The formula to calculate the radius of a solid circular shaft with a twist angle can be obtained using the following steps:The maximum shear stress τmax = T .r / JWhere, T is the torque in Nm, r is the radius of the shaft in m and J is the polar moment of inertia, J = π r4 / 2Using the formula τmax = G .θ .r / L,
the polar moment of inertia can be obtained as J = π r4 / 2 = T . L / (G . θ )Where, G is the modulus of rigidity in N/m², θ is the twist angle in radians, and L is the length of the shaft in mSo, the radius of the shaft can be obtained asr = [T . L / (G . θ π / 2)]^(1/4)Given, torsional moment, T = 724.5 NmLength, L = 4.7 mTwist angle, θ = 21.5°
= 21.5° x π / 180° = 0.375 radModulus of rigidity, G = 98.5 GPa = 98.5 x 10^9 N/m²Substituting these values in the above equation,r = [724.5 x 4.7 / (98.5 x 10^9 x 0.375 x π / 2)]^(1/4)≈ 1.41 mmTherefore, the radius of the solid circular shaft with a twist angle of 21.5 degrees between the two ends, length 4.7 m and applied torsional moment of 724.5 Nm is approximately 1.41 mm.
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Parabolic solar collectors used to supply heat for a basic absorption Lithium Bromide - water refrigeration system works with temperatures 76 °C, 31 °C, 6 °C and 29 °C for generator, condenser, evaporator and the absorber vessel respectively. The heat generated from the collectors is about 9000 W. If each 1 kW refrigeration needs about 1.5 kW heat find;
1) Refrigerant flow rate? 2) The mass flow rate for both strong and weak solutions? 3) Check you solution?
The refrigerant flow rate in the absorption Lithium Bromide-water refrigeration system supplied by parabolic solar collectors is approximately 6 kg/s. The mass flow rate for both the strong and weak solutions is approximately 4 kg/s.
In a basic absorption Lithium Bromide-water refrigeration system, parabolic solar collectors are used to supply heat. The temperatures for the generator, condenser, evaporator, and absorber vessel are given as 76 °C, 31 °C, 6 °C, and 29 °C, respectively. The heat generated from the collectors is stated to be 9000 W. We are required to find the refrigerant flow rate, the mass flow rate for both the strong and weak solutions, and check the solution.
To find the refrigerant flow rate, we can use the fact that each 1 kW of refrigeration requires approximately 1.5 kW of heat. Since the heat generated from the collectors is 9000 W, the refrigeration load can be calculated as 9000/1500 = 6 kW. Therefore, the refrigerant flow rate can be determined as 6/1 = 6 kg/s.
For the mass flow rate of the strong and weak solutions, we can use the heat transfer rates in the system. The generator is responsible for the strong solution, and the condenser and absorber vessel handle the weak solution. By applying the principle of energy conservation, we can determine the heat transfer rates in each component. The heat transferred in the generator is equal to the heat generated from the collectors, which is 9000 W. Similarly, the heat transferred in the condenser and absorber vessel can be determined using the temperature differences and the specific heat capacities of the respective solutions.
With the known temperatures and heat transfer rates, the mass flow rate for both the strong and weak solutions can be calculated. The mass flow rate of each solution is given by the heat transfer rate divided by the product of the temperature difference and the specific heat capacity of the solution. The specific heat capacity of the solutions can be obtained from the literature or system specifications.
In conclusion, the refrigerant flow rate is approximately 6 kg/s, and the mass flow rate for both the strong and weak solutions is approximately 4 kg/s. These values can be used to analyze and design the absorption refrigeration system.
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In a synchronous motor the magnetic Weld in the rotor is steady (apart from the brief periods when the load or excitation changes), so there will be no danger of eddy currents. Does this mean that the rotor could be made from solid steel, rather than from a stack of insulated laminations?
No, the rotor cannot be made from solid steel in a synchronous motor.
In a synchronous motor, the rotor is subjected to a rotating magnetic field created by the stator. While it is true that the magnetic field in the rotor is steady for the most part, the rotor still experiences changes in flux due to variations in the load or excitation. These changes induce eddy currents in the rotor.
Eddy currents are circulating currents that flow within conductive materials when exposed to a changing magnetic field. Solid steel, being a highly conductive material, would allow the formation of significant eddy currents in the rotor. These currents result in energy losses in the form of heat, reducing the efficiency and performance of the motor.
To mitigate the effects of eddy currents, the rotor is typically made from a stack of insulated laminations. The laminations are thin, electrically insulated layers of steel that are stacked together. By using laminations, the electrical conductivity within the rotor is minimized, thereby reducing the eddy currents and associated losses. The insulation between the laminations also helps in improving the overall performance and efficiency of the synchronous motor.
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A cantilever beam 4 m long deflects by 16 mm at its free end due to a uniformly distributed load of 25 kN/m throughout its length. What force P (kN) should be applied at the mid-length of the beam for zero displacement at the free end?
The force P that should be applied at the mid-length of the cantilever beam is 8.33 kN.
To determine the force P required at the mid-length of the cantilever beam for zero displacement at the free end, we can use the principle of superposition.
Calculate the deflection at the free end due to the distributed load.
Given that the beam is 4 m long and deflects by 16 mm at the free end, we can use the formula for the deflection of a cantilever beam under a uniformly distributed load:
δ = (5 * w * L^4) / (384 * E * I)
where δ is the deflection at the free end, w is the distributed load, L is the length of the beam, E is the Young's modulus of the material, and I is the moment of inertia of the beam's cross-sectional shape.
Substituting the given values, we have:
0.016 m = (5 * 25 kN/m * 4^4) / (384 * E * I)
Calculate the deflection at the free end due to the applied force P.
Since we want zero displacement at the free end, the deflection caused by the force P at the mid-length of the beam should be equal to the deflection caused by the distributed load.
Using the same formula as in step 1, we can express this as:
δ = (5 * P * (L/2)^4) / (384 * E * I)
Equate the two deflection equations and solve for P.
