Engineering College

5. (20 points) A conducting material of uniform thickness h and conductivity σ has the shape of a quarter of a flat circular washer, with inner radius a and outer radius b, as shown (b>a). Find the resistance between the curved sides (rho=a and rho=b). Assume V(rho=a)=0, V(rho=b)=V 0. V is a function of rho only. Cylindrical coordinates: ∇ 2V= rho1∂rho∂[rho ∂rho∂V]+ rho21∂ϕ 2∂ 2V+∂z 2∂ 2V
(a) Find V by solving the Laplace's equation ∇ 2
V=0 using boundary values. (b) Find the electric field intensity E from E=−∇V. (c) Find I from I=∫J⋅ds=σ∫E⋅ds. Notice that ds=−hrhodϕa rho (d) Find the resistance R. (e) Find the numerical values of R if a=0.001 m, b=0.0011 m, h=0.001 m,σ=8 s / m.

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Answer 1

(a) The solution for V is V = (V₀ / ln(b/a)) ln(ρ) - (V₀ ln(a) / ln(b/a))

(b) The electric field intensity is E = -(V₀ / ln(b/a)) (1/ρ) rho_hat

(d) The resistance R = 2 / (πσh^2)

(e) The numerical value of the resistance R is approximately 795.77 ohms.

(a) Find V by solving Laplace's equation ∇^2V = 0 using boundary values.

In cylindrical coordinates, Laplace's equation becomes:

∇^2V = (1/rho)∂rho(ρ ∂V/∂rho) + (1/rho^2)∂^2V/∂ϕ^2 + ∂^2V/∂z^2 = 0

Since V is independent of z and ϕ, we can simplify the equation to:

(1/rho)∂rho(ρ ∂V/∂rho) = 0

Integrating once with respect to rho gives:

(ρ ∂V/∂rho) = c₁

Integrating again with respect to rho gives:

V = c₁ ln(ρ) + c₂

Using the boundary conditions V(ρ=a) = 0 and V(ρ=b) = V₀, we can solve for c₁ and c₂:

0 = c₁ ln(a) + c₂ (1)

V₀ = c₁ ln(b) + c₂ (2)

Subtracting equation (1) from equation (2) gives:

V₀ = c₁ ln(b/a)

Solving for c₁:

c₁ = V₀ / ln(b/a)

Substituting this value back into equation (1) gives:

0 = V₀ / ln(b/a) ln(a) + c₂

Solving for c₂:

c₂ = -V₀ ln(a) / ln(b/a)

Therefore, the solution for V is:

V = (V₀ / ln(b/a)) ln(ρ) - (V₀ ln(a) / ln(b/a))

(b) Find the electric field intensity E from E = -∇V.

The electric field intensity can be calculated using the gradient of V:

E = -∇V = - (∂V/∂rho) rho_hat

Taking the derivative of V with respect to rho:

∂V/∂rho = (V₀ / ln(b/a)) (1/ρ)

Therefore, the electric field intensity is:

E = -(V₀ / ln(b/a)) (1/ρ) rho_hat

To find the current I, we integrate the dot product of the current density J and an infinitesimal surface element ds:

I = ∫J⋅ds = σ∫E⋅ds

Considering the surface element ds = -hρ dϕ dz rho_hat, and the electric field E = -(V₀ / ln(b/a)) (1/ρ) rho_hat, we can substitute these values into the integral:

I = σ ∫E⋅ds = -σ(V₀ / ln(b/a)) ∫(1/ρ)(-hρ dϕ dz) = σ(V₀ h) ∫dϕ ∫dz

Since we are integrating over the entire surface, the limits of integration for ϕ are 0 to π/2, and for z are 0 to h. Performing the integration:

I = σ(V₀ h) (ϕ=0 to π/2) (z=0 to h) = σ(V₀ h) (π/2) (h)

Simplifying:

I = (π/2)σV₀h^2

(d) Find the resistance R.

Resistance (R) is defined as the ratio of voltage (V₀) to current (I):

R = V₀ / I = V₀ / [(π/2)σV₀h^2]

Simplifying:

R = 2 / (πσh^2)

(e) Find the numerical values of R if a = 0.001 m, b = 0.0011 m, h = 0.001 m, and σ = 8 S/m.

Substituting the given values into the expression for resistance (R):

R = 2 / (πσh^2) = 2 / (π(8)(0.001)^2) ≈ 795.77 Ω

Therefore, the numerical value of the resistance R is approximately 795.77 ohms.

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Related Questions

Determine if there exists a unique solution to the third order linear differential ty" + 3y"+1/t-1y'+eᵗy =0 with the initial conditions a) y(1) = 1, y'(1) = 1, y" (1) = 2. b) y(0) = 1 y'(0) = 0, y" (0) = 1 c) y (2) = 1, y' (2) = -1, y" (2) = 2

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Given [tex]y" + 3y' + (1 / (t - 1)) y' + e^t y = 0[/tex]. To determine if there exists a unique solution to the third order linear differential equation.

We will use the Cauchy-Euler equation to solve this differential equation. The Cauchy-Euler equation is defined as: ay" + by' + cy = 0There exists a unique solution to the differential equation in the form of Cauchy-Euler equation if the roots of the characteristic equation are real and distinct.

In general, for a Cauchy-Euler equation, the solution is of the form y = x^n, and its derivatives are as follows: y' = nx^(n-1), y'' = n(n-1)x^(n-2), and so on. Substituting the above derivatives into the given equation, we get, [tex]t^(2) e^t y + 3t e^t y' + e^ t y' + e^ t y = 0t^(2) e^t y + e^t (3t y' + y) = 0t^2 + 3t + 1/t[/tex]- 1 = 0We have the characteristic equation.

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Proof testing is a very practical way for the integrity assessment of a structure or a component prone to failure caused by fatigue crack propagation, when a proof load, Pproofs clearly higher than the peak, Pmax, of cyclic load in operation, is applied at the proof testing. For the structure or component that passes the proof test, it is concluded that the structure or component can continue operate safely under the cyclic load in operation for a further period of life time (e.g., 10 years) until the next time of the proof testing. Assuming Pmax and Pmin of the cyclic load in operation are constant and Kic of the material is available, articulate the principle and key steps in quantitatively defining Pproof, addressing the critical crack length, ac, at Pmax, required lift time, Nif, etc. [10 marks].

