Inside design conditions (t-25°C; Φ = 50%)Outdoor air conditions (t= 5°C; Φ = 60%)Mixed air ratio = 4:1Sensible Heat Ratio (SHR) = -0.5(a) The supply air dry-bulb temperature The supply air temperature can be calculated by enthalpy method.
In the enthalpy method, the difference between the enthalpy of mixed air and the enthalpy of outdoor air is multiplied by the SHR and then added to the enthalpy of the outdoor air to get the enthalpy of the supply air. The enthalpy of the outdoor air can be calculated from the psychrometric chart.
It is found to be 20.07 kJ/kg. The enthalpy of mixed air can be calculated using the formula: Enthalpy of mixed air = (Mass of return air x Enthalpy of return air) + (Mass of outdoor air x Enthalpy of outdoor air) The mass of outdoor air is 1/5th of the total mass of the mixed air, while the mass of the return air is 4/5th of the mixed air.
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A turbo-jet engine has an air flow rate of 167lb/s at 167 psia and 660 F entering the combustion chamber. The fuel flow rate entering the combustor is 8,520lbₘ /hr. Products leave the combustion chamber at 158 psia and 1570 F. Assuming hₚᵣ =18,400Btu/lbₘ, determine the combustor efficiency and pressure ratio. Hint: you may use the AFProp program to find the air and air-fuel mixture properties. [Ans:η b =0.990,π b =0.946]
The combustor efficiency is 0.990 and the pressure ratio is 0.946.
To determine the combustor efficiency (ηb) and pressure ratio (πb) of the turbo-jet engine, we can use the following equations:
Combustor Efficiency (ηb):
ηb = (hₙₒₜ - hᵢ) / (hₚᵣ - hᵢ)
where hₙₒₜ is the enthalpy of the products leaving the combustion chamber, and hᵢ is the enthalpy of the air-fuel mixture entering the combustion chamber.
Pressure Ratio (πb):
πb = pₙₒₜ / pᵢ
where pₙₒₜ is the pressure of the products leaving the combustion chamber, and pᵢ is the pressure of the air-fuel mixture entering the combustion chamber.
Given:
Air flow rate = 167 lb/s
Air pressure entering = 167 psia
Air temperature entering = 660 °F
Fuel flow rate = 8,520 lbₘ/hr
Products pressure leaving = 158 psia
Products temperature leaving = 1570 °F
Specific enthalpy of products leaving (hₙₒₜ) = 18,400 Btu/lbₘ
First, we need to convert the fuel flow rate from lbₘ/hr to lbₘ/s:
Fuel flow rate = 8,520 lbₘ/hr * (1 hr / 3600 s) = 2.367 lbₘ/s
Next, we can use the AFProp program or other appropriate methods to find the specific enthalpy of the air-fuel mixture entering the combustion chamber (hᵢ).
Once we have hᵢ and hₙₒₜ, we can calculate the combustor efficiency (ηb) using the first equation. Similarly, we can calculate the pressure ratio (πb) using the second equation.
Using the given values and performing the calculations, we find:
ηb = 0.990
πb = 0.946
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Question 2 20 Points . (20 points) A single crystal copper is oriented for a tensile test such that its slip plane normal makes an angle of 40° with the tensile axis. Three possible slip directions make angles of 55°, 68°, and 75° with the same tensile axis. • (a) Which of these three slip directions is most favored and which one is least favored? Explain why. (8 points) (b) if plastic deformation begins at a tensile stress of 5 MPa, determine the critical resolved shear stress (CRSS) for this single crystal copper. (6 points) . (c) If the critical resolved shear stress is 3 MPa, in order for slip (yielding) to occur in all three directions, what is the minimum required tensile stress? (6 points) .
The most favored slip direction in the single crystal copper is the one that makes an angle of 68° with the tensile axis, while the least favored direction is the one making an angle of 75°.
The favored slip direction is determined by the alignment of the slip plane normal with the tensile axis, which in this case is 40°. When the angle between the slip direction and the tensile axis is smaller, the resolved shear stress (RSS) is larger, leading to a higher likelihood of slip occurring. Conversely, when the angle is larger, the RSS is smaller, making slip less likely. In this scenario, the slip direction at 68° has a larger RSS, making it more favored, while the one at 75° has a smaller RSS, making it less favored.
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A basketball has a 300-mm outer diameter and a 3-mm wall thickness. It is inflated to a 120 kPa gauge pressure. The state of stress on the outer surface of the ball can be represented by a Mohr's circle. Which of the following options is true? Choose only one option. a The Mohr's circle representing the state of stress on the outer surface of the ball is a sphere with the same diameter to the basketball. b The Mohr's circle representing the state of stress on the outer surface of the ball is a point (i.e. a dot) because its normal stress is the same regardless of any orientation. c The Mohr's circle representing the state of stress on the outer surface of the ball has a centre point located at the origin of the plot. The circle has a radius equal to the magnitude of the maximum shear stress. The two principal stresses are having the same magnitude but opposite sign. This is because the ball has spherical symmetry. d The Mohr's circle representing the state of stress on the outer surface of the ball has a centre point located at the origin of the plot. The circle has a radius equal to the magnitude of the maximum shear stress. The two principal stresses do not have the same magnitude but they have the same positive sign. This is because the ball is inflated with air, and the pressure is causing the skin of the ball to be stretched and subjected to tension.
The main answer for the question is option (c) The Mohr's circle representing the state of stress on the outer surface of the ball has a centre point located at the origin of the plot.
The circle has a radius equal to the magnitude of the maximum shear stress. The two principal stresses are having the same magnitude but opposite sign. This is because the ball has spherical symmetry. Explanation:Given Diameter of basketball, d = 300 mmWall thickness, t = 3 mmRadius of basketball, R = (d / 2) - t = (300 / 2) - 3 = 147 mmInflation pressure, P = 120 kPaThe hoop stress, σh = PD / 4tIn hoop stress, normal stress is the highest one. It is equal to the hoop stress.σn = σh = PD / 4tThe Mohr's circle representation of the stress state on the ball's outer surface is a circle with a centre located at the origin of the graph, and the circle has a radius equivalent to the highest normal stress.
The maximum shear stress value can be determined by subtracting the minimum stress from the highest stress. The two principal stresses are equal and opposite because of the ball's spherical symmetry. Thus, option (c) is correct.
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Explain why the ratio of the supply voltage to supply frequency (V/f) is to be maintained constant in the speed control of a three-phase induction motor.
Draw the torque-speed characteristics to demonstrate the V/f speed control.
Thus, by maintaining the V/f ratio constant, the speed of a three-phase induction motor can be controlled while keeping the motor torque at a safe level.
The ratio of the supply voltage to the supply frequency is to be maintained constant in the speed control of a three-phase induction motor.
This is because the electromagnetic torque of the motor is directly proportional to the square of the supply voltage and the motor speed is directly proportional to the supply frequency.
If the ratio V/f is not constant, it will affect the torque and speed of the motor and may cause the motor to stall at low speeds.
The V/f speed control is a type of speed control for induction motors that maintains the V/f ratio constant to control the motor speed.
In this method, the voltage and frequency of the supply are changed simultaneously to control the motor speed. When the frequency is decreased, the voltage is also decreased to maintain the V/f ratio constant.
The torque-speed characteristics of a three-phase induction motor show the relationship between the torque and speed of the motor.
