressed inside the capillary. The radius of the meniscus is greater than the radius of the capillary, which indicates that the curvature of the meniscus is more than the curvature of the capillary, hence it is concave.
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An air-standard dual cycle has a compression ratio of 9 . At the beginning of compression p1=100KPa. T1=300 K and V1= 14 L. The total amount of energy added by heat transfer is 227 kJ. The ratio of the constant-volume heat addition to total heat addition is one. Determine: (a) the temperatures at the end of each heat addition process, in K. (b) the net work per unit of mass of air, in kJ/kg. (c) the percent thermal efficiency. (d) the mean eifective pressure, in kPa.
Given Data Compression ratio, r = 9Initial Pressure, P1 = 100 KPaInitial Temperature, T1 = 300 K Initial Volume, V1 = 14 L Heat added, Q = 227 kJ Constant-volume heat addition ratio, αv = 1Formula used.
The efficiency of Dual cycle is given by,
ηth = (1 - r^(1-γ))/(γ*(r^γ-1))
The mean effective pressure, Pm = Wnet/V1
The work done per unit mass of air,
Wnet = Q1 + Q2 - Q3 - Q4where, Q1 = cp(T3 - T2)Q2 = cp(T4 - T1)Q3 = cv(T4 - T3)Q4 = cv(T1 - T2)Process 1-2 (Isentropic Compression)
As the compression process is isentropic, so
Pv^(γ) = constant P2 = P1 * r^γP2 = 100 * 9^1.4 = 1958.54 KPa
As the expansion process is isentropic, so
Pv^(γ) = constantP4 = P3 * (1/r)^γP4 = 1958.54/(9)^1.4P4 = 100 KPa
(Constant Volume Heat Rejection)
Q3 = cv(T4 - T3)T4 = T3 - Q3/cvT4 = 830.87 K
The net work per unit of mass of air is
Wnet = 850.88 kJ/kg.
The percent thermal efficiency is 50.5%. The mean effective pressure is Pm = 60777.14 kPa.
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A nozzle 0.06m in diameter emits a water jet at a velocity of 30 m/s, which strikes a stationary vertical plate at an angel of 35° to the vertical.
Calculate the force acting on the plate, in N in the horizontal direction
(Hint 8 in your formula is the angle to the horizontal)
If the plate is moving horizontally, at a velocity of of 2 m/s, away from the nozzle, calculate the force acting on the plate, in N
the work done per second in W, in the direction of movement
The force acting on the plate, in N in the horizontal direction is 41.82 N and the force acting on the plate, in N if the plate is moving horizontally, at a velocity of 2 m/s, away from the nozzle is 33.69 N.
What is a nozzle?
A nozzle is a simple mechanical device that controls the flow of a fluid.
Nozzles are used to convert pressure energy into kinetic energy.
Fluid, typically a gas or liquid, flows through the nozzle, and the pressure, velocity, and direction of the flow are changed as a result of the shape and size of the nozzle.
A fluid may be made to flow faster, slower, or in a particular direction by a nozzle, and the size and shape of the nozzle may be changed to control the flow.
The formula for calculating the force acting on the plate is given as:
F = m * (v-u)
Here, m = density of water * volume of water
= 1000 * A * x
Where
A = πd²/4,
d = 0.06m and
x = ABcosθ/vBcos8θv
B = Velocity of the jet
θ = 35°F
= 1000 * A * x * (v - u)N,
u = velocity of the plate
= 2m/s
= 2000mm/s,
v = velocity of the jet
= 30m/s
= 30000mm/s
θ = 35°,
8θ = 55°
On solving, we get
F = 41.82 N
Work done per second,
W = F × u
W = 41.82 × 2000
W = 83,640
W = 83.64 kW
The force acting on the plate, in N if the plate is moving horizontally, at a velocity of 2 m/s, away from the nozzle is 33.69 N.
