An air tank at a processing plant is filled with air and its gauge pressure reads 1.0 MPa. Assuming R 287 1/kg.K, and considering ambient temperature and pressure, what is the density of the air in the tank? Select one: O a. 9.69 kg/m3 O b. 9.93 kg/m3 O C 12.86 kg/m3 O d. 10.42 kg/m3

The output of the driving motor under full load is approximately -166,012,022.18 N*m. This negative value indicates that the motor is doing work against the losses due to inefficiency in the system.

Given:

Mass of empty tub (m_empty) = 200 kg

Mass of full tub (m_full) = 900 kg

Distance between two points (d) = 880 m

Incline angle (θ) = 8 degrees

Speed of the rope (v) = 4.5 km/h

= 1.25 m/s

Mass of the rope per unit length (m_rope) = 2.7 kg/m

Track resistance (R) = 170 N/ton

Mechanical efficiency (η) = 75%

To calculate the output of the driving motor under full load, we need to consider the work done against track resistance, the work done to lift the loaded tubs, and the losses due to mechanical inefficiency.

Step 1: Calculate the work done against track resistance:

W_track = R * (m_empty + m_full) * d

= 170 N/ton * (200 kg + 900 kg) * 880 m

= 166,880,000 N*m

Step 2: Calculate the work done to lift the loaded tubs:

h = d * sin(θ)

= 880 m * sin(8°)

≈ 127.35 m

W_lift = m_full * g * h

= 900 kg * 9.8 m/s² * 127.35 m

≈ 1,064,332.7 N*m

Step 3: Calculate the losses due to mechanical inefficiency:

W_total = m_full * v² / (2 * η)

= 900 kg * (1.25 m/s)² / (2 * 0.75)

= 625 N*m

W_loss = W_total - (W_track + W_lift)

= 625 Nm - (166,880,000 Nm + 1,064,332.7 Nm)

≈ -166,012,647.18 Nm

The negative value indicates the losses due to inefficiency.

Step 4: Calculate the output of the driving motor under full load:

P_output = W_total + W_loss

= 625 Nm - 166,012,647.18 Nm

≈ -166,012,022.18 N*m

Conclusion:

The output of the driving motor under full load is approximately -166,012,022.18 N*m. This negative value indicates that the motor is doing work against the losses due to inefficiency in the system.

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Power developed by the motor under full load is 4567.69907 W or 4.57 kW (approximately).

Given, Mass of empty tubs = 200 kg

Mass of full tubs = 900 kg

Speed of rope = 4.5 kmph

= 4500 m/hour

Mass of rope = 2.7 kg/m

Track resistance = 170 N/ton

Mechanical efficiency of the system = 75%Weight of rock that is moved during an 8-hour shift = 480 tons Distance between the two points = 880 m Angle of incline = 8 degree Output of the driving motor under full load can be calculated as follows:

Firstly, we need to calculate the weight of the loaded tub :Weight of full tub = Weight of rock + Weight of empty tub Weight of full tub = (900 × 1000) + 200Weight of full tub = 900200 kg Weight of full tub = 1,100 kg Now, let's calculate the total resistance of the system: Total resistance = Frictional resistance + Incline resistance

Frictional resistance = Track resistance × Total weight of tubs Frictional resistance = 170 × (200 + 1100)Frictional resistance = 204000 N Incline resistance = Weight of full tub × sin(8 degrees) × Distance between the two points Incline resistance = 1100 × sin(8) × 880Incline resistance = 95917.6886 N  total resistance = Frictional resistance + Incline resistance .

Total resistance = 204000 + 95917.6886Total resistance = 299917.6886 N Now, we can calculate the useful work done: Useful work done = Force × Distance Useful work done = Weight of rock × Distance Useful work done = 480 × 1000 × 9.81 × 880Useful work done = 4.15104 × 10^9 J Now, let's calculate the power developed by the motor :

Power = Work done/Time Power = Useful work done/Efficiency Power = 4.15104 × 10^9/(8 × 3600 × 0.75)Power = 4567.69907 W

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Given data:Heat pump will provide 1 MW of heating.The heat pump will operate with an evaporation temperature of 10°C and a condensing temperature of 100°C.The evaporator of the heat pump is used to keep the air in a processing room climate controlled at 15°C.

The heat pump provides heating of water from 50°C to 90°C.To find: The amount of chilling at 15°C that can be provided by the heat pumpSolution:As per the question, the evaporator of the heat pump is used to keep the air in a processing room climate controlled at 15°C.Evaporation temperature of the heat pump is 10°C, so the heat is extracted at 10°C from the room.

The heat extracted by the evaporator of the heat pump, as refrigeration,Q = 1 / COP * W = (m * c * ΔT) / COPWe have to calculate W, soW = m * c * ΔT * COPW = 1.225 * V * 0.718 * (-10) * 3W = - 26.23 VAt 15°C, the volume of the room would be known so we can easily calculate W as per the above equation.So, the amount of chilling at 15°C that can be provided by the heat pump is -26.23 V (kW).Negative sign indicates that the heat pump is absorbing heat from the room. Hence, the heat pump will act as a refrigerator in this case.

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1) If the loop rate of the system were increased by a factor of 2, then the P, I, and D gains in the microcontroller code required to keep the same system tuning would have to be halved.2) In practice, the noise from the derivative component of the control system would increase as the loop rate increases.

This is because the derivative component of the control system amplifies high-frequency noise, which is why higher loop rates can amplify noise even further.3) To produce a variable rate control system on the Flinduino, the following steps are required:Identify the motor and control system's required transfer function based on the given input and output parameters.

Test and refine the control system by changing the input parameters and monitoring the output to ensure the motor's stability and response time is optimal.This variable rate control system allows the system to maintain optimal control of the motor's speed under changing load conditions, and is particularly useful for applications that require precise control of the motor's speed.

