Compute the estimated actual endurance limit for SAE 4130 WQT 1300 steel bar with a rectangular cross section of 20.0 mm by 60 mm. It is to be machined and subjected to repeated and reversed bending stress. A reliability of 99% is desired

Answer:

Given

Decimal variable: currentSales

Decimal test value: 1000

Boolean variable: newCustomer

Boolean default value: true

The following code segment is written in Java

if(currentSales == 1000 || newCustomer)

{

//Some statements

}

The Program above tests two conditions using one of statement

The first condition is to check if currentSales is 1000

== Sign is s a relational operator used for comparison (it's different from=)

|| represents OR

The second condition is newCustomer, which is a Boolean variable

If one or both of the conditions is true, the statements within the {} will be executed

Meaning that, both conditions doesn't have to be true;

At least 1 condition must be true for the statement within the curly braces to be executed

Answer:

(a) 1.125 ft, Section factor = 22.78

(b) 42.75 ft

Explanation:

Hydraulic radius is given by [tex]R_{H} = \frac{A}{P}[/tex] Where

A = Cross sectional area of flow and

P = Perimeter  h

Since the cross section is a circle  then at depth 4 of 4.5 the perimeter

[tex]=2 \pi r-\frac{\theta }{360} *2 \pi r[/tex]  where r = 2.25 and θ = 102.1 °

perimeter = 10.1 ft and the  area = [tex]=\pi r^2-\frac{\theta }{360} * \pi r^2[/tex] =  11.39 ft²

Therefore [tex]R_{H} = \frac{11.39}{10.1} = 1.125 ft[/tex]

Section factor is given by for critical flow = Z = A×√D

= 11.39 ft² × √(4 ft) = 22.78

for normal flow Z =[tex]Z_{} ^{2} = \frac{A^{3}}{T}[/tex] = 22.78

(b) The alternate depth of flow is given by

for a given flow rate, we have from chart for flow in circular pipes

Alternative depth = 0.9×45 = 42.75 ft

Answer:

Negotiated speed should be lower. Perception/reaction time is too less than design values.

Explanation:

Given:

- The claimed safe speed V_1 = 65 mi/h

- Sight distance D = 510 ft

- The practical deceleration a = 11.2 ft/s   ... according to standards

Find:

Assuming practical stopping distance, comment on the student whether the claim is correct or not

Solution:

- Calculate the practical stopping distance:

                      d = V_1^2 / 2*a

                      d = ( 65 * 1.46 )^2 / 2*11.2 = 402.054 ft

- Solve for reaction distance d_r is as follows:

                     d_r = D - d = 510 - 402.054 = 107.945 ft

- The perception/time reaction is:

                   t_r = d_r/V_1 = 107.945 / 94.9

                  t_r = 1.17 sec

Answer: The perception/reaction time t_r = 1.17 s is well below the t = 2.3 s.

Hence, the safe speed should be lower.

Answer:True

Explanation:

Answer:

True

Explanation:

Input bias current:

It is a small current that flows in parallel with the input terminals of op-amp to bias the input transistors. This current gets converted into voltage and amplified which results in incorrect output results. This bias current Ib+ and Ib- flows in the positive and negative input terminals of the op-amp.  

Ib+ and Ib- create errors of opposite polarity. Therefore, bias current can be minimized by carefully adding a resistor in the positive input terminal.

Input offset current:

Unfortunately, a small error still remains due to the mismatch between input currents Ib+ and Ib-.

This input offset current error can be adjusted by adding a potentiometer and resistor in the negative input terminal.