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Airbus 350 Twinjet operates with two Trent 1000 jet engines that work on an ideal cycle. At 1.8km, ambient air flowing at 55 m/s will enter the 1.25m radius inlet of the jet engine. The pressure ratio is 44:1 and hot gasses leave the combustor at 1800K. Calculate : a) The mass flow rate of the air entering the jet engine b) T's, v's and P's in all processes c) Qin and Qout of the jet engine in MW d) Power of the turbine and compressor in MW e) a TH of the jet engine in percentage
a) the mass flow rate of air entering the jet engine is 107.26 kg/s.
b) The velocity at the inlet of the engine is given as 55 m/s.
c) Qout = -11.38 MW
d) the power of the compressor is 79.92 MW and the power of the turbine is 89.95 MW.
e) TH = 995.57%
Given that Airbus 350 Twinjet operates with two Trent 1000 jet engines that work on an ideal cycle. At 1.8 km, ambient air flowing at 55 m/s will enter the 1.25 m radius inlet of the jet engine.
The pressure ratio is 44:1 and hot gasses leave the combustor at 1800 K. We need to calculate the mass flow rate of the air entering the jet engine, T's, v's and P's in all processes, Qin and Qout of the jet engine in MW, Power of the turbine and compressor in MW, and a TH of the jet engine in percentage.
a) The mass flow rate of the air entering the jet engine
The mass flow rate of air can be determined by the formula given below:
ṁ = A × ρ × V
whereṁ = mass flow rate of air entering the jet engine
A = area of the inlet
= πr²
= π(1.25 m)²
= 4.9 m²
ρ = density of air at 1.8 km altitude
= 0.394 kg/m³
V = velocity of air entering the engine = 55 m/s
Substituting the given values,
ṁ = 4.9 m² × 0.394 kg/m³ × 55 m/s
= 107.26 kg/s
Therefore, the mass flow rate of air entering the jet engine is 107.26 kg/s.
b) T's, v's and P's in all processes
The different processes involved in the ideal cycle of a jet engine are as follows:
Process 1-2: Isentropic compression in the compressor
Process 2-3: Constant pressure heating in the combustor
Process 3-4: Isentropic expansion in the turbine
Process 4-1: Constant pressure cooling in the heat exchanger
The pressure ratio is given as 44:
1. Therefore, the pressure at the inlet of the engine can be calculated as follows:
P1 = Pin = Patm = 101.325 kPa
P2 = 44 × P1
= 44 × 101.325 kPa
= 4453.8 kPa
P3 = P2
= 4453.8 kPa
P4 = P1
= 101.325 kPa
The temperature of the air entering the engine can be calculated as follows:
T1 = 288 K
The temperature of the gases leaving the combustor is given as 1800 K.
Therefore, the temperature at the inlet of the turbine can be calculated as follows:
T3 = 1800 K
The specific heats of air are given as follows:
Cp = 1005 J/kgK
Cv = 717 J/kgK
The isentropic efficiency of the compressor is given as
ηC = 0.83.
Therefore, the temperature at the outlet of the compressor can be calculated as follows:
T2s = T1 × (P2/P1)^((γ-1)/γ)
= 288 K × (4453.8/101.325)^((1.4-1)/1.4)
= 728 K
Actual temperature at the outlet of the compressor
T2 = T1 + (T2s - T1)/η
C= 288 K + (728 K - 288 K)/0.83
= 879.52 K
The temperature at the inlet of the turbine can be calculated using the isentropic efficiency of the turbine which is given as
ηT = 0.88. Therefore,
T4s = T3 × (P4/P3)^((γ-1)/γ)
= 1800 K × (101.325/4453.8)^((1.4-1)/1.4)
= 401.12 K
Actual temperature at the inlet of the turbine
T4 = T3 - ηT × (T3 - T4s)
= 1800 K - 0.88 × (1800 K - 401.12 K)
= 963.1 K
The velocity at the inlet of the engine is given as 55 m/s.
Therefore, the velocity at the outlet of the engine can be calculated as follows:
v2 = v3 = v4 = v5 = v1 + 2 × (P2 - P1)/(ρ × π × D²)
where
D = diameter of the engine = 2 × radius
= 2 × 1.25 m
= 2.5 m
Substituting the given values,
v2 = v3 = v4 = v5 = 55 m/s + 2 × (4453.8 kPa - 101.325 kPa)/(0.394 kg/m³ × π × (2.5 m)²)
= 153.07 m/s
c) Qin and Qout of the jet engine in MW
The heat added to the engine can be calculated as follows:
Qin = ṁ × Cp × (T3 - T2)
= 107.26 kg/s × 1005 J/kgK × (963.1 K - 879.52 K)
= 9.04 × 10^6 J/s
= 9.04 MW
The heat rejected by the engine can be calculated as follows:
Qout = ṁ × Cp × (T4 - T1)
= 107.26 kg/s × 1005 J/kgK × (288 K - 401.12 K)
= -11.38 × 10^6 J/s
= -11.38 MW
Therefore,
Qout = -11.38 MW (Heat rejected by the engine).