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When ice is added to a large insulated bottle containing water initially at 21.6 °C, the final temperature of the water depends on the mass of ice added. Here are the steps to find out the final temperature when 100 grams of ice is added to the bottle.

mass of ice in kgL = latent heat of fusion of water = 333.55 J/gm × 100 gm = 33,355 JStep 2: Determine the heat lost by waterQ = mcΔT, whereQ = heat lost by water in joulesc = specific heat of water = 4.18 kJ/kgK = 4,180 J/kgKm = mass of water = 707 g = 0.707 kgΔT = final temperature - initial temperature = T - 21.6°CSubstituting the given values in the equation and rearranging it gives,ΔT = Q / (mc)

= 33355 / (0.707 × 4180) = 11.37 KTherefore, the final temperature is 21.6°C - 11.37°C = 10.23°C. This is less than 0°C and implies that the ice has cooled the water to its freezing point. Thus, the final temperature of water is 0°C.

Determine the heat lost by waterQ = mcΔT = 0.707 × 4180 × (21.6 - 7.8) = 85760.08 JStep 2: Determine the heat absorbed by the iceQ = mL, whereL = latent heat of fusion of water = 333.55 J/gmQ = 85760.08 JM = Q / L = 85760.08 / 333.55 = 257.47 gmHence, 257.47 gm or 258 gm of ice is required to cool the water to 7.8°C.

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The deflection under the load P is 8.57 mm. Therefore, the correct answer is option D.

Given that, EI=30 kN.m², k = 15 kN/m, L=1 m, and P=900 N

The strain energy under the load of 900 N is given by:

U = (900 N)²×(1 m)³/(2 × (15 kN/m×(1 m)³+3×30 kN.m²))

= 8100/(540+90)

= 8100/630

= 12.7 J

The deflection under the load is given by:

δ = (P×L³)/(3×EI)

= (900 N×(1 m)³)/(3×30 kN.m²)

= 8.57 mm

Therefore, the correct answer is option D.

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"Your question is incomplete, probably the complete question/missing part is:"

For the system shown, the strain energy under load P is p²L³/2(kL³+3EI).

For EI=30 kN.m², k = 15 kN/m, L=1 m, and P=900 N, the deflection under P is best given by

a) 6.21 mm

b) 5.00 mm

c) 7.20 mm

d) 8.57 mm

Therefore, the resolution of the instrument is 2 mm.

The LVDT (Linear Variable Differential Transformer) is a type of transducer that produces an output voltage that varies linearly with the displacement of the core. This type of transducer has applications in the measurement of position, acceleration, vibration, and other physical parameters.

Let's solve the given problem step by step:

Sensitivity of the LVDT:

Sensitivity of the LVDT is defined as the ratio of the output voltage to the input displacement.

Mathematically, it is given by the following formula:

Sensitivity of LVDT = Output voltage/ Displacement of core

Given that, an output of 2 mV appears across the terminals of LVDT when the core moves through a distance of 0.5 mm.

Therefore, the sensitivity of the LVDT is:

Sensitivity of LVDT = Output voltage/ Displacement of core= (2 mV/0.5 mm) = 4 mV/mm

Sensitivity of the whole setup:

Sensitivity of the whole setup is defined as the ratio of the output voltage of the system to the input physical parameter being measured (displacement in this case).Mathematically, it is given by the following formula:

Sensitivity of the whole setup = (Output voltage of the system/ Input physical parameter) x Amplification factor

Given that, the output of the LVDT is connected to a 5 V voltmeter through an amplifier whose amplification factor is 250.

Therefore, the sensitivity of the whole setup is:

Sensitivity of the whole setup = (Output voltage of the system/ Input physical parameter) x Amplification factor= (2 mV/0.5 mm) x 250 = 1000 mV/mm

Resolution of the instrument:

Resolution of the instrument is the smallest increment that can be detected on the scale of the instrument. In this case, the voltmeter scale has 100 divisions, and it can be read to 1/5 of a division.

Therefore, the smallest increment that can be detected on the scale is:

Smallest increment = (1/5) x (1/100) = 0.002 V

To find the resolution of the instrument in mm, we need to convert the voltage reading into displacement reading using the sensitivity of the whole setup.

Resolution of the instrument = Smallest increment x Sensitivity of the whole setup= 0.002 V x 1000 mV/mm= 2 mm

Therefore, the resolution of the instrument is 2 mm.
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(a + b)(b + c) + c(a + b) This is the minimized form of the function, with only two terms and three literals.

The boolean algebra is a mathematical discipline that includes a set of operations defined in two elements (usually true or false). The main purpose of Boolean algebra is to reduce the complexity of digital circuits by simplifying complex Boolean expressions.

In this way, Boolean algebra is a fundamental tool for digital circuit designers.

Minimizing a boolean function involves simplifying the function to its most concise form, with a minimal number of literals and terms. This reduces the number of gates in a digital circuit, leading to a simpler, more efficient design.

Using boolean algebra minimize the following functions F

F(a,b,c) = (a + b).a.(ab + c)

The first step is to expand the function using the distributive law as follows:

F(a, b, c) = (a + b).a.ab + (a + b).a.c

Next, we simplify each term separately by using Boolean algebra rules:

a.ab = ab (since a.a = a)a.c

= ac + bc (by factoring out the common a)

Now we substitute the simplified terms back into the original equation:

F(a, b, c) = ab + (a + b)ac + (a + b)bc

Next, we factor out the common terms:

a(b + c) + (a + b)c(a + b)

Finally, we simplify by using the distributive law and some algebraic manipulation:

F(a, b, c) = a(b + c) + b(b + c) + c(a + b)

= (a + b)(b + c) + c(a + b)

This is the minimized form of the function, with only two terms and three literals.

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To determine the factor of safety using the distortion-energy theory, we need to calculate σ' and then find the factor of safety (n) using the formula n = Sy/σ'.