Answer:

The answer to the question is

The heat transferred in the process is -274.645 kJ

Explanation:

To solve the question, we list out the variables thus

R-134a = Tetrafluoroethane

Intitial Temperaturte t₁ = 100 °C

Initial pressure = 3.5 bar = 350 kPa

For closed system we have m₁ = m₂ = m

ΔU = m×(u₂ - u₁) = ₁Q₂ -₁W₂

For constant pressure process we have

Work done = W = [tex]\int\limits^a_b P \, dV[/tex]  = P×ΔV = P × (V₂ - V₁) = P×m×(v₂ - v₁)

From the tables we have

State 1 we have h₁ = (490.48 +489.52)/2 = 490 kJ/kg

State 2 gives h₂ = 206.75 + 0.75 × 194.57= 352.6775 kJ/kg

Therefore Q₁₂ = m×(u₂ - u₁) + W₁₂ = m × (u₂ - u₁) + P×m×(v₂ - v₁)

= m×(h₂ - h₁) = 2.0 kg × (352.6775 kJ/kg - 490 kJ/kg) =-274.645 kJ

Answer:

P= 541.5 kW.

Explanation:

Given that

velocity of water after leaving the nozzle ,v= 95 m/s

The mass flow rate of the water , m= 120 kg/s

The power generated P is given as

[tex]P=\dfrac{1}{2}mv^2[/tex]

Now by putting the values in the above equation we get

[tex]P=\dfrac{1}{2}\times 120\times 95^2\ W[/tex]

P=541500  W

The  power in kW will be 541.5 kW.

Therefore the answer will be 541.5 kW

P= 541.5 kW.

The power generation potential of the water jet is 542.25 kW.

To determine the power generation potential of the water jet, we need to calculate the kinetic energy of the water jet and then convert it to power. The kinetic energy of the water jet can be calculated using the formula KE = 0.5 * m * v^2, where m is the mass flow rate of the water and v is the velocity of the water jet. Given that the flow rate is 120 kg/s and the velocity is 95 m/s, we can calculate the kinetic energy to be KE = 0.5 * 120 * 95^2 = 0.5 * 120 * 9025 = 542,250 J/s.

To convert the kinetic energy to power, we divide by the time taken to deliver the energy. Since the flow rate is given in kg/s, we can assume the time taken is 1 second. Therefore, the power generation potential of the water jet is 542,250 J/s, or 542.25 kW.

Answer:

Yes and no

Explanation:

The thermodynamic equation for the heat transfer in a constant volume process is the following:

[tex]Q=\Delta U=mC_V\Delta T[/tex]

where Q is the required heat, U is the internal energy, m the mass of the gas, C_V the heat capacity assuming consant volume and [tex]\Delta T[/tex] is the change in temperature.

If you assume the heat capacity doesn't change with temperature at which the gas is currently at then the heat transfer depends solely on the change in temperature. With this assumption the transfered heat would be the same in both cases.

In reality the heat capacity does change with respect to temperature. Depending on the type of gas. In reality there would be difference in heat transfered between 295/205 K and 245/255 . Only then you wouldn't use the [tex]\Delta T[/tex] expression since the integral would be different depending on the heat capacity in relation to temperature.

The work done by the gas is -940 kJ

Explanation:

In this process, we are told that the product of pressure and volume remains constant:

[tex]pV=const.[/tex]

so we can write

[tex]p_1 V_1 = p_2 V_2[/tex]

where

[tex]p_1 = 1.4 bar[/tex] is the initial pressure

[tex]p_2 = 6.8 bar[/tex] is the final pressure

[tex]V_1=4.25 m^2[/tex] is the initial volume

Solving for [tex]V_2[/tex], we find the final volume:

[tex]V_2=\frac{p_1V_1}{p_2}=\frac{(1.4)(4.25)}{6.8}=0.875 m^3[/tex]

Now by looking at the equation of state of an ideal gas:

[tex]pV=nRT[/tex] (1)

we notice that since [tex]pV=const.[/tex], this means that also the absolute temperature of the gas T remains constant (because the number of moles n does not change). Therefore this is an isothermal process: the work done in an isothermal process is given by

[tex]W=nRTln(\frac{V_2}{V_1})[/tex]

And by looking again at (1), we  can substitute (nRT) with (pV), so we get

[tex]W=p_1 V_1 ln (\frac{V_2}{V_1})[/tex]

Converting the pressure into SI units,

[tex]p_1 = 1.4 bar = 1.4\cdot 10^5 Pa[/tex]

So the work done is

[tex]W=(1.4\cdot 10^5)(4.25)ln(\frac{0.875}{4.25})=-9.4\cdot 10^5 J[/tex]

Which means -940 kJ. This value is negative since the work is done by the surroundings on the gas (because the gas is compressed).