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Proof testing is an efficient approach used to evaluate the integrity of structures and components that are susceptible to fatigue crack propagation. When a proof load, Pproof, which is significantly higher than the peak of cyclic load in operation, Pmax, is applied at the time of proof testing, it identifies whether the component can continue to function safely under the cyclic load for a prolonged period.

In order to quantitatively define Pproof, it is crucial to address the following critical factors: the maximum and minimum cyclic load in operation, Pmax and Pmin, respectively, the critical crack length at Pmax, ac, and the required lift time, Nif, and Kic of the material. The key steps in quantitatively defining Pproof are as follows:Step 1: Determine the range of Pmax and Pmin of the cyclic load in operation.Step 2: Select the maximum and minimum cyclic load among the Pmax and Pmin values.

Step 3: Calculate the stress intensity factor Kmax at the peak stress level of the cyclic load in operation.Step 4: Determine the critical crack length, ac, required for unstable crack growth using Kmax and Kic of the material.Step 5: Calculate the number of cycles, Nif, for unstable crack growth to reach ac.Step 6: Calculate Pproof based on the maximum allowable crack size and the calculated critical crack length and Pmax values. Thus, this is how the principle and key steps in quantitatively defining Pproof, addressing the critical crack length, ac, at Pmax, required lift time, Nif, etc. are articulated in the case of proof testing.

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What are the methods which used in Nano-composites preparations?

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Methods used in Nano-composites preparations: In-situ synthesis, Ex-situ blending.

Nano-composites are prepared using various methods to ensure the proper dispersion and integration of nanoparticles into a matrix material. These methods can be broadly categorized into in-situ synthesis and ex-situ blending. In-situ synthesis- involves synthesizing nanoparticles within the matrix material during composite preparation. Techniques like sol-gel, chemical vapor deposition, and electrochemical deposition are utilized to grow or deposit nanoparticles directly in the matrix, ensuring uniform distribution. Ex-situ blending- involves blending pre-synthesized nanoparticles with the matrix material. Techniques such as melt mixing, solution casting, and mechanical alloying are employed to disperse the nanoparticles within the matrix through mechanical or chemical means. Both in-situ synthesis and ex-situ blending methods have their advantages and limitations, and the choice of method depends on specific requirements, nanoparticle properties, and the matrix material used.

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A series circuit having a resistance of 75 ohms and a 1 microfarad capacitor which are connected to a 220-volt source.
Determine the impedance of the circuit in kiloohms.

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A series circuit having a resistance of 75 ohms and a 1 microfarad capacitor which are connected to a 220-volt source. The impedance of the circuit is 3.1856 kiloohms.

The impedance (Z) of the series circuit can be calculated by using the formula, Z = sqrt (R² + Xc²), where R is the resistance and Xc is the capacitive reactance which is given by the formula, Xc = 1/(2πfC), where f is the frequency and C is the capacitance.Given that the resistance (R) of the circuit is 75 ohms and the capacitance (C) is 1 microfarad. The frequency is not given, so we assume it to be 50 Hz (typical AC frequency).Using the formula for capacitive reactance,Xc = 1/(2πfC)Xc = 1/(2 × 3.14 × 50 × 10⁶ × 1 × 10⁻⁶)Xc = 3183.1 ohmsZ = sqrt(R² + Xc²)Z = sqrt(75² + 3183.1²)Z = 3185.6 ohms = 3.1856 kΩ Answer: The impedance of the circuit is 3.1856 kiloohms.

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An electric resistance heater works with a 245 V power-supply and consumes approximately 1.4 kW. Estimate the electric current drawn by this heater. Provide your answer in amperes rounded to three significant digits.

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The electric current drawn by this heater is 5.71 Amperes.

The formula for electric power is given by:

P = VI

where P is electric power,

V is voltage, and

I is the current

An electric resistance heater works with a 245 V power-supply and consumes approximately 1.4 kW.

We have to estimate the electric current drawn by this heater.We know that:

Power (P) = 1.4 kW

= 1400 W

Voltage (V) = 245 V

Substituting these values in the formula of electric power:

P = VI1400

= 245*I

= 1400/245I

= 5.71 Amperes

Therefore, the electric current drawn by this heater is 5.71 Amperes.

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An ASCII message is stored in memory, starting at address 1000h. In case this message is "BLG"
Write the H register state in the form FFh, otherwise a subroutine.

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An ASCII message is stored in memory, starting at address 1000h. In case this message is "BLG" then the H register state in the form FFh is 0C4h.

The ASCII code for B is 42h, L is 4Ch, and G is 47h. The three-character string BLG will be stored in memory locations 1000h, 1001h, and 1002h, respectively. The H register contains the high byte of the memory address of the last byte accessed in an operation.

In this scenario, when the computer accesses memory location 1002h, the H register will contain the high byte of 1002h, which is 10h. Thus, the H register state is 10h in this case.To convert the H register state to the form FFh, we'll add FFh to the number. In this example, FFh + 10h = 0C4h, which is the H register state in the form FFh. Therefore, the H register state in the form FFh for this scenario is 0C4h.

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A black thermocouple measures the temperature in a chamber with black walls. If the air
around the thermocouple is at 20 °C, the walls are at 100 °C, and the heat transfer
coefficient between the thermocouple and the air is 75 W / m2 K, what temperature will
the thermocouple read?
HINT: The heat convected away from the thermocouple by the air must exactly balance
that radiated to
it by the hot walls if the system is in steady state.

Answers

The black thermocouple measures the temperature in a chamber with black walls. If the air around the thermocouple is at 20°C and the walls are at 100°C, and the heat transfer coefficient between the thermocouple and the air is 75 W/m2K.

Then, the temperature that the thermocouple will read can be found by the following calculation. The convected heat away from the thermocouple by the air must exactly balance the radiated heat to it by the hot walls if the system is in steady state.According to the question, the wall's temperature is 100°C and the thermocouple's temperature is unknown.

Thus, assuming that the thermocouple's temperature is equal to the air's temperature, i.e., Tc = Ta. The rate of heat transfer from the black wall to the thermocouple is given by the following formula:q_conv = hA(Ta − Twall)Where q_conv is the heat transfer by convection, h is the convective heat transfer coefficient, A is the surface area, Ta is the air's temperature, and Twall is the wall's temperature.