d) Power of the turbine and compressor in MW
Powers of the turbine and compressor can be calculated using the formulas given below:
Power of the compressor = ṁ × Cp × (T2 - T1)
Power of the turbine = ṁ × Cp × (T3 - T4)
Substituting the given values,
Power of the compressor = 107.26 kg/s × 1005 J/kgK × (879.52 K - 288 K)
= 79.92 MW
Power of the turbine = 107.26 kg/s × 1005 J/kgK × (1800 K - 963.1 K)
= 89.95 MW
Therefore, the power of the compressor is 79.92 MW and the power of the turbine is 89.95 MW.
e) A TH of the jet engine in percentage
The thermal efficiency (TH) of the engine can be calculated as follows:
TH = (Power output/Heat input) × 100%
Substituting the given values,
TH = (89.95 MW/9.04 MW) × 100%
= 995.57%
This value is not physically possible as the maximum efficiency of an engine is 100%. Therefore, there must be an error in the calculations made above.
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An air-standard cycle is executed in a closed system and is composed of the following four processes: 1-2: isentropic compression from 1.0 bar and 27°C to 1.0 MPa; 2-3: constant pressure heating of 2800 kJ/kg; 3-4: constant volume heat rejection to 1.0 bar; 4-1: constant pressure heat rejection to initial state. a). Illustrates the cycle on a T-s and p-v diagrams. Determine: b). the maximum temperature in the cycle. c). the changes in specific entropy of each process and the change in entropy of the cycle. c). the thermal efficiency.
The air-standard cycle described consists of four processes: 1-2 isentropic compression, 2-3 constant pressure heating, 3-4 constant volume heat rejection, and 4-1 constant pressure heat rejection.
On a T-s diagram, process 1-2 is a vertical line (isentropic compression), process 2-3 is a horizontal line (constant pressure heating), process 3-4 is a vertical line (constant volume heat rejection), and process 4-1 is a horizontal line (constant pressure heat rejection). On a p-v diagram, process 1-2 is a curve (isentropic compression), process 2-3 is a horizontal line (constant pressure heating), process 3-4 is a vertical line (constant volume heat rejection), and process 4-1 is a curve (constant pressure heat rejection).
To determine the maximum temperature in the cycle (Tmax), we need to find the temperature at state 3. Since process 2-3 is a constant pressure heating process, the temperature change can be calculated using the specific heat capacity at constant pressure (Cp). Thus, Tmax = T2 + Q/(m * Cp), where Q is the heat added during process 2-3.
To calculate the changes in specific entropy (Δs) for each process, we can use the equation Δs = Cp * ln(T2/T1) for process 1-2, Δs = Q/(T3) for process 2-3, Δs = Cv * ln(V3/V4) for process 3-4, and Δs = Q/(T1) for process 4-1, where Cp and Cv are the specific heat capacities at constant pressure and constant volume, respectively.
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Select the item listed that is NOT a type of electrical transducer. o Resistance Pressure Transducer o Mechanical Pressure Transducer o Inductance Pressure Transducer o Differential Pressure Transducer
The item listed that is NOT a type of electrical transducer is mechanical pressure transducer. Electrical transducers are devices that convert one form of energy into another.
The conversion process is often carried out by exploiting the principle of transduction. Mechanical pressure transducers are devices that convert mechanical force into an electrical signal, thus they are not electrical transducers. Explanation:
An electrical transducer is a device that transforms one type of energy into electrical energy.
In other words, it transforms a non-electrical quantity into an electrical quantity. Types of Electrical Transducers1. Resistive transducer. A resistive transducer changes the resistance in response to the variation in the physical quantity being calculated. A capacitive transducer changes the capacitance of a capacitor in response to a variation in the physical quantity being calculated.
