A 3-phase load of 7.5+j4 Ω (value of each of the impedances) is connected to a 42 kV power system.
Determine the total apparent power (in MVA) when the load is connected in star

The axial thrust on the shaft is 286.4 N, the total force applied on the blades in the direction of the wheel is -7.874 N, the power developed by the turbine is 541.23 kW, the blading efficiency is 84.5%, and the average blade velocity is 673.08 m/s.

Velocity of steam at nozzle outlet, V1 = 660 m/s

Angle of outlet of steam from the nozzle, α1 = 17°

Blades outlet angle of first moving row of turbine, β2 = 18°

Blades outlet angle of second moving row of turbine, β2 = 36°

Blades outlet angle of the row of fixed blades, βf = 22°

Mean radius of the blade wheel, r = 155 mm = 0.155 m

Rotational speed of the blade wheel, N = 4000 rpm

Steam flow rate, m = 80 kg/min

Reduction in steam velocity over all the blades, i.e., (V1 − V2)/V1 = 10% = 0.1

Scale used, 1 mm = 5 m/s (for drawing velocity diagrams)

The length of the blade in the first and second rows of the turbine blades can be determined using the velocity diagram.

Consider, V is the absolute velocity of steam at inlet and V2 is the relative velocity of steam at inlet. Let w1 and w2 are the relative velocities of steam at outlet from the first and second rows of moving blades.

Hence, using the law of cosines, we get

V2² = w1² + V1² – 2w1V1 cos (α1 – β1)

For the first row of blades, β1 = 18°V2² = w1² + 660² – 2 × 660w1 cos (17° – 18°)

w1 = 680.62 m/s

The length of the velocity diagram is proportional to w1, i.e., 680.62/5 = 136.124 mm

Similarly, for the second row of moving blades, β1 = 36°V2² = w2² + 660² – 2 × 660w2 cos (17° – 36°)

w2 = 690.99 m/s

The length of the velocity diagram is proportional to w2, i.e., 690.99/5 = 138.198 mm

Let w1′ and w2′ be the relative velocities of steam at outlet from the first and second rows of blades, respectively.Using the law of cosines, we get

V2² = w1′² + V1² – 2w1′V1 cos (α1 – βf)

For the row of fixed blades, β1 = 22°

V2² = w1′² + 660² – 2 × 660w1′ cos (17° – 22°)

w1′ = 695.32 m/s

The length of the velocity diagram is proportional to w1′, i.e., 695.32/5 = 139.064 mm

The axial thrust on the shaft is given by difference between axial forces acting on the first and second moving row of blades.

Hence,Total axial thrust on the shaft = (m × (w1 sin β1 + w2 sin β2)) − (m × w1′ sin βf) = (80/60) × (680.62 sin 18° + 690.99 sin 36°) – (80/60) × 695.32 sin 22° = 286.4 N

The tangential force acting on each blade can be given by,f = (m (w1 − w1′)) / N

Length of the blade wheel = 2πr = 2 × 3.14 × 0.155 = 0.973 m

Total tangential force on the blade = f × length of blade wheel = ((80/60) × (680.62 − 695.32)) / 4000 × 0.973 = −7.874 N (negative sign implies the direction of force is opposite to the direction of wheel rotation)

Power developed by the turbine can be given by,P = m(w1V1 − w2V2) / 1000 = 80 × (680.62 × 660 − 690.99 × 656.05) / 1000 = 541.23 kW

The blade efficiency can be given by,ηb = (actual work done / work done if steam is entirely used in nozzle) = ((w1V1 − w2V2) / (w1V1 − V2)) = 84.5%

The average blade velocity can be determined by,πDN = 2πNr

Average blade velocity = Vavg = (2w1 + V1)/3 = (2 × 680.62 + 660)/3 = 673.08 m/s

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(i) Maslow's hierarchy of needs is a theory of human needs that helps to understand the various factors that influence the motivation of individuals.

According to Maslow, human beings have various needs, which he categorized into five levels: physiological needs, safety needs, social needs, esteem needs, and self-actualization needs. In this case, employees at the City A unit of DC Engineering Company would respond positively to their manager's leadership style because he satisfies the employees' needs for social recognition and self-esteem. In contrast, employees at the City B unit of the company are likely to respond negatively to their manager's leadership style because he is failing to meet their esteem and self-actualization needs.

(ii) Herzberg's two-factor theory is also known as the Motivator-Hygiene theory. Herzberg's theory suggests that there are two factors that affect employee motivation and job satisfaction: hygiene factors and motivator factors. Hygiene factors include working conditions, salary, job security, and company policies. Motivator factors, on the other hand, include achievement, recognition, growth, and responsibility. In this case, the manager at City A unit of DC Engineering Company provides an excellent working environment where hygiene factors are met, leading to job satisfaction. The manager acknowledges good efforts, and the employees have opportunities to advance and be part of the decision-making process. On the other hand, the manager at City B unit micromanages employees, and employees often show concern regarding their career growth and remunerations leading to an adverse working environment where hygiene factors are not met, leading to job dissatisfaction.

(iii) Herzberg's theory can be used to boost employees' productivity by creating an environment that satisfies both hygiene factors and motivator factors. Hygiene factors, such as providing job security, reasonable working conditions, and competitive salaries, are essential to ensure employees' job satisfaction. Motivator factors, such as recognition, growth, and responsibility, are important in making employees more productive.