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Which of the following is true?
A. BCC metals are more ductile than FCC metals
B. FCC metals are more ductile than HCP metals
C. HCP metals are more ductile than BCC metals
D. the crystal structure of a metal cannot affect the ductility of the metal

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Answer:Option B: FCC metals are more ductile than HCP metalsExplanation:In metallurgy, ductility refers to a material's capacity to deform plastically under tensile stress. The greater the amount of plastic deformation that occurs before failure, the more ductile a material is.

The ductility of metals varies according to their crystal structure. Metals can have one of three crystal structures: face-centered cubic (FCC), body-centered cubic (BCC), or hexagonal close-packed (HCP).The FCC metals, such as copper and aluminum, have a crystalline structure in which atoms are arranged in a cubic configuration with an atom at each corner and one at the center of each face.

Due to this regular atomic arrangement, FCC metals are more ductile than HCP metals, such as magnesium, which have a hexagonal arrangement of atoms. Therefore, option B: FCC metals are more ductile than HCP metals is true.

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A gas turbine cycle incorporating an intercooler is to be designed to cater to a power requirement of 180MW. The pressure ratio across each compressor stage is 5 . The temperature of air entering the first compressor is 295 K and that at the exit of the intercooler is 310 K. Note that the turbine comprises of a single stage. The temperature of the gases entering the turbine is 1650 K. A regenerator with a thermal ratio of 0.7 is also incorporated into the cycle. Assume isentropic efficiencies of the compressors and the turbine to be 87%. Taking the specific heat at constant pressure as 1.005 kJ/kg.K and the ratio of the specific heats as 1.4: (a) Draw the Temperature-Entropy (T-S) diagram for this process. (b) Calculate: (i) The temperature at the exit of each compressor stage.
(ii) The compressor total specific work. (iii) The net specific work output. (iv) The work ratio. (v) The mass flowrate of gases in kg/s. (vi) The temperature of the gases at the exit of regenerator before entering the combustion chamber. (vii) The cycle efficiency.

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(a) The Temperature-Entropy (T-S) diagram for the process is shown below.(b) (i) The temperature at the exit of each compressor stage is as follows:Stage 1: T2 = 295 × (5)^0.287 = 456.5 KStage 2: T3 = 456.5 × (5)^0.287 = 702 KStage 3: T4 = 702 × (5)^0.287 = 1079 K.

(ii) The compressor total specific work is given by,Wc = cp(T3 - T2) + cp(T4 - T3) + cp(T5 - T4)= 1.005 [(702 - 456.5) + (1079 - 702) + (1650 - 1079)]/0.87= 732.6 kW/kg(iii) The net specific work output is given by,Wnet = Wt - Wc= (cp(T5 - T6) - cp(T4 - T3))/0.87= (1.005 x (1650 - 861.6) - 1.005 x (1079 - 702))/0.87= 226.8 kW/kg(iv) The work ratio is given by,WR = Wc/Wt= 732.6/(226.8 + 732.6)= 0.763(v) The mass flow rate of gases is given by,mg = Wnet/[(cp(T5 - T6)) + (cp(T3 - T2))] = 226.8/[(1.005 x (1650 - 861.6)) + (1.005 x (702 - 456.5))] = 39.34 kg/s(vi) The temperature of gases at the exit of the regenerator before entering the combustion chamber is given by,T6 = T2 + (T5 - T4) x TR = 295 + (1650 - 1079) x 0.7 = 837.4 K(vii) The cycle efficiency is given by,ηcycle = Wnet/Qin= Wnet/(cp(T5 - T6) - cp(T3 - T2))= 226.8/[(1.005 x (1650 - 861.6)) - (1.005 x (702 - 456.5))] = 0.396 or 39.6%.Keywords: gas turbine cycle, intercooler, temperature, pressure ratio, compressors, thermal ratio, isentropic efficiencies, specific heat, ratio of specific heats, Temperature-Entropy (T-S) diagram, compressor stage, compressor total specific work, net specific work output, work ratio, mass flow rate of gases, temperature of gases, cycle efficiency.

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i) A pressure relief valve is to be used as a mechanical safety device on pressure vessel containing dry saturated steam. The relief valve is to be set to fully open at pressure of 26 bar. Using an approximate method, determine the nozzle throat radius of the pressure relief valve for a steam expansion mass flow rate of 0.29 kg/s. State all assumptions and show all calculation steps in your analysis.
ii) Explain the alternative graphical method to determine the critical pressure and area required at the throat of a nozzle flowing a condensable vapour. Use suitable diagrams, sketches, and equations in your answer.
iii) Briefly describe the behaviour of supersaturation for real high speed nozzle steam flows and discuss the implications of this phenomenon with an appropriate temperature – specific entropy diagram sketch.

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i) A pressure relief valve is to be used as a mechanical safety device on pressure vessel containing dry saturated steam. The relief valve is to be set to fully open at a pressure of 26 bar.

Using an approximate method, determine the nozzle throat radius of the pressure relief valve for a steam expansion mass flow rate of 0.29 kg/s. State all assumptions and show all calculation steps in your analysis.The basic equations used in determining the nozzle throat radius are the mass flow rate equation and the isentropic relation for choked flow.

The assumptions made are that the flow is adiabatic, steady-state, and fully developed, and that the pressure at the outlet of the nozzle is atmospheric. Here are the calculations for the nozzle throat radius:

r^2 = [A*(2/π)]^1/2

= [0.29/((26*(10^5))*(1.106))]^0.5

= 0.000177 m^2A

= πr^2

= π*(0.01331^2)

= 0.000556 m^2

Thus, the nozzle throat radius of the pressure relief valve for a steam expansion mass flow rate of 0.29 kg/s is 0.01331 m.

It is a chart that displays the enthalpy (h) and entropy (s) of a substance. The Mollier chart has a vertical axis of enthalpy and a horizontal axis of entropy.

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The total radiation energy leaving a surface per unit time and per unit area is known as Fill in the blank

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The total radiation energy leaving a surface per unit time and per unit area is known as radiant flux. Radiant flux, also referred to as radiant power, is a measure of the rate at which electromagnetic radiation is emitted or transmitted from a surface.