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Water is likely to leak out of the hole in the pipe when the average velocity is 5.0 m/s due to high dynamic pressure. With a slower velocity of 0.5 m/s, air might enter the pipe if the static pressure at the hole is less than atmospheric, but this requires additional details to confirm.

The question addresses the behavior of water within a pipe system under different flow conditions, involving principles of fluid dynamics. Specifically, it asks whether water will leak out of a small hole in a vertical pipe or if air will enter into the pipe through the hole given two different average velocities of water flow.

Case 1: Average velocity of 5.0 m/s

With an average velocity of 5.0 m/s, the dynamic pressure of the flowing water is considerable, and thus, water is likely to leak out of the hole due to the higher pressure inside the pipe compared to atmospheric pressure.

Case 2: Average velocity of 0.5 m/s

With a decreased velocity of 0.5 m/s, the dynamic pressure is significantly lower. If the static pressure at the hole's location is less than the atmospheric pressure, air might enter the pipe; however, if it is still higher than atmospheric, water would continue to leak out. The determination requires additional information, such as the height of the water column above the hole and any applied pressures at the water's source.

Generally, the behavior can be predicted using Bernoulli's principle and the continuity equation for incompressible flow, which together relate the velocities, pressures, and cross-sectional areas in different sections of a pipe.

Answer:

a= - 2.6 m/s².

Explanation:

u = 32 m/s

The speed after 6 s is half of u

[tex]v= \dfrac{32}{2}=16\ m/s[/tex]

t= 6 s

The average acceleration = a

We know v = u +at

v=final velocity

u=initial velocity

Now by putting the values in the above equation

16= 32 + a x 6

[tex]a=\dfrac{16-32}{6}\ m/s^2[/tex]

[tex]a=-2.6\ m/s^2[/tex]

Therefore the acceleration will be - 2.6 m/s².

a= - 2.6 m/s².

Negative indicates that velocity and acceleration is is opposite direction.

Answer:

a) -k* dT / dx = q_o

b) T(x) = -280*x + 80

c) T(L) = -4 C

Explanation:

Given:

- large plane wall of thickness L = 0.3 m

- thermal conductivity k = 2.5 W/m · °C

- surface area A =12 m2.

- left side of the wall at net heat flux q_o = 700 W/m2 @ x = 0

- temperature at that surface is measured to be T1 =80°C.

Find:

- Express the differential equation and the boundary conditions for steady one dimensional heat conduction through the wall.

- Obtain a relation for the variation of temperature in the wall by solving the differential equation

- Evaluate the temperature of the right surface of the wall at x = L.

Solution:

- The mathematical formulation of Rate of change of temperature is as follows:

                                    d^2T / dx^2 = 0

- Using energy balance:

                                    E_out = E_in

                                   -k* dT / dx = q_o

- Integrate the ODE with respect to x:

                                     T(x) = - (q_o / k)*x + C

- Use the boundary conditions, T(0) = T_1 = 80C

                                     80 = - (q_o / k)*0 + C

                                      C = 80 C

-Hence the Temperature distribution in the wall along the thickness is:

                                    T(x) = - (q_o / k)*x + 80

                                    T(x) = -(700/2.5)*x + 80

                                    T(x) = -280*x + 80

- Use the above relation and compute T(L):

                                     T(L) = -280*0.3 + 80

                                     T(L) = -84 + 80 = -4 C

Differential equation: [tex]\(\frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0\).[/tex]Temperature variation: [tex]\(T(x) = T_1\).[/tex]Temperature at x = L is [tex]\(T(L) = T_1\)[/tex].

a. The differential equation for steady one-dimensional heat conduction through the wall is given by Fourier's law:

[tex]\[ \frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0 \][/tex]

This equation states that the rate of change of heat flux with respect to distance ( x ) is constant and equal to zero in steady-state conditions.