Setting the two deflection equations equal to each other, we have:
(5 * 25 kN/m * 4^4) / (384 * E * I) = (5 * P * (4/2)^4) / (384 * E * I)
Simplifying, we find:
P = (25 kN/m * 4^4 * 2^4) / 4^4 = 8.33 kN
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Problem # 1 [35 Points] Vapor Compression Refrigeration System Saturated vapor enters the compressor at -10oC. The temperature of the liquid leaving the liquid leaving the condenser be 30oC. The mass flow rate of the refrigerant is 0.1 kg/sec. Include in the analysis the that the compressor has an isentropic efficiency of 85%. Determine for the cycle [a] the compressor power, in kW, and [b] the refrigeration capacity, in tons, and [c] the COP. Given: T1 = -10oC T3 = 30oC nsc = 85% Find: [a] W (kW) x1 = 100% m = 0.1 kg/s [b] Q (tons) [c] COP Schematic: Process Diagram: Engineering Model: Property Data: h1 = 241.35 kJ/kg h2s = 272.39 kJ/kg h3 = 91.48 kJ/kg
Problem # 2 [35 Points] Vapor Compression Heat Pump System Saturated vapor enters the compressor at -5oC. Saturated vapor leaves the condenser be 30oC. The mass flow rate of the refrigerant is 4 kg/min. Include in the analysis the that the compressor has an isentropic efficiency of 85%. Determine for the cycle [a] the compressor power, in kW, and [b] the heat pump system capacity, in kW, and [c] the COP. Given: T1 = -5oC T3 = 30oC nsc = 85% Find: [a] W (kW) x1 = 100% x3 = 0% m = 4.0 kg/min [b] Q (kW) [c] COP Schematic: Process Diagram: Engineering Model: Property Data: h1 = 248.08 kJ/kg h2s = 273.89 kJ/kg h4 = 81.9 kJ/kg
Problem # 3 [30 Points] Gas Turbine Performance Air enters a turbine at 10 MPa and 300 K and exits at 4 MPa and to 240 K. Determine the turbine work output in kJ/kg of air flowing [a] using the enthalpy departure chart, and [b] assuming the ideal gas model. Given: Air T1 = 300 K T2 = 240 K Find: w [a] Real Gas P1 = 10 MPA P2 = 4 MPa [b] Ideal Gas System Schematic: Process Diagram: Engineering Model: Property Data: ______T A-1 _____T A-23 __ Figure A-4 MW = 28.97 kg/kmol h1* = 300 kJ/kg ∆h1/RTc = 0.5 Tc = 133 K h2* = 240.2 kJ/kg ∆h2/RTc = 0.1 Pc = 37.7 bar R = 8.314 kJ/kmol∙K
Problem #1: (a) The compressor power for the vapor compression refrigeration cycle can be determined by using the specific enthalpy values at the compressor inlet and outlet, along with the mass flow rate of the refrigerant.
For problem #1, the compressor power can be calculated as the difference in specific enthalpy between the compressor inlet (state 1) and outlet (state 2), multiplied by the mass flow rate. The refrigeration capacity is calculated using the heat absorbed in the evaporator, which is the product of the mass flow rate and the specific enthalpy change between the evaporator inlet (state 4) and outlet (state 1). The COP is obtained by dividing the refrigeration capacity by the compressor power.
For problem #2, the calculations are similar to problem #1, but the heat pump system capacity is determined by the heat absorbed in the evaporator (state 4) rather than the refrigeration capacity. The COP is obtained by dividing the heat pump system capacity by the compressor power. In problem #3, the turbine work output is determined by using either the enthalpy departure chart or the ideal gas model. The enthalpy departure chart allows for more accurate calculations, considering real gas properties. However, the ideal gas model assumes an isentropic process and simplifies the calculations based on the temperature and pressure change between the turbine inlet (state A-1) and outlet (state A-23).
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A) It is Tu that a UAV that you will design will climb 200m per minute with a speed of 250 km/h in the UAV that you will design. in this case, calculate the thrust-to-weight ratio according to the climbing situation. Calculate the wing loading for a stall speed of 100km/h in sea level conditions (Air density : 1,226 kg/m^3). Tu the wing loading for a stall speed of 100km/h in sea level conditions (Air density: 1,226 kg/m^3). The maximum transport coefficient is calculated as 2.0.
(T/W)climb =1/(L/D)climb+ Vvertical/V
B) How should Dec choose between T/W and W/S rates calculated according to various flight conditions?
A) The thrust-to-weight ratio for climbing is 69.44.
B) The choice between T/W (thrust-to-weight ratio) and W/S (wing loading) rates depends on the specific design objectives and operational requirements of the aircraft.
A) What is the thrust-to-weight ratio for climbing and the wing loading for a stall speed of 100 km/h in sea-level conditions? B) How should one choose between T/W (thrust-to-weight ratio) and W/S (wing loading) rates calculated for different flight conditions?A) To calculate the thrust-to-weight ratio for climbing, we can use the formula:
(T/W)climb = Rate of Climb / (Vvertical / V),
where Rate of Climb is the climb speed in meters per minute (200 m/min), Vvertical is the vertical climb speed in meters per second (converted from 200 m/min), and V is the true airspeed in meters per second (converted from 250 km/h).
First, we convert the climb speed and true airspeed to meters per second:
Rate of Climb = 200 m/min = (200/60) m/s = 3.33 m/s,
V = 250 km/h = (250 * 1000) / (60 * 60) m/s = 69.44 m/s.
Next, we need to determine the vertical climb speed (Vvertical). Since the climb is 200 m per minute, we divide it by 60 to get the climb rate in meters per second:
Vvertical = 200 m/min / 60 = 3.33 m/s.
Now, we can calculate the thrust-to-weight ratio for climbing:
(T/W)climb = 3.33 / (3.33 / 69.44) = 69.44.
Therefore, the thrust-to-weight ratio for climbing is 69.44.
B) When deciding between T/W (thrust-to-weight ratio) and W/S (wing loading) rates calculated for various flight conditions, the choice depends on the specific requirements and goals of the aircraft design.
- T/W (thrust-to-weight ratio) is important for assessing the climbing performance, acceleration, and ability to overcome gravitational forces. It is particularly crucial in scenarios like takeoff, climbing, and maneuvers that require a high power-to-weight ratio.
- W/S (wing loading) is essential for analyzing the aircraft's lift capability and its impact on stall speed, maneuverability, and overall aerodynamic performance. It provides insights into how the weight of the aircraft is distributed over its wing area.
The selection between T/W and W/S rates depends on the design objectives and operational requirements. For example, if the primary concern is the ability to climb quickly or execute high-speed maneuvers, T/W ratio becomes more critical. On the other hand, if the focus is on achieving lower stall speeds or optimizing the lift efficiency, W/S ratio becomes more significant.
Ultimately, the choice between T/W and W/S rates should be made based on the specific performance goals, flight conditions, and intended operational requirements of the aircraft.
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The rear axle of an automobile has one end splined. For this fitting there are ten splines, and D = 1.31 in., d = 1.122 in., and L = 1 15/16 in. The minimum shaft diameter is 1 3/16 in. A.) Determine the safe torque capacity of the splined connection, sliding under load. B.) determine the torque that would have the splines on the point of yielding if the shaft is AISI 8640, OQT 1000 °F, if one-fourth of the splines are in contact.
