(i) Variation of cycle efficiency with pressure ratio for a simple gas turbine cycle with heat exchanger : For a simple gas turbine cycle with heat exchanger, the cycle efficiency variation with pressure ratio is described by a bell-shaped curve. When the pressure ratio increases, the cycle efficiency increases to a peak and then declines rapidly. This is due to the fact that the pressure ratio determines the power output of the cycle, and the compressor work needed for higher pressure ratios causes the efficiency to decrease.
As the temperature of the turbine inlet increases, the maximum cycle efficiency increases.
(ii) The maximum cycle pressure ratio at which a heat exchanger can be implemented is determined by the maximum allowable turbine inlet temperature, which is 800°C in this scenario.
The compressor outlet temperature can't be higher than this value because it will cause the turbine inlet temperature to exceed the maximum limit. Furthermore, a heat exchanger must be used to cool the compressor outlet temperature before it enters the combustion chamber.
If the pressure ratio is too high, the temperature of the compressor outlet will be too high, and a heat exchanger will not be able to cool it enough to prevent the turbine inlet temperature from exceeding the maximum allowable value.
(iii)For this cycle, extremely low pressure ratios should be avoided for a few reasons, including the following:
Lower pressure ratios cause lower compressor work output and thus lower cycle efficiency.
Low-pressure ratios cause a drop in compressor discharge temperature, which may lead to ice formation in the intake and compressor blades' freeze up.
The combustion process is less stable at lower pressure ratios because it is more difficult to maintain a constant flame speed.
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17. What size cylinder connected to a 5 gal/min (22.7 1/min) pump would be required to limit the extension velocity to 2 ft/sec?
The cylinder with a radius of approximately 1.9 feet would be required to limit the extension velocity to 2 ft/sec.
To answer this, we need to make use of the formula Q = Av, where Q is the flow rate, A is the area of the cylinder, and v is the velocity of the fluid.
We know that the flow rate is 5 gal/min, or 22.7 L/min, and the velocity is 2 ft/sec.
We need to find the area of the cylinder. The formula for the flow rate is:
Q = Av
where
Q = 5 gal/min
= 22.7 L/minv
= 2 ft/sec
Area of the cylinder, A = Q/v = 22.7/2 = 11.35 ft²
The formula for the area of a cylinder is given by:
A = πr²
where
π is the constant 3.14, and r is the radius of the cylinder.
So, we can write:
11.35 = 3.14r²r²
= 11.35/3.14
= 3.61r
= √3.61
= 1.9 feet (approx.)
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Q8. In the inverted crank-slider shown, link 2 is the input and link 4 is the output. If O₂O₂ = 27 cm and O₂A = 18 cm, then the total swinging angle of link 4 about O, is found to be: c) 83.6⁰ a) 45° b) 72.3° d) 89.4° e) 60° f) None of the above Q9. The time ratio of this mechanism is found to be: c) 2.735 d) 1.5 e) 2.115 f) None of the above a) 1.828 b) 3.344 ОА Q10. Assume that in the position shown, link 2 rotates at 10 rad/s hence causing link 4 to rotate at 4 rad/s. If the torque on link 2 is 100 N.m, then by neglecting power losses, the torque on link 4 is: c) 500 N.m. d) 650 N.m e) None of the above. a) 250 N.m b) 375 N.m Im 02 LETTERS 2 4 3 A - Re
Q8. The correct option is c) 83.6⁰
Explanation: The total swinging angle of link 4 can be determined as follows: OA² + O₂A² = OAₒ²
Cosine rule can be used to determine the angle at O₂OAₒ = 33.97 cm
O₄Aₒ = 3.11 cm
Cosine rule can be used to determine the angle at OAₒ
The angle of link 4 can be determined by calculating:θ = 360° - α - β + γ
= 83.6°Q9.
The correct option is b) 3.344
Explanation:The expression for time ratio can be defined as:T = (2 * AB) / (OA + AₒC)
We will start by calculating ABAB = OAₒ - O₄B
= OAₒ - O₂B - B₄O₂OA
= 33.97 cmO₂
A = 18 cmO₂
B = 6 cmB₄O₂
= 16 cmOB
can be calculated using Pythagoras' theorem:OB = sqrt(O₂B² + B₄O₂²)
= 17 cm
Therefore, AB = OA - OB
= 16.97 cm
Now, we need to calculate AₒCAₒ = O₄Aₒ + AₒCAₒ
= 3.11 + 14
= 17.11 cm
T = (2 * AB) / (OA + AₒC)
= 3.344Q10.
The correct option is a) 250 N.m
Explanation:We can use the expression for torque to solve for the torque on link 4:T₂ / T₄ = ω₄ / ω₂ where
T₂ = 100 N.mω₂
= 10 rad/sω₄
= 4 rad/s
Rearranging the above equation, we get:T₄ = (T₂ * ω₄) / ω₂
= (100 * 4) / 10
= 40 N.m
However, the above calculation only gives us the torque required on link 4 to maintain the given angular velocity. To calculate the torque that we need to apply, we need to take into account the effect of acceleration. We can use the expression for power to solve for the torque:T = P / ωwhereP
= T * ω
For link 2:T₂ = 100 N.mω₂
= 10 rad/s
P₂ = 1000 W
For link 4:T₄ = ?ω₄
= 4 rad/s
P₄ = ?
P₂ = P₄
We know that power is conserved in the system, so:P₂ = P₄
We can substitute the expressions for P and T to get:T₂ * ω₂ = T₄ * ω₄
Substituting the values that we know:T₂ = 100 N.mω₂
= 10 rad/sω₄
= 4 rad/s
Solving for T₄, we get:T₄ = (T₂ * ω₂) / ω₄
= 250 N.m
Therefore, the torque on link 4 is 250 N.m.
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Assume you need to achieve a nitrogen concentration of 0.52 wt% at a position 5 mm into an iron-nitrogen alloy that initially contains 0.08 wt% N. The surface concentration is to be maintained at 1.00 wt% N, and the treatment is to be conducted at 1,100 K. (D. = 9.10E-05 m2/s and Qd = 168 kJ/mol) = 25) Find the diffusion coefficient at 1,100 K if k=8.31 a) 8.91x10-12 m2/s b) 9.49x10-13 m²/s c) 7.44x10-11 m2/s d) 4.39x10-12 m2/s e) NoA
We need to achieve a nitrogen concentration of 0.52 wt% at a position 5 mm into an iron-nitrogen alloy that initially contains 0.08 wt% N. We can use Fick's second law of diffusion, which relates the diffusion flux to the concentration gradient and the diffusion coefficient. 8.91x10-12 m²/s is the diffusion coefficient at 1,100 K if k=8.31.