It represents the total energy transferred through radiation per unit time and per unit area. The radiant flux is typically measured in watts (W) and is used to quantify the amount of energy radiated by a surface. Radiant flux is an important concept in various fields, including physics, engineering, and thermal sciences. It is used to characterize the emission and transfer of thermal energy through radiation, which plays a significant role in heat transfer processes. By understanding the radiant flux, researchers and engineers can analyze and design systems involving radiative heat transfer, such as thermal insulation, solar energy devices, and radiative cooling systems. In summary, the term "radiant flux" refers to the total radiation energy leaving a surface per unit time and per unit area. It is a fundamental quantity in the study of radiative heat transfer and is crucial for analyzing and designing systems involving electromagnetic radiation.

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The probability density function for the diameter of a drilled hole in millimeters is 10e^(-10(x-5)) for x > 5 mm. Although the target diameter is 5 millimeters, vibrations, tool wear, and other nuisances produce diameters greater than 5 millimeters. a. Draw the probability distribution curve. b. Determine the probability that the hole diameter is 5 to 5.1mm c. Determine the expected diameter of the drilled hole. d. Determine the variance of the diameter of the holes. Determine the cumulative distribution function. e. Draw the curve of the cumulative distribution function. f. Using the cumulative distribution function, determine the probability that a diameter exceeds 5.1 millimeters.

Answers

a. To draw the probability distribution curve, we can plot the probability density function (PDF) over a range of values.

The probability density function for the diameter of a drilled hole is given by:

f(x) = 10e^(-10(x-5)), for x > 5

To plot the curve, we can choose a range of x-values, calculate the corresponding y-values using the PDF equation, and plot the points.

b. To determine the probability that the hole diameter is between 5 and 5.1 mm, we need to calculate the area under the probability distribution curve within that range. Since the PDF represents the probability density, we can integrate the PDF function over the given range to find the probability.

P(5 ≤ x ≤ 5.1) = ∫[5, 5.1] f(x) dx

c. To determine the expected diameter of the drilled hole, we need to calculate the expected value or the mean of the probability distribution. The expected value is given by:

E(X) = ∫[5, ∞] x * f(x) dx

d. To determine the variance of the diameter of the holes, we need to calculate the variance of the probability distribution. The variance is given by:

Var(X) = ∫[5, ∞] (x - E(X))^2 * f(x) dx

e. The cumulative distribution function (CDF) represents the probability that a random variable is less than or equal to a given value. To draw the curve of the CDF, we need to calculate the cumulative probability for different x-values.

CDF(x) = ∫[5, x] f(t) dt

f. Using the CDF, we can determine the probability that a diameter exceeds 5.1 millimeters by subtracting the CDF value at 5.1 from 1:

P(X > 5.1) = 1 - CDF(5.1)

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MCQ: In triangulation surveys: A. The area is divided into triangular figures B. Control stations are located from which detailed surveys are carried out C. Sides are not measured excepting the base line D. All the above 11- The necessary geometrical condition for triangulation adjustments is: A. The sum of the angles around a station should be 360° B. The sum of the three angles of a plane triangle should be 180° C. The sum of the eight angles of a braced quadrilateral should be 360° D. All the above 12- Triangulation surveys are carried out for providing: A. Planimetric control B. Height control C. Both planimetric and height control D. None of these 1- If the magnetic bearing of the sun at place at noon in southern hemisphere is 167°, the magnatic declination at that place is: A. 77° N B. 23° S C. 13° E D. 13° W 2-Triangulation surveys are carried out for locating: A. Control points for surveys of large areas. B. Control points for photogrammetric surveys. C. Engineering works, i.e. terminal points of long tunnel, bridge abutments, etc. D. All above 3- The horizontal angle between the true meridian and magnetic meridian at a place is called: A. azimuth B. declination C. local attraction D. magnetic bearing 4- If the reduced bearing of a line AB is N60°W and length is 100 m, then the latitude and departure respectively of the line AB will be: A. +50 m, +86.6 m B. +86.6 m, -50 m C. +50 m, -86.6 m D. +70.7 m, -50 m 5- The smaller horizontal angle between the true meridian and a survey line, is known: A. Declination B. Bearing C. Azimuth D. Dip 6- For a line AB: A. the foreword bearing of AB and back bearing of AB differ by 180° B. the foreword bearing of AB and back bearing of BA differ by 180° C. both (A) and (B) are correct D. none is correct 7- There are two stations A and B. which of the following statements is correct? A. The fore bearing of AB is AB B. The back bearing of AB is BA D. All the above C. The fore and back bearings of AB differ by 180° 8- Intersection method of detailed plotting is most suitable for: B. urban areas A. forests C. hilly areas D. plains 9- If in aclosed traverse, the sum of the north latitudes is more than the sum of the south latitudes. And alse the sum of west departures is more than the sum of the east departures, the bearing of the closing line is in the: A. NE quadrant B. SE quadrant C. NW quadrant D. SW quadrant

Answers

A.Triangulation surveys are conducted for dividing the area into triangular figures, locating control stations, and measuring sides to establish planimetric and height control.

B. In triangulation surveys, the area is divided into triangular figures, which helps in achieving accurate measurements and calculations for mapping purposes.Control stations are strategically located to serve as reference points from which detailed

What topics are covered in the provided set of multiple-choice questions?

The provided multiple-choice questions cover various aspects of triangulation surveys and related concepts in surveying.

Triangulation surveys involve dividing the area into triangular figures and locating control stations for detailed surveys. The necessary geometrical condition for triangulation adjustments is that the sum of angles around a station should be 360° or the sum of the three angles of a plane triangle should be 180°.Triangulation surveys are carried out for providing planimetric control, height control, or both.The magnetic declination at a place can be determined based on the given magnetic bearing of the sun.5. Triangulation surveys are conducted for locating control points for surveys of large areas, photogrammetric surveys, and engineering works. The horizontal angle between the true meridian and magnetic meridian at a place is called declination. The reduced bearing of a line AB can be used to determine the latitude and departure of the line. The smaller horizontal angle between the true meridian and a survey line is known as the azimuth.For a line AB, the forward bearing of AB and back bearing of BA differ by 180°.The fore and back bearings of AB differ by 180°.The intersection method of detailed plotting is most suitable for urban areas.If the sum of north latitudes is greater than the sum of south latitudes and the sum of west departures is greater than the sum of east departures in a closed traverse, the bearing of the closing line is in the NW quadrant.

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Beceiving current is high in case of a) No load) 2 by Full load Resistive load d) Inductive load 2. If the transmission line is folle loaded the voltage at the receiving end compared with the Sending and is: a) Greater b) Smaller c) Equal d) None of the above 3. The transmission line require (a) Active power in no-load operation. b) Reactive e) Apparent d) None of the above In case of matched load only the -power is transmitted. a) Active> b) Reactive c) Apparent d) None of the above

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1. Receiving current is high in the case of d) Inductive load.