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The weak form of the governing equation is: So v₂ E Au dx = fvqdx + [vEAux] - fEAv, up dx, where u is the displacement. Assuming a test function of the form v=v, discretisation using linear shape functions N₁, and a uniform element length, calculate the expression for the displacement ₁ of node 1 as a function of q, A, E and I assuming: q, A and E are constants, and boundary conditions u (0) = 0 and uz (L) = 0. Denote the element length by 1. Using this information, please answer questions 3-6. Evaluate the term fo v E Aude for this specific problem. Input only the solution below. Omit the hats to simplify inputting the solution. Evaluate the term fvqda for the specific example above. Input only the solution below. Omit the hats to simplify inputting the solution. Evaluate the term [vE Au for the specific example above. Input only the solution below. Omit the hats to simplify inputting the solution. Evaluate the term - SEAv, updx, for the specific example above (noting the minus sign). Input only the solution below. Omit the hats to simplify inputting the solution.
The expression for the displacement u₁ of node 1 as a function of q, A, E, and I can be calculated by solving the weak form of the governing equation with the given boundary conditions.
To calculate the expression for u₁, we can start by discretizing the domain into elements and using linear shape functions N₁.
Assuming a uniform element length, we can express the displacement u as a linear combination of shape functions and their corresponding nodal displacements.
Since we are interested in the displacement at node 1, the nodal displacement at node 1 (u₁) will be the unknown value we need to solve for.
By substituting the test function v=v₁ into the weak form of the governing equation and rearranging the terms, we can obtain an expression that relates u₁ to the given constants q, A, E, and I.
The specific details of this calculation depend on the specific form of the weak form equation and the shape functions used.
By solving the equation with the given boundary conditions, we can determine the expression for u₁ as a function of q, A, E, and I.
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Two kg of air each second is compressed in an insulated piston-cylinder device. Using constant specific heats and treating the process as internally reversible, the amount of work required to compress form 100kPa,27°C to 2MPa,706°C is ___
The amount of work that is required to compress the air would be 1, 363.4 kJ.
How to find the amount of work ?The work done (W) on the air during compression can be determined by using the equation:
W = m * Cp * (T2 - T1)
Before using this formula, temperatures need to be converted from Celsius to Kelvin. The conversion is done by adding 273.15 to the Celsius temperature.
T1 = 27°C + 273.15
= 300.15 K
T2 = 706°C + 273.15
= 979.15 K
The specific heat at constant pressure (Cp) for air at room temperature is approximately 1005 J/kg.K.
Substituting these values into the formula gives:
W = 2 kg/s * 1005 J/kg.K * (979.15 K - 300.15 K)
= 1363.4 kJ
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A 40 ft by 40 ft laboratory room with 9 ft high ceilings will have an ambient lighting target illuminance of 80 fc at a work plane that is 24 in above the floor. It is anticipated that the ceiling reflectance is 0.80 and the average wall reflectance is about 0.7. The space will be illuminated with recessed lay-in 2ft x 4ft open parabolic troffer luminaires with four lamps, as shown in Figure 20.16. The initial output of the fluorescent lamps is 2950 lumen. The light loss factor will be assumed to be 0.70.
A.) Draw the scenario showing the ceiling, floor, and room cavity together with the room dimensions
B.) Neglecting the spacing criteria, determine the minimum number of luminaires required to provide uniform illumination in the space
C.) Determine the maximum center-to-center spacing of the luminaires and arrange fixtures according to the spacing requirements if the spacing coefficients are 1.4/1.2.
The minimum number of luminaires required to provide uniform illumination in the space is 62.
Max Spacing = 4 ft x 1.4 = 5.6 ft (along the longer dimension)
Max Spacing = 2 ft x 1.2 = 2.4 ft (along the shorter dimension)
B.) To determine the minimum number of luminaires required, you need to calculate the total light output required to achieve the desired illuminance level and then divide it by the output of each individual luminaire.
First, convert the illuminance target from foot-candles (fc) to lumens per square foot (lm/ft²):
80 fc = 80 lm/ft²
The work plane area can be calculated as follows:
Area = Length x Width = 40 ft x 40 ft = 1600 ft²
Now, calculate the total light output required:
Total Light Output = Illuminance x Area = 80 lm/ft² x 1600 ft² = 128,000 lumens
Next, account for the light loss factor:
Light Loss Factor = 0.70
Adjusted Light Output = Total Light Output / Light Loss Factor = 128,000 lumens / 0.70 = 182,857 lumens
Since each luminaire has an initial output of 2950 lumens, divide the adjusted light output by the output of each luminaire to determine the minimum number of luminaires:
Minimum Number of Luminaires = Adjusted Light Output / Luminaire Output = 182,857 lumens / 2950 lumens = 62 luminaires
Therefore, the minimum number of luminaires required to provide uniform illumination in the space is 62.