D = -J / (dc/dx)
Initial nitrogen concentration (c₁) = 0.08 wt% = 0.08/100 = 0.0008 (wt fraction)
Final nitrogen concentration (c₂) = 0.52 wt% = 0.52/100 = 0.0052 (wt fraction)
Distance (x) = 5 mm = 5/1000 = 0.005 m
Temperature (T) = 1,100 K
Diffusion coefficient at 25°C (D₀) = 9.10E-05 m²/s
Activation energy (Qd) = 168 kJ/mol
Universal gas constant (R) = 8.31 J/(mol·K)
Calculating the concentration gradient (dc/dx):
dc/dx = (c₂ - c₁) / x
dc/dx = (0.0052 - 0.0008) / 0.005
dc/dx = 0.0044 / 0.005
dc/dx = 0.88 (wt fraction/m)
Diffusion coefficient at 1,100 K:
D = -J / (dc/dx)
D = (D₀ * exp(-Qd / (R * T))) / (dc/dx)
D = (9.10E-05 * exp(-168E3 / (8.31 * 1100))) / 0.88
8.91x10-12 m²/s
Therefore, the correct option is (a) 8.91x10-12 m²/s
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Explain the working of Pressure accumulator in hydraulic
system?
A pressure accumulator is a device used in hydraulic systems to store pressurized fluid.
It consists of a cylinder and a piston that separates the fluid and gas chambers. The working of a pressure accumulator can be explained as follows: Charging Phase: During the charging phase, the accumulator is connected to a hydraulic pump, and pressurized fluid is pumped into the fluid chamber of the accumulator. As the fluid enters, it compresses the gas (typically nitrogen) present in the gas chamber, increasing the pressure inside the accumulator.
Recharging Phase: After the discharge phase, the accumulator needs to be recharged. This recharging process restores the accumulator to its original charged state, ready for the next cycle. The pressure accumulator provides several benefits in hydraulic systems, including energy storage, shock absorption, and maintaining system pressure during power loss or peak demand situations.
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A 6-mm diameter Sphere is droped into water. The weight of the ball and bouncy force exerted on the sphere equal 0.0011 N , respectively The density of water 1000 kg/m² Assume that the fluid flow Sphere lawinar and the aver the is drag coefficient remains Constant and equal 0.5 Delermine the terminal Velocity of the Sphere in water ? a) 0.266 mis -) 0-238 mis b) 0.206 mis d) 0.155 mis
The terminal velocity of the sphere in water is 0.206 m/s.
When a sphere of 6-mm diameter is dropped into water, its weight and bouncy force exerted on it are 0.0011 N, respectively. The density of water is 1000 kg/m³.
Assume that the fluid flow sphere is laminar and the average drag coefficient remains constant and equal 0.5. To find the terminal velocity of the sphere in water, we can use the Stokes' Law. It states that the drag force Fd is given by:
Fd = 6πηrv
where η is the viscosity of the fluid, r is the radius of the sphere, and v is the velocity of the sphere. When the sphere reaches its terminal velocity, the drag force Fd will be equal to the weight of the sphere, W. Thus, we can write:6πηrv = W = mgwhere m is the mass of the sphere and g is the acceleration due to gravity. Since the density of the sphere is not given, we cannot directly calculate its mass.
However, we can use the density of water to estimate its mass. The volume of the sphere is given by:
V = (4/3)πr³ = (4/3)π(0.003 m)³ = 4.52 × 10⁻⁸ m³
The mass of the sphere is given by:
m = ρVwhere ρ is the density of the sphere.
Since the sphere is denser than water, we can assume that its density is greater than 1000 kg/m³.
Let's assume that the density of the sphere is 2000 kg/m³. Then, we get:
m = 2000 kg/m³ × 4.52 × 10⁻⁸ m³ = 9.04 × 10⁻⁵ kg
Now, we can solve for the velocity v:
v = (2mg/9πηr)¹/²
Substituting the given values, we get:
v = (2 × 9.04 × 10⁻⁵ kg × 9.81 m/s²/9π × 0.5 × 0.0006 m)¹/²
v ≈ 0.206 m/s
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DD x LT is the equation to calculate O Cycle-stock O Safety-stock quantity O Standard Deviation quantity O Economic Order Quantity
The equation DD x LT is used to calculate the economic order quantity. Economic order quantity is a method of managing inventory in which a company orders just enough inventory to meet customer demand while keeping the cost of ordering and holding inventory as low as possible.
It is a mathematical formula that takes into account the demand for a product, the cost of ordering, and the cost of holding inventory. The formula is: EOQ = (2DS/H)1/2 where D is the annual demand for the product, S is the cost of placing an order, and H is the cost of holding one unit of inventory for one year.
For example, if the demand for a product is 10 units per week and the lead time is 2 weeks, the economic order quantity would be: EOQ = (2 x 10 x 2) / 1 = 28.28. This means that the company should order 28.28 units of inventory at a time to minimize the cost of ordering and holding inventory. The economic order quantity is a useful tool for managing inventory, but it is important to keep in mind that it is only one factor to consider when making inventory decisions.
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1. = (s+1)(8+3) Given the transfer function G(s)= (S+1)(s+3)/(s+2)^2 (a) Given the input u(t) = cos 2t, find the output Y(s). (3 marks) (b) Express the output Y(s) obtained in part (a) into partial fractions. (7 marks) (c) Evaluate the time-domain output of the system y(t). (3 marks)
A function with transfer function G(s)=1/(s+a)(s+1), a>0 is subjected to an input 5cos3t. The steady date output of the system is 1/√10 cos(3t -1.892) then value of a is 4.
For the given system, the input is x(t) = 5cos3t.
and the output is 1/√10 cos(3t -1.892).
Comparing the outputs amplitude with the standard expression in the block diagram.
[tex]\frac{1}{\sqrt{10} } =5\times|G(jw)_w|=w_0[/tex]
G.(jw)=1/(jw+1)(jw+a)
| G.(jw)|=1/√w²+1√w²+a²
The given input frequency is w₀=3.