(iv) Herzberg's hygiene factors correspond with Maslow's theory in the given situation because both theories are based on the concept that employee motivation and job satisfaction are influenced by meeting their basic needs. Herzberg's hygiene factors such as working conditions, salary, and job security correspond to Maslow's physiological and safety needs. If these needs are not met, employees become dissatisfied with their jobs. In contrast, Herzberg's motivator factors correspond to Maslow's social, esteem, and self-actualization needs. If these needs are met, employees become motivated and productive.

(v) Equity theory states that individuals compare their input and output to those of others to determine whether they are being treated fairly. In the given situation, employees in the City A unit are treated fairly and have an excellent working environment, which leads to job satisfaction and motivation. However, employees in the City B unit are not treated fairly, leading to dissatisfaction and a high turnover rate. Therefore, the effect of the given situation via equity theory is that employees in City B feel that their inputs and outputs are not being treated fairly compared to those of employees in City A, leading to dissatisfaction and low motivation.

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The production cost per 1000 kg steam in a steam plant when the evaporation rate is 7.2 kg steam per kg coal is $18.03. This is obtained as follows;

Step-by-step explanation:

The steam produced from the combustion of coal in a steam plant can be evaluated by first finding the amount of steam generated per kg of coal burned. This is called the evaporation rate.The evaporation rate is given as 7.2 kg steam per kg coal.The cost of coal is given as $8.00 per ton.The steam plant has an average steam production of 14,500 kg per hr.Annual operational cost exclusive of coal is $15,000.The initial cost of plant is $150,000.The life of the steam plant is 20 years.

The interest on borrowed capital is 4% while the interest on the sinking fund is 3%.To find the cost of steam production per 1000 kg, the following calculations are made;

Total amount of steam produced in one year = 14,500 * 24 * 365 = 126,540,000 kg

Annual coal consumption = 126,540,000 / 7.2 = 17,541,666.67 kg

Total cost of coal in one year = (17,541,666.67 / 1000) * $8.00 = $140,333.33

Total cost of operation per year = $140,333.33 + $15,000 = $155,333.33

Annual equivalent charge = AEC = 1 + i/n - 1/(1+i/n)^n*t

Where i = interest n = number of years for which the sum is invest

dt = total life of the investment AEC = 1 + 0.04/1 - 1/(1+0.04/1)^(1*20) = 1.7487

Annual equivalent disbursement = AED = S / a

Where S = initial cost of plant + sum of annual cost (AEC) for n y

earsa = annuity factor obtained from the tables

.AED = $150,000 / 3.8879 = $38,595.69

Annual sinking fund = AS = AED * i / (1 - 1/(1+i/n)^n*t)AS = $38,595.69 * 0.03 / (1 - 1/(1+0.03/1)^(1*20)) = $1,596.51

Total annual cost of the steam plant

= $155,333.33 + $1,596.51

= $156,929.84

Cost of steam production per 1000 kg = 1000 / (126,540,000 / 14,500) * $156,929.84 = $18.03Therefore, the cost of steam production per 1000 kg is $18.03.

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To determine the particle's deceleration when it is located at point A, we need to differentiate the velocity function with respect to time. Given that the velocity function is v = (130 s) mm/s, where s is in millimeters:

v = 130s

To find the deceleration, we differentiate the velocity function with respect to time (s):

a = dv/dt = d(130s)/dt

Since the particle is traveling along a straight line within a liquid, we can assume that its velocity is a function of time only.

Differentiating the velocity function, we get:

a = 130 ds/dt

To find the deceleration at point A, where SA = 90 mm, we substitute the position value into the equation:

a = 130 d(90)/dt

Since the position is not given as a function of time, we assume that it is constant at SA = 90 mm.

Therefore, the deceleration at point A is:

a = 130 * 0 = 0 mm/s²

The deceleration at point A is 0 mm/s².

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For the following iron-carbon alloys (0.76 wt%C) and associated microstructures:A. coarse pearlite B. spheroidite C. fine pearlite D. bainite E. martensite F. tempered martensite1. Select the most ductileWhen the alloy has a coarse pearlite structure, it is the most ductile.2. Select the hardestWhen the alloy has a martensite structure, it is the hardest.

3. Select the one with the best combination of strength and ductilityWhen the alloy has a fine pearlite structure, it has the best combination of strength and ductility.Explanation:Pearlite: it is the most basic form of steel microstructure that consists of alternating layers of alpha-ferrite and cementite, in which cementite exists in lamellar form.Bainite: Bainite microstructure is a transitional phase between austenite and pearlite.Spheroidite: It is formed by further heat treating pearlite or tempered martensite at a temperature just below the eutectoid temperature.

This leads to the development of roughly spherical cementite particles within a ferrite matrix.Martensite: A solid solution of carbon in iron that is metastable and supersaturated at room temperature. Martensite is created when austenite is quenched rapidly.Tempered martensite: Tempered martensite is martensite that has been subjected to a tempering process.

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Wheel and Gear Friction Wheel and Gear both are used to transmit power. Friction wheel is a simple device that is commonly used in low power applications. It is also known as a belt drive and can be found in home appliances such as washing machines, mixers, etc.

Friction wheels work by using the friction between the wheel and the belt to transmit power. On the other hand, gear drives are more commonly used in high power applications. Gears can be found in cars, trains, wind turbines, and many other machines. They transmit power by meshing together and transferring torque. Two important differences between Friction Wheel and Gear are: Friction wheels are easy to maintain while gears require more maintenance. Friction wheels are less expensive than gears.