When we compare the inductive load to the resistive load, we notice that the receiving current is high in the case of the inductive load. Inductive loads can create power factor problems because the current and voltage waveforms are out of phase. When compared to resistive loads, inductive loads produce more waste energy and thus demand more current.

2. The voltage at the receiving end compared with the sending end is b) Smaller when the transmission line is fully loaded. When a transmission line is fully loaded, the receiving end voltage is smaller than the sending end voltage because voltage is lost due to line resistance and inductive reactance.

3. The transmission line requires b) Reactive power in no-load operation. When there is no load, the transmission line requires reactive power.

4. In the case of matched load, only the a) Active power is transmitted. When the load is matched, there is no reactive power. As a result, only the active power is transmitted and not the reactive power.

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Question 1 a) Develop the activity sequence model and determine the normal time for the following work activities: 1. A worker sitting on chair stands up walk 7 steps toward an old cabinet. He opens the old drawer and he face some resistance because the drawer is kind stuck. 2. He collects 8 screws from the drawer and returns back to his chair which 7 steps away, sit down and hold the screws. 3. He inserts with adjustments the 8 screws in each hole in laptop in front of him. 4. He picks up the screwdriver laying aside next to him and turns each screw X times using wrist action. After he is done, he performs Y(unknown) body motion put the screwdriver next to the laptop. b) If the total TMU is 1640 which one of the following is true. "Hint chose the answer that is close to your value if not exact" (Show your work) X= 9 turns and Y is no body motion X= 3 turns and Y is sitting down X= 3 turns and Y is standing up X=5 turns and Y is standing up c) Given that the performance rating is 110% for the process above and the PFD allowance is 20%. Calculate the standard time for the process.

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The given problem pertains to work measurement and time study, which are aspects of industrial engineering.

The answer depends on the specific times assigned to the actions listed. For part c, standard time is computed by multiplying the normal time by the sum of 1, the performance rating, and the allowance factor. In a more detailed sense, the normal time for work activities can be computed using predetermined motion time systems (PMTS) or time study. Given the task sequence, you'd need to assign each action a time value based on the complexity and duration. In this context, we need information on TMU (time measurement unit) values for actions such as walking, opening a drawer, picking screws, sitting, and screwing. For part b, we'd compare the total TMU with each option. The option with TMU closest to 1640 is correct. In part c, standard time = normal time x (1 + performance rating/100 + allowance factor), assuming normal time includes rest and delay allowances.

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A centrifugal pump may be viewed as a vortex, where the 0.15m diameter impeller, rotates within a 0.65m diameter casing at a speed of 150 rpm. The outer edge of the vortex may NOT be considered infinite.
Determine
The circumferential velocity, in m/s at a radius of 0.225 m
The angular velocity, in rad/s at a radius of 0.055;
The circumferential velocity, in m/s at a radius of 0.04 m
The angular velocity, in rad/s s at a radius of 0.225 m

Answers

The circumferential velocity at a radius of 0.225 m is approximately 23.56 m/s for centrifugal pump. The angular velocity at a radius of 0.055 m is approximately 686.68 rad/s.

The circumferential velocity can be calculated using the formula:

V = π * d * n

where V is the circumferential velocity, d is the diameter, and n is the rotational speed in revolutions per minute (rpm). Substituting the given values, we have:

V = π * 0.15 m * 150 rpm = 70.69 m/s

To find the circumferential velocity at a specific radius, we can use the following formula:

V_ r = V * (r_ impeller / r_ radius)

where V_ r is the circumferential velocity at the desired radius, r_ impeller is the radius of the impeller (0.15 m), and r_ radius is the desired radius. Substituting the given values, we get:

V_ r = 70.69 m/s * (0.15 m / 0.225 m) = 47.13 m/s

Thus, the circumferential velocity at a radius of 0.225 m is approximately 47.13 m/s.

The angular velocity can be calculated using the formula:

ω = 2π * n

where ω is the angular velocity in radians per second and n is the rotational speed in revolutions per minute (rpm). Substituting the given values, we have:

ω = 2π * 150 rpm = 942.48 rad/s

To find the angular velocity at a specific radius, we can use the following formula:

ω_r = ω * (r_ impeller / r_ radius)

where ω_r is the angular velocity at the desired radius, r_ impeller is the radius of the impeller (0.15 m), and r_ radius is the desired radius. Substituting the given values, we get:

ω_r = 942.48 rad/s * (0.15 m / 0.225 m) = 628.32 rad/s

Thus, the angular velocity at a radius of 0.055 m is approximately 628.32 rad/s.

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In the armature of a dc machine running at 1750 rpm and having four poles The flux per pole is 25 mWb, and the armature is lap-wound with 728 conductors. Determine: a. the induced voltage. b. the speed at which it is to be driven to produce the same emf, if it is wave winding.

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The question involves a DC machine with four poles and an armature that is lap-wound with 728 conductors. The machine is running at a speed of 1750 rpm, and the flux per pole is given as 25 mWb. The task is to determine the induced voltage and calculate the speed required to produce the same electromotive force (emf) if the armature is wave-wound.

In this scenario, the DC machine operates at a speed of 1750 rpm and has four poles. The armature is lap-wound, meaning it has 728 conductors. The flux per pole is provided as 25 mWb. Part (a) asks for the calculation of the induced voltage, which is the voltage generated in the armature due to the interaction with the magnetic field. By using the given information and applying the appropriate formulas and equations for DC machines, we can determine the induced voltage.

In part (b), we are required to find the speed at which the machine must run with a wave-wound armature to produce the same electromotive force (emf). The wave winding configuration differs from the lap winding, and the change in winding style affects the relationship between speed speed and emf. By analyzing the characteristics of wave winding and and emf. By analyzing the characteristics of wave winding and considering the impact on the electrical output, we can calculate the required speed to achieve the desired emf.

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1) The induced voltage in dc machine : E = 667.78 V

2) The speed at which it is to be driven to produce the same emf, if it is wave winding : N = 328.13 RPM

Given,

N = 1750RPM

Flux per pole = 25mWb

a. Induced voltage:

The induced voltage of a DC machine is given by the equation; E = ΦNZP / 60A, where E is the induced voltage, Φ is the flux per pole, N is the speed of rotation in revolutions per minute (RPM), Z is the total number of armature conductors, P is the number of poles on the machine, and A is the number of parallel paths in the armature winding.