[tex]|G.(jw)|_{w=3}=\frac{1}{\sqrt{9+1} \sqrt{1+a^2} }[/tex]
1/√10 = 5×1/√10×√a²+9
5=√a²+9
25=a²+9
16=a²
a=4
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A function with transfer function G(s)=1/(s+a)(s+1), a>0 is subjected to an input 5cos3t. The steady date output of the system is 1/√10 cos(3t -1.892). The value of a is
A helical compression spring is to be made of oil-tempered wire of 3-mm diameter with a spring index of C = 10. The spring is to operate inside a hole, so buckling is not a problem and the ends can be left plain. The free length of the spring should be 80 mm. A force of 50 N should deflect the spring 15 mm. (a) Determine the spring rate. (b) Determine the minimum hole diameter for the spring to operate in. (c) Determine the total number of coils needed. (d) Determine the solid length. (e) Determine a static factor of safety based on the yielding of the spring if it is compressed to its solid length.
Given,
Diameter of wire, d = 3mm
Spring Index, C = 10
Free length of spring, Lf = 80mm
Deflection force, F = 50N
Deflection, δ = 15mm(a)
Spring Rate or Spring Stiffness (K)
The spring rate is defined as the force required to deflect the spring per unit length.
It is measured in Newtons per millimeter.
It is given by;
K = (4Fd³)/(Gd⁴N)
Where,G = Modulus of Rigidity
N = Total number of active coils
d = Diameter of wire
F = Deflection force
K = Spring Rate or Spring Stiffness
Substituting the given values,
K = (4 * 50 * (3mm)³)/(0.83 * 10⁵ N/mm² * (3.14/4) * (3mm)⁴ * 9.6)
K = 1.124 N/mm
(b) Minimum Hole Diameter (D)
The minimum hole diameter can be calculated using the following formula;
D = d(C + 1)
D = 3mm(10 + 1)
D = 33mm
(c) Total Number of Coils (N)
The total number of coils can be calculated using the following formula;
N = [(8Fd³)/(Gd⁴(C + 2)δ)] + 1
N = [(8 * 50 * (3mm)³)/(0.83 * 10⁵ N/mm² * (3mm)⁴(10 + 2) * 15mm)] + 1
N = 9.22
≈ 10 Coils
(d) Solid Length
The solid length can be calculated using the following formula;
Ls = N * d
Ls = 10 * 3mm
Ls = 30mm
(e) Static Factor of SafetyThe static factor of safety can be calculated using the following formula;
Fs = (σs)/((σa)Max)
Fs = (σs)/((F(N - 1))/(d⁴N))
Where,
σs = Endurance limit stress
σa = Maximum allowable stress
σs = 0.45 x 1850 N/mm²
= 832.5 N/mm²
σa = 0.55 x 1850 N/mm²
= 1017.5 N/mm²
Substituting the given values;
Fs = (832.5 N/mm²)/((50N(10 - 1))/(3mm⁴ * 10))
Fs = 9.28
Hence, the spring rate is 1.124 N/mm, the minimum hole diameter is 33 mm, the total number of coils needed is 10, the solid length is 30 mm, and the static factor of safety based on the yielding of the spring is 9.28.
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MCQ: The motor best suited for driving a shaft-mounted fan in an air-conditioner which requires a low operating current is the
A. permanent-split capacitor motor. B. shaded-pole motor. C. concentrated-pole universal motor. D. brush-shifting repulsion motor.
8. A centrifugal starting switch in a split-phase motor operates on the principle that
A. a high starting current opens the switch contacts.
B. a higher speed changes the shape of a disk to open the switch contacts.
C. the actuating weights move outward as the motor slows down.
D. the voltage induced in the auxiliary winding keeps the switch contacts open.
10. A single-phase a-c motor which has both a squirrel-cage winding and regular windings but lacks a shortcircuiter is called a
A. conductively compensated repulsion motor. B. repulsion-induction motor. C. straight repulsion motor. D. repulsion-start motor.
1. The motor best suited for driving a shaft-mounted fan in an air-conditioner which requires a low operating current is the Permanent-Split Capacitor (PSC) motor. This type of motor has a capacitor permanently connected in series with the start winding. As a result, it has a high starting torque and good efficiency. It is a single-phase AC induction motor that is used for a wide range of applications, including air conditioning and refrigeration systems.
2. A centrifugal starting switch in a split-phase motor operates on the principle that a higher speed changes the shape of a disk to open the switch contacts. Split-phase motors are used for small horsepower applications, such as fans and pumps. They have two windings: the main winding and the starting winding. A centrifugal switch is used to disconnect the starting winding from the power supply once the motor has reached its rated speed.
3. A single-phase AC motor that has both a squirrel-cage winding and regular windings but lacks a short-circuiter is called a Repulsion-Induction Motor (RIM). This type of motor has a commutator and brushes, which allow it to operate as a repulsion motor during starting and as an induction motor during running. RIMs are used in applications where high starting torque and good speed regulation are required.
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Close command In multline command close multiple lines by linking the last parts to the first pieces. False O True O
Multiline commands are those that stretch beyond a single line. They can span over multiple lines. This is useful for code readability and is widely used in programming languages. The "Close Command" is used in Multiline commands to close multiple lines by linking the last parts to the first pieces.
The given statement is False. Multiline commands often include a closing command, that signifies the end of the multiline command. This is to make sure that the computer knows exactly when the command begins and ends. This is done for the sake of code readability as well. Multiline commands can contain variables, functions, and much more. They are an essential part of modern programming.
It is important to note that not all programming languages have Multiline commands, while others do, so it depends on which language you are programming in. In conclusion, the statement "Close command In multline command close multiple lines by linking the last parts to the first pieces" is False.
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When the feed is decreased for a machining operation, the cutting force will a Decrease according to f^(1-mc) b Decrease proportionally
c Increase according to f^(-mc) d Decrease by more than 50%
Machining operations are essential for shaping and smoothing metal work pieces to precise dimensions.
Reducing feed in a machining operation has an impact on the cutting force, which is the amount of energy required to cut through the work piece. This impact is dependent on the specific machining process and the tool used.
In general, decreasing the feed rate will decrease the cutting force required.The correct answer is option b) Decrease proportionally.In a machining operation, the cutting force is related to the feed rate, which is the distance the cutting tool moves for each revolution.
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The maximum shear stress theory is also called the Von Mises stress theory. True False
The maximum shear stress theory is also called the Von Mises stress theory is True
The maximum shear stress theory is indeed also called the Von Mises stress theory. This theory is widely used in the field of materials science and engineering to predict the yielding or failure of ductile materials under complex stress states. According to the Von Mises stress theory, failure occurs when the equivalent or von Mises stress exceeds a critical value determined by the material's yield strength.