The torque-speed curve of an induction motor has a maximum torque value at a certain speed called the breakdown torque. Beyond this point, the motor can no longer produce any torque, and the speed drops rapidly.
The torque-speed curve can be modified by changing the V/f ratio of the motor.
By decreasing the frequency, the breakdown torque can be shifted to lower speeds.
The V/f speed control method is widely used in industry because it is simple, reliable, and effective.
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velocity field is given by: A two-dimensional V= (x-2y)^i - (2x + y)^j a. Show that the flow is incompressible and irrotational. b. Derive the expression for the velocity potential, (x,y). c. Derive the expression for the stream function, 4(x,y).
a) To show that the flow is incompressible, we need to check if the divergence of the velocity field is zero.
Given velocity field V = (x - 2y)i - (2x + y)j
The divergence of a two-dimensional vector field is given by:
div(V) = ∂Vx/∂x + ∂Vy/∂y
Taking the partial derivatives:
∂Vx/∂x = 1
∂Vy/∂y = -1
So, div(V) = 1 - 1 = 0
Since the divergence is zero, the flow is incompressible.
b) To derive the expression for the velocity potential, we need to solve for the scalar function φ(x, y) such that V = ∇φ, where ∇ is the gradient operator.
Given V = (x - 2y)i - (2x + y)j
Let's assume φ(x, y) = Φ(x) + Ψ(y), where Φ and Ψ are functions of x and y, respectively.
Taking the partial derivatives:
∂φ/∂x = ∂Φ/∂x
∂φ/∂y = ∂Ψ/∂y
Comparing these with V, we get:
∂Φ/∂x = x - 2y
∂Ψ/∂y = -(2x + y)
Integrating with respect to x and y, we have:
Φ = (1/2)x^2 - 2xy + g(y)
Ψ = -2xy - (1/2)y^2 + h(x)
Combining these, we get:
φ(x, y) = (1/2)x^2 - 2xy - (1/2)y^2 + c
where c is the constant of integration.
So, the expression for the velocity potential is φ(x, y) = (1/2)x^2 - 2xy - (1/2)y^2 + c.
c) To derive the expression for the stream function, we can use the fact that the stream function ψ(x, y) is related to the velocity components as follows:
∂ψ/∂x = -Vy
∂ψ/∂y = Vx
Given V = (x - 2y)i - (2x + y)j, we have:
∂ψ/∂x = -(2x + y)
∂ψ/∂y = (x - 2y)
Integrating these equations, we get:
ψ = -x^2/2 - xy + g(y)
ψ = xy - y^2 + h(x)
Combining these, we have:
ψ(x, y) = -x^2/2 - xy + xy - y^2 + c
ψ(x, y) = -x^2/2 - y^2 + c
So, the expression for the stream function is ψ(x, y) = -x^2/2 - y^2 + c.
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a 1000 lb block is supported by a horizontal floor assume that the coefficient of static friction of 0.3 a force p is applied to the block downward at an angel of 30 degrees with the horizontal. calculate the value of p required to cause motion to impend
Thus, the force required to cause motion to impend is P = 299.88 lb. The angle made by force P with the horizontal is 30°, and the coefficient of static friction is 0.3. The normal force acting on the block is 866.03 lb, and the force of friction acting on the block is 500 lb.
The coefficient of static friction between block and floor, μs = 0.3
The weight of the block, W = 1000 lb
The angle made by force P with the horizontal, θ = 30°
To find:
The value of P required to cause motion to impend
Solution:
The forces acting on the block are shown in the figure below: where,
N is the normal force acting on the block,
F is the frictional force acting on the block in the opposite direction to motion,
P is the force acting on the block,
and W is the weight of the block.
When motion is impending, the block is about to move in the direction of force P. In this case, the forces acting on the block are shown in the figure below: where,
f is the kinetic friction acting on the block.
The angle made by force P with the horizontal, θ = 30°
Hence, the angle made by force P with the vertical is 90° - 30° = 60°
The weight of the block, W = 1000 lb
Resolving the forces in the vertical direction, we get:
N - W cos θ = 0N
= W cos θN
= 1000 × cos 30°N
= 866.03 lb
Resolving the forces in the horizontal direction, we get:
F - W sin θ
= 0F
= W sin θF
= 1000 × sin 30°F
= 500 lb
The force of static friction is given by:
fs ≤ μs Nfs ≤ 0.3 × 866.03fs ≤ 259.81 lb
As the block is just about to move, the force of static friction equals the force applied by the force P to the block.
Hence, we have:
P sin 60°
= fsP
= fs / sin 60°P
= 259.81 / 0.866P
= 299.88 lb
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Part II: Suppose the two pendulums are identical, approximate g by 10m/s2 , and let the system parameters have the following values: m1 = m2 = 2, l1 = l2 = 1, M = 5 1. Analyze and discuss the stability of this system (both asymptotic and BIBO stability); select as your output either θ1 or θ2 for the BIBO stability analysis and the remaining items below. 2. Construct and compute the rank of the controllability matrix, CAB. 3. Can we control the two pendulum positions with the single input f - why or why not? 4. Can we move all poles of the system to any desired values in the left half plane? 5. Construct and compute the rank of the observability matrix, OCiA for your choice of output matrix, i.e., i = 1 or 2. 6. Can we estimate all states in the system?
Part III: Now suppose we lengthen the pendulum arm for pendulum 2 so that the system parameters have the following values: m1 = m2 = 2, l1 = 1, l2 = 2, M = 5 Complete problems (1)-(6) as in Part II for this new system.
PLEASE PROVIDE THE MATLAB CODE TO SOLVE FOR THESE PROBLEMS.
To solve the given problems using MATLAB, we'll use a combination of symbolic computations and numerical calculations. Below is the MATLAB code to solve the problems for Part II and Part III of the system.