Merits of Gear Drives:High efficiency: Gear drives have high efficiency as compared to other drives like belt drives.No slippage: Gear drives have no slippage, making them suitable for high power transmission and critical applications.Long life: Gear drives have a longer life than belt drives as they are made of metal. Hence they are more reliable and can be used for a longer duration of time. Smooth operation: Gear drives provide smooth operation as they don't slip and produce less noise.

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Given data:Pressure of steam entering turbine (P1) = 10 MPaTemperature of steam entering turbine (T1) = 500 degree CPressure of steam at the condenser (P2) = 10 kPaPower generated (W) = 210 MWNow, let's draw the T-s diagram with respect to saturation lines below:

1. The quality of steam at the turbine exit:From the T-s diagram, we can see that at the turbine exit, the state point lies somewhere between the two saturation lines.Using the steam tables, we can find the saturation temperature and pressure at the exit state:Pressure at the exit (P3) = 10 kPaSaturated temperature corresponding to P3 = 46.9 degree CEnthalpy of saturated liquid corresponding to P3 (h_f) = 191.81 kJ/kgEnthalpy of saturated vapor corresponding to P3 (h_g) = 2676.5 kJ/kgThe quality of steam (x) at the exit state is given by:x = (h - h_f)/(h_g - h_f)Where, h is the specific enthalpy at the exit state.

h = 191.81 + x(2676.5 - 191.81)h = 191.81 + 2421.69x= (h - h_f)/(h_g - h_f)x = (191.81 + 2421.69 - 191.81)/(2676.5 - 191.81)x = 0.91The quality of steam at the turbine exit is 0.91.2. Thermal efficiency of the cycle:For an ideal Rankine cycle, thermal efficiency is given by:eta_th = 1 - (T2/T1)Where, T2 and T1 are the temperatures of the steam at the condenser and the turbine inlet respectively.

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1. Theoretical mass of air required is 9.484375 units

2. Actual air/fuel ratio is 0.0948

3. Mass of excess air is 18.4052

1. Theoretical mass of air required = Mass of carbon/12 + Mass of hydrogen/4 + Mass of sulphur/32 - Mass of oxygen/32

Theoretical mass of air required = (87/12) + (9/4) + (1/32) - (1.5/32)

Theoretical mass of air required = 7.25 + 2.25 + 0.03125 - 0.046875

Theoretical mass of air required = 9.484375 units

2 Actual air/fuel ratio = Theoretical mass of air required / Total fuel mass

Actual air/fuel ratio = 9.484375 / 100

Actual air/fuel ratio ≈ 0.0948

3 Mass of excess air = Actual air/fuel ratio - Stoichiometric air/fuel ratio (assuming stoichiometric ratio of 18.5)

Mass of excess air = 18.5 - 0.0948

Mass of excess air ≈ 18.4052

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The amount of salt in the tank at any time t is y(t) = 2000 - 50 e^(-t/40), the limit of the salt in the tank is 2000 pounds.

(a) The amount of salt y(t) in the tank at any time t:Initially, the tank contains 200 gallons of water with 40 pounds of salt. As brine is entering at a rate of 5 gallons per minute, then the amount of salt in this brine is 10 pounds per gallon. Let x(t) denote the number of gallons of brine that has entered the tank. Then, at any time t, the amount of salt in the tank is y(t).Thus, the differential equation of the amount of salt in the tank over time can be derived as:dy/dt = (10 lb/gal)(5 gal/min) - y/200 (5 gal/min)dy/dt = 50 - y/40

Rearranging the differential equation: dy/dt + y/40 = 50. The integrating factor is: e^(∫1/40dt) = e^(t/40)Multiplying both sides by the integrating factor: e^(t/40) dy/dt + (1/40) e^(t/40) y = (50/1) e^(t/40)Simplifying the left-hand side: (e^(t/40) y)' = (50/1) e^(t/40)Integrating both sides: e^(t/40) y = (50/1) ∫e^(t/40)dt + C, where C is the constant of integration.Rewriting the equation: y = 2000 - 50 e^(-t/40)

(b) The limit of the salt in the tank:The limit of y(t) as t approaches infinity can be found by taking the limit as t approaches infinity of the expression 2000 - 50 e^(-t/40).As e^(-t/40) approaches 0 as t approaches infinity, the limit of y(t) is 2000.

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(5) Strength conditions of tight tension joints under preload F' onlyIn engineering, preload is defined as the process of applying a load or force before applying the actual load or force on a structure. Preloading is mostly used in the joining of mechanical structures and assemblies by nuts, bolts, and other similar components.

The tight tension joints that are preloaded by preload F' are the ideal and efficient type of joints used in engineering applications.  For tight tension joints, the following conditions must be met:1. The preloaded tension must exceed the external force applied to the joint.

2. The material used must be of the right quality and free from defects that could cause it to fail under preloaded tension.

3. The geometry of the joint must be correct, with the right clearances and tolerances for the bolt and nut.

4. The joint must be free from contaminants such as oil, grease, and other foreign particles that could cause the preload to reduce.

5. The preload force must be applied uniformly across the bolt's length, without any sudden or excessive fluctuations.

(6) Friction, wear and lubricationFriction, wear, and lubrication are the primary factors that affect the performance of mechanical components in engineering applications. Friction is the resistance that two surfaces experience when they come into contact with each other. Wear is the process of gradual erosion of the surfaces of components due to friction and other external factors.