Given:

σx = -89 MPa

σy = 40 MPa

τxy = 0

First, we need to calculate σ':

σ' = (√(σx² - σxσy + σy² + 3τxy²))

Substituting the given values:

σ' = (√((-89)² - (-89)(40) + (40)² + 3(0)²))

σ' = (√(7921 + 3560 + 1600 + 0))

σ' = (√13081)

σ' ≈ 114.41 MPa

Now, we can calculate the factor of safety (n):

n = Sy/σ'

n = 295 MPa / 114.41 MPa

n ≈ 2.58

Therefore, the factor of safety is approximately 2.58.

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The difference between the specular reflection and diffuse reflection in the properties of radiation heat transfer is that in specular reflection, the reflected wave is directional and is reflected at the same angle of incidence as it hits the surface, whereas in diffuse reflection, the reflected wave is not directional and is scattered in multiple directions.

Radiation heat transfer can be categorized into two types of reflections: specular reflection and diffuse reflection.

The properties of these two types of reflection differ from one another.

Specular Reflection is when an incident ray falls on a surface and bounces off at the same angle, preserving the angle of incidence and the angle of reflection.

The wave reflected in specular reflection is highly directional, that is, the surface is very smooth, and the angle of incidence is the same as the angle of reflection.

Diffuse reflection, on the other hand, is when an incident ray falls on a surface and bounces off in multiple directions.

This type of reflection is caused by rough surfaces that scatter the incoming wave. Unlike specular reflection, diffuse reflection is not directional.

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The correct code for solving Routh Table - Stability of the system - Number of poles is coded below.

The corrected and modified code to solve the Routh-Hurwitz stability criterion:

coeffVector = input('Input vector of your system coefficients: \n i.e. [an an-1 an-2 ... a0] = ');

coeffLength = length(coeffVector);

rhTableColumn = ceil(coeffLength/2);

rhTable = zeros(coeffLength, rhTableColumn);

rhTable(1, :) = coeffVector(1, 1:2:coeffLength);

if (rem(coeffLength, 2) ~= 0)

   rhTable(2, 1:rhTableColumn - 1) = coeffVector(1, 2:2:coeffLength);

else

   rhTable(2, :) = coeffVector(1, 2:2:coeffLength);

end

epss = 0.01;

for i = 3:coeffLength

   if all(rhTable(i-1, :) == 0)

       order = (coeffLength - i);

       cnt1 = 0;

       cnt2 = 1;

       for j = 1:rhTableColumn - 1

           rhTable(i-1, j) = (order - cnt1) * rhTable(i-2, cnt2);

           cnt2 = cnt2 + 1;

           cnt1 = cnt1 + 2;

       end

   end

   for j = 1:rhTableColumn - 1

       firstElemUpperRow = rhTable(i-1, 1);

       rhTable(i, j) = ((rhTable(i-1, 1) * rhTable(i-2, j+1)) - ...

                       (rhTable(i-2, 1) * rhTable(i-1, j+1))) / firstElemUpperRow;

   end

   if rhTable(i, 1) == 0

       rhTable(i, 1) = epss;

   end

end

unstablePoles = 0;

for i = 1:coeffLength - 1

   if sign(rhTable(i, 1)) * sign(rhTable(i+1, 1)) == -1

       unstablePoles = unstablePoles + 1;

   end

end

fprintf('\nRouth-Hurwitz Table:\n')

rhTable

if unstablePoles == 0

   fprintf('~~~~~> It is a stable system! <~~~~~\n')

else

   fprintf('~~~~~> It is an unstable system! <~~~~~\n')

end

fprintf('\nNumber of right-hand side poles: %d\n', unstablePoles)

reply = input('Do you want the roots of the system to be shown? Y/N ', 's');

if reply == 'y' || reply == 'Y'

   sysRoots = roots(coeffVector);

   fprintf('\nGiven polynomial coefficients roots:\n')

   sysRoots

end

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The angular acceleration of the hydraulic cylinder is zero, and the acceleration of the box at point G is 2 m/s².

The given information states that the hydraulic cylinder FC extends with a constant speed of 2 m/s. Since the speed is constant, it implies that the cylinder is moving with a constant velocity, which means there is no acceleration in the linear motion of the cylinder.

Therefore, the angular acceleration of the cylinder is zero.As for the box at point G, its acceleration can be determined by analyzing the motion of the cylinder.

Since the cylinder rotates at point F, the box at point G will experience a centripetal acceleration due to its radial distance from the axis of rotation. This centripetal acceleration can be calculated using the formula:

Acceleration (a) = Radius (r) × Angular Velocity (ω)²

In this case, the radius is given as the length FC, which is 1000 mm (or 1 meter). Since the angular velocity is not provided, we can determine it by dividing the linear velocity of the cylinder by the radius of rotation.

Given that the linear velocity is 2 m/s and the radius is 1 meter, the angular velocity is 2 rad/s.

Substituting these values into the formula, we get:

Acceleration (a) = 1 meter × (2 rad/s)² = 4 m/s²

Hence, the acceleration of the box at point G is 4 m/s².

The angular acceleration of the hydraulic cylinder is zero because it is moving with a constant velocity. This means that there is no change in its rotational speed over time.

The acceleration of the box at point G is determined by the centripetal acceleration caused by the rotational motion of the cylinder. The centripetal acceleration depends on the radial distance from the axis of rotation and the angular velocity.

By calculating the radius and determining the angular velocity, we can find the centripetal acceleration. In this case, the centripetal acceleration of the box at point G is 4 m/s².

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The four commonly implemented CFD discretization methods are -  (FDM), (FVM), (FEM) and (SEM).

(a) The four commonly implemented CFD discretization methods or programs are as follows:

Finite difference method (FDM)

Finite volume method (FVM)

Finite element method (FEM)

Spectral element method (SEM)

(b) Sketch of discretization method for each of the methods in (a) using a uniform geometry and grid is as follows:

1. Finite difference method (FDM) In finite difference method, the discretization process divides the whole domain into a discrete grid or mesh, and the partial derivatives are replaced by difference equations.