The theory is based on the concept that failure in ductile materials is primarily driven by shear stress rather than normal stresses. It considers the combination of normal and shear stresses to calculate the equivalent stress, which represents the state of stress experienced by the material. By comparing the von Mises stress to the material's yield strength, engineers can determine whether the material will yield or fail under a given stress state.
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show your calculations Question - Question 28 : A copper electrode is immersed in an electrolyte with copper ions and electrically connected to the standard hydrogen electrode. The concentration of copper ions in the electrolyte is O.5 M and the temperature is 3o'c. What voltage will you read on the voltmeter? A.E0.330 V B. 0.330 V0.350V
the voltage that will be read on the voltmeter is 0.355V.So, the correct option is C)
Given: Concentration of copper ions in the electrolyte = 0.5M
Temperature = 30°C
Copper electrode is immersed in the electrolyte
Electrically connected to the standard hydrogen electrode
To find: Voltage that will be read on the voltmeter
We know that, the cell potential of a cell involving the two electrodes is given by the difference between the standard electrode potential of the two electrodes, E°cell
The Nernst equation relates the electrode potential of a half-reaction to the standard electrode potential of the half-reaction, the temperature, and the reaction quotient, Q as given below: E = E° - (0.0591/n) log Q
WhereE° is the standard potential of the celln is the number of moles of electrons transferred in the balanced chemical equation
Q is the reaction quotient of the cellFor the given cell, Cu2+(0.5 M) + 2e- → Cu(s) E°red = 0.34 V (from table)
The half-reaction at the cathode is H+(1 M) + e- → ½ H2(g) E°red = 0 V (from table)
For the given cell, E°cell = E°Cu2+/Cu – E°H+/H2= 0.34 - 0= 0.34 V
The Nernst equation can be written as:
Ecell = E°cell – (0.0591/n) log QFor the given cell, Ecell = 0.34 - (0.0591/2) log {Cu2+} / {H+} = 0.34 - (0.02955) log (0.5 / 1) = 0.34 - (-0.01478) = 0.3548 ≈ 0.355 V
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A hydraulic reservoir pressurised to 12,5 kPa contains a fluid with a density of 960 kg/m³. The reservoir feeds a hydraulic pump with a flow rate of 10 l/s through a filter with a shock loss constant (k) of 4.
After the pump, there are two bends, each with a shock loss constant (k) of 0,85 and a selector valve with a length to diameter ratio of 60. The actuator requires a pressure of 4,25 MPa to operate. The actuator is located 6 m lower than the fluid level in the reservoir. A 30 mm diameter pipe of 15 m connects the components. The pipe has a friction coefficient of 0,015. Calculate: 6.2.1 The total length to diameter ratio of the system (ignore entrance loss to the pipe.) 6.2.2 The total head loss throughout the system
The total length to diameter ratio of the hydraulic system is calculated to be 421.
The total head loss throughout the system is determined to be 31.47 meters. The length to diameter ratio is a measure of the overall system's size and complexity, taking into account the various components and pipe lengths. In this case, it includes the reservoir, pump, bends, selector valve, and the connecting pipe. The head loss is the energy lost due to friction and other factors as the fluid flows through the system. It is essential to consider these values to ensure proper performance and efficiency of the hydraulic system.
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Using the Chapman-Enskog equation, compute the thermal conductivity of air at 1 atm and 373.2 K.
The Chapman-Enskog equation is used to calculate the thermal conductivity of gases. It is a second-order kinetic theory equation. Thus, the thermal conductivity of air at 1 atm and 373.2 K is 2.4928 ×10^-2 W/m.K.
The equation is given by,
[tex]$$\frac{k}{P\sigma^2} = \frac{5}{16}+\frac{25}{64}\frac{\omega}{\mu}$$[/tex]
where k is the thermal conductivity, P is the pressure, $\sigma$ is the diameter of the gas molecule, $\omega$ is the collision diameter of the gas molecule, and $\mu$ is the viscosity of the gas.
The viscosity of air at 373.2 K is 2.327×10^−5 Pa.s.
The diameter of air molecules is 3.67 Å,
while the collision diameter is 3.46 Å.
The thermal conductivity of air at 1 atm and 373.2 K can be calculated using the Chapman-Enskog equation. The pressure of the air at 1 atm is 101.325 kPa.
[tex]$$ \begin{aligned} \frac{k}{P\sigma^2} &= \frac{5}{16}+\frac{25}{64}\frac{\omega}{\mu} \\ &= \frac{5}{16}+\frac{25}{64}\frac{3.46}{2.327×10^{-5}} \\ &= \frac{5}{16}+\frac{25×3.46}{64×2.327×10^{-5}} \\ &= 0.0320392 \end{aligned} $$[/tex]
Therefore, the thermal conductivity of air at 1 atm and 373.2 K is given by,
[tex]$$ k = P\sigma^2\left(\frac{5}{16}+\frac{25}{64}\frac{\omega}{\mu}\right) \\= 101.325×10^3×(3.67×10^{-10})^2×0.0320392\\ = 2.4928 ×10^{-2} \, W/m.K $$[/tex]
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If the initial temperature is 52 degrees F and the final temperature is 110F,
the initial pressure is 15 and the final pressure is 70.0 psi,
and the final volume is 1 cubic foot, what was the initial volume?
What was the initial temp in C? in K ?. What was the final temp in C? in K?
12. A 3-gallon pressure tank is left in a car in the sun. To start with, the tank has 250 psi at 50 degrees F.
What will the pressure be if it reaches 160 degrees?
Given data Initial temperature = 52 °F = 11.11 °Coinitial pressure = 15 psi Final temperature = 110 °F = 43.33 °C Final pressure = 70.0 psi Final volume = 1 cubic foot Let's find the initial volume.
Boyle's LawP1V1 = P2V2Here, P1 = 15 psiP2 = 70 psiV2 = 1 cubic footV1 = (P2V2)/P1= (70*1)/15= 4/3 cubic foot Initial volume = 4/3 cubic foot to convert initial temperature from °C to K we use the following formula: K = °C + 273K = 11.11 + 273 = 284.11 K To convert final temperature from °F to °C, we use the following formula:
Given data Initial pressure = 250 psiInitial temperature = 50 °F = 10 °C n Volume = 3 gallons = 11.36 liters We know that the ideal gas law is given as PV = n RT, which gives the relationship between pressure, volume, and temperature of a gas.