Substituting the given values into the formula, we have:

E = ΦNZP / 60A

E = (25 x 728 x 4 x 1750) / (60 x 2)

E = 40,066.67 / 60

E = 667.78 V

Therefore, the induced voltage of the machine is 667.78 V.

b. Speed required for the same emf, if it is wave winding:

For a wave winding, the formula for induced voltage is given as E = ΦNZP / 60, where N is the speed of rotation in RPM.

Substituting the given values into the formula, we have:

E = ΦNZP / 60

667.78 = (25 x 728 x 4 x N) / 60

N = (667.78 x 60) / (25 x 728 x 4)

N = 328.13 RPM

Therefore, the machine must be driven at 328.13 RPM to produce the same induced voltage with a wave winding.

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12. 2 points Capacitive susceptance decreases as frequency increases O a. True O b. False 13. 2 points The amplitude of the voltage applied to a capacitor affects its capacitive reactance. O a. True O b. False 14. 2 points For any given ac frequency a 10 μF capacitor will have more capacitive reactance than a 20 μF capacitor. O a. True
O b. False 15. 2 points In a series capacitive circuit, the smallest capacitor has the largest voltage drop. O a. True O b. False 16. 2 points In a parallel capacitive circuit all capacitors store the same amount of charge O a. True O b. False

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12. False 13. False 14. FALSE 15. true 16. true are the answers

12. False

Capacitive susceptance is the reciprocal of the capacitive reactance, and it varies with frequency. The higher the frequency of the AC, the lower the capacitive reactance.

13. False

Capacitive reactance is determined by the capacitance and frequency of the applied voltage, and it is not influenced by the voltage level.

14. False

Capacitive reactance varies with the capacitance and frequency of the applied voltage. A capacitor with a capacitance of 20 μF has less capacitive reactance than a capacitor with a capacitance of 10 μF.

15. True

The capacitive reactance is inversely proportional to the capacitance of the capacitor in a series capacitive circuit, so the capacitor with the lowest capacitance will have the largest voltage drop across it.

16. True

In a parallel capacitive circuit, all capacitors receive the same voltage because they are linked across the same voltage source, and they all store the same amount of charge.

Q = CV is the equation used to calculate the amount of charge stored in a capacitor,

where Q is the charge stored in coulombs, C is the capacitance in farads, and V is the voltage across the capacitor in volts.

Since the voltage across each capacitor is the same in a parallel circuit, all capacitors store the same amount of charge.

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If a line-to-line fault occurs across "b" and "c" and Ea = 230 V/0°, Z₁ = 0.05 +j 0.292, Zn = 0 and Zf = 0.04 + j0.3 02, find: a) the sequence currents la1 and laz fault current If b) c) the sequence voltages Vǝ1 and Va2 d) sketch the sequence network for the line-to-line fault.

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Line-to-line fault across "b" and "c". Ea = 230 V/0°.Z₁ = 0.05 +j 0.292,Zn = 0.f = 0.04 + j0.302.

(a) The sequence currents: Sequence currents la1 and laz fault current are calculated by using the following formulae:

la1 = (-2/3)[(0.05 + j0.292) / (0.05 + j0.292 + 0.04 + j0.302)] * (230 / √3)la1 = (-2/3)[0.05 + j0.292 / 0.0896 + j0.594] * 230la1

= -28.7 + j51.5A

Let us use the below formula to calculate the fault current: if = 3 * la1if

= 3 * (-28.7 + j51.5)if = -86.1 + j154.5

A(b) The sequence voltages :Sequence voltages Vǝ1 and Va2 are calculated using the following formulae: For voltage

Vǝ1:(Vǝ1 / √3) = Ea / √3Vǝ1 = Ea = 230V/0

°For voltage Va2:Va2 = 0

(As the fault is a line-to-line fault, the phase voltages are equal in magnitude but opposite in direction, and they are canceled out due to phase shifting in a balanced system.

Hence, the zero sequence voltage is zero.) (c) The sequence diagram can be shown as follows: Sequence Network The sequence network for the line-to-line fault is shown below: Sequence Network for the line-to-line fault.

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A simple ideal Brayton Cycle is modified to use a two stage turbine with reheating, while keeping constant: the maximum cycle temperature, the boiler pressure, the condenser pressure, and the steam mass flow rate. Sketch the process with and without reheating in a T-s plot with these constraints.
How do the following quantities change if reheating is used (compared to the simple cycle)?
Cycle Thermal Efficiency
Select one:
a. Unanswerable b. Increases c. Decreases
d. No effect Heat Addition
Select one: a. Decreases b. No effect c. Increases
d. Unanserable Turbine Outlet Quality
Select one: a. Unanswerable b. No effect c. Decreases d. Increases Turbine Work
Select one: a. No effect b. Unanswerable c. Decreases d. Increases Pump Work
Select one: a. No effect b. Unanswerable c. Decrescer

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The thermal efficiency and turbine work increase, while heat addition, turbine outlet quality, and pump work remain unaffected.

When a simple ideal Brayton Cycle is modified to use a two-stage turbine with reheating while keeping the maximum cycle temperature, boiler pressure, condenser pressure, and steam mass flow rate constant, the T-s (temperature-entropy) plot will show the following changes:
Cycle Thermal Efficiency: b. Increases
The addition of reheating improves the thermal efficiency of the cycle. Reheating allows for additional heat addition at a higher temperature, resulting in increased work output and improved overall efficiency.
Heat Addition: c. Increases
With reheating, additional heat is added to the cycle at a higher temperature after the first expansion in the turbine. This increases the overall heat input into the cycle and allows for more work extraction.
Turbine Outlet Quality: b. No effect
The use of reheating does not directly affect the quality (or dryness fraction) of the steam at the turbine outlet. The quality of the steam depends on the condenser pressure and the turbine efficiency, which remain constant in this scenario.
Turbine Work: d. Increases
The introduction of reheating increases the total work output of the turbine. After the first expansion in the high-pressure turbine, the partially expanded steam is reheated before entering the second-stage turbine. This reheating process allows for additional expansion and work extraction, resulting in increased turbine work output.
Pump Work: a. No effect
The use of reheating does not have a direct effect on the pump work. The pump work is determined by the pressure difference between the condenser pressure and the boiler pressure, which remains constant in this case.
Hence, when a two-stage turbine with reheating is added to the Brayton Cycle while keeping the specified constraints constant, the cycle thermal efficiency increases, heat addition increases, turbine outlet quality remains unchanged, turbine work increases, and pump work remains unaffected.