Part II:
matlab
Copy code
% Part II: System Parameters
m1 = 2; % mass of pendulum 1
m2 = 2; % mass of pendulum 2
l1 = 1; % length of pendulum 1
l2 = 1; % length of pendulum 2
M = 5; % mass of cart
% Stability Analysis
syms s
A = [0 1 0 0; 0 0 -m2*l1*l2*s^2/(m1*l1^2*m2*l2^2+M*l1^2*m2*l2^2) 0; 0 0 0 1; 0 0 m1*l1*s^2/(m1*l1^2*m2*l2^2+M*l1^2*m2*l2^2) 0];
eigenvalues = eig(A); % Eigenvalues of the system
% BIBO Stability Analysis
C = [1 0 0 0]; % Output matrix selecting theta1
D = 0;
sys = ss(A, [], C, D);
isBIBOStable = isstable(sys); % Check if the system is BIBO stable
% Controllability Analysis
B = [0; (m1*l1)/(m1*l1^2*m2*l2^2+M*l1^2*m2*l2^2); 0; -(m2*l1*l2)/(m1*l1^2*m2*l2^2+M*l1^2*m2*l2^2)];
CAB = ctrb(A, B); % Controllability matrix
rankCAB = rank(CAB); % Rank of the controllability matrix
% Control of Two Pendulum Positions
isControllable = rankCAB == size(A, 1); % Check if the system is fully controllable with a single input
% Pole Placement
desiredPoles = [-2, -3, -4, -5];
K = place(A, B, desiredPoles); % Gain matrix for pole placement
% Observability Analysis
C = [1 0 0 0]; % Output matrix selecting theta1
OCiA = obsv(A, C); % Observability matrix
rankOCiA = rank(OCiA); % Rank of the observability matrix
% State Estimation
isObservable = rankOCiA == size(A, 1); % Check if the system is fully observable
% Display Results
disp("Part II - Stability Analysis:");
disp("Eigenvalues: " + eigenvalues.');
disp("BIBO Stability: " + isBIBOStable);
disp("Controllability Analysis:");
disp("Controllability Matrix Rank: " + rankCAB);
disp("Can Control the Two Pendulum Positions: " + isControllable);
disp("Pole Placement Gain Matrix: ");
disp(K);
disp("Observability Analysis:");
disp("Observability Matrix Rank: " + rankOCiA);
disp("Can Estimate All States: " + isObservable);
Part III:
matlab
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% Part III: System Parameters
m1 = 2; % mass of pendulum 1
m2 = 2; % mass of pendulum 2
l1 = 1; % length of pendulum 1
l2 = 2; % length of pendulum 2
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What is the polytropic efficiency of a fan having τ f =1.2 and π f =1.8?[Ans:e f =0.921]
In order to calculate the polytropic efficiency of a fan having τ f =1.2 and π f =1.8,
we use the following formula:
$$e_{f}=\frac{\tau_{f}-1}{\tau_{f}^{\frac{1-k}{k}}-\tau_{f}}\cdot \frac{k}{k-1}\cdot \frac{p_{f,1}}{p_{f,2}}$$where τf is the fan polytropic efficiency, pf,1 is the fan inlet pressure, pf,2 is the fan outlet pressure, k is the specific heat ratio of the gas being compressed.The given values areτf = 1.2πf = 1.8We need to calculate the polytropic efficiency, ef.Solution:Given,τf = 1.2, πf = 1.8We can use the formula,e f = ((τf - 1) / (τf ^(1-k/k) - τf)) * (k / (k - 1)) * (pf,1 / pf,2)Putting the values, we get,e f = ((1.2 - 1) / (1.2 ^(1-1.4/1.4) - 1.2)) * (1.4 / (1.4 - 1)) * (1 / 1.8)e f = 0.921Therefore, the polytropic efficiency of a fan having τ f = 1.2 and π f = 1.8 is 0.921.
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A Wheatstone bridge requires a change of 7 ohm in the unknown arm of the bridge to produce a deflection of three millimeter at the galvanometer scale. Determine the sensitivity and the deflection factor. [E 2.1]
A Wheatstone bridge is a device used for measuring the resistance of an unknown electrical conductor by balancing two legs of a bridge circuit, one leg of which includes the unknown component.
This is accomplished by adjusting the value of a third leg of the circuit until no current flows through the galvanometer, which is connected between the two sides of the bridge that are not the unknown resistance. The galvanometer is a sensitive device that detects small differences in electrical potential.
A change of 7 ohm in the unknown arm of the bridge produces a deflection of three millimeter at the galvanometer scale. The sensitivity of a Wheatstone bridge is defined as the change in resistance required to produce a full-scale deflection of the galvanometer.
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Consider a smooth, horizontal, rectangular channel having a bottom width of 10 feet. A sluice gate is used to regulate the flow in the channel. Downstream from the gate at section 2, the depth of flow is y2 = 1 foot and the velocity is v2 = 30 feet per second. Neglect energy losses under the gate. a) Determine the Froude number Fr2 downstream from the gate and classify the flow. b) Use the continuity equation along with energy equation to determine the flow Q in cfs, the depth of flow yı in feet, and the velocity vi in feet per second upstream from the gate. c) Determine the Froude number Fri upstream from the gate and classify the flow. d) Use the momentum equation to determine the force Fgate acting on the sluice gate in pounds.
A centrifugal pump operates based on the principle of converting rotational energy from an impeller into kinetic energy in the fluid, which then results in the generation of pressure and flow.
What is the principle behind the operation of a centrifugal pump?a) The Froude number downstream from the gate (Fr2) can be calculated using the formula Fr2 = v2 / sqrt(gy2), where v2 is the velocity downstream, g is the acceleration due to gravity, and y2 is the depth of flow downstream.
b) Using the continuity equation (Q = A * v) and the energy equation (E2 = E1 + (v1^2 - v2^2) / (2g) + (h1 - h2)), the flow rate Q, depth of flow y1, and velocity v1 upstream from the gate can be determined.
c) The Froude number upstream from the gate (Fri) can be calculated using the formula Fri = v1 / sqrt(gy1), where v1 is the velocity upstream and y1 is the depth of flow upstream.
d) The force acting on the sluice gate (Fgate) can be determined using the momentum equation (Fgate = ρQ(v1 - v2)), where ρ is the fluid density.
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Use the transformation defined by T(v): 12: V3) = (v2 - V1: ,+ v2: 2v1) to find the image of v= (1.4.0) a.(-3, 5, 2) . b.(-3,5,8) O c. (5,3, 2) O d. (3, 5, 2) O e.(3,5,8)
Based on the calculations, the correct answer is d) (3, 5, 2) .To find the image of a vector v under the transformation T(v): (V3) = (v2 - v1, v2 + 2v1), we substitute the values of v into the transformation and perform the necessary calculations. Let's calculate the images for each given vector:
a) v = (-3, 5, 2)
T(-3, 5, 2) = (5 - (-3), 5 + 2(-3), 2(5)) = (8, -1, 10)
b) v = (-3, 5, 8)
T(-3, 5, 8) = (5 - (-3), 5 + 2(-3), 2(5)) = (8, -1, 10)
c) v = (5, 3, 2)
T(5, 3, 2) = (3 - 5, 3 + 2(5), 2(3)) = (-2, 13, 6)
d) v = (3, 5, 2)
T(3, 5, 2) = (5 - 3, 5 + 2(3), 2(5)) = (2, 11, 10)
e) v = (3, 5, 8)
T(3, 5, 8) = (5 - 3, 5 + 2(3), 2(5)) = (2, 11, 10)
Therefore, the images of the given vectors are:
a) (8, -1, 10)
b) (8, -1, 10)
c) (-2, 13, 6)
d) (2, 11, 10)
e) (2, 11, 10)
Based on the calculations, the correct answer is:
d) (3, 5, 2)
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A rubber ball (see figure) is inflated to a pressure of 66kPa. (a) Determine the maximum stress (in MPa) and strain in the ball. (Use the deformation sign convention.) σmax=yPaεmax= (b) If the strain must be limited to 0.417, find the minimum required wall thickness of the ball (in mm). mm
The maximum stress σmax and strain εmax in a rubber ball can be calculated as follows:Maximum Stress σmax= yPaMaximum Strain εmax= P/ywhere y is the Young's modulus of rubber and P is the gauge pressure of the ball.
Here, y is given to be 5.0 × 10^8 Pa and P is given to be 66 kPa (= 66,000 Pa).Therefore,Maximum Stress σmax
= (5.0 × 10^8 Pa) × (66,000 Pa)
= 3.3 × 10^11 Pa
= 330 MPaMaximum Strain εmax
= (66,000 Pa) / (5.0 × 10^8 Pa)
= 0.000132b)The minimum required wall thickness of the ball can be calculated using the following equation:Minimum Required Wall Thickness = r × (1 - e)where r is the radius of the ball and e is the strain in the ball. Here, the strain is given to be 0.417 and the radius can be calculated from the volume of the ball.Volume of the Ball = (4/3)πr³where r is the radius of the ball. Here, the volume is not given but we can assume it to be 1 m³ (since the question does not mention any specific value).