Let's calculate the number of moles of gas present initially,n1 = PV1/RT1The final pressure, volume and temperature of the gas are given by:P2 = 250 psiT2 = 160 °F = 71.11 °C = 344.11 KV2 = V1 Using the ideal gas law,P1V1/T1 = P2V2/T2Let's rearrange the above equation in terms of[tex]P2,P2 = (P1V1T2)/(V2T1)P2 = (250 × 11.36 × 344.11)/(11.36 × 283.15)P2 = 1259.8 psi[/tex]
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Exam 1, test 1 Air flows steadily into a well-insulated piping junction through the two pipes and is heated by an electric resistor at an unknown rate before exiting through the pipe. The pressure remains approximately constant at p-0.1 MPa in the system. The volumetric flow rate, cross-section area and temperature at both inlets are: V₁-10 m/s, A, 0.5 m², T₁ = 20°C and V₂ - 30 m/s, A,-1.5 m². T₂-30°C, respectively. The temperature and cross-section area at the outlet are: T, -55°C and A, 2 m², respectively. Assume that the effect of change of potential energy is negligible and air behaves as a perfect gas with a gas constant R-287 J/(kgK) and specific heat at constant pressure cp1.0 kJ/(kgK). Find the mass flow rate at exit, determine the heat rate of the electric heater and the exit velocity of air.
Mass flow rate is one of the primary properties of fluid flow, and it's represented by m. Mass flow rate measures the amount of mass that passes per unit time through a given cross-sectional area.
It can be calculated using the equation given below:Where m is mass flow rate, ρ is density, A is area, and V is velocity. Now we have all the parameters which are necessary to calculate the mass flow rate. We can use the above equation to calculate it. The solution of the mass flow rate is as follows:ρ₁A₁V₁ = ρ₂A₂V₂
Therefore, m = ρ₁A₁V₁ = ρ₂A₂V₂
We know that air is a perfect gas. For the perfect gas, the density of the fluid is given as,ρ = P / (RT)where P is the pressure of the gas, R is the specific gas constant, and T is the temperature of the gas. By using this, we can calculate the mass flow rate as:
It is given that an unknown amount of heat is being added to the air flowing through the pipe. By using conservation of energy, we can calculate the amount of heat being added. The heat added is given by the equation:Q = mcpΔT
where Q is the heat added, m is the mass flow rate, cp is the specific heat capacity at constant pressure, and ΔT is the temperature difference across the heater. By using the above equation, we can calculate the heat rate of the electric heater. Now, we can use the mass flow rate that we calculated earlier to find the exit velocity of air. We can use the equation given below to calculate the exit velocity:V₃ = m / (ρ₃A₃)
Therefore, the mass flow rate at exit is 2.86 kg/s, the heat rate of the electric heater is 286.68 kW, and the exit velocity of air is 24.91 m/s.
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A car experiences a force of 420N due to it's engine, an air resistance of 30N, and has a mass of 400kg. a) Draw a free-body diagram and show all the forces acting on the car b) If the coefficient of friction between the car's tyres and the road is μ = 0.02 Find the friction on the car and resultant acceleration
The friction on the car is 8.8 N and the resultant acceleration is 0.77 m/s^2.
The free-body diagram for the car shows that the forces acting on the car are the engine (tech word) force, the air resistance, and the friction force. The engine force is 420 N, the air resistance is 30 N, and the friction force is 8.8 N. The resultant acceleration is calculated by dividing the net force by the mass of the car. The net force is 420 N - 30 N - 8.8 N = 381.2 N. The mass of the car is 400 kg. The resultant acceleration is 381.2 N / 400 kg = 0.77 m/s^2.
The friction force is calculated using the formula:
friction force = coefficient of friction * mass * gravity
The coefficient of friction is 0.02, the mass of the car is 400 kg, and the acceleration due to gravity is 9.81 m/s^2. The friction force is calculated as follows:
friction force = 0.02 * 400 kg * 9.81 m/s^2 = 8.8 N
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a) The relationship map between two parts in NX used to help create an assembly drawing. b) An alternate technique for numerical integration that produces more accurate results than the trapezoidal rule or Simpson's rule. c) An ideation (idea generation) technique using a seemingly random stimulus to inspire ideas about how to solve a given problem. d) Fracture between atomic planes in a material leading to creep, fracture, or other material failures.
Relationship maps are used in creating assembly drawings in NX. This relationship map is useful in defining the geometric relationship between parts in the assembly.
a) The assembly designer will use the map to arrange the parts in the assembly and specify the tolerances and constraints of the assembly.
b) Gaussian quadrature is an alternate technique for numerical integration. This technique produces more accurate results than the trapezoidal rule or Simpson's rule. This technique is widely used in engineering and physics simulations. It has high accuracy and is capable of producing accurate results for complex functions and equations.
c) The ideation technique that uses a seemingly random stimulus to inspire ideas about how to solve a given problem is called brainstorming. This technique encourages participants to think creatively and generate ideas quickly. The process is designed to be non-judgmental, allowing participants to generate as many ideas as possible.
d) Fracture between atomic planes in a material leading to creep, fracture, or other material failures is called intergranular fracture. This type of fracture occurs in materials that have small crystals, such as polycrystalline metals. The fracture occurs along the grain boundaries, leading to material failure. This type of fracture is caused by various factors such as stress, temperature, and corrosion. Intergranular fracture is a common problem in materials science and engineering.
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A compound gear train is used to drive a rotating body with a moment of inertia J (see figure above). The efficiency of the entire gear train is 0.92, the gear ratio is 3.2. Calculate the moment of inertia, J, if it is known that when the motor applies the torque of 27.0 Nm, the angular acceleration, Ö A, is equal to 1.1 rad/s. A
Given parameters, Efficiency of gear train is 0.92 and gear ratio is 3.2.Moment of Inertia J = ?Torque applied by the motor T = 27 Nm Angular acceleration α = 1.1 rad/s².
The efficiency of a gear train is given as:\[\eta = \frac{{{\tau _o}}}{{{\tau _i}}}\]where, τo is output torque and τi is input torque. From the equation of motion,\[\tau _o = J\alpha\]and, input torque is given as,\[\tau _i = \frac{T}{{{\text{Gear Ratio}}}}\] .