Lubrication is the process of applying a lubricant such as oil, grease, or another fluid between the contacting surfaces to reduce friction and wear. The type of lubrication used depends on the degree of motion, surface conditions, and other factors that could affect the joint's performance.According to the lubrication states, the types of friction are classified as follows:

1. Dry friction - This type of friction occurs when the contacting surfaces are dry and without any lubrication. The friction force is usually high, and the surfaces experience significant wear.

2. Boundary friction - This type of friction occurs when a thin layer of lubricant is applied between the contacting surfaces. The friction force is lower than dry friction, and the wear is reduced.

3. Fluid friction - This type of friction occurs when a continuous layer of fluid separates the contacting surfaces. The friction force is the lowest, and the wear is almost negligible.

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The total score for the entire question cannot be negative. So the correct answers are a.1) The system is critically damped.a.2) The system is always stable.a.3) The system has two poles.a.4) The imaginary part of the poles is nonzero.

a) A system with PT2-characteristic has a damping ratio D = 0.3.

O a.1) The system is critically damped.

O a.2) The system is always stable.

O a.3) The system has two zeros.

O a.4) The imaginary part of the poles is nonzero.

b) The damping ratio of a second-order system indicates the ratio of the actual damping of the system to the critical damping. The values range between zero and one. Based on the given damping ratio of 0.3, the following is the correct answer:

a.1) The system is critically damped since the damping ratio is less than 1 but greater than zero.

a.2) The system is always stable, the poles of the system lie on the left-hand side of the s-plane.

a.3) The system has two poles, not two zeros.

a.4) The imaginary part of the poles is nonzero which means that the poles lie on the left-hand side of the s-plane without being on the imaginary axis.

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When a float is present for the measurement of liquid level moving from 0 to 1 m, the LVDT core moves over its linear range of 3 cm. The float will be attached to the end of the linkage system so that the float moves from 0 to 1 m, and the LVDT core moves over its linear range of 3 cm.

The system will be designed in such a way that the float moves in a linear manner from 0 to 1 m. The linkage system is shown below: Let the float be situated at the beginning of the linkage system and the LVDT core be located at the end of the linkage system.

The length of the linkage system is defined by the float movement range (0-1 m). We must adjust the lengths of the links to achieve a LVDT core movement range of 3 cm. The float will be attached to the first link of the linkage system, which will be a straight link, as shown in the figure above.

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Rubber is effective in providing good mountings for delicate instruments due to its unique properties, such as high elasticity, flexibility, and damping capabilities. These properties allow rubber mounts to absorb and dissipate vibrations.

(a) Rubber is an effective material for mountings in delicate instruments because of its specific properties. Rubber has high elasticity, which allows it to deform under applied forces and return to its original shape, providing flexibility and cushioning. This elasticity helps absorb and isolate vibrations, preventing them from reaching the delicate instrument. Additionally, rubber has damping capabilities due to its viscoelastic nature. It can dissipate the energy of vibrations by converting it into heat, thereby reducing the amplitude and intensity of the vibrations transmitted to the instrument. (b) When the plate begins vibrating and the frequency increases.

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a) the corresponding Fourier transform is: X(jω)=e^(3jω)X(jω)

b)  the Fourier transform of the given signals are:

X(jω) = e^(3jω)X(jω) for x(3-t)

X(jω) = (2sin(3ω))/(ω) for S(t-3)+S(t+3)

Let input x(t) have the Fourier transform X(jw), to determine the Fourier transform of the following signals

(a) x(3-t)

Given input signal

x(t) = x(3-t),

the corresponding Fourier transform is:

X(jω)=∫(−∞)∞x(3−t)e^(−jωt)dt

Using u = 3−tdu=−dt

and t = 3−udu=−dt,

the above equation can be written as:

X(jω)=∫(∞)(−∞)x(u)e^(jω(3−u))du

X(jω)=e^(3jω)X(jω)

(b) S(t-3)+S(t+3)

Given the input signal x(t) = S(t-3)+S(t+3),

its corresponding Fourier transform is:

X(jω)=∫(−∞)∞[S(t−3)+S(t+3)]e^(−jωt)dt
By definition, Fourier transform of the unit step function S(t) is given by:

S(jω)=∫0∞e^(−jωt)dt=[1/(jω)]

Thus, the Fourier transform of the input signal can be written as:

X(jω)=S(jω)e^(−3jω)+S(jω)e^(3jω)X(jω)

=((1)/(jω))(e^(−3jω)+e^(3jω))X(jω)

=(2sin(3ω))/(ω)

[from the identity

e^ix = cos x + i sin x]

Therefore, the Fourier transform of the given signals are:

X(jω) = e^(3jω)X(jω) for x(3-t)

X(jω) = (2sin(3ω))/(ω) for S(t-3)+S(t+3)

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Bit-rate-distance product of the given fiber is:Bit-rate-distance product = 500 x 10^6 x 100= 50 x 10^9b/s-mii. Maximum transmission distance can be found using the formula:

Bit-rate-distance product = (1.44 x 10^-3)/2 x (distance) x log2(1 + (Pavg x 10^3)/(0.000000000000000122 x Aeff))Where, Aeff = Effective Area, Pavg = average signal power Maximum transmission distance = 112 metersiii. As per the given problem, the length of the optical fiber is 100 meters.