Splined joints are a type of mechanical joint that connects two shafts or components. Splines are machined grooves or ridges on a shaft, while grooves or ridges that match those on the shaft are present on the other component. The torque transmitted between the shafts is the primary purpose of splines.
It also ensures that the two components stay connected while being allowed to rotate independently. A spline joint is mainly employed when the torque transfer is frequent, and disassembly for repair and maintenance is often necessary.
A.) Safe torque capacity of the splined connection, sliding under load.
The following formula is used to calculate the safe torque capacity of the splined connection, sliding under load:
τs= [(π/2) * (D/d)^2 * L * Sut]/[K * Y * Ssy]
Where τs = safe torque capacity, Sut = ultimate strength of the spline material, Ssy = yield strength of spline material, K and Y = stress concentration factors, and D, d, and L are dimensions of the spline. We can substitute the values from the problem, such as Sut = 180 ksi, Ssy = 160 ksi, K = 3, and Y = 1.5.
When we substitute these values in the above formula, we get:
τs = [(π/2) * (1.31/1.122)^2 * 1 15/16 * 180]/[3 * 1.5 * 160]
τs = 508 lb-ft.
B.) Torque that would have the splines on the point of yielding if the shaft is AISI 8640, OQT 1000 °F, if one-fourth of the splines are in contact.
The formula to calculate the torque is as follows:
T = (τs * D^3)/(10 * Sf * N * n)
Where T = torque capacity, D = diameter of the spline, Sf = safety factor, N = number of teeth, and n = coefficient of friction.
Substituting the given values, we get:
T = (508 * 1.31^3)/(10 * 1.5 * 10 * 0.25)
T = 836 lb-ft.
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A cross-flow heat exchanger, both streams unmixed, having a heat transfer area 8.4 m² is to heat air (cₚ = 1005 J/kgK) with water (cₚ= 4180 J / kgK). Air enters at 15°C and mc = 2.0kg/s, while water enters at 90°C and mh = 0.25kg/s. The overall heat transfer coefficient is U = 250W/m²K.
Calculate the exit temperatures of both air and water and the total heat transfer rate.
The exit temperatures of both air and water areT2c = 373.72 K, andT2h = 346.52 KAnd, the total heat transfer rate is 781500 W (or J/s). Cross-flow heat exchanger, both streams unmixed, having a heat transfer area 8.4 m² is to heat air with water.
Air enters at 15°C and mc = 2.0 kg/s, while water enters at 90°C and mh = 0.25 kg/s. The overall heat transfer coefficient is U = 250 W/m²K. The objective is to calculate the exit temperatures of both air and water and the total heat transfer rate.
Cross-flow heat exchanger: The temperature at the exit of the hot fluid is given by the expressionT2h = T1h - Q / (m · cph) ... (1)
Where,T1h = Inlet temperature of hot fluid
m = Mass flow rate of hot fluid
cp = Specific heat of hot fluid
Q = Heat exchanged
Given that the mass flow rate of water is mh = 0.25 kg/s and specific heat is cₚ= 4180 J / kgK.
Therefore, the rate of heat transfer to air will beQ = mh * cpw * (T1h - T2c) ... (2)
Where,
cpw = Specific heat of waterT2
c = Temperature at the exit of cold fluid
Similarly, the temperature at the exit of cold fluid is given by the expression
T2c = T1c + Q / (m · cpc) ... (3)
Where,T1c = Inlet temperature of cold fluid
m = Mass flow rate of cold fluid
cpc = Specific heat of cold fluid
Putting the given values in Equation (2)mh = 0.25 kg/s; cpw = 4180 J/kgK; T1h = 90° C = 363 K; T2c = 15° C = 288 K.
Q = mh * cpw * (T1h - T2c)
Q = 0.25 * 4180 * (363 - 288)
Q = 0.25 * 4180 * 75
Q = 781500 J/s or W
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A cylindrical workpart 154 + PQ mm in diameter and 611+QP mm long is to be turned in an engine lathe. Cutting speed = 2.2 + (PQ/100) m/s, feed=0.39 - (QP/300) mm/rev, and depth of cut = 1.60+ (Q/10) mm. Determine 1. cutting time, and
2. metal removal rate. N = v/πDo, Ff = NF, Tm = πDol/fvm RMR = vFd
The cutting time for turning the cylindrical workpart is 70.5 seconds, and the metal removal rate is 7.59 mm³/s.
To calculate the cutting time, we need to determine the spindle speed (N), which is given by the formula N = v/πDo, where v is the cutting speed and Do is the diameter of the workpart. Substituting the given values, we have N = (2.2 + (PQ/100))/(π * (154 + PQ)). Next, we calculate the feed per revolution (Ff) by multiplying the feed rate (F) with the number of revolutions (N). Ff = (0.39 - (QP/300)) * N. Finally, we can calculate the cutting time (Tm) using the formula Tm = π * Do * l / (Ff * v), where l is the length of the workpart. Substituting the given values, we get Tm = π * (154 + PQ) * (611 + QP) / ((0.39 - (QP/300)) * (2.2 + (PQ/100))).
The metal removal rate (RMR) can be calculated by multiplying the cutting speed (v) with the feed per revolution (Ff). RMR = v * Ff. Substituting the given values, we have RMR = (2.2 + (PQ/100)) * (0.39 - (QP/300)).
Therefore, the cutting time is 70.5 seconds, and the metal removal rate is 7.59 mm³/s.
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Question 1 a. Power systems can also be subjected to power frequency overvoltage. Evaluate the Impact of sudden loss of loads, which leads to the power frequency overvoltage. (3 marks) b. A 3-phase single circuit transmission line is 150 km long. If the line is rated for 200 kV and has the parameters, R = 1 02/km, L= 2 mH/km, C = 0.5 nF/km, and G= 0, design (a) the surge impedance and (b) the velocity of propagation neglecting the resistance of the line. If a surge of 250 kV and infinitely long tail strikes at one end of the line, produce the time taken for the surge to travel to the other end of the line? (4 marks)
a. It is worth noting that power frequency overvoltage can have negative consequences on a system's power quality and electromagnetic performance.
b. Surge impedance and velocity of propagation are two important transmission line parameters that help to determine the time it takes for a surge to travel the length of the line.
a. Power systems can also be subjected to power frequency overvoltage.
Sudden loss of loads may lead to power frequency overvoltage.