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Refrigerant −134 a expands through a valve from a state of saturated liquid (quality x =0) to a pressure of 100kpa. What is the final quality? Hint: During this process enthalpy remains constant.

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The given scenario involves Refrigerant-134a expanding through a valve from a state of saturated liquid (quality x = 0) to a pressure of 100 kPa. The question asks for the final quality of the refrigerant, considering that the enthalpy remains constant during this process.

We use the quality-x formula for determining the final quality of the liquid after expanding it through the valve.

The quality-x formula is defined as follows:

x2 = x1 + (h2 - h1)/hfgwhere x1 is the initial quality of the liquid, which is zero in this case; x2 is the final quality of the liquid; h1 is the enthalpy of the liquid at the initial state; h2 is the enthalpy of the liquid at the final state; and hfg is the enthalpy of vaporization.

It is mentioned that the enthalpy remains constant. So, h1 = h2 = h. Now, the formula becomes:x2 = x1 + (h - h1)/hfgBut h = h1.

Therefore, the above formula can be simplified as:x2 = x1 + (h - h1)/hfgx2 = 0 + 0/hfgx2 = 0.

This implies that the final quality of the refrigerant is zero. Hence, the final state of the refrigerant is saturated liquid.

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Write a function to implement Left Hand Rectangular Integration from a to b of the given function, . Use N linearly spaced values for x. With a function signature of: function integ= Ih_rect_int(a, b, g, N)

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The function Ih_rect_int implements the Left Hand Rectangular Integration method to approximate the integral of a given function g over the interval [a, b] using N linearly spaced values for x.

The function returns the value of the computed integral. In the Left Hand Rectangular Integration method, the area under the curve is approximated by dividing the interval into N subintervals of equal width. The height of each rectangle is determined by evaluating the function at the left endpoint of each subinterval, and the width of each rectangle is the width of the subinterval. The sum of the areas of all the rectangles gives an approximation of the integral. The function Ih_rect_int takes the parameters a and b to define the integration interval, the function g to be integrated, and N to determine the number of subintervals. It uses a loop to calculate the value of the integral by summing the areas of the rectangles. The width of each rectangle is determined by the spacing between the linearly spaced x values. Finally, the computed integral value is returned as the output of the function.

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what is the best option for saving money (electrical heater or kitchen LPG system? As an expert in energy, What you always advice your family and friends in this regard?

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An electrical heater and kitchen LPG system are two popular options for heating and cooking. The choice between the two depends on several factors that you need to consider before making a final decision. I advise my family and friends to consider the following factors before deciding which option is best for them.

1. Energy efficiency: Energy efficiency is the primary factor you should consider when choosing between an electrical heater and kitchen LPG system. The kitchen LPG system is generally more energy-efficient than electrical heaters. LPG gas can heat up a pot or pan faster than an electric heating element, which saves energy.2. Cost: Cost is another important factor you should consider when choosing between an electrical heater and a kitchen LPG system. LPG gas is generally cheaper than electricity in most parts of the world. However, the cost of LPG varies depending on your location, so it's important to do some research to find out the price of LPG in your area.3. Safety: Safety is also an essential factor you should consider when choosing between an electrical heater and a kitchen LPG system. Both options have their unique risks and safety concerns. For example, LPG is highly flammable, while electrical heaters can pose an electrocution hazard.

Availability is another essential factor you should consider when choosing between an electrical heater and a kitchen LPG system. LPG is not available in some areas, while electricity is readily available in most parts of the world. You need to consider the availability of the energy source in your area before deciding which option is best for you.In conclusion, both an electrical heater and kitchen LPG system have their unique benefits and drawbacks. The best option for saving money depends on several factors such as energy efficiency, cost, safety, convenience, and availability. I always advise my family and friends to consider these factors carefully before making a final decision.

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A contractor manufacturing company purchased a production equipment for $450,000 to meet the specific needs of a customer that had awarded a 4-year contract with the possibility of extending the contract for another 4 years. The company plans to use the MACRS depreciation method for this equipment as a 7-year property for tax purposes. The combined income tax rate for the company is 24%, and it expects to have an after-tax rate of return of 8% for all its investments. The equipment generated a yearly revenue of $90,000 for the first 4 years. The customer decided not to renew the contract after 4 years. Consequently, the company decided to sell the equipment for $220,000 at the end of 4 years. Answer the following questions, (a) Show before tax cash flows (BTCF) from n= 0 to n=4 (b) Calculate depreciation charges (c) Compute depreciation recapture or loss (d) Find taxable incomes and income taxes (e) Show after-tax cash flows (ATCF). (f) Determine either after tax NPW or after-tax rate of return for this investment and indicate if the company obtained the expected after-tax rate of retum

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a) Before-tax cash flows (BTCF) from n= 0 to n=4Year

RevenueDepreciationBTCF0-$450,000-$450,0001$90,000$57,144$32,8562$90,000$82,372$7,6283$90,000$59,013$30,9874$90,000$28,041$61,959

b) Depreciation charges

Using the MACRS depreciation method, the annual depreciation expenses are as follows:Year

Depreciation rate Depreciation charge1 14.29% $64,215.002 24.49% $110,208.753 17.49% $78,705.754 12.49% $56,216.28Therefore, the total depreciation charge over 4 years is $309,345.75.

c) Depreciation recapture or loss

After 4 years, the equipment was sold for $220,000. The adjusted basis of the equipment is the initial cost minus the accumulated depreciation, which is:$450,000 - $309,345.75 = $140,654.25Therefore, the depreciation recapture or loss is:$220,000 - $140,654.25 = $79,345.75The depreciation recapture is positive and hence, the company must report this as ordinary income in the current tax year.

d) Taxable incomes and income taxesYearRevenueDepreciationBTCFTaxable IncomeTax1$90,000$64,215.00$25,785.00$6,187.60(24% x $25,785.00)2$90,000$110,208.75-$20,208.75-$4,850.10(24% x -$20,208.75)3$90,000$78,705.75$11,294.25$2,710.22(24% x $11,294.25)4$90,000$56,216.28$33,783.72$8,107.69(24% x $33,783.72)

The total income taxes paid over 4 years is $21,855.61.e) After-tax cash flows (ATCF)YearBTCFTaxIncome TaxATCF0-$450,000-$450,0001$32,856$6,188$26,6692$7,628$4,850$2,7793$30,987$2,710$28,2774$61,959$8,108$53,851The total ATCF over 4 years is $110,576.f)

After-tax NPW or After-tax rate of return (ARR) for this investmentAfter-tax NPW = -$450,000 + $110,576(P/A,8%,4 years)= -$450,000 + $110,576(3.3121)= -$28,128.04Since the NPW is negative, the company did not obtain the expected after-tax rate of return.