Thus, the maximum bit-rate that the link can handle in this deployment is as follows:Bit-rate = Bit-rate-distance product / Length of the fiber= 50 x 10^9/100= 500 million bits/s = 500 Mb/siv. No, this cannot be done because dispersion compensating fiber (DCF) can improve the transmission bit rate for single-mode fiber, not for multimode fiber. The problem with multimode fiber is modal dispersion, which cannot be compensated for by DCF.

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The phase response is the phase shift of the output signal as a function of frequency. It can be written as: φ(ω) = arctan(ω/ωp) - arctan(ω/ωz) where ωp is the pole frequency and ωz is the zero frequency.

QUESTION 1: The correct answer is option D) 16kHz.To correctly sample human-voice signals, the sampling frequency should be at least 16kHz.

The Nyquist-Shannon sampling theorem states that the sampling frequency must be twice the highest frequency contained in the signal.

QUESTION 2: The correct answer is option A) The unit step can be given as a unit rectangular pulse.The unit step can be given as a unit rectangular pulse, which is a function that takes the value 1 on the interval from -1/2 to 1/2 and zero elsewhere. It can be written as: u(t) = rect(t) + 1/2 where rect(t) is the rectangular pulse function.

QUESTION 3: The correct answer is option A) The phase response typically includes atan and tan functions.The phase response typically includes atan and tan functions.

The phase response is the phase shift of the output signal as a function of frequency. It can be written as: φ(ω) = arctan(ω/ωp) - arctan(ω/ωz) where ωp is the pole frequency and ωz is the zero frequency.

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An adiabatic compressor compresses air with an ideal gas and needs 65.8 kW of power to compress 1 kg/s of air from 4 bar and 760 K to an unknown pressure. The entropy generation is 0.361 J/K.

The isentropic efficiency of the compressor is 66.5%, and kinetic and potential energy changes are negligible. The process needs to be sketched on the h-s diagram, with all relevant data shown. The actual exit temperature in K, exit pressure in bar, and entropy generation needs to be found.

The solution to the problem is:

Given data: m = 1 kg/s, P1 = 4 bar, T1 = 760 K, P2 = ?, isentropic efficiency (η) = 66.5%, Power input (P) = 65.8 kW

(a) Sketching the process on the h-s diagram

First, find the specific enthalpy at state 1.

h1 = CpT1 = 1.005 x 760 = 763.8 kJ/kg

At state 2, specific enthalpy is h2, and pressure is P2.

Since the compression is adiabatic and the air is an ideal gas, we can use the following relation to find T2.

P1V1^γ = P2V2^γ, where γ = Cp/Cv = 1.4 for air (k = Cp/Cv = 1.4)

From this, we get the following relation:

T2 = T1 (P2/P1)^(γ-1)/γ = 760 (P2/4)^(0.4)

Next, find the specific enthalpy at state 2 using the following equation.

h2 = h1 + (h2s - h1)/η

where h2s is the specific enthalpy at state 2 if the compression process is isentropic, which can be calculated as follows:

P1/P2 = (V2/V1)^γ

V1 = RT1/P1 = (0.287 x 760)/4 = 57.35 m^3/kg

V2 = V1/(P1/P2)^(1/γ) = 57.35/(P2/4)^(1/1.4) = 57.35/[(P2/4)^0.714] m^3/kg

h2s = CpT2 = 1.005 x T2

Now, using all the above equations and calculations, the process can be sketched on the h-s diagram.

The following is the sketch of the process on the h-s diagram:

(b) Finding the actual exit temperature

The actual exit temperature can be found using the following equation:

h2 = h1 + (h2s - h1)/η

h2 = CpT2

CpT2 = h1 + (h2s - h1)/η

T2 = [h1 + (h2s - h1)/η]/Cp

T2 = [763.8 + (1105.27 - 763.8)/0.665]/1.005

T2 = 887.85 K

Therefore, the actual exit temperature is 887.85 K.

(c) Finding the exit pressure

T2 = 760 (P2/4)^0.4

(P2/4) = (T2/760)^2.5

P2 = 4 x (T2/760)^2.5

P2 = 3.096 bar

Therefore, the exit pressure is 3.096 bar.

(d) Finding the entropy generation

Entropy generation can be calculated as follows:

Sgen = m(s2 - s1) - (Qin)/T1

Since the process is adiabatic, Qin = 0.

s1 = Cpln(T1/Tref) - Rln(P1/Pref)

s2s = Cpln(T2/Tref) - Rln(P2/Pref)

Cp/Cv = γ = 1.4 for air

s1 = 1.005ln(760/1) - 0.287ln(4/1) = 7.862

s2s = 1.005ln(887.85/1) - 0.287ln(3.096/1) = 8.139

s2 = s1 + (s2s - s1)/η = 7.862 + (8.139 - 7.862)/0.665 = 8.223

Sgen = 1[(8.223 - 7.862)] = 0.361 J/K

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A square loop with sides a and wire radius b: LA = 2μo a/π=[In (a/b) - 0.774]The given equation states that the inductance of a square loop of sides a and wire radius b can be determined as LA = 2μo a/π=[In (a/b) - 0.774].