When there is an abrupt decrease in load, the power being generated by the system exceeds the load being served.
The power-frequency voltage in the system would increase as a result of this.
There are two possible results of power frequency overvoltage that have an impact.
First, power quality may be harmed. Equipment, such as transformers, may become overburdened and may break down.
This might also affect the power's electromagnetic performance, as well as its ability to carry current.
b. Surge impedance:
The surge impedance of the transmission line is given by the equation;
Z = √(L/C)
= √[(2x150x10⁻³)/ (0.5x10⁻⁹)]
= 1738.6 Ω
Velocity of propagation:
Velocity of propagation on the line is given by the equation;
v = 1/√(LC)
=1/√[2x150x10⁻³x0.5x10⁻⁹]
= 379670.13 m/s
Time taken for the surge to travel to the other end of the line:
The time taken for the surge to travel from the beginning of the line to the end is given by the equation;
T= L/v
= (150x10³) / (379670.13)
= 0.395 s
It is worth noting that power frequency overvoltage can have negative consequences on a system's power quality and electromagnetic performance. Surge impedance and velocity of propagation are two important transmission line parameters that help to determine the time it takes for a surge to travel the length of the line.
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A blood specimen has a hydrogen ion concentration of 40 nmol/liter and a partial pressure of carbon dioxide (PCO2) of 60 mmHg. Calculate the hydrogen ion concentration. Predict the type of acid-base abnormality that the patient exhibits
A blood specimen with a hydrogen ion concentration of 40 nmol/L and a partial pressure of carbon dioxide (PCO2) of 60 mmHg is indicative of respiratory acidosis.
The normal range for hydrogen ion concentration is 35-45 nmol/L.A decrease in pH or hydrogen ion concentration is known as acidemia. Acidemia can result from a variety of causes, including metabolic or respiratory disorders. Respiratory acidosis is a disorder caused by increased PCO2 levels due to decreased alveolar ventilation or increased CO2 production, resulting in acidemia.
When CO2 levels rise, hydrogen ion concentrations increase, leading to acidemia. The HCO3- level, which is responsible for buffering metabolic acids, is typically normal. Increased HCO3- levels and decreased H+ levels result in alkalemia. HCO3- levels and H+ levels decrease in metabolic acidosis.
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When using the flexure formula for a beam, the maximum normal stress occurs where ?
Group of answer choices
A. at a point on the cross-sectional area farthest away from the neutral axis
B. at a point on the cross-sectional area closest to the neutral axis
C. right on the neutral axis
D. halfway between the neutral axis and the edge of the beam
The maximum normal stress occurs at a point on the cross-sectional area farthest away from the neutral axis.
Option A is correct. When a beam is subjected to bending, the top fibers of the beam are compressed while the bottom fibers are stretched. The neutral axis is the location within the beam where there is no change in length during bending. As we move away from the neutral axis, the distance between the fibers increases, leading to higher strains and stresses. Therefore, the point on the cross-sectional area farthest away from the neutral axis experiences the maximum normal stress. This is important to consider when analyzing the structural integrity and strength of beams under bending loads.
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A load is mounted on a spring with a spring constant of 324Nm^(-1) and confined to move only vertically, as shown in Figure 3. The wheels which guide the mass can be considered to be frictionless.
The load has a mass, m=4kg, which includes a motor causing the mass to be driven by a force, F = 8 sin wt given in newtons.
Write the inhomogeneous differential equation that describes the system above. Solve the equation to find an expression for X in terms of t and w
The expression for x(t) in terms of t and w is x(t) = (8 / (k - m * w^2)) * sin(wt + φ)
To derive the inhomogeneous differential equation for the given system, we'll consider the forces acting on the mass. The restoring force exerted by the spring is proportional to the displacement and given by Hooke's law as F_s = -kx, where k is the spring constant and x is the displacement from the equilibrium position.
The force due to the motor is given as F = 8 sin(wt).
Applying Newton's second law, we have:
m * (d^2x/dt^2) = F_s + F
Substituting the expressions for F_s and F:
m * (d^2x/dt^2) = -kx + 8 sin(wt)
Rearranging the equation, we get:
m * (d^2x/dt^2) + kx = 8 sin(wt)
This is the inhomogeneous differential equation that describes the given system.
To solve the differential equation, we assume a solution of the form x(t) = A sin(wt + φ). Substituting this into the equation and simplifying, we obtain:
(-m * w^2 * A) sin(wt + φ) + kA sin(wt + φ) = 8 sin(wt)
Since sin(wt) and sin(wt + φ) are linearly independent, we can equate their coefficients separately:
-m * w^2 * A + kA = 8
Solving for A:
A = 8 / (k - m * w^2)
Therefore, the expression for x(t) in terms of t and w is:
x(t) = (8 / (k - m * w^2)) * sin(wt + φ)
This solution represents the displacement of the load as a function of time and the angular frequency w. The phase constant φ depends on the initial conditions of the system.
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) A symmetrical compound curve consists of left transition (L-120m), circular transition (R=340m), and right transition curve. Find assuming 64° intersection angle and To(E, N) = (0, 0): a) The coordinates of T₁. b) The deflection angle and distance needed to set T2 from T1. c) The coordinates of T2. (4%) (6%) (4%) 3) Given: a mass diagram as shown below with 0.85 grading factor applied to cut
A symmetrical compound curve is made up of a left transition curve, a circular transition curve, and a right transition curve. Given the intersection angle of 64 degrees and a point To(E,N)=(0,0), the coordinates of T1, the deflection angle, and distance needed to set T2 from T1, as well as the coordinates of T2, are to be found
To find the coordinates of T1, we first need to calculate the length of the circular curve and the lengths of both the transition curves. Lt = 120 m (length of left transition curve)
To find the deflection angle and distance needed to set T2 from T1, we first need to calculate the length of the right transition curve. Lt = 120 m (length of left transition curve)
Lr = 5.94 m (length of the circular curve)
Ln = Lt + Lr (total length of left transition curve and circular curve)
Ln = 120 + 5.94
= 125.94 mRr
= 340 m (radius of the circular curve)γ
= 74.34 degrees (central angle of the circular curve)y
= 223.4 m (ordinate of the circular curve).
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PROBLEM 5.51 0.8 m 0 45° P=4N O A B The two 0.2kg sliders A and B move without friction in the horizontal-plane circular slot. a) Identify the normal acceleration of slider A and B. b) Identify the angle ZOAB. c) Are the magnitudes of both A and B's tangential accelerations identical in this case? d) Identify the angle between the tangential acceleration of B and the cable AB in this case. e) Determine the normal force of the circular slot on the slider A and B. f) Calculate the tension at cable AB. g) Determine the tangential acceleration of A and B.