The boundary conditions are:

1. At  x = 0 : [tex]\( q = q_0 \)[/tex], [tex]\( T = T_1 \)[/tex]

2. At  x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex], as there is no heat flux across the right surface of the wall.

b. To solve the differential equation, integrate it twice:

[tex]\[ k \frac{{dT}}{{dx}} = C_1 \][/tex]

[tex]\[ \frac{{dT}}{{dx}} = \frac{{C_1}}{{k}} \][/tex]

[tex]\[ T = \frac{{C_1}}{{k}} x + C_2 \][/tex]

Apply the boundary conditions:

At  x = 0 : [tex]\( T = T_1 \)[/tex]

[tex]\[ C_2 = T_1 \][/tex]

At  x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex]

[tex]\[ \frac{{C_1}}{{k}} = 0 \][/tex]

[tex]\[ C_1 = 0 \][/tex]

Therefore, the temperature variation in the wall is given by:

[tex]\[ T(x) = T_1 \][/tex]

c. The temperature of the right surface of the wall at  x = L  is equal to [tex]( T(L) = T_1 \)[/tex], as there is no variation in temperature along the wall according to the solution obtained in part b.

Answer:

The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.

Xₚբᵣ = 0.632

X꜀ₘբᵣ = 0.5

Xₚբᵣ > X꜀ₘբᵣ

Explanation:

From the reaction rate coefficient, it is evident the reaction is a first order reaction

Performance equation for a CMFR for a first order reaction is

kτ = (X)/(1 - X)

k = reaction rate constant = 0.05 /day

τ = Time constant or holding time = V/F₀

V = volume of reactor = 280 m³

F₀ = Flowrate into the reactor = 14 m³/day

X = conversion

k(V/F₀) = (X)/(1 - X)

0.05 × (280/14) = X/(1 - X)

1 = X/(1 - X)

X = 1 - X

2X = 1

X = 1/2 = 0.5

For the PFR

Performance equation for a first order reaction is given by

kτ = In [1/(1 - X)]

The parameters are the same as above,

0.05 × (280/14) = In (1/(1-X)

1 = In (1/(1-X))

e = 1/(1 - X)

2.718 = 1/(1 - X)

1 - X = 1/2.718

1 - X = 0.3679

X = 1 - 0.3679

X = 0.632

The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.

To determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste stream, we compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation r = kC. In a CMFR, the volume is constant, while in a PFR, the volume varies. Therefore, a PFR may be more efficient depending on the reactor design.

To determine whether a CMFR (Continuous Mixed Flow Reactor) or a PFR (Plug Flow Reactor) is more efficient in removing a reactive compound from the waste stream, we need to compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation: r = kC, where r is the reaction rate, k is the reaction rate coefficient, and C is the concentration of the reactive compound.

In a CMFR, the volume is constant, so the reactor volume (280 m3) is equal to the product of the flow rate (14 m3·day−1) and the residence time (t), which is the time it takes for the fluid to pass through the reactor. Therefore, t = V/Q = 280/14 = 20 days.

In a PFR, the volume varies along the length of the reactor, and the residence time is defined as the integral of the volume divided by the flow rate. Using the equation t = ∫V/Q, we can calculate the residence time for a PFR.

Since the residence time for a CMFR is fixed at 20 days, and the residence time for a PFR can be longer or shorter depending on the reactor design, a PFR may be more efficient in removing the reactive compound from the waste stream under steady-state conditions with a first-order reaction.