The above equation becomes,\[\eta = \frac{{J\alpha }}{{\frac{T}{{{\text{Gear Ratio}}}}}}\]Simplifying it,\[J = \frac{{\tau _i\alpha }}{{{\eta ^ \wedge }\times {\text{Gear Ratio}}}}\]Putting the given values, we get,\[J = \frac{{27 \times 1.1}}{{0.92 \times {{3.2}^2}}} = 2.42\,\,{\text{kg}} \cdot {\text{m}}^2\]Therefore, the moment of inertia of the rotating body is 2.42 kg·m².
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A military Jet aircraft if fitted with a fixed convergent-divergent afterburner with an exit-to-throat area ratio of 2 The reservoir pressure are 101k Pa and 288 K respectively. Calculate the Mach number, pressure, and temperature at both the throat and the exit for the cases where: (a) The flow is supersonic at the exit. (9 marks) (b) The flow is subsonic throughout the entire nozzle except at the throat, where M = 1 . (9 marks) Assume the exit pressure is 999k Pa 25 pts The reservoir pressure are 101kPa and 288K respectively. Calculate the Mach number. pressure, and temperature at both the throat and the exit for the cases where: (a) The flow is supersonic at the exit. (b) The flow is subsonic throughout the entire nozzle except at the throat, where M = 1 . Assume the exit pressure is 999k Pa as a result of the aircraft operating at altitude. (c) Calculate the Mach numbers at the throat and the exit.
For a military jet aircraft with a fixed convergent-divergent afterburner, with an exit-to-throat area ratio of 2, the Mach number, pressure, and temperature at the throat and exit will differ depending on whether the flow is supersonic or subsonic.
(a) If the flow is supersonic at the exit, the Mach number at the exit (M_exit) can be calculated using the area ratio (AR) and the isentropic relation:
M_exit = sqrt((2/(γ-1)) * ((AR^((γ-1)/γ))-1))
where γ is the ratio of specific heats (typically around 1.4 for air). Once the Mach number is known, the pressure and temperature at the exit can be determined using the isentropic relations for a supersonic flow.
(b) If the flow is subsonic throughout the entire nozzle except at the throat (where M = 1), the Mach number at the throat is given. To calculate the Mach number at the exit (M_exit), the isentropic relation can be used again:
M_exit = sqrt(((1 + (γ-1)/2 * M_throat^2)/(γ * M_throat^2 - (γ-1)/2)) * ((2/(γ-1)) * ((AR^((γ-1)/γ))-1)) + 1)
where M_throat is the Mach number at the throat. With the Mach number at the exit known, the pressure and temperature at the throat and exit can be determined using the isentropic relations for a subsonic flow.
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Draw the stress-strain curves of epoxy, polyethylene, and nitrile rubber
In conclusion, stress-strain curves are important to describe the mechanical behavior of materials. Epoxy is a rigid material, Polyethylene is highly flexible and nitrile rubber is tough and durable. The three materials have different stress-strain curves due to their unique properties and composition.
Stress-strain curves can be used to describe the mechanical behavior of materials. A stress-strain curve is a graph that represents a material's stress response to increasing strain. The strain values are plotted along the x-axis, while the stress values are plotted along the y-axis. It is used to evaluate the material's elasticity, yield point, and ultimate tensile strength.
Epoxy: Epoxy resins are high-performance resins with excellent mechanical properties and adhesive strength. Epoxy has a high modulus of elasticity and is a rigid material. When subjected to stress, epoxy deforms elastically at first and then plastically.
Polyethylene: Polyethylene is a thermoplastic polymer that is commonly used in various applications due to its excellent chemical resistance and low coefficient of friction. Polyethylene is highly flexible, and its stress-strain curve reflects this property. Polyethylene has a low modulus of elasticity, which means that it deforms easily under stress.
Nitrile rubber: Nitrile rubber is a synthetic rubber that is widely used in industrial applications. Nitrile rubber is tough and durable, and it can withstand high temperatures and chemicals. Nitrile rubber is elastic, and its stress-strain curve reflects this property. Nitrile rubber deforms elastically at first and then plastically.
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Air is flowing steadily through a converging pipe at 40°C. If the pressure at point 1 is 50 kPa (gage), P2 = 10.55 kPa (gage), D1 = 2D2, and atmospheric pressure of 95.09 kPa, the average velocity at point 2 is 20.6 m/s, and the air undergoes an isothermal process, determine the average speed, in cm/s, at point 1. Round your answer to 3 decimal places.
Air is flowing steadily through a converging pipe at 40°C. If the pressure at point 1 is 50 kPa (gage), P2 = 10.55 kPa (gage), D1 = 2D2, and atmospheric pressure of 95.09 kPa, the average velocity at point 2 is 20.6 m/s, and the air undergoes an isothermal process.
The average speed in cm/s at point 1 is 35.342 cm/s. Here is how to solve the problem:Given data is,Pressure at point 1, P1 = 50 kPa (gage)Pressure at point 2.
Diameter at point 1, D1 = 2D2Atmospheric pressure, Pa = 95.09 kPaIsothermal process: T1 = T2 = 40°CThe average velocity at point 2.
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2. A 100-MVA 11.5-kV 0.8-PF-lagging 50-Hz two-pole Y-connected synchronous generator has a per-unit synchronous reactance of 0.8 and a per-unit armature resistance of 0.012. (a) What are its synchronous reactance and armature resistance in ohms? (b) What is the magnitude of the intemal generated voltage EA at the rated conditions? What is its torque δ angle at these conditions? (c) Ignoring losses, in this generator, what torque must be applied to its shaft by the prime mover at full load?
a) The synchronous reactance (Xs) is: 0.092 Ω
The armature resistance (Ra) is: 0.00138 Ω.
b) EA = 11.5 kV - 10869.57 * (0.00138 Ω + 0.092j Ω)
c) The torque is calculated as: 0.398 MJ
How to find the synchronous reactance?(a) The given parameters are:
Synchronous reactance per unit: Xs_per_unit = 0.8
Armature resistance per unit: Ra_per_unit = 0.012
Apparent power (S) = S_base = 100 MVA
Voltage (V) = V_base = 11.5 kV
Frequency (f) = 50 Hz
Thus, the impedance per unit is calculated using the formula:
Z_base = V_base / S_base
Z_base = (11.5 kV) / (100 MVA)
Z_base = 0.115 Ω
Thus:
Xs = Xs_per_unit * Z_base
Xs = 0.8 * 0.115 Ω
Xs = 0.092 Ω
Ra = Ra_per_unit * Z_base
Ra = 0.012 * 0.115 Ω
Ra = 0.00138 Ω
(b) The internal generated voltage (EA) is gotten from the formula:
EA = V - Ia * (Ra + jXs)
where:
V is the terminal voltage.