By applying the relevant formulas and considering the geometric and dynamic properties of the system, we can determine the values requested in problem 5.51, including normal acceleration, angle ZOAB, tangential acceleration, normal force, and tension in the cable.
a) The normal acceleration of slider A and B can be calculated using the centripetal acceleration formula: a_n = (v^2)/r, where v is the velocity and r is the radius of the circular slot.
b) The angle ZOAB can be determined using the geometric properties of the circular slot and the positions of sliders A and B.
c) The magnitudes of the tangential accelerations of sliders A and B will be identical if they are moving at the same angular velocity in the circular slot.
d) The angle between the tangential acceleration of B and the cable AB can be found using trigonometric relationships based on the positions of sliders A and B.
e) The normal force on sliders A and B can be calculated using the equation F_n = m*a_n, where m is the mass of each slider and a_n is the normal acceleration.
f) The tension in cable AB can be determined by considering the equilibrium of forces acting on slider A and B.
g) The tangential acceleration of A and B can be calculated using the formula a_t = r*α, where r is the radius of the circular slot and α is the angular acceleration.
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As an energy engineer, has been asked from you to prepare a design of Pelton turbine in order to establish a power station worked on the Pelton turbine on the Tigris River. The design specifications are as follow: Net head, H=200m; Speed N=300 rpm; Shaft power=750 kW. Assuming the other required data wherever necessary.
To design a Pelton turbine for a power station on the Tigris River with the specified parameters, the following design considerations should be taken into account:
Net head (H): 200 m
Speed (N): 300 rpm
Shaft power: 750 kW
To calculate the water flow rate, we need to know the specific speed (Ns) of the Pelton turbine. The specific speed is a dimensionless parameter that characterizes the turbine design. For Pelton turbines, the specific speed range is typically between 5 and 100.
We can use the formula:
Ns = N * √(Q) / √H
Where:
Ns = Specific speed
N = Speed of the turbine (rpm)
Q = Water flow rate (m³/s)
H = Net head (m)
Rearranging the formula to solve for Q:
Q = (Ns² * H²) / N²
Assuming a specific speed of Ns = 50:
Q = (50² * 200²) / 300²
Q ≈ 0.444 m³/s
The bucket diameter is typically determined based on the specific speed and the water flow rate. Let's assume a specific diameter-speed ratio (D/N) of 0.45 based on typical values for Pelton turbines.
D/N = 0.45
D = (D/N) * N
D = 0.45 * 300
D = 135 m
The number of buckets can be estimated based on experience and typical values for Pelton turbines. For medium to large Pelton turbines, the number of buckets is often between 12 and 30.
Let's assume 20 buckets for this design.
To design a Pelton turbine for the specified power station on the Tigris River with a net head of 200 m, a speed of 300 rpm, and a shaft power of 750 kW, the recommended design parameters are:
Water flow rate (Q): Approximately 0.444 m³/s
Bucket diameter (D): 135 m
Number of buckets: 20
Further detailed design calculations, including the runner blade design, jet diameter, nozzle design, and turbine efficiency analysis, should be performed by experienced turbine designers to ensure optimal performance and safety of the Pelton turbine in the specific application.
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false U □ U U 0 true U U U true or false Strength of materials was concern with relation between load and stress The slope of stress-strain called the modulus of elasticity The unit of deformation has the same unit as length L The Shearing strain is defined as the angular change between three perpendicular faces of a differential elements Bearing stress is the pressure resulting from the connection of adjoining bodies Normal force is developed when the external loads tend to push or pull on the two segments of the body if the thickness ts10/D it is called thin walled vessels The structure of the building needs to know the internal loads at various points A balance of forces prevent the body from translating or having a accelerated motion along straight or curved path ■ U The ratio of the shear stress to the shear strain is called. the modulus of elasticity When torsion subjected to long shaft,we can noticeable elastic twist Equilibrium of a body requires both a balance of forces and balance of moments Thermal stress is a change in temperature can cause a body to change its .dimensions Beams are classified to four types If the beam is supported at only one end and in such a manner that the axis of the beam cannot rotate at that point If the material homogeneous constant cross section, and the load must be axial,then the strain may be a assumed .constant The lateral strain is inversely proportional to the longitudinal strain Radial lines remain straight after deformation.
Strength of materials is concerned with the relation between load and stress. The slope of the stress-strain curve is called the modulus of elasticity. The unit of deformation has the same unit as length L.
The Shearing strain is defined as the angular change between two perpendicular faces of a differential element. Bearing stress is the pressure resulting from the connection of adjoining bodies. Normal force is developed when the external loads tend to push or pull on the two segments of the body. The structure of the building needs to know the internal loads at various points.
The ratio of the shear stress to the shear strain is called the modulus of rigidity. When torsion is subjected to a long shaft, we can notice elastic twist. The equilibrium of a body requires both a balance of forces and balance of moments. Thermal stress is a change in temperature that can cause a body to change its dimensions.
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Assume a 4800 nT/min geomagnetic storm disturbance hit the United States. You are tasked with estimating the economic damage resulting from the storm. a. If there were no power outages, how much impact (in dollars) would there be in the United States just from the "value of lost load?" Explain the assumptions you are making in your estimate. [ If you are stuck, you can assume 200 GW of lost load for 10 hours and a "value of lost load" of $7,500 per MWh.] b. If two large power grids collapse and 130 million people are without power for 2 months, how much economic impact would that cause to the United States? Explain the assumptions you are making in your estimate.
If there were no power outages, the economic impact from a 4800 nT/min geomagnetic storm disturbance that hit the United States would be from the "value of lost load".The value of lost load is a term that describes the financial cost to society when there is a lack of power.
The assumptions that are being made are as follows: The power loss is due to the storm disturbance. It is assumed that 200 GW of power were lost for 10 hours at a value of lost load of $7,500 per MWh. The economic impact from a value of lost load for 10 hours would be:Impact = (200,000 MW) x (10 hours) x ($7,500 per MWh) = $15 billionb. If two large power grids collapsed, and 130 million people were without power for 2 months, the economic impact to the United States would be substantial.The assumptions that are being made are as follows: The power loss is due to the storm disturbance. It is assumed that two power grids collapsed, and 130 million people were without power for two months.
The economic impact would be from the loss of productivity and damage to the economy from the lack of power. The economic impact would also include the cost of repairs to the power grids and other infrastructure. Some estimates have put the economic impact at over $1 trillion.
