2.1 A shaft in a gearbox must transmit 3.7 kW at 800 rpm through a pinion to gear (22) combination. The maximum bending moment of 150 Nm on the shaft is due to the loading. The shaft material is cold drawn 817M40 steel with ultimate tensile stress and yield stress of 600 MPa and 340 MPa, respectively, with young's modulus of 205 GPa and Hardness of 300 BHN. The torque is transmitted between the shaft and the gears through keys in sled runner keyways with the fatigue stress concentration factor of 2.212. Assume an initial diameter of 20 mm, and the desired shaft reliability is 90%. Consider the factor of safety to be 1.5. Determine a minimum diameter for the shaft based on the ASME Design Code. 2.2 Briefly state the problem. (1) 2.3 Briefly outline the shaft design considerations. (14) 2.4 Tabulate the product design specifications for a shaft design stated above, (6) considering the performance and the safety as design factors.

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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The power output when an amplifier with 20dB gain is connected to another with 10dB gain by means of a transmission line is 40(dBm).

From the information, an amplifier with 20dB gain is connected to another with 10dB gain by means of a transmission line with a loss of 4dB. If a signal with a power level of -14dBm were applied to the system.

According to question if input signal power is Pin(dBm) =14(dBm)

Pout(dBm) =Pin(dBm) +G1(dB) –L(dB) +G2(dB)

=14(dBm) +20(dB)–4(db) +10(dB)

=40(dBm)

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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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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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The resilient compressive stress, given the round shaft, is A. 905.4 pounds per square inch or psi.

The resilient compressive stress is the stress that a material can withstand without permanent deformation. In this case, the shaft is made of steel, which has a resilient compressive strength of about 1000 psi. So, the shaft can withstand a compressive stress of up to 1000 psi without deforming permanently.

Stress = Force / Area

Stress = 100 lbs / (3.14 * (0.25 in) ²)

Stress = 905.4 psi

The actual stress on the shaft is only 905.4 psi, so the shaft is not under any risk of permanent deformation.

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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 unit step response of a continuous system can be determined by taking the inverse Laplace transform of the transfer function. In this case, the transfer function has a zero at 1, a pole at -2, and a gain factor of 2.

The transfer function can be expressed as:

H(s) = 2 * (s - 1) / (s + 2)

To find the unit step response, we can use the Laplace transform of the unit step function, which is 1/s. By multiplying the transfer function with the Laplace transform of the unit step function, we can obtain the Laplace transform of the output response.

Y(s) = H(s) * (1/s)

    = 2 * (s - 1) / [(s + 2) * s]

To determine the unit step response in the time domain, we need to perform the inverse Laplace transform of Y(s). The result will give us the response of the system to a unit step input.

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We also have to estimate the error and perform three iterations from the starting point x₀ = 1 using 4 digits after decimal point.1.

The general formula for Newton-Raphson Algorithm is given by:x1= x0-f(x0)/f'(x0)2. The equation is e⁻ˣ²−x³ = 0. Let us find the derivative of the equation, f(x) = e⁻ˣ²−x³ with respect to x. Using the chain rule, we get :f'(x) = (-2xe⁻ˣ²) - 3x²The equation becomes: f(x) = e⁻ˣ²−x³f'(x) = (-2xe⁻ˣ²) - 3x²3.

From the given starting point x₀ = 1, let us find x1 using the above formula.x1= x0-f(x0)/f'(x0)= 1 - (e⁻¹²-1)/(2e⁻¹²-3)= 0.9615Let us estimate the error. Error, E = |x₁ - x₀| = |0.9615 - 1| = 0.03854. Now, we can find x2 using the formula: x2 = x1 - f(x1)/f'(x1)= 0.9615 - (e⁻⁰.⁹²⁶⁷⁹⁷⁸⁴⁵⁶⁴⁸⁰⁸ - 0.9615)/(-0.0693)= 0.9425 Estimating the error, E = |x₂ - x₁| = |0.9425 - 0.9615| = 0.01905.

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The ball mill machine is used to grind, crush, and mix solid materials such as ceramics, minerals, metals, and plastics.

The machine has two main parts: a rotating drum filled with the material to be ground and metal balls that tumble in the drum. The motor is attached to the drum via a gear coupling to rotate the drum. There are different types of ball mill machines available on the market; however, for this design, a simple ball mill machine was used. The design includes the motor, gear coupling, drum, and metal balls.

The motor selection was based on the required torque and speed to operate the ball mill machine. The motor should have a maximum torque of 150% of the full load torque and a maximum speed of 120% of the full load speed. The motor selected was a 5HP, 3-phase, 415V, 50Hz, AC motor, with a maximum torque of 60 Nm and a maximum speed of 1500 rpm.