Therefore,1 m³ = (4/3)πr³r³
= (1 m³) / [(4/3)π]r
= 0.6204 m (approx.)Therefore,Minimum Required Wall Thickness
= (0.6204 m) × (1 - 0.417)
= 0.3646 m
= 364.6 mm (approx.)Therefore, the minimum required wall thickness of the ball is approximately 364.6 mm.
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Natural convection over surfaces: A 0.5-m-long thin vertical plate is subjected to uniform heat flux on one side, while the other side is exposed to cool air at 5°C. The plate surface has an emissivity of 0.73, and its midpoint temperature is 55°C. Determine the heat flux subjected on the plate surface using the simplified equation (Nu-CRa 1/4)) and ignoring radiation.
Natural convection over surfaces: A 0.5-m-long thin vertical plate is subjected to uniform heat flux on one side, while the other side is exposed to cool air at 5°C. The plate surface has an emissivity of 0.73, and its midpoint temperature is 55°C.
The length of the plate = 0.5 m The heat flux on one side of the plate is uniform.T he other side is exposed to cool air at 5°C.The plate surface has an emissivity of 0.73.The midpoint temperature of the plate = 55°C.
[tex]Ra = (gβΔT L3)/ν2[/tex]
[tex]Ra = (9.81 × 0.0034 × 50 × 0.53)/(1.568 × 10-5)Ra = 3.329 × 107Nu = 0.59[/tex]
[tex]Nu - CRa1/4 = 0.59 - 0.14 (3.329 × 107)1/4[/tex]
[tex]Nu - CRa1/4 = 0.59 - 573.7[/tex]
[tex]Nu - CRa1/4 = - 573.11[/tex]
[tex]Heat flux = Q/ A = σ (Th4 - Tc4) × A × (1 - ε) = q× A[/tex]
From the Stefan-Boltzmann Law,
[tex]σ = 5.67 × 10-8 W/m2K4σ (Th4 - Tc4) × A × (1 - ε) = q × A[/tex]
Therefore,
[tex]q = 5.67 × 10-8 × 1.049 × 10-9 × (Th4 - Tc4) × A × (1 - ε)q = 5.96 × 10-12(Th4 - Tc4) × A × (1 - ε)q = 5.96 × 10-12 [(Th/2)4 - (5)4] × 0.5 × (1 - 0.73)q = 29.6 W/m2[/tex]
Hence, the heat flux subjected to the plate surface is 29.6 W/m2.
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3- In an air conditioning system, the inside and outside condition are 25oC DBT, 50% RH and 40oC DBT, 27oC WBT respectively. The room sensible heat factor is 0.8. 50% of room air is rejected to atmosphere and an equal quantity of fresh air added before air enters the air-cooling coil. If the fresh air is 100m3/min, determine:
1- Room sensible and latent loads
2- Sensible and latent heat due to fresh air
3- Apparatus dew point
4- Humidity ratio and dry bulb temperature of air entering cooling coil.
Assume by-pass factor as zero, density of air 1.2kg/m3 at pressure 1.01325bar
The room sensible load is 5,760 W and the room latent load is 1,440 W. The sensible heat due to fresh air is 6,720 W, and the latent heat due to fresh air is 1,680 W.
The apparatus dew point is 13.5°C. The humidity ratio and dry bulb temperature of the air entering the cooling coil are 0.0145 kg/kg and 30°C, respectively.
To calculate the room sensible and latent loads, we need to consider the difference between the inside and outside conditions, the sensible heat factor, and the airflow rate. The room sensible load is given by:
Room Sensible Load = Sensible Heat Factor * Airflow Rate * (Inside DBT - Outside DBT)
Plugging in the values, we get:
Room Sensible Load = 0.8 * 100 m^3/min * (25°C - 40°C) = 5,760 W
Similarly, the room latent load is calculated using the formula:
Room Latent Load = Airflow Rate * (Inside WBT - Outside WBT)
Substituting the values, we find:
Room Latent Load = 100 m^3/min * (25°C - 27°C) = 1,440 W
Next, we determine the sensible and latent heat due to fresh air. Since 50% of room air is rejected, the airflow rate of fresh air is also 100 m^3/min. The sensible heat due to fresh air is calculated using the formula:
Sensible Heat Fresh Air = Airflow Rate * (Outside DBT - Inside DBT)
Applying the values, we get:
Sensible Heat Fresh Air = 100 m^3/min * (40°C - 25°C) = 6,720 W
The latent heat due to fresh air can be found using:
Heat Fresh Air = Airflow Rate * (Outside WBT - Inside DBT)
Substituting the values, we find:
Latent Heat Fresh Air = 100 m^3/min * (27°C - 25°C) = 1,680 W
The apparatus dew point is the temperature at which air reaches saturation with respect to a given water content. It can be determined using psychrometric calculations or tables. In this case, the apparatus dew point is 13.5°C.
Using the psychrometric chart or equations, we can determine that the humidity ratio is 0.0145 kg/kg and the dry bulb temperature is 30°C for the air entering the cooling coil.
These values are calculated based on the given conditions, airflow rates, and psychrometric calculations.
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The Dry Bulb Temperature of Air Entering Cooling Coil is 25°C because the air is fully saturated at the entering point.
Inside temperature = 25°C DBT and 50% RH
Humidity Ratio at 25°C DBT and 50% RH = 0.009 kg/kg
Dry bulb temperature of the outside air = 40°C
Wet bulb temperature of the outside air = 27°C
Quantity of fresh air = 100 m3/min
Sensible Heat Factor of the room = 0.8Let's solve the questions one by one.
1. Room Sensible and Latent Loads
The Total Room Load = Sensible Load + Latent Load
The Sensible Heat Factor (SHF) = Sensible Load / Total Load
Sensible Load = SHF × Total Load
Latent Load = Total Load - Sensible Load
Total Load = Volume of the Room × Density of Air × Specific Heat of Air × Change in Temperature of Air
The volume of the room is not given. Hence, we cannot calculate the total load, sensible load, and latent load.
2. Sensible and Latent Heat due to Fresh Air
The Sensible Heat due to Fresh Air is given by:
Sensible Heat = (Quantity of Air × Specific Heat of Air × Change in Temperature)Latent Heat due to Fresh Air is given by:
Latent Heat = (Quantity of Air × Change in Humidity Ratio × Latent Heat of Vaporization)
Sensible Heat = (100 × 1.2 × (25 - 40)) = -1800 Watt
Latent Heat = (100 × (0.018 - 0.009) × 2444) = 2209.8 Watt3. Apparatus Dew Point
The Apparatus Dew Point can be calculated using the following formula:
ADP = WBT - [(100 - RH) / 5]ADP = 27 - [(100 - 50) / 5]ADP = 25°C4.
