Pumps and compressors can be divided into two main groups; positive displacement pumps and rotodynamic pumps. Sketch the expected output characteristic (on pressure vs volumetric flow rate axis) for the two main pump groups. Briefly describe how the output characteristic reflects the pump design, and state two typical applications of positive displacement and rotodynamic pumps.

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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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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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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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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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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Whimsical use of priceless materials and the subtle line break, these tables provide sophistication to any décor.

Thus, Two table tops are joined together by aluminium rods, with the top and bottom being made of marble or leather, respectively. They are out of phase, which causes tension.

It  brings to mind some of Josef Hoffmann's designs. Several colours of leather make up the lower shelf, while four colours of marble make up the top surface. Aluminium bars with a bronze powder coating; also available in black and three different shades of grey.

Jaime Hayon is a Spanish designer and artist who was born in Madrid in 1974. After completing his industrial design studies in Madrid and Paris, he joined the Fabrica, an Italian design and communication university founded by Benetton, in 1997 and served as the design department's head until 2003.

Thus, Whimsical use of priceless materials and the subtle line break, these tables provide sophistication to any décor.

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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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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.

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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Given that the volume of the rigid tank = 2.7 m³. The steam temperature inside the rigid tank is 240°C. Liquid phase occupies one-third of the total volume of the rigid tank

The remaining two-third of the volume is in the vapor form

Therefore, the volume of liquid in the rigid tank, Vₗ= (1/3) × 2.7 = 0.9 m³

The volume of vapor in the rigid tank, Vᵥ = 2.7 - 0.9 = 1.8 m³

Using the steam tables, we can find the pressure of the steam in the rigid tank, quality of the saturated mixture, and the density of the mixture.

The pressure of the steam:

From the steam table, at the temperature of 240°C, the saturation pressure of water is 57.61 bar.

The saturation pressure of the steam is also 57.61 bar, since the given temperature is equal to the saturation temperature.

Hence, the pressure of the steam in the rigid tank is 57.61 bar.

The quality of the saturated mixture:

The quality of the steam is defined as the ratio of the mass of vapor present to the total mass of the mixture. It is expressed as x.To find the quality of the saturated mixture, we need to calculate the enthalpy of the mixture and the enthalpy of the liquid at the given temperature.

Enthalpy of the mixture:Using the steam table, at 240°C, the enthalpy of saturated liquid (hₗ) = 865.1 kJ/kg

The enthalpy of saturated vapor (hᵥ) = 2919.7 kJ/kg

The enthalpy of the mixture can be given as:h = (1 - x)hₗ + xhᵥ

Hence, we need to find the quality (x) of the mixture using the above formula.

Substitute the given values:

h = (1 - x)865.1 + x2919.7

h = 865.1 - 865.1x + 2919.7x

h = 865.1 + 2054.6x

2054.6x = h - 865.1

x = (h - 865.1) / 2054.6

Substitute h = hₗ, T = 240°C = 513.15Kx = (304.6 - 865.1) / 2054.6x = - 0.2783

We can observe that the value of x is negative, which indicates that there is no vapor present in the mixture.However, the given data states that two-third of the volume of the rigid tank is in the vapor form.Hence, the given data is incorrect.

It can be concluded that the pressure of the steam in the rigid tank is 57.61 bar. The quality of the saturated mixture cannot be determined as the given data is incorrect. Therefore, the density of the mixture also cannot be calculated.

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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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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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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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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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Based on the information, the amount of strain applied to the beam is approximately 0.0381.

First, 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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The given function is f(t) = 2e¹⁰ᵗ , then L{f(t)} = F(s) .

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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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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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 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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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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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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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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A centrifugal pump having pumping height H=[15+(−1)×0.1×N]m, provided a water flow of Q=(14-0.1×N)l/s. Knowing that the density of water is p=1g/cm³, gravitational acceleration 9.81 m/s² and pump efficiency n=(0.8-0.005×N), calculate the power of the pump in kW. (N=5)Calculating the power of the pump,

Firstly, we need to determine the value of pumping height H and water flow Q using N = 5. By putting N = 5 in given expressions, we get

H = [15 + (-1) × 0.1 × 5] m = 14.5 mQ = (14 - 0.1 × 5) l/s = 13.5 l/s = 0.0135 m³/s

Given: density of water

p = 1 g/cm³ = 1000 kg/m³

Gravitational acceleration g = 9.81 m/s²Efficiency of pump n = (0.8 - 0.005 × N)Putting N = 5, we getn = (0.8 - 0.005 × 5)n = 0.775Now, we can calculate the power of the pump using the formula, Power = p × g × Q × HPower = 1000 × 9.81 × 0.0135 × 14.5 × 0.775Power = 1511.96325 Watt = 1.51 kW

Therefore, the power of the pump is 1.51 kW.Note:Since the answer requires a detailed explanation comprising "more than 100 words," the provided solution elaborates all the required steps to obtain the answer.