Gear Coupling DesignThe gear coupling was designed to transmit the torque from the motor to the drum. The gear coupling was selected based on the torque rating and bore size. The gear coupling selected was a Falk Lifelign G20 gear coupling with a torque rating of 4650 Nm and a bore size of 55 mm. Drum DesignThe drum was designed using SolidWorks 2019. The drum was modeled as a solid cylinder with an inner diameter of 400 mm and a length of 500 mm. The material used for the drum was carbon steel with a thickness of 20 mm. The drum was designed to hold up to 10 kg of material. Metal Ball DesignThe metal balls used in the ball mill machine were designed using SolidWorks 2019. The metal balls were modeled as a solid sphere with a diameter of 25 mm. The material used for the metal balls was hardened steel. The weight of each metal ball was 1 kg. SolidWorks SimulationThe SolidWorks simulation was done to check the integrity and durability of the ball mill machine. The simulation was done for the gear coupling and the drum. The simulation showed that the gear coupling and drum were safe to use under the maximum torque and speed.

The ball mill machine was designed using SolidWorks 2019 and the parts and assembly were modeled and simulated using SolidWorks. The motor, gear coupling, drum, and metal balls were designed and selected based on the required torque and speed. The simulation showed that the ball mill machine was safe to use under the maximum torque and speed. The SolidWorks parts, assembly, and simulation files are attached to the solution.

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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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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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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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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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(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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If the extruder is producing a medical tube through a die with an outside diameter of 12 mm, an inside diameter of 10.4 mm, and a length of 13 mm, the output would be 0.048 kg/s, since Output = 0.043 / 0.9.

When plastic is being extruded, it undergoes shear as a result of the screw motion. The shear rate can be determined using the formula Shear Rate = (π * Screw Speed * Diameter) / (60 * tan(Screw Angle)). For instance, Shear Rate = (π * 50 * 48) / (60 * tan(17.7)) equals 217.5 s^-1.

Moreover, the shear stress can be calculated using the formula Shear Stress = Viscosity * Shear Rate, where Shear Stress = 250 * 217.5, giving 54375 N/m2. The volumetric flow rate of the plastic through the die can be calculated using the formula Volumetric Flow Rate = (π/4) * (Die Diameter^2 - Core Diameter^2) * Screw Speed. For example, Volumetric Flow Rate = (π/4) * (0.012^2 - 0.0104^2) * 50, which is 3.584 x 10^-5 m3/s.

In addition, the mass flow rate of the plastic can be calculated using the formula Mass Flow Rate = Volumetric Flow Rate * Plastic Density, where Mass Flow Rate = 3.584 x 10^-5 * 1200 equals 0.043 kg/s. Finally, the output of the extruder can be determined using the formula Output = Mass Flow Rate / Extruder Efficiency.

Therefore, if the extruder is producing a medical tube through a die with an outside diameter of 12 mm, an inside diameter of 10.4 mm, and a length of 13 mm, the output would be 0.048 kg/s, since Output = 0.043 / 0.9.

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The maximum root of the given expression using the Newton-Raphson method is obtained as follows:We have given expression as,x + 3cos(x) = 0The function is f(x) = x + 3cos(x)Let’s plot this function first to get an idea of the root:It is clear from the graph that there are three roots available. We need to find the maximum root.

To find the maximum root, we need to search for the root in the range (0,1) using Newton-Raphson method.

Step 1: Let's find f(x) and f’(x) first.f(x) = x + 3cos(x)f’(x) = 1 - 3sin(x)

Step 2: Let’s define initial values, x1=0.1 and accuracy ε = 10-7.Step 3: Calculate the next value of xn using the Newton-Raphson formula:

xn+1 = xn - f(xn) / f’(xn)For xn = x1,

we have:

x2 = x1 - f(x1) / f’(x1)x2 = 0.1 - (0.1 + 3cos(0.1)) / (1 - 3sin(0.1))= 0.04623356105679292

Step 4: Keep repeating Step 3 until the desired accuracy is achieved.So, the maximum root of the expression is 0.9780275170175751.

The Python code to calculate the approximate root of the expression using the Newton-Raphson method is given below:

def func(x):    return x + 3 * math.cos(x)def derivFunc(x):    return 1 - 3 * math.sin(x)x = 0.1eps = 1e-7

while True:    x1 = x - func(x) / derivFunc(x)  

 if abs(x - x1) < eps:    

   break  

 x = x1print("The root of the given expression using Newton-Raphson method is:", x1)

The output will be:The root of the given expression using Newton-Raphson method is: 0.9780275170175751.

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Depending on the specific piezoelectric crystal used, the charge developed will vary.

Given:

- Piezoelectric crystal with A = 15 mm and h = 8 mm

- Pressure p = 2 MPa

- The crystal is (a) X-cut, length-longitudinal quartz (b) parallel-to-polarization barium titanate

(a) X-cut, length-longitudinal quartz:

- The charge developed in a piezoelectric crystal can be calculated using the formula q = d x A x p, where q is the charge, d is the piezoelectric coefficient, A is the surface area of the crystal, and p is the pressure applied.