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A gas contained within a piston-cylinder assembly undergoes two processes, A and B, between the same end states, 1 and 2, where P1 = 10 bar, V1 0.1m³, U1 = 400 kJ and P2 = 1 bar, V2 = 1.0 m³, U2 = 200 kPa: Process A: Process from 1 to 2 during which the pressure-volume relation is PV = constant. Process B: Constant-volume process from state 1 to a pressure of 1 bar, followed by a linear pressure-volume process to state 2. Kinetic and potential energy effects can be ignored. For each of the processes A and B. (a) evaluate the work, in kJ, and (b) evaluate the heat transfer, in kJ. Enter the value for Process A: Work, in kJ. Enter the value for Process A: Heat Transfer, in kJ. Enter the value for Process B: Work, in kJ. Enter the value for Process B: Heat Transfer, in kJ.

Answers

The values of work and heat transfer for the given processes are given below:

Process A:Work = -5.81 kJ

Heat Transfer = 0kJ

Process B:Work = 0.45 kJ

Heat Transfer = -199.55 kJ.

Initial state: P1 = 10 bar, V1 = 0.1 m³, U1 = 400 kJ

Final state: P2 = 1 bar, V2 = 1.0 m³, U2 = 200 kJ

Process A:Pressure-volume relation is PV = constant

Process B:Constant-volume process from state 1 to a pressure of 1 bar,

followed by a linear pressure-volume process to state 2(a) Evaluate the work, in kJ for process A:

For process A, pressure-volume relation is PV = constant

So, P1V1 = P2V2 = C
Work done during process A is given as,W = nRT ln(P1V1/P2V2)

Here, n = number of moles,

R = gas constant,

T = temperature.

For an ideal gas,

PV = mRT

So, T1 = P1V1/mR and

T2 = P2V2/mR

T1/T1 = T2/T2

W = mR[T2 ln(P1V1/P2V2)]

= mR[T2 ln(P1V1/P2V2)]/1000W

= (1/29)(1/0.29)[1.99 ln(10/1)]

= -5.81 kJ(b)

Evaluate the heat transfer, in kJ for process A:

Since it is an adiabatic process, so Q = 0kJ

(a) Evaluate the work, in kJ for process B:For process B, V1 = 0.1 m³, V2 = 1.0 m³, P1 = 10 bar and P2 = 1 bar.

For the process of constant volume from state 1 to a pressure of 1 bar: P1V1 = P2V1

The work done in process B is given as,The initial volume is constant, so the work done is 0kJ for the constant volume process.

The final process is a linear process, so the work done for the linear process is,

W = area of the trapezium OACB Work done for linear process is given by:

W = 1/2 (AC + BD) × ABW

= 1/2 (P1V1 + P2V2) × (V2 - V1)W

= 1/2 [(10 × 0.1) + (1 × 1.0)] × (1.0 - 0.1)W = 0.45 kJ

(b) Evaluate the heat transfer, in kJ for process B:Heat transfer, Q = ΔU + W

Here, ΔU = U2 - U1= 200 - 400 = -200 kJ

For process B, heat transfer is given by:Q = -200 + 0.45

= -199.55 kJ

So, the values of work and heat transfer for the given processes are given below:

Process A:Work = -5.81 kJ

Heat Transfer = 0kJ

Process B:Work = 0.45 kJ

Heat Transfer = -199.55 kJ.

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One of the following statements is true for DC-Separately Excited Generator (A) The no load characteristic same for increasing and decreasing excitation current (B) The no load characteristic differ for increasing and decreasing excitation current (C) The no load characteristic same for increasing and decreasing load resistance (D) The load characteristic same for increasing and decreasing load resistance

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DC-separately excited generator A separately excited DC generator is the one in which the field coils are excited separately from the armature coils by a separate DC supply or a battery.

The armature is connected to the load and the separate supply is used to energize the field coils. As the field coils are excited with a separate source of DC supply, the generator is called a separately excited generator.The no load characteristic differ for increasing and decreasing excitation current.

in DC-separately excited generator. The graph between open circuit voltage (V) and field current (If) is known as open circuit characteristics. If field current is increased keeping armature current and speed fixed, the open-circuit voltage also increases.

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Consider the following plane stress state: Ox=12 kpsi, Oy = 6 kpsi, Txy = 4 kpsi cw Calculate the following: 1. The coordinates of the center of the Mohr's circle C The location of the center of the Mohr's circle Cis ( 2. Principal normal stresses (01, 02) The principal normal stresses are 0₁ = 3. Maximum shear stress (T) The maximum shear stress is 4. The angle from the x axis to 01 (pl The angle from the x axis to 01 (p) is 5. The angle from the x axis to T (Ps) The angle from the x axis to 7 (s) is 6. The radius of the Mohr's circle The radius of the Mohr's circle is kpsi.

Answers

The radius of the Mohr's circle (R) is 5 kpsi

To calculate the coordinates of the center of the Mohr's circle (C), we can use the following formulas:

Center of Mohr's circle (C) = ((σx + σy) / 2, 0)

Given the stress state: σx = 12 kpsi, σy = 6 kpsi, and τxy = 4 kpsi (cw),

Substituting the values into the formula, we get:

Center of Mohr's circle (C) = ((12 + 6) / 2, 0) = (9 kpsi, 0)

Therefore, the coordinates of the center of the Mohr's circle (C) are (9 kpsi, 0).

To calculate the principal normal stresses (σ1, σ2), we can use the following formulas:

σ1 = ((σx + σy) / 2) + √(((σx - σy) / 2)^2 + τxy^2)

σ2 = ((σx + σy) / 2) - √(((σx - σy) / 2)^2 + τxy^2)