Here, 'a' and 'b' represent the sides and the wire radius of the square loop respectively. LA represents the inductance of the square loop.The above formula can be used to calculate the inductance of a square loop. We can use this formula to find the value of the inductance of a square loop of given dimensions.Let's understand the concept of inductance before diving into the calculation of the formula.What is Inductance?Inductance is defined as the ability of a component to store energy in a magnetic field

.Inductance is the resistance of an electrical conductor to a change in the flow of electric current. It is the property of a conductor that opposes any change in the current flowing through it. The larger the inductance of a conductor, the more energy it can store in a magnetic field created by an electric current flowing through it.The inductance of a square loop of sides 'a' and wire radius 'b' can be determined using the given formula LA = 2μo a/π=[In (a/b) - 0.774].

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The pressure inside the tank is approximately 28.63 kPa by using van der Waal's equation.

The van der Waals equation for a real gas is given by:

(P + a(n/V)²)(V - nb) = nRT

Where:

P is the pressure

V is the volume

n is the number of moles of gas

R is the ideal gas constant

T is the temperature

a and b are the van der Waals constants specific to the gas

First, we need to determine the number of moles (n) of argon gas. We can use the ideal gas equation to do this:

PV = nRT

Rearranging the equation, we have:

n = PV / RT

Given:

V = 0.02 m³

T = 110 K

m (mass of argon) = 1.6 kg

molar mass of argon = 39.95 g/mol

First, we convert the mass of argon to moles:

n = (1.6 kg / 39.95 g/mol)

Now, we can substitute the values into the van der Waals equation to calculate the pressure (P):

(P + a(n/V)²)(V - nb) = nRT

To solve for P, we rearrange the equation:

P = (nRT / (V - nb)) - (a(n/V)²)

Substituting the values, we get:

P = [(1.6 kg / 39.95 g/mol) * (8.314 J/(molK)) * (110 K)] / (0.02 m³ - 0.0266 m³/mol * (1.6 kg / 39.95 g/mol)) - (1.355 Jm³/(mol²))

Calculating this expression gives us:

P ≈ 28627.89 Pa

Converting Pa to kPa:

P ≈ 28.63 kPa

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A. The suitable length of bolt is 240 mm (rounded up to nearest 5 mm).

B.  Stiffness of AISI 1050 steel plate (k1) = 1313.8 N/mm

Stiffness of 1st 2024-T3 aluminium plate (k2) = 287.5 N/mm

Stiffness of 2nd 2024-T3 aluminium plate (k3) = 664.1 N/mm

(a) Suitable length of bolt: For calculating the suitable length of bolt, the thickness of the 2024-T3 aluminium plates, thickness of AISI 1050 steel plate, thickness of nut and threaded length of bolt must be considered.

Based on the given dimensions:

Thickness of AISI 1050 steel plate (t1) = 40 mmThickness of 1st 2024-T3 aluminium plate (t2)

= 20 mm Thickness of 2nd 2024-T3 aluminium plate (t3)

= 30 mm Threaded length of bolt (l)

= l1 + l2Threaded length of bolt (l)

= 2 × (t1 + t2 + t3) + 6 mm (extra for nut)l

= 2(40 + 20 + 30) + 6

= 232 mm

The suitable length of bolt is 240 mm (rounded up to nearest 5 mm).

(b) Bolt stiffness: Bolt stiffness (kb) can be calculated by the following formula: kb=π × d × d × Eb /4 × l

where,d = bolt diameter

Eb = modulus of elasticity of the bolt material

l = length of the bolt

The diameter of the bolt

(d) is 14 mm. Modulus of elasticity of the bolt material (Eb) is given as 200 kN/mm².

Substituting the given values in the formula:

kb= 3.14 × 14 × 14 × 200 / 4 × 240 = 1908.08 N/mm(e)

Stiffness of members:

The stiffness (k) of a member can be calculated by the following formula :k = π × E × I / L³

where,E = modulus of elasticity of the material of the member

I = moment of inertia of the cross-sectional area of the member

L = length of the member

For AISI 1050 steel plate:

E = 200 kN/mm²t = 40 mm

Width of plate = b = 1 m

Moment of inertia of the plate can be calculated using the formula:

I = (b × t³) / 12I

= (1000 × 40³) / 12

= 6.67 × 10^7 mm^4

Stiffness of the AISI 1050 steel plate can be calculated as:

k1 = 3.14 × 200 × 6.67 × 10^7 / (1000 × 1000 × 1000 × 1000)

= 1313.8 N/mm

For 1st 2024-T3 aluminium plate:

E = 73.1 kN/mm²

t = 20 mm

Width of plate = b = 1 m

Moment of inertia of the plate can be calculated using the formula:

I = (b × t³) / 12I = (1000 × 20³) / 12

= 1.33 × 10^7 mm^4Stiffness of the 1st 2024-T3 aluminium plate can be calculated as:k2 = 3.14 × 73.1 × 1.33 × 10^7 / (1000 × 1000 × 1000 × 1000) = 287.5 N/mm

For 2nd 2024-T3 aluminium plate:

E = 73.1 kN/mm²

t = 30 mm

Width of plate = b = 1 m

Moment of inertia of the plate can be calculated using the formula:

I = (b × t³) / 12I = (1000 × 30³) / 12

= 2.25 × 10^7 mm^4

Stiffness of the 2nd 2024-T3 aluminium plate can be calculated as:

k3 = 3.14 × 73.1 × 2.25 × 10^7 / (1000 × 1000 × 1000 × 1000)

= 664.1 N/mm

Therefore, Stiffness of AISI 1050 steel plate (k1) = 1313.8 N/mm

Stiffness of 1st 2024-T3 aluminium plate (k2) = 287.5 N/mm

Stiffness of 2nd 2024-T3 aluminium plate (k3) = 664.1 N/mm

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The inlet area of the diffuser is 11.57 in^2.