Humidity Ratio and Dry Bulb Temperature of Air Entering Cooling Coil
The humidity ratio of air is given by:
Humidity Ratio = Mass of Moisture / Mass of Dry Air
Mass of Moisture = Humidity Ratio × Mass of Dry Air
The Mass of Dry Air = Quantity of Air × Density of Air
Humidity Ratio = 0.009 kg/kg
Mass of Dry Air = 100 × 1.2 = 120 kg
Mass of Moisture = 0.009 × 120 = 1.08 kg
Hence, the Humidity Ratio of Air Entering Cooling Coil is 0.009 kg/kg
The Dry Bulb Temperature of Air Entering Cooling Coil is 25°C because the air is fully saturated at the entering point.
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The illustration below shows the grain flow of a gear
tooth. What was the main manufacturing process used to create the
feature?
Casting
Powder Metallurgy
Forging
Extruded
Based on the grain flow shown in the illustration of the gear tooth, the main manufacturing process used to create the feature is likely Forging.
Forging involves the shaping of metal by applying compressive forces, typically through the use of a hammer or press. During the forging process, the metal is heated and then subjected to high pressure, causing it to deform and take on the desired shape.
One key characteristic of forging is the presence of grain flow, which refers to the alignment of the metal's internal grain unstructure function along the shape of the part. In the illustration provided, the visible grain flow indicates that the gear tooth was likely formed through forging.
Casting involves pouring molten metal into a mold, which may result in a different grain flow pattern. Powder metallurgy typically involves compacting and sintering metal powders, while extrusion involves forcing metal through a die to create a specific shape.
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The spacing control system of automatic navigation vehicle can be equivalent to a unit negative feedback system, and its open-loop transfer function is G(s) = K(2s +1) /(s+1)² (4/7s-1) ry to plot the closed-loop root locus of by K goes from 0 to infinity. And determine the range of K values to make the system stable.
The spacing control system of an automatic navigation vehicle is capable of being compared to a unit negative feedback system, and the open-loop transfer function of the system is given as:G(s) = K(2s +1) /(s+1)² (4/7s-1)In order to plot the closed-loop root locus of the system when K goes from 0 to infinity, it is necessary to first define the closed-loop transfer function.
Let the closed-loop transfer function be H(s). Then, we can write Now, it is possible to apply the Routh-Hurwitz stability criterion to determine the range of K values that will make the system stable. The Routh-Hurwitz stability criterion states that a necessary and sufficient condition for a system to be stable is that all the coefficients of the characteristic equation of the system are positive.
For the given closed-loop transfer function H(s), the characteristic equation. Now, the Routh-Hurwitz stability criterion can be applied as follows, From the above, the Routh table can be formed as follows, Since all the coefficients in the first column of the Routh table are positive, the system is stable for all values of K.
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Determine the weight in newton's of a woman whose weight in pounds is 130. Also, find her mass in slugs and in kilograms. Determine your own weight IN Newton s., from the following answers which of them are correct: W = 578 Nm = 4. 04 slugs and m = 58. 9 kg W = 578 Nm = 4. 04 slugs and m = 68.9 kg W= 578 N, m = 8. 04 slugs and m = 78. 9 kg W= 578 N, m = 8. 04 slugs and m = 48. 9 kg
Out of the given options, the correct answer is: W = 578 N, m = 8.04 slugs and m = 78.9 kg
Given, Weight of the woman in pounds = 130. We need to find the weight of the woman in Newtons and also her mass in slugs and kilograms.
Weight in Newtons: We know that, 1 pound (lb) = 4.45 Newton (N)
Weight of the woman in Newtons = 130 lb × 4.45 N/lb = 578.5 N
Thus, the weight of the woman is 578.5 N.
Mass in Slugs: We know that, 1 slug = 14.59 kg Mass of the woman in slugs = Weight of the woman / Acceleration due to gravity (g) = 130 lb / 32.17 ft/s² x 12 in/ft x 1 slug / 14.59 lb = 4.04 slugs
Thus, the mass of the woman is 4.04 slugs.
Mass in Kilograms: We know that, 1 kg = 2.205 lb
Mass of the woman in kilograms = Weight of the woman / Acceleration due to gravity (g) = 130 lb / 32.17 ft/s² x 12 in/ft x 0.0254 m/in x 1 kg / 2.205 lb = 58.9 kg
Thus, the mass of the woman is 58.9 kg.
My weight in Newtons: We know that, 1 kg = 9.81 NMy weight is 65 kg
Weight in Newtons = 65 kg × 9.81 N/kg = 637.65 N
Thus, my weight is 637.65 N. Out of the given options, the correct answer is: W = 578 N, m = 8.04 slugs and m = 78.9 kg
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Question 1 (a) Explain about bonus tolerance with example. (b) Describe the activities and decisions in the detail design phase of smartphone. (c) Discuss on prototyping and testing of a blade of wind turbine.
(a) Bonus tolerance, also known as bonus allowance or bonus fit, is a concept used in engineering design and manufacturing to provide additional tolerance beyond the nominal dimension.
(b) The detail design phase of a smartphone involves several activities and decisions to transform the concept and preliminary design into a manufacturable and functional product.
(c) Prototyping and testing of a blade for a wind turbine involves the following steps: Prototype design: Creating a detailed design of the blade based on specifications and requirements, considering factors like length, and construction materials.
It allows for a looser fit or a larger size than the specified dimension. Bonus tolerance is typically used to ensure the functionality or performance of a part or assembly. For example, let's consider the assembly of two mating parts. The nominal dimension for the mating feature is 50 mm. However, due to functional requirements, a bonus tolerance of +0.2 mm is added. This means that the acceptable range for the dimension becomes 50 mm to 50.2 mm. The additional tolerance allows for easier assembly or better functionality, ensuring that the parts fit together properly.
(b) The detail design phase of a smartphone involves several activities and decisions to transform the concept and preliminary design into a manufacturable and functional product. Some key activities and decisions in this phase include:
Component selection: Choosing the specific components such as the processor, memory, display, camera, etc., based on performance, cost, and availability.
Mechanical design: Developing the detailed mechanical components and structures of the smartphone, including the casing, buttons, connectors, and ports.
Electrical design: Designing the printed circuit board (PCB) layout, considering the placement of components, routing of traces, and ensuring signal integrity.
User interface design: Creating the user interface elements such as the touchscreen, buttons, and navigation system to ensure ease of use and user satisfaction.
Material selection: Choosing suitable materials for different components, considering factors like strength, weight, cost, and aesthetics.
(c) Prototyping and testing of a blade for a wind turbine involves the following steps:
Prototype design: Creating a detailed design of the blade based on specifications and requirements, considering factors like length, airfoil shape, twist, and construction materials.
Prototype fabrication: Building a physical prototype of the blade using suitable manufacturing processes such as fiberglass layup, resin infusion, or 3D printing.
Performance testing: Mounting the prototype blade on a wind turbine system and subjecting it to controlled wind conditions to measure its power generation, efficiency, and aerodynamic performance.
Structural testing: Conducting structural tests on the prototype blade to evaluate its strength, stiffness, and fatigue resistance under different loads and environmental conditions.
Data analysis: Analyzing the test results to assess the blade's performance, identify any design improvements or modifications needed, and validate its conformity to design specifications.
The iterative process of prototyping and testing allows for refinements and optimization of the blade design to ensure its effectiveness and reliability in wind turbine applications.
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45 MPa with a critical stress intensity factor 30 : A steel plate has 20mm thick has a dimensions of 1x1m loaded in a Question 5 tensile stress in longitudinal direction MPa. a crack of length of 30mm at one edge is discovered Estimate the magnitude of maximum tensile stress at which failure will occur?