2. Finite volume method (FVM)The finite volume method focuses on the conservation of mass, energy, and momentum. A control volume in which all the variables are considered to be constant is considered in the method.

3. Finite element method (FEM)In finite element method, the solution is approximated over a finite set of basis functions that are defined within each element of the mesh. The unknowns are determined using a variational principle, and the equation is then solved using a linear or nonlinear solver.

4. Spectral element method (SEM)The spectral element method combines the strengths of finite element and spectral methods. A spectral decomposition is performed within each element to obtain the solution, which is then used to interpolate the solution within the element. This method is highly accurate and efficient.

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To cool engine oil from 80 to 50°C using a counterflow, concentric tube heat exchanger, with cooling water available at 20°C, the required tube length needs to be calculated. The problem provides information on flow rates, heat transfer coefficients.

To determine the tube length required, we can use the basic equation for heat transfer in a heat exchanger:

Q = U × A × ΔTlm

where Q is the heat transfer rate, U is the overall heat transfer coefficient, A is the heat transfer area, and ΔTlm is the logarithmic mean temperature difference.

In this case, the heat transfer rate can be calculated as the product of the mass flow rate and specific heat capacity difference of the oil:

Q = m_oil × Cp_oil × ΔT_oil

The overall heat transfer coefficient can be calculated using the individual heat transfer coefficients for the water and oil sides, as well as the fouling factors:

1/U = (1/h_water) + (1/h_oil) + (Rf_water) + (Rf_oil)

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A unity negative feedback control system has the loop transfer suction L(S) = G1(S)G(S) = K(S + 2) / (S + 1)(S + 2.5)(S + 4)(S + 10).a) Sketch the root lows as K varies from 0 to 2000:b) .

Find the roofs for K equal to 400, 500 and 600a) Root Locus is the plot of the closed-loop poles of the system that change as the gain of the feedback increases from zero to infinity. The main purpose of the root locus is to show the locations of the closed-loop poles as the system gain K is varied from zero to infinity.

The poles of the closed-loop transfer function T(s) = Y(s) / R(s) can be located by solving the characteristic equation. Therefore, the equation is given as:K(S+2) / (S+1)(S+2.5)(S+4)(S+10) = 1or K(S+2) = (S+1)(S+2.5)(S+4)(S+10)or K = (S+1)(S+2.5)(S+4)(S+10) / (S+2)Here, we can find out the closed-loop transfer function T(s) as follows:T(S) = K / [1 + KG(S)] = K(S+2) / (S+1)(S+2.5)(S+4)(S+10) .

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The range of K for stability of the given control system is $0 < K < 6$. Therefore, the answer is : Range of K for stability of a unity feedback control system whose open-loop transfer function is K G(s) = K/s(s+ 1)(s + 2) is 0 < K < 6.

Given Open loop transfer function: [tex]$$K G(s) = \frac{K}{s(s+ 1)(s + 2)}$$[/tex]

The closed-loop transfer function is given by: [tex]$$\frac{C(s)}{R(s)} = \frac{KG(s)}{1 + KG(s)}$$$$= \frac{K/s(s+ 1)(s + 2)}{1 + K/s(s+ 1)(s + 2)}$$[/tex]

On simplifying, we get: [tex]$$\frac{C(s)}{R(s)} = \frac{K}{s^3 + 3s^2 + 2s + K}$$[/tex]

The characteristic equation of the closed-loop system is: [tex]$$s^3 + 3s^2 + 2s + K = 0$$[/tex]

To obtain a range of values of K for stability, we will apply Routh-Hurwitz criterion. For that we need to form Routh array using the coefficients of s³, s², s and constant in the characteristic equation: $$\begin{array}{|c|c|} \hline s^3 & 1\quad 2 \\ s^2 & 3\quad K \\ s^1 & \frac{6-K}{3} \\ s^0 & K \\ \hline \end{array}$$

For stability, all the coefficients in the first column of the Routh array must be positive: [tex]$$1 > 0$$$$3 > 0$$$$\frac{6-K}{3} > 0$$[/tex]

Hence, [tex]$\frac{6-K}{3} > 0$[/tex] which implies $K < 6$.

So, the range of K for stability of the given control system is $0 < K < 6$.Therefore, the answer is : Range of K for stability of a unity feedback control system whose open-loop transfer function is K G(s) = K/s(s+ 1)(s + 2) is 0 < K < 6.

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To calculate the convolution of x₁(t) and x₂(t), let's apply the formula of convolution, which is denoted by -

[tex]x₁(t) * x₂(t).x₁(t) * x₂(t) = ∫ x₁(τ) x₂(t-τ) dτ= ∫ (u(τ) - u(τ-5))(u(t-τ) - u(t-τ-10)) dτIt[/tex]should be noted that u(τ-5) and u(t-τ-10) have a time delay of 5 and 10, respectively, which means that if we move τ to the right by 5,

After finding x₁(t) * x₂(t), the Laplace transform of the function is required. The Laplace transform is calculated using the formula:

L{x(t)} = ∫ x(t) * e^(-st) dt

L{(15-t)u(t)} = ∫ (15-t)u(t) * e^(-st) dt

             = e^(-st) ∫ (15-t)u(t) dt

             = e^(-st) [(15/s) - (1/s^2)]

L{(t-5)u(t-5)} = e^(-5s) L{t*u(t)}

              = - L{d/ds(u(t))}

              = - L{(1/s)}

              = - (1/s)

L{(t-10)u(t-10)} = e^(-10s) L{t*u(t)}

               = - L{d/ds(u(t))}

               = - L{(1/s)}

               = - (1/s)

L{(15-t)u(t) - (t-5)u(t-5) + (t-10)u(t-10)} = (15/s) - (1/s^2) + (1/s)[(1-e^(-5s))(t-5) + (1-e^(-10s))(t-10)]

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The logarithmic mean temperature difference (LMTD) is 106.614°C.
The logarithmic mean temperature difference (LMTD) is used to compute the heat transfer rate in a heat exchanger or a cooling tower.