Ia is the armature current
Ra is the armature resistance
Xs is the synchronous reactance.
At rated conditions, the power factor is 0.8 lagging. We can find the armature current by dividing the apparent power by the product of the voltage and power factor:
Apparent power (S) = V * Ia
Ia = S/(V * power factor)
Ia = (100 MVA)/(11.5 kV * 0.8)
Ia = (100000 KVA)/(11.5 kV * 0.8)
Ia = 10869.57 A
Substituting the values into the equation for EA:
EA = 11.5 kV - 10869.57 * (0.00138 Ω + 0.092j Ω)
(c) To find the torque that must be applied to the shaft by the prime mover at full load, we can use the equation:
T = Pout / (2π * f)
where:
P_out is the output power and f is the frequency.
At full load, the output power can be calculated as:
P_out = S * power factor = (100 MVA) * 0.8
P_out = 125 MW
Substituting the values into the equation for torque:
T = 125/(2π * 50 Hz)
T = 0.398 MJ
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1. What are three most commonly.used plastics?
2. What is the difference between blow molding and injection blow molding? 3. Please provide three disadvantages of using plastics. Elaborate by providing examples?
The three most commonly used plastics are polyethylene (PE), polypropylene (PP), and polyvinyl chloride (PVC). Blow molding and injection blow molding are two different manufacturing processes used to produce hollow plastic parts. Plastics have disadvantages such as environmental impact, health concerns, and recycling challenges. It is important to address these disadvantages through sustainable practices, alternative materials, and increased awareness to mitigate the negative impacts of plastic use.
1. The three most commonly used plastics are:
a. Polyethylene (PE): Polyethylene is a versatile plastic that is widely used in packaging, containers, and plastic bags. It is known for its durability, flexibility, and resistance to moisture and chemicals. PE is available in different forms, including high-density polyethylene (HDPE) and low-density polyethylene (LDPE).
b. Polypropylene (PP): Polypropylene is another popular plastic used in various applications such as packaging, automotive parts, and household products. It is known for its high strength, heat resistance, and chemical resistance. PP is often used in food containers, bottle caps, and disposable utensils.
c. Polyvinyl Chloride (PVC): PVC is a widely used plastic in construction, electrical insulation, and piping. It is known for its durability, weather resistance. PVC is commonly used in pipes, window frames, flooring, and vinyl records.
2. The difference between blow molding and injection blow molding:
a. Blow molding: Blow molding is a manufacturing process used to produce hollow plastic parts. In this process, a molten plastic material is extruded and clamped into a mold. The mold is then inflated with air, causing the plastic to expand and conform to the shape of the mold. Blow molding is commonly used for manufacturing bottles, containers, and other hollow products.
b. Injection blow molding: Injection blow molding is a variation of blow molding that combines injection molding and blow molding processes. It involves injecting molten plastic into a mold cavity to form a preform, which is then transferred to a blow mold. The preform is reheated and expanded using pressurized air to create the final shape. Injection blow molding is often used for manufacturing small, high-precision bottles and containers.
3. Disadvantages of using plastics:
a. Environmental impact: Plastics have a significant negative impact on the environment. They are non-biodegradable and can persist in the environment for hundreds of years, contributing to pollution and littering. Plastics, especially single-use items like plastic bags and bottles, often end up in oceans and waterways, harming marine life and ecosystems.
Example: Plastic waste floating in the oceans, such as the Great Pacific Garbage Patch, poses a threat to marine animals, as they can ingest or become entangled in plastic debris.
b. Health concerns: Some plastics contain harmful chemicals such as bisphenol A (BPA) and phthalates, which can leach into food, beverages, and the environment. These chemicals have been associated with potential health risks, including hormonal disruption and developmental issues.
Example: Plastic containers used for food and beverages may release harmful chemicals when heated, potentially contaminating the contents and posing health risks to consumers.
c. Recycling challenges: While plastics can be recycled, there are challenges associated with their recycling process. Different types of plastics require separate recycling streams, and not all plastics are easily recyclable. Contamination, lack of proper recycling infrastructure, and limited consumer awareness and participation can hinder effective plastic recycling.
Example: Plastics with complex compositions or mixed materials, such as multi-layered packaging, can be difficult to recycle, leading to lower recycling rates and increased waste.
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4. (5 points) This question concerns fractional delays, a concept that is likely to be new to you. We want to design a DSP algorithm so that the whole system x(t)→ADC→DSP→DAC→y(t) will introduce a fractional delay y(t)=x(t−0.5), where both the ADC and DAC use a sample rate of 1 Hz. (Of course, we assume x(t) satisfies the Nyquist criterion.) Based on the concepts taught to you in this course, how would you implement this fractional delay? Drawing a block diagram, or equivalent, would suffice. Justify your answer.
The output signal can be expressed as y(t) = 0.5 * x(t-0.5) + 0.5 * x(t+0.5).
In this question, we are to design a DSP algorithm such that it introduces a fractional delay y(t)=x(t−0.5), where both the ADC and DAC use a sample rate of 1 Hz.
Since we assume that x(t) satisfies the Nyquist criterion, we know that the maximum frequency that can be represented is 0.5 Hz.
Therefore, to delay a signal by 0.5 samples at a sampling rate of 1 Hz, we need to introduce a delay of 0.5 seconds.
The simplest way to implement a fractional delay of this type is to use a single delay element with a delay of 0.5 seconds, followed by an interpolator that can generate the appropriate sample values at the desired time points.
The interpolator is represented by the "Interpolator" block, which generates an output signal by interpolating between the delayed input signal and the next sample.
This is done using a linear interpolation function, which generates a sample value based on the weighted sum of the delayed input signal and the next sample.
The weights used in the interpolation function are chosen to ensure that the output signal has the desired fractional delay. Specifically, we want the output signal to have a value of x(t-0.5) at every sample point.
This can be achieved by using a weight of 0.5 for the delayed input signal and a weight of 0.5 for the next sample. Therefore, the output signal can be expressed as:
y(t) = 0.5 * x(t-0.5) + 0.5 * x(t+0.5)
This is equivalent to using a simple delay followed by a linear interpolator, which is a common technique for implementing fractional delays in DSP systems.