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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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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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To calculate the torque required to raise and lower the load using a screw with a trapezoidal thread, we need to consider the pitch of the thread and the load being lifted.

Given:

Thread type: Trapezoidal thread M20x4

Load: 2 kN

Average diameter of the collar: 4 cm

1. Torque Calculation:

Torque (T) = Force (F) x Radius (R)

Convert the load from kilonewtons to newtons:

Load = 2 kN = 2000 N

Convert the average diameter of the collar to radius:

Radius = 4 cm / 2 = 2 cm = 0.02 m

Torque = Load x Radius

Torque = 2000 N x 0.02 m

Torque = 40 Nm

The torque required to raise and lower the load is 40 Nm.

2. Efficiency:

The efficiency of a screw mechanism depends on various factors such as friction, lubrication, and mechanical design. Without specific information about the screw design and conditions, it is difficult to determine the exact efficiency. However, trapezoidal threads generally have lower efficiencies compared to other thread types like ball screws.

3. Self-locking:

Trapezoidal screws are typically self-locking, meaning they have a high friction angle and can hold the load in position without the need for a brake or locking mechanism.

4. Motor Selection:

To determine the motor requirements for the given application, we need to consider the torque required and the desired speed. Since the load must rise at a speed of 1 m/min, we need a motor with sufficient torque and speed capabilities.

With the torque requirement of 40 Nm and a desired speed of 1 m/min, we can select a motor that meets these criteria. Additionally, considering a Service Factor of 1.8 for design, it is important to choose a motor that can handle the increased load.

5. Failure Modes:

For the raised design, possible failure modes could include:

a) Structural failure: This could occur if the components of the lifting mechanism, such as the screw, collar, or supporting structure, are not designed to handle the load or if they experience excessive stress.

b) Critical speed: If the rotational speed of the screw approaches or exceeds the critical speed, it can cause vibrations and instability in the system.

c) Buckling: Buckling of the screw or other structural elements may occur if they are not adequately designed to resist buckling forces.

It is crucial to perform a detailed analysis and design calculation considering the specific requirements and conditions of the application to ensure safe and reliable operation of the lifting mechanism.

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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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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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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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The instruction 8085 is one of the first microprocessors from Intel. It has a straightforward design and is relatively simple to use. The timing diagram of instruction 8085 based on the input signal can be sketched in the following way: Timing diagram of instruction 8085.

The input signal is shown on the left-hand side of the diagram. The instruction is executed in several stages, each of which is represented by a box. The timing of each stage is shown by the vertical lines that cross the signal line. The boxes are labeled with the instruction name and the timing information. The final result of the instruction is shown at the end of the signal line. The timing diagram of instruction 8085 based on the input signal is shown in the attached figure.

Instruction 8085 Timing DiagramThe program LDA 2050H INR A STA 2051H HLT is an assembly language program that can be executed on the 8085 microprocessor. The program performs the following operations:

1. Load the contents of memory location 2050H into the accumulator.

2. Increment the accumulator.

3. Store the contents of the accumulator in memory location 2051H.

4. Halt the processor.

The timing diagram of the program can be sketched by combining the timing diagrams of the individual instructions. The program timing diagram is shown in the attached figure. Program Timing Diagram.

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The formula used to calculate the maximum deflection of a cantilever beam with a concentrated load P applied at the free end of a beam with length L is PL³/3El.

Hence, the correct option is b) PL³/3El.

What is a cantilever beam?

A cantilever beam is a type of beam that is fixed at one end and is free at the other.

This type of beam is common in many engineering structures, including bridges and buildings.

Due to its simple design, it is often used in a wide range of applications.

Cantilever beams are used in a variety of applications, including cranes, bridges, and even diving boards.

How to calculate the maximum deflection of a cantilever beam?

The maximum deflection of a cantilever beam can be calculated using the formula PL³/3El,

where

P is the load applied,

L is the length of the beam,

E is the elastic modulus of the material, and I is the moment of inertia of the beam cross-section.

This formula is based on the Euler-Bernoulli beam theory, which is commonly used to calculate the deflection of beams.

The formula is only valid if the load is applied perpendicular to the axis of the beam, and the beam is homogeneous and isotropic.

In addition, the beam must be long enough so that its deflection is negligible compared to its length, and the load must be concentrated at a single point.

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