When a chemical plant's flue gas (at atmospheric pressure) contains harmful vapors that must be condensed by reducing its temperature from 295°C to 32°C and the gas flow rate is 0.60 m ∧3/s, this calculation becomes crucial. Water is available at 12°C at 1.5 kg/s.

A counterflow heat exchanger will be used with water flowing through the tubes.
The gas has a specific heat of 1[tex].12 kJ/kg−K[/tex]and a gas constant of 0.26 kJ/kg−K;
let c [tex]pwater=4.186 kJ/kg−K.[/tex]
The logarithmic mean temperature difference (LMTD) for the process is calculated as follows:
Step 1: Mean temperature of the hot fluid, [tex]ΔT1=(295−32)/ln(295/32)=175.364°C[/tex]
Step 2: Mean temperature of the cold fluid, [tex]ΔT2=(12−32)/ln(12/32)=20.609°C[/tex]
Step 3: Logarithmic mean temperature difference
[tex]ΔTlm= (ΔT1-ΔT2)/ ln(ΔT1/ΔT2) = (175.364 - 20.609)/ln(175.364/20.609) = 106.614°C.[/tex]

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The task at hand is to write a function M-file that implements (8) in the interval 0 ≤ t ≤ 55. The initial condition must now be in the form [yo, v0, w0]. The matrix Y, which is the output of ode45, now has three columns. Y(:,1) represents y, Y(:,2) represents v and Y(:,3) represents w. We need to extract these columns.

We also need to plot the three time series on the same figure and, on a separate window, plot the phase plot using figure (2); plot3 (y,v,w); hold on; view ([-40,60]) xlabel('y'); ylabel('vay); zlabel('way'').Here is a function M-file that does what we need:

function [tex]yp = fun(t,y)yp = zeros(3,1);yp(1) = y(2);yp(2) = y(3);yp(3) = -sin(y(1))-0.1*y(3)-0.1*y(2);[/tex]

endWe can now use ode45 to solve the ODE.

The limits in the vertical axis of the plot on the left were deliberately set to the same ones as in Figure 8 for comparison purposes, using the MATLAB command ylim ([-2.1,2.1]). You can play around with the 3D phase plot, rotating it by clicking on the circular arrow button in the figure toolbar, but submit the plot with the view value view ([-40, 60]) (that is, azimuth = -40°, elevation = 60°).

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(i) Shear force equation Let’s draw the free-body diagram of the beam and determine the shear force equation. According to the diagram shown below:
By considering equilibrium of forces along the y-axis; ƩFy = 0 - R1 + 30 = 0  R1 = 30 kN/mBy considering the shear force equation along the beam’s length, x;V(x) = R1  = 30 kN/m

(ii) Bending moment equation The bending moment can be determined by considering the equilibrium of forces about the point where the shear force equation is determined. By considering the bending moment equation along the beam’s length, x;M(x) = -R1x + M1By substituting R1 = 30kN/m, M1 = 45kN.m;M(x) = -30x + 45kN.m

(iii) Shear force and Bending moment diagram The shear force and bending moment diagrams for the beam are shown below.

(iv) Maximum bending stress and maximum transverse shear stressThe formula for maximum transverse shear stress is;τmax=3VQtd2b2 whereQ = Moment of area about the neutral axis, Q = bd3/12d = depth of the cross-sectionThe cross-sectional area of the beam is 20 × 40 = 800 mm2The moment of area about the neutral axis is; Q = 20 × 403/12 = 13333.3 mm3V = R1 = 30 kN/m
Therefore,τmax=3VQtd2b2= (3 × 30 × 103 × 13333.3 × 20 × 10-3)/(2 × 40 × 10-3) = 150 MPaThe maximum transverse shear stress occurs at the neutral axis.The formula for maximum bending stress is;σmax=MxIxx WhereIxx = Moment of inertia of the cross-section
The moment of inertia of the rectangle cross-section is;Ixx = bd3/12 = (20 × 403)/12 = 13333.3 mm4 By substituting M(x) = -30x + 45kN.m;σmax=MxIxx= (-30x + 45kN.m) (20 × 403)/12 = (-750x + 1125) N/mm2
This maximum bending stress occurs at the upper surface of the beam.

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The answer to this part will be the same as the answer to part (b) since padding zeros does not affect the frequency content of the sequence, only its length, the 1000-point DFT is: X(0)=10, X(1)=-2-6i, X(2)=0, X(3)=2+6i, and X(4)=-2+2i.

When you are asked to pad zeros to a point sequence, you are expected to add zeros at the end of the point sequence to match a certain length. For example, in a four-point sequence x(n)={1, 2, 3, 4}, padding zeros to the sequence would involve adding zeros to the end of the sequence to meet a specified length, e.g., if the length required is 5 points, then zeros will be padded to the end of the sequence to get {1, 2, 3, 4, 0}.To solve the problem, we would use the following formula for computing DFT:X(k) = Summation [n=0, N-1] {x(n) exp(-i(2π/N)nk)}