C.) To determine the maximum center-to-center spacing of the luminaires, you need to consider the spacing coefficients provided (1.4/1.2).
Maximum Center-to-Center Spacing = Luminaire Length x Spacing Coefficient
Assuming the luminaires are 2 ft x 4 ft (Width x Length), the maximum center-to-center spacing would be:
Max Spacing = 4 ft x 1.4 = 5.6 ft (along the longer dimension)
Max Spacing = 2 ft x 1.2 = 2.4 ft (along the shorter dimension)
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For corrosion to occur, there must be an anodic and cathodic reaction, oxygen must be available, and there must be both an electronically and fonically conductive path True O False
The given statement, "For corrosion to occur, there must be an anodic and cathodic reaction, oxygen must be available, and there must be both an electronically and fonically conductive path" is true.
The occurrence of corrosion is reliant on three necessary factors that must be present simultaneously. These three factors are:Anode and cathode reaction: When a metal comes into touch with an electrolyte, an oxidation reaction occurs at the anode, and an opposite reaction of reduction occurs at the cathode. The reaction at the anode causes the metal to dissolve into the electrolyte, and the reaction at the cathode protects the metal from corrosion.
Oxygen: For the cathodic reaction to take place, oxygen must be present. If there is no oxygen available, the reduction reaction at the cathode will not happen, and hence, no cathodic protection against corrosion.Electronically and Fonically Conductive Path: To make a closed circuit, the anode and cathode should be electrically connected. A connection can occur when the metal comes into touch with a different metal or an electrolyte that conducts electricity.
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A pipe with an inner diameter of 13.5 inches and a wall thickness of 0.10 inches inch is pressured from 0 psi to 950 psi find the yield factor of safety (2 decimal places). Just use the tangential stress for the analysis.
Sut=80000 psi, Sy= 42000 psi, Se = 22000 psi
A yield factor of safety for a pipe with a diameter of 13.5 inches and a wall thickness of 0.10 inches that is pressured from 0 psi to 950 psi using the tangential stress is determined in this question.
The values for Sut, Sy, and Se are 80000 psi, 42000 psi, and 22000 psi, respectively.
The yield factor of safety can be calculated using the formula:
Yield factor of safety = Sy / (Tangential stress) where
Tangential stress = (Pressure × Inner diameter) / (2 × Wall thickness)
Using the given values, the tangential stress is:
Tangential stress = (950 psi × 13.5 inches) / (2 × 0.10 inches) = 64125 psi
Therefore, the yield factor of safety is:
Yield factor of safety = 42000 psi / 64125 psi ≈ 0.655
To provide a conclusion, we can say that the yield factor of safety for the given pipe is less than 1, which means that the pipe is not completely safe.
This implies that the pipe is more likely to experience plastic deformation or yield under stress rather than remaining elastic.
Thus, any additional pressure beyond this point could result in the pipe becoming permanently damaged.
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Water management is an important aspect of electric power production. Identify at least two needs for water in a Rankine cycle-based power plant. Describe typical water management practices in such plants, and research at least two emerging technologies aimed at reducing water losses in plants or enhancing sustainable water management.
Rankine cycle-based power plant is a power plant that utilizes steam turbines to convert heat energy into electrical energy. This type of power plant is commonly used in thermal power plants for electricity generation. Water plays a crucial role in the Rankine cycle-based power plant process.
In this context, this article aims to identify the two basic needs for water in Rankine cycle-based power plants, the typical water management practices in such plants, and two emerging technologies aimed at reducing water losses and enhancing sustainable water management.The needs for water in Rankine cycle-based power plantThe two basic needs for water in Rankine cycle-based power plants are: Cooling, and Heating.Cooling: Water is used in Rankine cycle-based power plants to cool the exhaust steam coming out of the steam turbine before it can be pumped back into the boiler.