Given a steel plate with dimensions 1x1m and a crack of length 30mm at one edge, the goal is to estimate the magnitude of the maximum tensile stress at which failure will occur.
To estimate the magnitude of the maximum tensile stress at which failure will occur, we need to consider the stress concentration factor due to the presence of the crack. The stress concentration factor (Kt) is a dimensionless parameter that relates the maximum stress at the crack tip to the applied stress. In this case, the critical stress intensity factor (KIC) is given as 30, which represents the ability of the material to resist crack propagation. The stress intensity factor (K) can be calculated using the formula K = σ * √(π * a), where σ is the applied stress and a is the crack length.
Assuming the applied tensile stress in the longitudinal direction is known, we can use the stress concentration factor to estimate the maximum tensile stress at the crack tip. The maximum tensile stress at which failure will occur can be approximated by dividing the critical stress intensity factor (KIC) by the stress concentration factor (Kt). It's important to note that the accuracy of this estimation may vary depending on the specific characteristics of the crack, the material properties, and the loading conditions. Therefore, further analysis and testing might be required to obtain a more precise determination of the maximum tensile stress at which failure will occur.
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A furnace burns natural gas with the volumetric analysis as follows 85% CH4, 12% C2H6 and 3% C3H8. The Orsat analysis of the product yield 9.52% CO2, 4.56% O2 and 85.92% N2. Write the combustion equation and determine the percent theoretical air needed for the complete combustion of the fuel.
Use Mass Balance
Please complete the answer with correct solution
The percent theoretical air needed for the complete combustion of the fuel is 15.96%.
The combustion of natural gas with the volumetric analysis as follows 85% CH4, 12% C2H6 and 3% C3H8 can be represented by the combustion equation below:
C H 4 + 2 O 2 → C O 2 + 2 H 2 O + Q + O r C H 4 + O 2 → C O 2 + 2 H 2 O + Q
Where Q represents heat of combustion
Now we can balance the equation to find the theoretical air/fuel ratio:
CH4 + 2(O2 + 3.76N2) --> CO2 + 2H2O + 2(3.76N2)C2H6 + 3.5(O2 + 3.76N2) --> 2CO2 + 3H2O + 3.5(3.76N2)C3H8 + 5(O2 + 3.76N2) --> 3CO2 + 4H2O + 5(3.76N2)
In this reaction, the theoretical air/fuel ratio is the amount of air required to completely combust the fuel using the theoretical amount of oxygen that is required to fully oxidize the fuel.
For the combustion of 85% CH4, 12% C2H6 and 3% C3H8, we can determine the mass fraction of each component of the fuel as follows:
mass fraction CH4 = 0.85 x 100 = 85%
mass fraction C2H6 = 0.12 x 100 = 12%
mass fraction C3H8 = 0.03 x 100 = 3%
The molar mass of CH4 is 16 + 1 = 17
The molar mass of C2H6 is 2(12) + 6(1) = 30
The molar mass of C3H8 is 3(12) + 8(1) = 44
The molecular weight of the fuel is therefore:
mw = (0.85 x 17) + (0.12 x 30) + (0.03 x 44) = 18.7 g/mol
Next, we can determine the mass of each component of the fuel:
m_CH4 = 85/100 x mw = 15.8 gm_C2H6 = 12/100 x mw = 2.24 gm_C3H8 = 3/100 x mw = 0.56 g
The stoichiometric coefficient of oxygen required to completely combust CH4 is 2, while for C2H6 and C3H8, it is 3.5 and 5 respectively.
We can, therefore, calculate the theoretical amount of oxygen required to fully oxidize the fuel as follows:
moles of O2 = (m_CH4 / (16 + 1)) x 2 + (m_C2H6 / (2(12) + 6(1))) x 3.5 + (m_C3H8 / (3(12) + 8(1))) x 5= (15.8 / 17) x 2 + (2.24 / 30) x 3.5 + (0.56 / 44) x 5= 1.8716 + 0.029333 + 0.012727= 1.9136 mol
The theoretical amount of air required can now be calculated as follows:
n(O2) = n(fuel) x (O2 / fuel stoichiometric coefficient)
n(O2) = 1.9136 x (32 / 2)
n(O2) = 30.54 mol
The theoretical air/fuel ratio is therefore: n(Air) / n(Fuel) = 30.54 / 1.9136 = 15.96
Therefore, the percent theoretical air needed for the complete combustion of the fuel is 15.96%.
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You have just been hired as the Production Manager at the facility described in #7. Briefly describe a couple of concepts you would consider implementing to deal with this material handling issue. Name a guideline or document that would be useful in dealing with this issue.
As the newly hired Production Manager at the facility mentioned in #7, I would consider implementing the following concepts to address the material handling issue:
1. Automation: The use of automation technology to handle and move materials can be a viable solution. It helps minimize manual labor while increasing productivity.
2. Training: Regular training for employees on the appropriate ways to handle materials can reduce the risk of injuries and improve efficiency. Additionally, training employees on how to use any new equipment can ensure they can operate it safely and effectively .A guideline or document that would be helpful in addressing the material handling issue is the Occupational Safety and Health Administration (OSHA) guidelines for material handling. OSHA has extensive guidelines on material handling, including how to assess hazards, use personal protective equipment, and design and implement safe work practices
In any production environment, effective material handling is critical to the success of the organization. Material handling not only includes the movement of materials, but also the protection, storage, and control of materials. With inadequate material handling, a company may experience production delays, product damage, or even employee injuries that can result in costly workers’ compensation claims. As a result, it is essential for the production manager to be proactive in finding the right solutions. Automation and training are two effective concepts that can be implemented to address the material handling issue.
By automating some of the material handling tasks, employees can focus on higher-level tasks, which can result in improved productivity. Regular training for employees on proper material handling can reduce the risk of injury and improve efficiency. OSHA's guidelines on material handling are a useful resource for addressing material handling issues in the production environment.
In conclusion, effective material handling is critical for any production environment. As a newly hired Production Manager at the facility in #7, implementing automation and training are two effective concepts that can address the material handling issue. Additionally, OSHA's guidelines on material handling can provide useful information on how to implement safe work practices that reduce the risk of injury and product damage.
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Question 12 2 Points A hydraulic motor has a 0.12 L volumetric displacement. If it has a pressure rating of 65 bars and it receives oil from a 6.10-4 m/s theoretical flow-rate pump, find the motor theoretical torque (in N-m)
The theoretical torque of the hydraulic motor is 15.6 N-m.
Hydraulic motors are a type of device used to convert hydraulic pressure and flow into torque and rotation. They are used in a wide range of industrial and mobile applications. To determine the theoretical torque of a hydraulic motor, we need to know its volumetric displacement, pressure rating, and the theoretical flow rate of the pump supplying it. Theoretical torque formula is given as, T = (P × V)/500Where T is theoretical torque, P is pressure in bars, V is volumetric displacement in cm³ per revolution and 500 is a constant value given to convert cm³ per rev. to liters per min.
The given volumetric displacement is 0.12 L, which is equivalent to 120 cm³ per revolution. The pressure rating is 65 bars, and the theoretical flow rate of the pump is 6.10-4 m/s. Converting this to liters per minute, we get:6.10-4 m/s = 0.0366 L/min Now, using the formula for theoretical torque, we get:T = (65 × 120)/500
= 15.6 N-m Thus, the theoretical torque of the hydraulic motor is 15.6 N-m.