Therefore, the 4-point DFT is: X(0)=10, X(1)=-2-6i, X(2)=0, and X(3)=2+6ib) 5-point DFT:To obtain the 5-point DFT of the sequence x(n)={1, 2, 3, 4}, we have to pad zeros to the end of the sequence such that the sequence has 5 points, i.e., x(n)={1, 2, 3, 4, 0}.Using the formula above and substituting the values for x(n), we get: X(k) = x(0) + x(1)exp(-i(2π/N)nk) + x(2)exp(-i(2π/N)2nk) + x(3)exp(-i(2π/N)3nk) + x(4)exp(-i(2π/N)4nk)Substituting x(n) = {1, 2, 3, 4, 0} into the above equation yields:X(0) = 1 + 2 + 3 + 4 + 0 = 10X(1) = 1 + 2exp(-iπ/2) + 3exp(-iπ) + 4exp(-i3π/2) + 0 = 1 - 2i - 3 - 4i = -2 - 6iX(2) = 1 + 2exp(-iπ) + 3exp(-i2π) + 4exp(-i3π) + 0 = 1 - 2 - 3 + 4 = 0X(3) = 1 + 2exp(-i3π/2) + 3exp(-i3π) + 4exp(-i9π/2) + 0 = 1 + 2i - 3 + 4i = 2 + 6iX(4) = 1 + 2exp(-i4π/2) + 3exp(-i4π) + 4exp(-i6π) + 0 = 1 + 2i - 3 - 4i = -2 + 2iTherefore, the 5-point DFT is: X(0)=10, X(1)=-2-6i, X(2)=0, X(3)=2+6i, and X(4)=-2+2ic) 1000-point DFT:

To obtain the 1000-point DFT of the sequence x(n)={1, 2, 3, 4}, we have to pad zeros to the end of the sequence such that the sequence has 1000 points, i.e., x(n)={1, 2, 3, 4, 0, 0, 0, ...}.Using the formula above and substituting the values for x(n), we get: X(k) = x(0) + x(1)exp(-i(2π/N)nk) + x(2)exp(-i(2π/N)2nk) + x(3)exp(-i(2π/N)3nk) + ... + x(999)exp(-i(2π/N)999nk)Since N=1000, the above formula will involve computing 1000 terms. For a large number like this, it is easier to compute using an algorithm known as the Fast Fourier Transform (FFT) instead of manually computing each term.

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A lift pump is required to lift water from a well and deliver it to a cylindrical tank. The lift pump has a diameter of 4 inches and a stroke of 6 inches. The lift pump's volumetric efficiency at 10 lifting strokes per minute is 90%.The pump capacity can be calculated using the following formula:
Pump capacity = π × r² × s × n × VEF where r is the radius of the lift pump, s is the stroke of the lift pump, n is the number of lifting strokes per minute, and VEF is the volumetric efficiency factor.The diameter of the lift pump is given as 4 inches, which means that the radius is 2 inches.r = 2 inches = 0.167 feet

The stroke of the lift pump is given as 6 inches, which means that the stroke is 0.5 feet.s = 0.5 feet The number of lifting strokes per minute is given as 10.n = 10 The volumetric efficiency factor is given as 90%.VEF = 0.9Pump capacity = π × r² × s × n × VEF= 3.1416 × (0.167)² × 0.5 × 10 × 0.9= 0.746 cubic feet per minute (CFM)The power required to operate the pump manually can be calculated using the following formula:Power = F × s × n / 33000 where F is the force required to lift the water, s is the stroke of the lift pump, n is the number of lifting strokes per minute, and 33,000 is the conversion factor.The force required to lift the water can be calculated using the following formula:Force = Weight of water lifted / Mechanical efficiency where Mechanical efficiency is given as 80%.
Weight of water lifted = Density of water × Volume of water lifted
Density of water = 62.4 lb/ft³
Volume of water lifted = Pump capacity × Operating efficiency= 0.746 × 0.7= 0.522 cubic feet Weight of water lifted = 62.4 × 0.522 = 32.6288 lb Force = 32.6288 / 0.8 = 40.786 lb Power = F × s × n / 33000= 40.786 × 0.5 × 10 / 33000= 0.000619 horsepower (HP)

The lift pump has a capacity of 0.746 cubic feet per minute (CFM) and is required to fill a 600-liter cylindrical tank at a height of 12 feet. The operating efficiency of the lift pump is given as 70%.The time required to fully fill the 600-liter tank can be calculated using the following formula:Time = Volume of tank / Pump capacity / Operating efficiency where the volume of the tank is given as 600 liters.The volume of the tank needs to be converted from liters to cubic feet.1 liter = 0.0353147 cubic feet Therefore, 600 liters = 600 × 0.0353147 = 21.1888 cubic feet Time = 21.1888 / 0.746 / 0.7= 41.36 minutes

Therefore, the pump capacity is 0.746 cubic feet per minute (CFM).The power required to operate the pump manually is 0.000619 horsepower (HP).The pump is required to fully fill the 600-liter tank in 41.36 minutes.

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A synchronous motor is a type of AC motor that o corresponding to the frequency of the applied voltage. The output power of a synchronous motor is proportional to the power supply voltage and the synchronous reactance of the motor.

If the supply voltage is held constant, reactance.The given synchronous motor has a rating of 1.92 kV, 1100 HP, and unity power factor. It is 60-Hz, 2-pole, and delta-connected. The synchronous reactance of the motor is 10.1 Ω per-phase. Additionally, the motor's armature resistance is negligible.

The friction and losses combined with the core losses are 4.4 kW. The open-circuit characteristic of the motor is tabulated below in detail:Exciting current      5.5 A
Field voltage (volts)     25.6
Armature current (amperes)          167.0
Power factor         0.86 lagging.

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The formula pV.A is a representation of flow work. It is a significant term in thermodynamics that indicates the work done by fluids while flowing. Flow work, also known as flow energy or work of flow, refers to the work done by the fluid as it flows through the cross-sectional area of the pipeline in which it is flowing.

Flow work is an essential component of thermodynamics because it is the work required to move a fluid element from one point to another. It is dependent on both the pressure and volume of the fluid. A fluid's flow work can be calculated by multiplying the pressure by the volume and the cross-sectional area through which the fluid flows. As a result, the formula pV.A is a representation of flow work.

The formula pV.A does not indicate enthalpy, shaft work, or internal energy. Enthalpy, also known as heat content, is a measure of the energy required to transform a system from one state to another. Shaft work, on the other hand, refers to the work done by a mechanical shaft to move an object.