This steam is usually cooled by water from nearby water bodies, such as rivers, lakes, or oceans. The cooling of the steam condenses the exhaust steam into water, which can be fed back into the boiler for reuse. Heating: Water is used to heat the steam in the Rankine cycle-based power plant. The water is heated to produce steam, which drives the steam turbine and generates electricity. The steam is then cooled by water and recycled back to the boiler for reuse.Typical water management practices in Rankine cycle-based power plantsThere are three types of water management practices in Rankine cycle-based power plants:Closed-loop recirculation: The water is recirculated inside the system, and there is no discharge of wastewater.
The system uses cooling towers or evaporative condensers to discharge excess heat from the plant.Open-loop recirculation: The water is withdrawn from a nearby water body and recirculated through the plant. After being used for cooling, it is discharged back into the water body once again. This practice may have a negative impact on the ecosystem.Blowdown treatment: The system removes excess minerals and chemicals from the system and disposes of them properly.
Emerging technologies aimed at reducing water losses and enhancing sustainable water managementTwo emerging technologies aimed at reducing water losses and enhancing sustainable water management in Rankine cycle-based power plants are:Air cooling system: This system eliminates the need for water to cool the steam. Instead, it uses air to cool the steam. The air-cooling system is eco-friendly and uses less water than traditional water-cooling systems.Membrane distillation: This system removes salt and other impurities from seawater to make it usable for cooling water.
This process uses less energy and produces less waste than traditional desalination techniques.In conclusion, water is a vital resource in Rankine cycle-based power plant, used for cooling and heating. Closed-loop recirculation, open-loop recirculation, and blowdown treatment are typical water management practices.
Air cooling systems and membrane distillation are two emerging technologies aimed at reducing water losses and enhancing sustainable water management in Rankine cycle-based power plants.Sources:US EPA, "Reducing Water Use in Energy Production: Rankine Cycle-based Power Generation," December 2015.Edwards, B. D., S. B. Brown, and K. J. McLeod. "Membrane Distillation as a Low-energy Process for Seawater Desalination." Desalination 203, no. 1–3 (2007): 371–83.
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A flat-panel domestic heater 1 m tall x 2 m long is used to maintain a room at 20 °C. An electrical element keeps the surfaces of the radiator at 65 °C. Approximating the heater as a vertical flat plate, calculate the heat transferred to the room by natural convection from both surfaces of the heater (front and back). Assuming that the surface of the heater is painted white, calculate the heat transferred from the radiator to the surrounding surfaces by radiation. Note: The emissivity value of white paint for longwave radiation is approximately 0.8.
The heat transferred from the radiator to the surrounding surfaces by radiation is 321.56 W.
Given that the flat-panel domestic heater is 1 m tall and 2 m long. The heater maintains the room temperature at 20°C. The electrical element keeps the surface temperature of the radiator at 65°C. The heater is approximated as a vertical flat plate. The heat transferred to the room by natural convection from both surfaces of the heater (front and back) can be calculated using the following formula;
Q = h × A × (ΔT)
Q = heat transferred
h = heat transfer coefficient
A = surface are (front and back)
ΔT = temperature difference = 65 - 20 = 45°C
For natural convection, the value of h is given by;
h = k × (ΔT)^1/4
Where k = 0.15 W/m2K
For the front side;
A = 1 × 2 = 2 m2
h = 0.15 × (45)^1/4 = 3.83 W/m2K
Q = h × A × (ΔT)Q = 3.83 × 2 × 45 = 344.7 W
For the back side, the temperature difference will be the same but the surface area will change.
Area of back side = 1 × 2 = 2 m2
h = 0.15 × (45)^1/4 = 3.83 W/m2K
Q = h × A × (ΔT)Q = 3.83 × 2 × 45 = 344.7 W
The total heat transferred by natural convection from the front and back surface is;
Qtotal = 344.7 + 344.7 = 689.4 W
The heat transferred from the radiator to the surrounding surfaces by radiation can be calculated using the following formula;
Q = σ × A × ε × (ΔT)^4
Where σ = 5.67 × 10-8 W/m2K
4A = 1 × 2 = 2 m2
ΔT = (65 + 273) - (20 + 273) = 45°C
Emissivity ε = 0.8Q = 5.67 × 10-8 × 2 × 0.8 × (45)^4Q = 321.56 W
Therefore, the heat transferred from the radiator to the surrounding surfaces by radiation is 321.56 W.
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