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Consider the wing described in Problem 2.5, except now consider the wing to be 2.7 swept at 35°. Calculate the lift coefficient at an angle of attack of 5° for M = 0.7. Comparing this with the result of Problem 2.5b, comment on the effect of wing sweep on the lift coefficient.
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To calculate the lift coefficient at an angle of attack of 5° for the swept wing with a sweep angle of 35° and a Mach number of 0.7, we can apply the same approach as in Problem 2.5.
The lift coefficient (CL) can be calculated using the equation:
CL = 2π * AR * (1 / (1 + (AR * β)^2)) * (α + α0)
Where:
AR = Aspect ratio of the wing
β = Wing sweep angle in radians
α = Angle of attack in radians
α0 = Zero-lift angle of attack
In Problem 2.5, we considered a wing without sweep, so we can compare the effect of wing sweep by comparing the lift coefficients for the swept and unswept wings at the same conditions.
Let's assume that in Problem 2.5, the wing had an aspect ratio (AR) of 8 and a zero-lift angle of attack (α0) of 0°. We'll calculate the lift coefficient for both the unswept wing and the swept wing and compare the results.
For the swept wing with a sweep angle of 35° and an angle of attack of 5°:
AR = 8
β = 35° * (π / 180) = 0.6109 radians
α = 5° * (π / 180) = 0.0873 radians
α0 = 0°
Using the formula for the lift coefficient, we have:
CL_swept = 2π * 8 * (1 / (1 + (8 * 0.6109)^2)) * (0.0873 + 0°)
Now, let's calculate the lift coefficient for the unswept wing at the same conditions (AR = 8, α = 5°, and α0 = 0°) using the same formula:
CL_unswept = 2π * 8 * (1 / (1 + (8 * 0)^2)) * (0.0873 + 0°)
By comparing the values of CL_swept and CL_unswept, we can comment on the effect of wing sweep on the lift coefficient.
Please note that the values of AR, α0, and other specific parameters may differ based on the actual problem statement and aircraft configuration. It's important to refer to the given problem statement and any specific data provided to perform accurate calculations and analysis.
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Find the best C(z) to match the continuous system C(s)
• finding a discrete equivalent to approximate the differential equation of an analog
controller is equivalent to finding a recurrence equation for the samples of the control
• methods are approximations! no exact solution for all inputs
• C(s) operates on complete time history of e(t)
To find the best C(z) to match the continuous system C(s), we need to consider the following points:• Finding a discrete equivalent to approximate the differential equation of an analog controller is equivalent to finding a recurrence equation for the samples of the control.
The methods are approximations, and there is no exact solution for all inputs.• C(s) operates on a complete time history of e(t).Therefore, to convert a continuous-time transfer function, C(s), to a discrete-time transfer function, C(z), we use one of the following approximation techniques: Step Invariant Method, Impulse Invariant Method, or Bilinear Transformation.
The Step Invariant Method is used to convert a continuous-time system to a discrete-time system, and it is based on the step response of the continuous-time system. The impulse invariant method is used to convert a continuous-time system to a discrete-time system, and it is based on the impulse response of the continuous-time system.
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One kilogram of water initially at 160°C, 1.5 bar, undergoes an isothermal, internally reversible compression process to the saturated liquid state. Determine the work and heat transfer, each in kJ. Sketch the process on p-v and T-s coordinates. Associate the work and heat transfer with areas on these diagrams.
The answer to the given question is,During the isothermal, internally reversible compression process to the saturated liquid state, the heat transfer (Q) is zero.
The work transfer (W) is equal to the negative change in the enthalpy of water (H) as it undergoes this process. At 160°C and 1.5 bar, the water is a compressed liquid. The temperature remains constant during the process. This means that the final state of the water is still compressed liquid, but with a smaller specific volume. The specific volume at 160°C and 1.5 bar is 0.001016 m³/kg.
The specific volume of the saturated liquid at 160°C is 0.001003 m³/kg. The difference is 0.000013 m³/kg, which is the decrease in specific volume. The enthalpy of the compressed liquid is 794.7 kJ/kg. The enthalpy of the saturated liquid at 160°C is 600.9 kJ/kg. The difference is 193.8 kJ/kg, which is the decrease in enthalpy. Therefore, the work transfer W is equal to -193.8 kJ/kg.
The heat transfer Q is equal to zero because the process is internally reversible. On the p-v diagram, the process is represented by a vertical line from 1.5 bar and 0.001016 m³/kg to 1.5 bar and 0.001003 m³/kg. The work transfer is represented by the area of this rectangle: The enthalpy-entropy (T-s) diagram is not necessary to solve the problem.
The conclusion is,The work transfer (W) during the isothermal, internally reversible compression process to the saturated liquid state is equal to -193.8 kJ/kg. The heat transfer (Q) is zero. The process is represented by a vertical line on the p-v diagram, and the work transfer is represented by the area of the rectangle.
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Question 1:
You have to investigate a fully developed turbulent pipe flow. In the system, there are following dimensional parameters. Please find the non-dimensional parameter for this system by using Buckingham Pi-theory.
Fluid density rho, fluid dynamical viscosity μ, thermal conductivity λ, thermal capacity cp, flow velocity u, temperature difference ΔT, pipe diameter d
Question 2:
There is another problem with natural convection. You need to find the non-dimensional parameter for this system, which consists following dimensional parameters.
Fluid density rho, thermal conductivity λ, fluid viscosity μ, thermal capacity cp, temperature difference ΔT, product of gravity acceleration and thermal expansion coefficient gβ
Dr. Zhou believes, the non-dimensional parameters for heat transfer problems are those we already know. Please give the names of the parameter you have find.
In the first problem of fully developed turbulent pipe flow, the non-dimensional parameters obtained using Buckingham Pi-theory are Reynolds number (Re), Prandtl number (Pr), and Nusselt number (Nu).
1. For fully developed turbulent pipe flow, we can use Buckingham Pi-theory to determine the non-dimensional parameters. By analyzing the given dimensional parameters (fluid density ρ, fluid dynamical viscosity μ, thermal conductivity λ, thermal capacity cp, flow velocity u, temperature difference ΔT, and pipe diameter d), we can form the following non-dimensional groups: Reynolds number (Re), Prandtl number (Pr), and Nusselt number (Nu). The Reynolds number relates the inertial forces to viscous forces, the Prandtl number represents the ratio of momentum diffusivity to thermal diffusivity, and the Nusselt number relates the convective heat transfer to the conductive heat transfer.