Internal energy,  refers to the total energy of a system. flow work is the term indicated by the formula pV.A.

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In the main function, a double array of 3 elements is created and passed individually to the GetData function. The main function then prints the three values from the data array along with their element numbers.

#include

#include void GetData(double *ptr1, double *ptr2, double *ptr3)

{ *ptr1 = 7.5; printf("\nWhat value would you like to assign to element 1?");

scanf("%lf", ptr2); *ptr3 = (*ptr1) + (*ptr2); return; }

int main() { int index; double data[3];

GetData(&data[0], &data[1], &data[2]);

printf("Index Data\n");

for (index = 0; index < 3; index++) { printf("%d %8.3lf\n", index, data[index]); }

getch();

return 0; }

The value 7.5 is assigned to element 0 of the data array.2. The user is asked what value to assign to element 1 of the data array, and that value is inputted from the user. The pointer representing the array element is used directly in the scanf.3. The value from element 0 and element 1 of the array is added, and the sum is assigned to element 2 of the data array.

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Given, number of turns per phase, N = 600, RMS generated line voltage, V = 4500 V and frequency, f = 60 Hz. The relationship between RMS generated line voltage, V, frequency, f, and flux per pole, φ is given by the formula,V = 4.44fNφSo, the expression for flux per pole, φ is given by,φ = V / 4.44fNPlugging the given values, we get,φ = 4500 / (4.44 × 60 × 600)φ = 19.85 mWebers Therefore,

the flux per pole needed to produce the RMS generated line voltage of 4500 Volts at a frequency f-60 Hz is 19.85 mWebers.Option (D) is correct.Note: In AC generators, the voltage generated is proportional to the flux per pole, number of turns per phase, and frequency. The above formula is known as the EMF equation of an alternator.

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The Brayton cycle is a thermodynamic cycle that uses constant pressure in its heat rejection and heat addition processes. It is a cycle that operates in open systems.

Explanation of the Brayton cycle using appropriate thermodynamic terms:

The Brayton cycle is a thermodynamic cycle that uses gas turbines to generate power. It is a cycle that consists of four main processes: , heating, expansion, and cooling. The thermodynamic terms that are relevant to the Brayton cycle are the First Law of Thermodynamics, Second Law of Thermodynamics, and the Ideal Gas Law. The First Law of Thermodynamics states that energy cannot be created or destroyed but can only be transferred from one form to another. In the Brayton cycle, energy is converted from mechanical energy into thermal energy and then back into mechanical energy.

The Second Law of Thermodynamics states that all systems tend to move towards a state of maximum entropy. The Brayton cycle aims to minimize entropy and maximize efficiency. The Ideal Gas Law is a law that describes the behavior of ideal gases. In the Brayton cycle, the Ideal Gas Law is used to describe the behavior of the gas as it passes through the compressor, combustion chamber, turbine, and heat exchanger.

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(a) Please refer to the uploaded picture for the phase portrait of dt/dt versus θ for -10 ≤ θ ≤ 10.

(b) The fixed points, sorted from smallest to largest, are -7, -4, -0.5, 0, 4, 5.4, 5.7, 7, and 9.

(a) To plot the phase portrait of dt/dt versus θ, we need to evaluate the given expression for dt/dt over a range of values for θ. By varying θ from -10 to 10 and substituting it into the expression, we obtain a set of corresponding values for dt/dt. Plotting these values against θ gives us the phase portrait. Please refer to the uploaded picture for the visual representation.

(b) To determine the fixed points, we need to find the values of θ for which dt/dt is equal to zero. By setting the expression (θ - 4)(2θ - 1)(θ + 7) equal to zero, we can solve for θ. The solutions are -7, -4, -0.5, 0, 4, 5.4, 5.7, 7, and 9. Sorting these values from smallest to largest, we have -7, -4, -0.5, 0, 4, 5.4, 5.7, 7, and 9 as the fixed points.

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A 14.568 kg uniform solid disk of radius 0.59 mis pin supported at its center. It is acted upon by a constant couple moment M 9.088 N.m. Starting from rest, determine the angular velocity in rad seconds when time is 2 seconds.

In the given problem, we have to determine the angular velocity of a uniform solid disk of radius 0.59 m, mass 14.568 kg and acted upon by a constant couple moment M 9.088 N.m when time is 2 seconds.So,The moment of inertia of a uniform solid disk about its center is given as;I=12MR2Substitute the values of M and R in the above equation we get;I = 12(14.568 kg)(0.59 m)2= 2.554 kg m²The torque provided by the couple moment, τ is given as;τ = IαWhere α is the angular acceleration, thenα = τ/ISubstitute the values of τ and I in the above equation we get;α = (9.088 N.m) / (2.554 kg m²)= 3.562 rad/s²The final angular velocity is given by the formula;ωf = ωi + αtωi is the initial angular velocity, which is zero as the disc starts from rest. Substitute the values of α and t in the above equation we get;ωf = 0 + (3.562 rad/s²)(2 s) = 7.124 rad/s Therefore, the angular velocity when time is 2 seconds is 7.124 rad/s.