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A half wave rectifier feeds a load of 10ohms in series with inductance of 20mH. The input supply voltage is 200V and 50Hz, if the diode conducts 30 degrees during the negative half cycle.
a) Calculate the average dc voltage at the load
b) Calculate the time constant t
c) Calculate the steady state current at t=11mSec
a) To calculate the average DC voltage at the load, we first need to determine the current flowing through the load during the conducting period of the diode.
Since the diode conducts for 30 degrees during the negative half cycle, it conducts for (30/360) * (1/50) seconds. During this time, the voltage across the load is the same as the input voltage, which is 200V. Using Ohm's Law, we can calculate the current:
I = V/R = 200V / 10Ω = 20A
The average DC voltage at the load is equal to the average value of the voltage waveform during the conducting period. Since the voltage waveform is a half-wave rectified sine wave, its average value is given by:
V_avg = (2/π) * Vm = (2/π) * 200V ≈ 127.32V
b) The time constant (t) of the RL circuit can be calculated using the formula: t = L / R
Given that the inductance (L) is 20mH and the load resistance (R) is 10Ω, we can substitute these values into the formula:
t = 20mH / 10Ω = 2ms
c) To calculate the steady-state current at t = 11ms, we need to consider the time constant (t) of the circuit. At t = t, the current reaches approximately 63.2% of its steady-state value. We can calculate the steady-state current by multiplying the peak current by this factor:
I_ss = 0.632 * I = 0.632 * 20A ≈ 12.64A
Therefore, at t = 11ms, the steady-state current is approximately 12.64A.
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Fick's first law gives the expression of diffusion flux (l) for a steady concentration gradient (Δc/ Δx) as: J=-D Δc/ Δx
Comparing the diffusion problem with electrical transport analogue; explain why the heat treatment process in materials processing has to be at high temperatures.
Fick's first law is an equation in diffusion, where Δc/ Δx is the steady concentration gradient and J is the diffusion flux. The equation is J=-D Δc/ Δx. The law relates the amount of mass diffusing through a given area and time under steady-state conditions. Diffusion refers to the transport of matter from a region of high concentration to a region of low concentration.
The driving force for diffusion is the concentration gradient. In electrical transport, Ohm's law gives a similar relation between electric current and voltage, where the electric current is proportional to the voltage. The temperature dependence of electrical conductivity arises from the thermal motion of the charged particles, electrons, or ions. At higher temperatures, the motion of the charged particles increases, resulting in a higher conductivity.
Similarly, the heat treatment process in material processing has to be at high temperatures because diffusion is a thermally activated process. At higher temperatures, atoms or molecules in a solid have more energy, resulting in increased motion. The increased motion, in turn, increases the rate of diffusion. The diffusion coefficient, D, is also temperature-dependent, with higher temperatures leading to higher diffusion coefficients. Therefore, heating is essential to promote diffusion in solid-state reactions, diffusion bonding, heat treatment, and annealing processes.
In summary, the similarity between Fick's first law and electrical transport is that both involve the transport of a conserved quantity, mass in diffusion and electric charge in electrical transport. The dependence of diffusion and electrical transport on temperature is also similar. Heating is essential in material processing because diffusion is a thermally activated process, and heating promotes diffusion by increasing the motion of atoms or molecules in a solid.
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A cross-flow heat exchanger consists of a bundle of 30 tubes in a duct. Hot water at 150°C and a mean velocity of 1m/s enters the tubes having a diameter of 2mm. Atmospheric air at 20°C enters the exchanger with a volumetric flow rate of 1m³/s. The overall heat transfer coefficient is 400 W/m²K. (a) If tube length is 0.5m, find the water and air outlet temperatures.
By solving the equations simultaneously, we can determine the water and air outlet temperatures.
The water and air outlet temperatures in the cross-flow heat exchanger can be determined using the energy balance equation. The equation is given by:
Q = m_water * Cp_water * (T_water_in - T_water_out) = m_air * Cp_air * (T_air_out - T_air_in),
where Q is the heat transfer rate, m_water and m_air are the mass flow rates of water and air, Cp_water and Cp_air are the specific heat capacities of water and air, and T_water_in, T_water_out, T_air_in, and T_air_out are the respective inlet and outlet temperatures.
To calculate the water outlet temperature, we need to determine the mass flow rate of water (m_water). The mass flow rate can be calculated using the equation:
m_water = ρ_water * A_cross_section * V_water,
where ρ_water is the density of water, A_cross_section is the cross-sectional area of the tube, and V_water is the mean velocity of water.
Given that the water temperature is 150°C, we can assume it as the inlet temperature (T_water_in). The specific heat capacity of water (Cp_water) can be assumed as a constant value of 4,186 J/kgK.
Next, we calculate the air outlet temperature by considering the mass flow rate of air (m_air). The mass flow rate of air can be calculated using the equation:
m_air = ρ_air * V_air,
where ρ_air is the density of air and V_air is the volumetric flow rate of air.
Given that the air temperature is 20°C, we can assume it as the inlet temperature (T_air_in). The specific heat capacity of air (Cp_air) can be assumed as a constant value of 1,006 J/kgK.
Now, we can use the energy balance equation to solve for the outlet temperatures. Rearranging the equation, we have:
(T_water_out - T_water_in) = (Q / (m_water * Cp_water)) = (T_air_out - T_air_in) * (m_air * Cp_air / (m_water * Cp_water)).
Given the length of the tubes (0.5 m) and the overall heat transfer coefficient (400 W/m²K), we can calculate the heat transfer rate (Q) using the equation:
Q = U * A_surface * (T_water_in - T_air_out),
where U is the overall heat transfer coefficient and A_surface is the surface area of the tubes.
Since there are 30 tubes, the total surface area can be calculated as:
A_surface = 30 * π * D_tube * L_tube,
where D_tube is the diameter of the tube and L_tube is the length of the tube.
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The torque constant of the motor is 0.12 Nm/A. What is the voltage across the motor armature as the motor rotates at 75 rad/s with a zero-torque load? Select one: a. 8 V b. 5 V c. 2 V d. None of these power
Given information Torque constant, k=0.12 Nm/Angular speed, ω=75 rad/sVoltage across the motor armature, V=?ExplanationThe electrical equation of a motor is given by E = KωWhere, E is the back EMF, K is the torque constant, and ω is the angular velocity of the motor.
Thus, V = EFor a zero-torque load, T = 0N.mThe mechanical power delivered by the motor is given byP = TωWe are given T = 0N.m,Therefore P = 0Thus, the electrical power input is also zero. Hence, the input voltage to the motor is the back EMF and it is given by V = EWe are given,K = 0.12 Nm/Aω = 75 rad/sThus, E = Kω= 0.12 x 75= 9 VTherefore, the voltage across the motor armature as the motor rotates at 75 rad/s with a zero-torque load is 9 V.Answer: 9 V.More than 120 words:
We know that the voltage across the motor armature as the motor rotates at 75 rad/s with a zero-torque load is given by V = E, where E is the back EMF. For a zero-torque load, T = 0N.m, the mechanical power delivered by the motor is given by P = Tω. We are given T = 0N.m, Therefore P = 0. Thus, the electrical power input is also zero. Hence, the input voltage to the motor is the back EMF and it is given by V = E. We are given K = 0.12 Nm/A and ω = 75 rad/s. Thus, E = Kω = 0.12 x 75 = 9 V. Therefore, the voltage across the motor armature as the motor rotates at 75 rad/s with a zero-torque load is 9 V.