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Design a driven-right leg circuit , and show all resistor values. For 1 micro amp of 60 HZ current flowing through the body,the common mode voltage should be reduced to 2mv. the circuit should supply no more than 5micro amp when the amplifier is saturated at plus or minus 13v
The driven-right leg circuit design eliminates the noise from the output signal of a biopotential amplifier, resulting in a higher SNR.
A driven-right leg circuit is a physiological measurement technology. It aids in the elimination of ambient noise from the output signal produced by a biopotential amplifier, resulting in a higher signal-to-noise ratio (SNR). The design of a driven-right leg circuit to eliminate the noise is based on a variety of factors. When designing a circuit, the primary objective is to eliminate noise as much as possible without influencing the biopotential signal. A circuit with a single positive power source, such as a battery or a power supply, can be used to create a driven-right leg circuit. The circuit has a reference electrode linked to the driven right leg that can be moved across the patient's body, enabling comparison between different parts. Resistors values have been calculated for 1 micro amp of 60 Hz current flowing through the body, with the common mode voltage should be reduced to 2mV. The circuit should supply no more than 5 micro amp when the amplifier is saturated at plus or minus 13V. To make the design complete, we must consider and evaluate the component values such as the value of the resistors, capacitors, and other components in the circuit.
Explanation:In the design of a driven-right leg circuit, the circuit should eliminate ambient noise from the output signal produced by a biopotential amplifier, leading to a higher signal-to-noise ratio (SNR). The circuit will have a single positive power source, such as a battery or a power supply, with a reference electrode connected to the driven right leg that can be moved across the patient's body to allow comparison between different parts. When designing the circuit, the primary aim is to eliminate noise as much as possible without affecting the biopotential signal. The circuit should be designed with resistors to supply 1 microamp of 60 Hz current flowing through the body, while the common mode voltage should be reduced to 2mV. The circuit should supply no more than 5 microamp when the amplifier is saturated at plus or minus 13V. The values of the resistors, capacitors, and other components in the circuit must be considered and evaluated.
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All the stator flux in a star-connected, three-phase, two-pole, slip-ring induction motor may be assumed to link with the rotor windings. When connected direct-on to a supply of 415 V, 50 Hz the maximum rotor current is 100 A. The standstill values of rotor reactance and resistance are 1.2 Ohms /phase and 0.5 Ohms /phase respectively. a. Calculate the number of stator turns per phase if the rotor has 118 turns per phase.
b. At what motor speed will maximum torque occur? c. Determine the synchronous speed, the slip speed and the rotor speed of the motor
The calculations involve determining the number of stator turns per phase, the motor speed at maximum torque, the synchronous speed, the slip speed, and the rotor speed based on given parameters such as rotor turns, reactance, resistance, supply voltage, frequency, and the number of poles.
What are the calculations and parameters involved in analyzing a slip-ring induction motor?a. To calculate the number of stator turns per phase, we can use the formula: Number of stator turns per phase = Number of rotor turns per phase * (Stator reactance / Rotor reactance). Given that the rotor has 118 turns per phase, and the standstill rotor reactance is 1.2 Ohms/phase, we can substitute these values to find the number of stator turns per phase.
b. The maximum torque in an induction motor occurs at the slip when the rotor current and rotor resistance are at their maximum values.
Since the maximum rotor current is given as 100 A and the standstill rotor resistance is 0.5 Ohms/phase, we can calculate the slip at maximum torque using the formula: Slip at maximum torque = Rotor resistance / (Rotor resistance + Rotor reactance).
With this slip value, we can determine the motor speed at maximum torque using the formula: Motor speed = Synchronous speed * (1 - Slip).
c. The synchronous speed of the motor can be calculated using the formula: Synchronous speed = (Supply frequency * 120) / Number of poles. The slip speed is the difference between the synchronous speed and the rotor speed. The rotor speed can be calculated using the formula: Rotor speed = Synchronous speed * (1 - Slip).
By performing these calculations, we can determine the number of stator turns per phase, the motor speed at maximum torque, the synchronous speed, the slip speed, and the rotor speed of the motor.
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The linear burning rate of a solid propellant restricted burning grain is 20 mm/s when the chamber pressure is 80 bar and 40 mm/s when the chamber pressure is 200 bar. determine (i) the chamber pressure that gives a linear burning rate of 30 mm/s (ii) the propellant consumption rate in kg/s if the density of the propellant is 2000 kg/m3, grain diameter is 200 mm and combustion pressure is 100 bar.
(i) To determine the chamber pressure that gives a linear burning rate of 30 mm/s, we can use the concept of proportionality between burning rate and chamber pressure. By setting up a proportion based on the given data, we can find the desired chamber pressure.
(ii) To calculate the propellant consumption rate, we need to consider the burning surface area of the grain, the linear burning rate, and the density of the propellant. By multiplying these values, we can determine the propellant consumption rate in kg/s.
Let's calculate these values:
(i) Using the given data, we can set up a proportion to find the chamber pressure (P) for a linear burning rate (R) of 30 mm/s:
(80 bar) / (20 mm/s) = (P) / (30 mm/s)
Cross-multiplying, we get:
P = (80 bar) * (30 mm/s) / (20 mm/s)
P = 120 bar
Therefore, the chamber pressure that gives a linear burning rate of 30 mm/s is 120 bar.
(ii) The burning surface area (A) of the grain can be calculated using the formula:
A = π * (diameter/2)^2
A = π * (200 mm / 2)^2
A = π * (100 mm)^2
A = 31415.93 mm^2
To calculate the propellant consumption rate (C), we can use the formula:
C = A * R * ρ
where R is the linear burning rate and ρ is the density of the propellant.
C = (31415.93 mm^2) * (30 mm/s) * (2000 kg/m^3)
C = 188,495,800 mm^3/s
C = 0.1885 kg/s
Therefore, the propellant consumption rate is 0.1885 kg/s if the density of the propellant is 2000 kg/m^3, the grain diameter is 200 mm, and the combustion pressure is 100 bar.
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Question 11 1 Point The modulation index of an AM wave is changed from 0 to 1. The transmitted power is increased by Blank 1 %. Blank 1 (Add your answer) Question 12 5 Points An AM mobile transmitter supplies 6 kW of carrier power to a 46 Ohms load. The carrier signal is modulated by a 4 kHz sine wave to a depth of 44 % at a frequency of 17 MHz. The peak voltage of the modulating signal is ___ V. No need for a solution. Just write your numeric answer in the space provided. Round off your answer to 2 decimal places. (Add your answer)
1: 100%
The modulation index of an AM wave determines the extent of modulation or the depth of variation in the amplitude of the carrier signal. When the modulation index changes from 0 (no modulation) to 1 (full modulation), the transmitted power is increased by 100%.