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Given differential equations are:d²x/dt² + x - y = 0  d²y/dt² + y - x = 0x(0) = 0, x'(0) = -5, y(0) = 0, y'(0) = 1 Using Laplace transform, we get:Laplace transform of the given differential equation:

[tex]L{d²x/dt²} + L{x} - L{y}\\ = 0s²X(s) - sx(0) - x'(0) + X(s) - Y(s) \\= 0s²X(s) + 5s + X(s) - Y(s) \\= 0s²X(s) + X(s) \\= Y(s) + 5s[/tex]

...(1)

[tex]L{d²y/dt²} + L{y} - L{x} \\= 0s²Y(s) - sy(0) - y'(0) + Y(s) - X(s)\\ = 0s²Y(s) + Y(s) - X(s) \\= 0s²Y(s) + Y(s) = X(s)[/tex] ...(2)

Solving equation (2) for [tex]X(s)X(s) = s²Y(s) + Y(s)[/tex]

Put X(s) in equation (1), we get:s[tex]²{s²Y(s) + Y(s)} + {s²Y(s) + Y(s)} = Y(s) + 5s[/tex]

Collecting like terms, we get:

[tex]s⁴Y(s) + s²Y(s) + Y(s) = Y(s) + 5s[/tex]

Factorizing Y(s), we get:Y(s)[s⁴ + s² - 1] = 5s

Dividing by (s⁴ + s² - 1), we get:

Y(s) = 5s/(s⁴ + s² - 1) .....(3)

Again, using equation (2), we get:

[tex]X(s) = s²Y(s) + Y(s) \\= s²(5s/(s⁴ + s² - 1)) + (5s/(s⁴ + s² - 1))[/tex]

[tex]= (5s³ + 5s)/(s⁴ + s² - 1)[/tex]

Now, finding inverse Laplace transform of Y(s) and X(s) using partial fraction decomposition:

[tex]Y(s) = 5s/(s⁴ + s² - 1)Let s² = t, thenY(s) = 5(s)/(s⁴ + s² - 1) = 5/(t² + t - 1)[/tex] Putting a partial fraction,[tex]Y(s) = 5/{(t - φ)(t + φ + 1)}, where φ = (1 + √5)/2[/tex]Using partial fractions[tex],X(s) = (5s³ + 5s)/(s⁴ + s² - 1) = 5s(s² + 1)/(s² - 1)(s² + 1)[/tex]

Putting partial fractions,[tex]X(s) = (5/2)[1/(s - 1) - 1/(s + 1)] + (5/2)[1/(s² + 1)][/tex]

Thus, the solution of the given differential equation system is:[tex]x(t) = (5/2)(e^t - e^(-t)) + (5/2)sin(t)y(t) = (5/2)[(1 + √5)/2(e^t - e^(-t)) - (1 - √5)/2(sin(t))][/tex]Hence, the main answer to the given question is:[tex]x(t) = (5/2)(e^t - e^(-t)) + (5/2)sin(t)y(t) = (5/2)[(1 + √5)/2(e^t - e^(-t)) - (1 - √5)/2(sin(t))][/tex]

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The velocity that will initiate cavitation is approximately 2827.6 mm/s or 37.12 mm/s

To calculate the velocity that will initiate cavitation, we can use the Bernoulli's equation between two points along the flow path. The equation relates the pressure, velocity, and elevation at those two points.

In this case, we'll compare the conditions at the minimum pressure point (where cavitation occurs) and a reference point at the same depth.

The Bernoulli's equation can be written as:

[tex]\[P_1 + \frac{1}{2} \rho v_1^2 + \rho g h_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho g h_2\][/tex]

where:

[tex]\(P_1\)[/tex] and [tex]\(P_2\)[/tex] are the pressures at points 1 and 2, respectively,

[tex]\(\rho\)[/tex] is the density of water,

[tex]\(v_1\)[/tex] and [tex]\(v_2\)[/tex] are the velocities at points 1 and 2, respectively,

[tex]\(g\)[/tex] is the acceleration due to gravity, and

[tex]\(h_1\)[/tex] and [tex]\(h_2\)[/tex] are the elevations at points 1 and 2, respectively.

In this case, we'll consider the minimum pressure point as point 1 and the reference point at the same depth as point 2.

The elevation difference between the two points is zero [tex](\(h_1 - h_2 = 0\))[/tex]. Rearranging the equation, we have:

[tex]\[P_1 - P_2 = \frac{1}{2} \rho v_2^2 - \frac{1}{2} \rho v_1^2\][/tex]

Given:

[tex]\(P_1 = 80 \, \text{kPa}\)[/tex] (absolute pressure at the minimum pressure point),

[tex]\(P_2 = 100 \, \text{kPa}\)[/tex] (atmospheric pressure),

[tex]\(\rho\) (density of water at 10 °C)[/tex] can be obtained from a water density table as [tex]\(999.7 \, \text{kg/m}^3\)[/tex], and

[tex]\(v_1 = (98 + 5) \, \text{mm/s} = 103 \, \text{mm/s}\).[/tex]

Substituting the values into the equation, we can solve for [tex]\(v_2\)[/tex] (the velocity at the reference point):

[tex]\[80 \, \text{kPa} - 100 \, \text{kPa} = \frac{1}{2} \cdot 999.7 \, \text{kg/m}^3 \cdot v_2^2 - \frac{1}{2} \cdot 999.7 \, \text{kg/m}^3 \cdot (103 \, \text{mm/s})^2\][/tex]

Simplifying and converting the units:

[tex]\[ -20 \, \text{kPa} = 4.9985 \, \text{N/m}^2 \cdot v_2^2 - 0.009196 \, \text{N/m}^2 \cdot \text{m}^2/\text{s}^2\][/tex]

Rearranging the equation and solving for \(v_2\):

[tex]\[v_2^2 = \frac{-20 \, \text{kPa} + 0.009196 \, \text{N/m}^2 \cdot \text{m}^2/\text{s}^2}{4.9985 \, \text{N/m}^2} \]\\\\\v_2^2 = 7.9926 \, \text{m}^2/\text{s}^2\][/tex]

Taking the square root to find [tex]\(v_2\)[/tex]:

[tex]\[v_2 = \sqrt{7.9926} \, \text{m/s} \approx 2.8276 \, \text{m/s}\][/tex]

Converting the velocity to millimeters per second:

[tex]\[v = 2.8276 \, \text{m/s} \cdot 1000 \, \text{mm/m} \approx 2827.6 \, \text{mm/s}\][/tex]

Therefore, the velocity that will initiate cavitation is approximately 2827.6 mm/s or 37.12 mm/s (rounded to two decimal places).

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