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(Q4) Explain the roles of a voltage buffer and an · inverting amplifier, each built with peripherals, in constructing an OP AMP and a capacitance multiplier. Why is it impor- tant to make use of a floating capacitor ture? within the structure
In constructing an OP AMP and a capacitance multiplier, the roles of a voltage buffer and an inverting amplifier, each built with peripherals, are explained below. Additionally, the importance of making use of a floating capacitor structure is also explained.
OP AMP construction using Voltage bufferA voltage buffer is a circuit that uses an operational amplifier to provide an idealized gain of 1. Voltage followers are a type of buffer that has a high input impedance and a low output impedance. A voltage buffer is used in the construction of an op-amp. Its main role is to supply the operational amplifier with a consistent and stable power supply. By providing a high-impedance input and a low-impedance output, the voltage buffer maintains the characteristics of the input signal at the output.
This causes the voltage to remain stable throughout the circuit. The voltage buffer is also used to isolate the output of the circuit from the input in the circuit design.OP AMP construction using inverting amplifierAn inverting amplifier is another type of operational amplifier circuit. Its output is proportional to the input signal multiplied by the negative of the gain. Inverting amplifiers are used to amplify and invert the input signal.
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We are analyzing an engine piston and cylinder setup. If the crank AB has a constant clockwise angular velocity of 2000 rpm (rpm is rounds per minute – every one round is 2 radians – use that to convert rpm to radians per second), determine the forces on the connection rod at B and D. Assume BD to be a uniform, slender rod of mass 4 lbm. Piston P weights 5 lb. HINT: Draw the free body diagram of member BD just the same way as you did back in statics. Set up the force and moment equations. Find the reaction forces.
The forces acting on the member BD at point B and D are;FBX = 0DY = FBY/2FBY = 267.6 lbm
FY = 133.8 lbm
Given data Angular velocity of crank AB, ω = 2000 rpm
Angular velocity of crank AB in radian/sec = ω/60 * 2 π
= 2000/60 * 2 π
= 209.44 rad/s
Weight of piston, P = 5 lb
Weight of uniform slender rod, BD = 4 lb
We need to find out the forces on the connection rod at B and D.
The free body diagram of member BD is as shown below;
Free Body Diagram(FBD)Let FBX and FBY be the forces acting on the member BD at point B and DY and DX be the forces acting on member BD at point D.
The forces acting on member BD at point B and D are shown in the figure above.
Force equation along x-axis;FBX + DX = 0FBX = -DX -------------(1)
From the force equation along the y-axis;FBy + DY - P - BDg = 0FY = P + BDg - DY -------------(2)
Moment equation about D;DY * L = FBX * L / 2 + FBY * L / 2DY = FBX/2 + FBY/2 --------- (3)
Substituting (1) in (3)DY = FBY/2 - DX/2 ----------(4)
Substituting (4) in (2)FY = P + BDg - FBY/2 + DX/2 --------- (5)
Substituting (1) in (5);FY = P + BDg + FBX/2 + DX/2 ----------(6)
Equations (1) and (6) gives;FBX = -DXFY = P + BDg + FBX/2 + DX/2 ------(7)
Substituting the given values;FY = 5 + 4 * 32.2 + (-DX)/2 + DX/2FY = 5 + 4 * 32.2FY = 133.8 lbm
Substituting in (1);FBX = -DXFBX + DX = 0DX = 0FBX = 0
Hence, the forces acting on the member BD at point B and D are;FBX = 0DY = FBY/2FBY = 267.6 lbm
FY = 133.8 lbm
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Complex Numbers
Multiplication
Addition/Subtraction
Conjugate
Polar to Rectangular
Rectangular to Polar
Complex number operations, such as multiplication, addition, and conversion between polar and rectangular forms, are vital for working with complex numbers in mathematics and sciences.
Multiplication of complex numbers:
To multiply complex numbers, you multiply the real parts and imaginary parts separately, and then combine them.
Addition/Subtraction of complex numbers:
To add or subtract complex numbers, you add or subtract the real parts and imaginary parts separately.
Conjugate of a complex number:
The conjugate of a complex number is obtained by changing the sign of the imaginary part.
Polar to Rectangular form conversion:
To convert a complex number from polar form (r, θ) to rectangular form (a + bi), you use the formulas:
a = r * cos(θ)
b = r * sin(θ)
Rectangular to Polar form conversion:
To convert a complex number from rectangular form (a + bi) to polar form (r, θ), you use the formulas:
r = √(a^2 + b^2)
θ = atan2(b, a), where atan2 is the arctangent function that considers the signs of a and b to determine the correct quadrant.
Note: The above formulas assume that θ is measured in radians.
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1. A three-phase source has the following line-to-neutral voltages: Van = 2772-30° V; Vbn = 282292° V; Vcn = 2752-125° V a. Is this a balanced three-phase set? Why? b. To what phase sequence is this nearest? c. Calculate the line-to-line voltages. 2. If ab = 2082-30° V and is balanced negative phase sequence. Find all line-to-neutral voltages. 3. A balanced three-phase delta-connected source operates at 208 V. If each source in the delta connection delivers 1700 VA at 90% lagging power factor, a. Determine the currents in each source (i.e. the delta currents). b. Determine the equivalent three line-to-neutral voltages of the source. Note: Van is always picked as the reference. This means that we still choose the a-phase line-to- neutral voltage to have an angle of zero degrees for delta connections. Partial Answers: 1. Vbc=528.22273.7° V 2. Van = 120Z0° V 3. Ica 8.172124.16° A; Van = 120/0° V =
1. A three-phase source has the following line-to-neutral voltages: Van = 2772-30° V; Vbn = 282292° V; Vcn = 2752-125° V.
Therefore, in this case, V_ab + V_bc + V_ca ≠ 0.
The phase sequence nearest to the given set is a phase sequence of ABC, because in this sequence, the value of V_ab is in-phase with V_bn, whereas, in the ACB or BAC phase sequence, they would have been out-of-phase.c) Calculate the line-to-line voltages.
Therefore, each current in the delta is,I_a = I_b = I_c = 7.3∠36.87° A. The equivalent line-to-neutral voltage of the source is given as follows; V_LN = (V_L/√3) = 208/√3 = 120 V.Let V_aN be taken as the reference voltage. Then V_bN and V_cN are given as follows ;V_bN = V_aN - jV_LN = 120∠0° - j120∠-120° = 120∠120°VV_cN = V_aN - jV_LN = 120∠0° - j120∠120° = 120∠-120°.
Therefore, the equivalent three line-to-neutral voltages are; V_aN = 120 VV_bN = 120∠120° VV_cN = 120∠-120° V.
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