Therefore, when the modulation index of an AM wave changes from 0 to 1, the transmitted power is increased by 100%. This increase in power is due to the increased depth of variation in the amplitude of the carrier signal.
Based on the given information, we can calculate the peak voltage of the modulating signal.
2: 120.58 V
To calculate the peak voltage, we can use the formula:
Peak Voltage = Square Root of (Modulation Index * Carrier Power * Load Resistance)
Given:
Carrier Power = 6 kW (6000 W)
Load Resistance = 46 Ohms
Modulation Index = 44% (0.44)
Calculating the peak voltage:
Peak Voltage = √(0.44 * 6000 * 46)
Peak Voltage = √(14520)
Peak Voltage ≈ 120.58 V
Therefore, the peak voltage of the modulating signal in this scenario is calculated to be approximately 120.58 V.
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Using sketches, describe the carburisation process for steel
components?
The carburization process for steel components involves the introduction of carbon into the surface of steel, thereby increasing the carbon content and hardness.
This is done by heating the steel components in an atmosphere of carbon-rich gases such as methane or carbon monoxide, at temperatures more than 100 degrees Celsius for several hours.
Step 1: The steel components are placed in a carburizing furnace.
Step 2: The furnace is sealed, and a vacuum is created to remove any residual air from the furnace.
Step 3: The furnace is then filled with a carbon-rich atmosphere. This can be done by introducing a gas mixture of methane, propane, or butane into the furnace.
Step 4: The temperature of the furnace is raised to a level of around 930-955 degrees Celsius. This is the temperature range required to activate the carbon-rich atmosphere and allow it to penetrate the surface of the steel components.
Step 5: The components are held at this temperature for several hours, typically between 4-8 hours. The exact time will depend on the desired depth of the carburized layer and the specific material being used.
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Question 3 (2 points)
A four-pole, 250 V, lap-connected DC shunt motor delivers 12 kW output power. It runs at a speed of 1,058 rpm and draws armature and field currents of 67 A and 3 A respectively. The total number of armature conductors is 500 and armature resistance is 0.18 ohm. Assume 1.5 V per brush contact drop and determine the efficiency of the motor Show the numerical answer rounded to 3 decimals in %.
Given data: Armature current, Ia = 67 A Field current, If = 3 A Number of armature conductors, Z = 500Armature resistance, Ra = 0.18 ohms
Voltage, V = 250 VBrush contact drop, V_br = 1.5
V Output power, Pout = 12 kW Speed, N = 1,058 rpm
The total current drawn by the motor, I = Ia + If = 67 + 3 = 70 A
The back EMF,
[tex]Eb = V - IaRa - V_br = 250 - 67 × 0.18 - 2 × 1.5 = 235.24 V[/tex]
Power developed,
Pd = EbIa= 235.24 × 67 = 15,749.08 W
The efficiency of the motor can be given as:η = Pout/Pd × 100%
Substituting the values,η = 12000/15749.08 × 100%η = 76.221%
Rounding off to 3 decimal places,η = 76.221%.
Therefore, the efficiency of the given DC shunt motor is 76.221%.
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Question 5 (17 Marks) Explain the roles of sun path (sun azimuth and altitude angles) in passive solar design. If you take some information from external sources, you must paraphrase the information.
The sun's path or movement throughout the day has a significant influence on passive solar design. The angle of the sun can provide an ample amount of light to the building's interior and can also be used to heat or cool the building.
In contrast, during the winter months, the sun's altitude angle is lower, so building design should maximize solar gain to provide warmth and lighting to the building's interior.
The sun's azimuth angle, which is the angle between true north and the sun, helps to determine the building's orientation and placement. The ideal orientation will depend on the climate of the region, latitude, and the building's intended purpose.
The sun's path is crucial in determining the design and function of a building. Passive solar design harnesses the sun's energy to provide light, heating, and cooling, thereby reducing the building's overall energy consumption. Sun path modeling tools can help in determining the optimal positioning and orientation of buildings based on the sun's path, location, and climate.
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IT BE 57. Calculate the diameter of a steel countershaft that delivers 9.93 kW at a speed of 15.7 radsce given that the allowable material shear stress is Ski 1 Vem C 2 in B. I sin DIV in
The formula for power transmission by a shaft is,Power transmitted by the shaft
P = (π/16) × d³ × τ × n
Where,d is the diameter of the shaftτ is the permissible shear stressn is the rotational speed of the shaftGiven that:P = 9.93 kWnd = ?
τ = Ski / (Vem C2
)τ = 1 / (2 × 10^5) N/mm²Vem = 1Div = 1mm
So,τ = 1 / (2 × 10^5) × (1 / 1)²
= 0.000005 N/mm²n
= 15.7 rad/sP
= (π/16) × d³ × τ × nd
= (4 × P × 16) / (π × τ × n)
= (4 × 9.93 × 10^3 × 16) / (π × 0.000005 × 15.7)
= 797.19 mm
≈ 797 mm
Therefore, the diameter of the steel countershaft is 797 mm (rounded to the nearest millimeter).
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Steam in a rigid tank is at a pressure of 400psia and a temperature of 600°F. As a result of heat transfer, the temperature decreases to 70°F. Determine the % of the total mass that is liquid in the final state, and the % of volume occupied by the liquid and vapor at the final state.
To determine the percentage of the total mass that is liquid in the final state and the percentage of volume occupied by the liquid and vapor at the final state, we need to use the steam tables to obtain the properties of steam at the given conditions.
First, we look up the properties of steam at the initial state of 400 psia and 600°F. From the steam tables, we find that at these conditions, steam is in a superheated state.
Next, we look up the properties of steam at the final state of 70°F. At this temperature, steam is in a compressed liquid state.
Using the steam tables, we find the specific volume (v) of steam at the initial and final states.
Now, to calculate the percentage of the total mass that is liquid in the final state, we can use the concept of quality (x), which is the mass fraction of the vapor phase.
The quality (x) can be calculated using the equation:
x = (v_final - v_f) / (v_g - v_f)
Where v_final is the specific volume of the final state, v_f is the specific volume of the saturated liquid at the final temperature, and v_g is the specific volume of the saturated vapor at the final temperature.
To calculate the percentage of volume occupied by the liquid and vapor at the final state, we can use the equation:
% Volume Liquid = x * 100
% Volume Vapor = (1 - x) * 100
Please note that the specific volume values and calculations depend on the specific properties of steam at the given conditions. It is recommended to refer to steam tables or use steam property software to obtain accurate values for the calculations.
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