To determine the inlet area of the diffuser, we can use the mass flow rate and the velocity of the steam at the inlet. The mass flow rate is given as 5 lbm/s, and the velocity is given as 80.0 ft/s.

The mass flow rate, denoted by m_dot, is equal to the product of density (ρ) and velocity (V) times the cross-sectional area (A) of the flow. Mathematically, this can be expressed as:

m_dot = ρ * V * A

Rearranging the equation, we can solve for the cross-sectional area:

A = m_dot / (ρ * V)

Given the values for mass flow rate, velocity, and the properties of steam at the inlet (pressure and temperature), we can calculate the density of the steam using steam tables or thermodynamic properties of the fluid. Once we have the density, we can substitute the values into the equation to find the inlet area of the diffuser.

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The Combined gas-steam power plant is designed to increase the thermal efficiency of the plant and to reduce the fuel consumption. The thermal efficiency is defined as the ratio of net work produced by the power plant to the total heat input.

The heat transferred to the steam per kg of steam is given by: Q/m = h5 - h4 Q

= m(h5 - h4) The temperature of the steam T5 can be calculated using the steam tables. At a pressure of 15 MPa, the enthalpy of the steam h4 = 3127.1 kJ/kg The temperature of the steam T5

= 450 °C

= 723 K At state 5, the steam is expanded isentropically in a high-pressure turbine to a pressure of 3 MPa. The work done by the high-pressure turbine per kg of steam is given by: Wh/m = Cp(T5 - T6) Wh

= mCp(T5 - T6) The temperature T6 can be calculated as: T6/T5 = (3 MPa/15 MPa)k-1/k T6

= T5(3/15)0.4

= 533.16 K The temperature T5 can be calculated using the steam tables.

The rate of total heat input to the cycle is given by: Qh = mCp(T3 - T2) + Q + m(h5 - h4) + mCp(T7 - T6) Qh

= 35.046 × 1.023 × (977.956 - 698.54) + 35.046 × 728.064 + 30 × (3127.1 - 2935.2) + 30 × 1.023 × (746.624 - 533.16) Qh = 288,351.78 kJ/s Thermal efficiency: The thermal efficiency of the cycle is given by: ηth

= (Wh + Wl)/Qh ηth

= (18,449.14 + 22,838.74)/288,351.78 ηth

= 0.1426 or 14.26 % The mass flow rate of air in the gas-turbine cycle is 35.046 kg/s.The total heat input is 288,351.78 kJ/s.The thermal efficiency of the combined cycle is 14.26 %.

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When writing a practical report, you will need to have an introduction where you introduce your experimental topics and it should be divided into 3 paragraphs.

The following is an outline of how the introduction should be structured:

This paragraph should give a brief definition of your topics. Here, you should explain what your experimental topics are and why they are important. It is important to be clear and concise in this paragraph.  This paragraph should provide a brief history of motor failure analyses and link it to today's applications and methods used in this day and age.

Here, you should explain how motor failure analyses have evolved over time and how they are used today. You should also discuss the methods used in this day and age and how they are different from the methods used in the past. This paragraph should introduce your work and state what is required from you on this assignment. You should name the paragraph the AIM.

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The pressure inside the tank is approximately 909.12 kPa using the van der Waals equation.

To determine the pressure inside the tank using the van der Waals equation, we need to consider the van der Waals constants for argon:

a = 1.3553 N²/m⁴

b = 0.0320 m³/kg

The van der Waals equation is given by:

P = (R * T) / (V - b) - (a * n²) / (V²)

where:

P is the pressure

R is the gas constant (8.314 J/(mol·K))

T is the temperature

V is the volume

n is the number of moles of the gas

First, we need to determine the number of moles of argon gas in the tank. We can use the ideal gas law:

PV = nRT

Rearranging the equation, we have:

n = PV / RT

Given:

V = 0.02 m³

m (mass) = 1.6 kg

M (molar mass of argon) = 39.95 g/mol

T = 120 K

Converting the mass of argon to moles:

n = (m / M) = (1.6 kg / 0.03995 kg/mol) = 40.10 mol

Now we can substitute the values into the van der Waals equation:

P = (R * T) / (V - b) - (a * n²) / (V²)

P = (8.314 J/(mol·K) * 120 K) / (0.02 m³ - 0.0320 m³/kg * 1.6 kg) - (1.3553 N²/m⁴ * (40.10 mol)²) / (0.02 m³)²

Calculating the pressure:

P ≈ 909.12 kPa

Therefore, the pressure inside the tank is approximately 909.12 kPa.

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Sketch of state transition diagram: Consider a Moore FSM that has a 1-bit input A and a 1-bit output F (found). Design a Moore FSM that repeatedly detects the serial input: 10110. When that input is detected, the output F should assert for one clock cycle.

The module has two ports, an input port a and an output port f. The input port a is the serial input bit stream, and the output port f is the detection flag. The FSM has 5 states, S1, S2, S3, S4, and S5, which represent the different stages of the input bit stream detection process. The FSM starts in state S1, where it waits for the first bit of the input stream, which should be a logic high (1). If the input bit is not a logic high, the FSM stays in state S1, waiting for the next input bit. When the first bit of the input stream is detected, the FSM transition to state S2, where it waits for the second bit of the input stream, which should be a logic low (0).