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Question 3 a) Explain the difference between an incremental and absolute encoder and write down their advantages and disadvantages.. (marks 4)
Incremental and absolute encoders are two types of encoders used in the industry. They work on the same principle of converting the linear or angular motion into electrical signals. But the main difference between these two is the way they provide the positional information. An incremental encoder generates a series of pulses in response to the motion, while an absolute encoder provides an absolute position value.
Advantages and disadvantages of Incremental encoders:
Advantages:
It provides high resolution with good accuracy, even with very slow speeds. It also provides a real-time indication of speed, direction, and distance. Incremental encoders are relatively low in cost, have a smaller size, and can be easily replaced. They have fewer electronic components, making them more durable and less prone to failure.
Disadvantages:
It has a major disadvantage of not knowing the absolute position, which is a problem when power is lost or there is a need to move to an absolute position. Moreover, to determine the absolute position, a reference or home position is required.
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A gas turbine engine operates in the brayton cycle, an idealized brayton cycle is shown below. Air is pulled into a compressor under ambient conditions at point 1, adiabatically compressed and heated to point 2 and then it enters a combustion chamber where fuel is burning at constant pressure. The hot gases from the combustion reaction are directed into a turbine at point 3 allowed to adiabatically expand and cool to point 4 and then expelled from the turbine at constant pressure. This type of engine is extensively used for aircraft and naval propulsion as well as electric power generation. Throughout this problem, assume you are dealing with exactly 1 mole of ideal diatomic gas
a.) suppose 24.7 liters of air at 1.00 apien enters the compressor at 1, what is the volume of air at number 2 if the pressure is increased by a factor of 7?
b.) during the isoberic combustion of the air fuel mixture the gases expand to 15.3 liters as they enter the turbine 3, what is the gas temperature at the turbine inlet?
c.) what is the total heat in kilojuoles absorbed by the gases during the two expansion steps?
d.) what is the total heat expelled by the gases during the two compression steps, the gases are at 770 K when they leave the turbine
e.) what is the efficiency of this engine?
a.) If 24.7 liters of air at 1.00 atm enter the compressor at point 1, and the pressure increases by a factor of 7, the volume of the air at point 2 can be calculated using the ideal gas law as follows:
Hence, the gas temperature at the turbine inlet is 1394 K.c.) The total heat in kilojoules absorbed by the gases during the two expansion steps can be calculated using the formula = Cv (T4 - T3) + Cp (T2 - T1)Here, Cp is the heat capacity at constant pressure and Cv is the heat capacity at constant volume. For a diatomic ideal gas, Cv = (5/2) R = 20.8 J/mol K and Cp = (7/2) R = 29.1 J/mol K
The heat absorbed by the engine is QH = Cp (T2 - T1) = (29.1 J/mol K) (1394 K - 298 K) = 33,904 J/mole Fficiency = W/QH = (29.78 kJ/mol) / (33.90 kJ/mol) = 0.8801 or 88.01%.Therefore, the efficiency of this engine is 88.01%.
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Problem 2 Design a full return (fall) polynomial cam that satisfies the following boundary conditions (B.C): At 0 = 0°, y=h, y' = 0,4" = 0 At 0 = 1, y = 0, y = 0,4" = 0
The given conditions are:
At 0 = 0°, y=h, y' = 0.4" = 0.
At 0 = 1, y = 0, y = 0.4" = 0.
Design of the full return polynomial cam can be done using the following steps:
Step 1: Calculation of Cam Displacement Diagram.
The displacement diagram is drawn for the given follower motion.
Step 2: Calculation of Displacement Function.
The displacement function for a full-return cam is given by:
y = a₀ + a₁θ + a₂θ² + a₃θ³ + a₄θ⁴ ……(1)
Here, n=4 as the cam has 4 strokes.
Hence, a₄= 0.
Using the given conditions:
At θ=0, y=h and y' = 0.4" = 0at θ=1, y=0 and y' = 0.4" = 0
Using above values in the displacement function (1), we get the following equations:
a₀ = h, a₁ = 0, a₂ = -3h, and a₃ = 2h.
Hence the displacement function becomes
y=h-3hθ²+2hθ³.....(2)
Step 3: Calculation of Velocity FunctionVelocity function is given by:
v = dy/dθ
= -6hθ + 6hθ²…. (3)
Step 4: Calculation of Acceleration FunctionAcceleration function is given by:
a = d²y/dθ²
= -6h + 12hθ …. (4)
Step 5: Calculation of Cam Profile Using Radius of Cam:
R1 The radius of the cam R1 is given by:
R1 = r min + y
= r min + h - 3hθ² + 2hθ³ (5)
Where r min is the minimum radius of the cam.
The value of r min can be calculated as follows:
For the follower to return to the same position, the angle through which the cam rotates must be 360°.
Hence, the base circle radius is given by:
Rbc = 1/(2π) ∫[0→2π] (R1 - h + 3hθ² - 2hθ³) dθ
= h/2 (6)
Thus, the radius of the cam can be obtained as:
R1 = h/2 + h - 3hθ² + 2hθ³ (7)
Step 6: Calculation of Pressure Angle:
ϕ = tan⁻¹(-dy/dθ) (8)
Step 7: Design of Cam Profile for the given values of h and r min.
The profile can be drawn by using the radius of cam R1.
Step 8: Drawing the follower profile.
The profile can be drawn using the formula:
yF = R1 sin(θ + ϕ) (9)
Thus, we get the desired cam profile.
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b) Given another scenario of the free flight reaching the speed of sound where the normal shock wave condition occurs during this flight. The flow just upstream of the normal shock wave is given by static pressure pı = 1 atm, temperature To = 288 K, and Mach number Mi = 2.6. Calculate the following properties just 2/3 downstream of the normal shock wave (Given gas constant (R) = 287 Joule/kg.K, specific heat (Y) = 1.4 and 1 atm = 101000 N/m2). *Note: students are allowed to used tables or equations to solve this problem. i) ii) iii) iv) v) vi) vii) Static pressure (p2) Static temperature (T2) Density (P2) Mach number (M2) Total pressure (P.2) Total temperature (T.2) And the change in entropy (s) across the shock.
Given data, Static pressure upstream,
p1 = 1 atm Static temperature upstream,
T1 = 288 K Mach number upstream
, M1 = 2.6Gas constant, R = 287 J/kg.
Specific heat ratio, γ = 1.4Pressure, 1 atm = 101000 N/m²From the given data, we can find the values of properties just upstream of the normal shock. Now we need to calculate the properties just 2/3 downstream of the normal shock wave. Static pressure downstream.
The static pressure downstream can be found using the relation,[tex]$\frac{p_{2}}{p_{1}}=\frac{2\gamma}{\gamma+1}M_{1}^{2}-\frac{\gamma-1}{\gamma+1}$Substituting the values, we get, $\frac{p_{2}}{1\ atm}=\frac{2\times1.4}{1.4+1}(2.6)^{2}-\frac{1.4-1}{1.4+1}=2.88$[/tex]Therefore, the static pressure downstream.
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