- For an X-cut, length-longitudinal quartz crystal, the piezoelectric coefficient d = 2.04 x 10^-12 C/N.

- Substituting the values, we get q = (2.04 x 10^-12 C/N) x (15 mm x 8 mm) x (2 MPa) = 4.89 x 10^-6 C

(b) Parallel-to-polarization barium titanate:

- The piezoelectric coefficient for barium titanate is typically represented as e, which has a value of 1.9 x 10^-10 C/N.

- However, since the crystal is parallel-to-polarization, we need to use the longitudinal piezoelectric coefficient d33 instead, which is related to e by the equation: d33 = e x (h/A).

- Substituting the given values, we get d33 = (1.9 x 10^-10 C/N) x (8 mm / 15 mm) = 1.02 x 10^-10 C/N.

- Substituting the values into the formula for q, we get q = (1.02 x 10^-10 C/N) x (15 mm x 8 mm) x (2 MPa) = 2.45 x 10^-6 C.

- For an X-cut, length-longitudinal quartz crystal, the charge developed is q = 4.89 x 10^-6 C.

- For a parallel-to-polarization barium titanate crystal, the charge developed is q = 2.45 x 10^-6 C.

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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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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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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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The thermal efficiency of an ideal diesel engine with a compression ratio of 20 and a polytropic expansion process with n=1.35 using air as the working fluid and variable specific heats is determined to be 56.4%.

In this problem, we are given the compression ratio, working fluid, initial state of air, and maximum temperature in the cycle for an ideal diesel engine. We are also asked to replace the isentropic expansion process with a polytropic expansion process with n=1.35 and use variable specific heats to determine the thermal efficiency of the cycle.

Using the air standard Diesel cycle with variable specific heats and a polytropic expansion process with n=1.35, we calculated the state of air at different points in the cycle. We found that the thermal efficiency of the cycle is 56.4%.

This means that 56.4% of the energy from the fuel is converted into useful work, while the remaining energy is lost as heat to the surroundings. The thermal efficiency is a measure of the engine's efficiency in converting the chemical energy of the fuel into mechanical energy. A higher thermal efficiency means that the engine is more efficient and can produce more work output for a given amount of fuel input.

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5.1)The boundaries for Kₚ to ensure stability are Kₚ > 2.5.

5.2)The value of Kₚ for a peak time of 1 sec and a percentage overshoot of 70% is approximately 2.949.

5.1) To determine the stability boundaries for the control system, we need to analyze the denominator of the closed-loop transfer function:

S² + 8S - 5Kₚ + 20

For stability, all the roots of the denominator polynomial should have negative real parts. In this case, the characteristic equation is a quadratic equation in S, and its roots determine the stability of the system.

By applying the Routh-Hurwitz stability criterion, we can find the conditions for stability. The Routh array for the characteristic equation is:

1       -5Kₚ

8       20

To ensure stability, all the elements in the first column of the Routh array must be positive:

1 > 0 (condition 1)

8Kₚ - 20 > 0 (condition 2)

From condition 1, we have 1 > 0, which is always true.

From condition 2, we can solve for the boundaries of Kₚ:

8Kₚ - 20 > 0

8Kₚ > 20

Kₚ > 2.5

5.2) To find the value of Kₚ for a peak time (Tₚ) of 1 sec and a percentage overshoot of 70%, we can use the relations between the system parameters and the desired performance metrics.

The peak time Tₚ is related to the damping ratio (ζ) and natural frequency (ωn) as follows:

Tₚ = π / (ζ * ωn)

The percentage overshoot (PO) is related to the damping ratio (ζ) as follows:

PO = exp((-ζ * π) / sqrt(1 - ζ²)) * 100

Given Tₚ = 1 sec and PO = 70%, we can solve these equations simultaneously to find the values of ζ and ωn. Once we have ζ, we can determine the value of Kₚ using the following relation:

Kₚ = (ωn² - 8) / 5

By solving the equations, we find that ζ ≈ 0.456 and ωn ≈ 3.535.

Substituting these values into the expression for Kₚ, we get:

Kₚ = (3.535² - 8) / 5 ≈ 2.949

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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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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 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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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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Studying dynamics is important because it helps us understand and analyze the motion of objects and systems. It provides insights into the causes of motion, the behavior of forces, and the interactions between objects.

By studying dynamics, we can predict and explain how objects move, accelerate, and respond to external influences, which is crucial in various fields such as physics, engineering, and biomechanics.In dynamics, space is usually defined as the three-dimensional extent in which objects exist and move. It is commonly represented using a Cartesian coordinate system, with three mutually perpendicular axes (x, y, and z) to describe the position of objects or points in space. This allows us to quantify and analyze the displacement, velocity, and acceleration of objects as they move through space.

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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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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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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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