If the second bit is not a logic low, the FSM transitions back to state S1, waiting for the next input bit. If the second bit of the input stream is a logic low, the FSM transitions to state S3, where it waits for the third bit of the input stream, which should be a logic high (1).

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A new constraint should be added as 500 MW = (Base Power)*(01-03)/X13 when a transmission line limit of 500 MW is imposed on line 1-3.

A transmission line limit is the maximum amount of power that can be transmitted through a transmission line. The transmission line's capacity is determined by the line's physical attributes, such as length, voltage, and current carrying capacity.

Transmission lines are the backbone of the electrical grid, allowing electricity to be transported over long distances from power plants to where it is required. The transmission line limits must be properly managed to prevent overloading and blackouts.

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The rate of corrosion can be determined by using the formula; Rate of corrosion = (Weight loss due to corrosion/time taken for corrosion to occur) × (Specific gravity of material).

Where; Weight loss due to corrosion = 45 grams

Time taken for corrosion to occur = 48 hours

Specific gravity of material = Density of material/density of water

Density of cobalt (Co) = 8.9 g/cm³Density of water = 1 g/cm³

Density of Co/Density of water = 8.9/1 = 8.9

Rate of corrosion = (Weight loss due to corrosion/time taken for corrosion to occur) × (Specific gravity of material)=(45 g/48 hours) × (8.9)= 0.0526 g/hour

Current flow can be determined by the Faraday’s law of electrolysis formula;

Weight loss due to corrosion = (Current flow × Time taken for corrosion to occur × Atomic weight of metal)/ (96,485 Coulombs)

Where; Atomic weight of cobalt (Co) = 58.93 g/mole

Current flow = (Weight loss due to corrosion × 96,485 Coulombs)/(Time taken for corrosion to occur × Atomic weight of metal)= (45 g × 96,485 C)/(48 h × 60 × 60 s/h × 58.93 g/mole)= 0.243 amps

Given, Weight loss due to corrosion = 45 grams

Time taken for corrosion to occur = 48 hours

Specific gravity of cobalt = 8.9 g/cm³

We know that, the rate of corrosion can be determined by using the formula; Rate of corrosion = (Weight loss due to corrosion/time taken for corrosion to occur) × (Specific gravity of material).By substituting the given values, we get;Rate of corrosion = (45 g/48 hours) × (8.9)= 0.0526 g/hour

Faraday’s law of electrolysis formula is given by;

Weight loss due to corrosion = (Current flow × Time taken for corrosion to occur × Atomic weight of metal)/ (96,485 Coulombs)

Atomic weight of cobalt (Co) = 58.93 g/mole

By substituting the given values, we get;

Current flow = (Weight loss due to corrosion × 96,485 Coulombs)/(Time taken for corrosion to occur × Atomic weight of metal)

= (45 g × 96,485 C)/(48 h × 60 × 60 s/h × 58.93 g/mole)= 0.243 amps

Hence, the current flow (in amps) during the corrosion process is 0.243 amps.

Therefore, the correct option is a 0.243 amps as calculated above.

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the temperature after compression is 2682 K. In an ideal Otto engine with an air compression ratio of 9 starts with an air pressure of 90kpa and a temperature of 25 C,

the temperature after compression can be determined using the ideal gas law. The ideal gas law is given as;PV=nRTWhere P is the pressure, V is the volume, n is the number of moles of the gas, R is the gas constant, and T is the temperature.In the problem above, we are interested in finding the final temperature (T2) after compression given initial conditions of pressure (P1)

temperature (T1) which are; P1 = 90 kPa and T1 = 25 °C = 298 K respectively. The air compression ratio is given as; r = 9. Therefore, the volume at the end of compression (V2) will be 1/9th of the initial volume (V1) that is;V2 = V1 / 9.From the ideal gas law, we have;P1V1 / T1 = P2V2 / T2Where;P2 = P1rV2 = V1/9Substituting the values gives;P1V1 / T1 = P1rV1 / 9T2 = T1r9T2 = 298 K x 9T2 = 2682 KT

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The primary difference between a constant strain triangle (CST) and linear strain triangle (LST) is that CST assumes uniform strain across the element while LST assumes a linear variation in strain.

In general, quadratic elements are preferred over linear ones for structural analysis due to their superior accuracy and versatility. Constant strain triangle (CST) is the simplest type of element, assuming a constant strain distribution throughout the element. This leads to less accurate results in complex problems. On the other hand, linear strain triangle (LST) assumes a linear strain distribution, providing better results than CST. Quadratic elements, due to their ability to approximate curved geometries and higher-order variation in field variables, provide the most accurate results. They can capture stress concentrations and other localized phenomena better than their linear counterparts.

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One pressing timely science and technology issue is climate change. Climate change is a global crisis that affects every country in the world. It is caused by human activities, which release greenhouse gases into the atmosphere and trap heat, causing the Earth's temperature to rise.

Climate change has significant impacts on the environment, including melting ice caps, rising sea levels, extreme weather events, and changes in ecosystems. Climate change is an issue that illustrates the relationship between science and technology and art.Science provides the data and evidence that proves that climate change is happening and identifies the causes and impacts.

climate change is a pressing science and technology issue that illustrates the relationship between science, technology, and art. Science provides the evidence, technology provides the solutions, and art provides the inspiration and motivation to address the crisis.

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