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Answer:Name resolution traffic goes to the single zone

Administrative burden

Submit

Centralized Management

Explanation:DNS(Domain name system) is a term used in the internet which Describes the conversion of alphabetical names into Numerical representations,he a large Organisation as stated which spans through different Geographical areas continue with a single Domain name system it will lead to the following.

Name resolution traffic will increase which might delay the execution of tasks

Administrative burden will be Increased as it is carrying out a wide range of activities.

Centralized management which may affect the flow of work.

Answer: Check the attached

Explanation:

Answer:

attached below

Explanation:

Answer:

(A) architectural sheet metal roofing

Explanation:

 By the name itself we can judge that the 'Architectural sheet metal roofing' is a kind of metal roofing.

And these type of metal roofing is primarily used for small and big houses, small buildings and as well as in a building that is for commercial use they can be totally flat as well as little bit sloped.  

And the words similarly like batten and standing seam, and flat seam all tells us that these are the types of architectural sheet metal roofing.

Answer:

59.9%

Explanation:

R^2 =(-0.774)^2  = 0.599

59.9% of fuel is accounted for

Answer:

the time, in hours = 4.07hrs

Explanation:

The detailed step by step derivation and appropriate integration is as shown in the attached files.

Answer:

Explanation:

Code used will be like

using System;

using System.Collections.Generic;

using System.Linq;

using System.Text;

using System.Threading.Tasks;

namespace PaintingWall

{

class Room

{

public int length, width, height,Area,Gallons;

public Room(int l,int w,int h)

{

length = l;

width = w;

height = h;  

}

private int getLength()

{

return length;

}

private int getWidth()

{

return width;

}

private int getHeight()

{

return height;

}

public void WallAreaAndNumberGallons()

{

Area = getLength() * getHeight() * getWidth();

if (Area < 350)

{

Gallons = 1;

}

else if (Area > 350)

{

Gallons = 2;

}    

Console.WriteLine ("The area of the Room is " + Area);

Console.WriteLine("The number of gallons paint needed to paint the Room is " + Gallons);

}

}

class PaintingDemo

{

static void Main(string[] args)

{

int l, w, h;

Room[] r = new Room[8];

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room "+(i+1));

Console.Write("Enter Length : ");

l = Convert.ToInt32(Console.ReadLine() );

Console.Write("Enter Width : ");

w = Convert.ToInt32(Console.ReadLine());

Console.Write("Enter Height : ");

h= Convert.ToInt32(Console.ReadLine());

r[i] = new Room(l,w,h);

Console.WriteLine();

}

for (int i = 0; i <= 7; i++)

{

Console.WriteLine("Room " + (i + 1));

r[i].WallAreaAndNumberGallons();

}

Console.ReadKey();  

}

}

}

Answer:

1) Glass

2) Rock sheet

3) Lexan

4) Masonite

b) k = 8.75 W/m.K

Explanation:

Given:

The thermal conductivity of certain materials as follows:

-Sheet Rock: k = 0.43 W/(m*K)

-Masonite: k = 0.047 W/(m*K)

-Glass: k = 0.72 W/(m*K)

-Lexan: k = 0.19 W/(m*K)

Data Given:

- Q = 100 W

- A = 8 m^2

- dT = 10 C

- L = 7 m

Find:

a) list the materials in order from most conductive to least conductive

b) calculate the thermal conductivity using Fourier's Equation

Solution:

- We know from Fourier's Law the relation between Heat transfer and thermal conductivity as follows:

                                   Q = k*A*dT / L

- From the relation above we can see that rate of heat transfer is directly proportional to thermal conductivity k.

- Hence, the list in order of decreasing conductivity is as follows:

- The list of materials in the decreasing order of thermal conductivity k is:

           1) Glass                 k = 0.72 W/m.K        

           2) Rock sheet      k = 0.43 W/m.K

           3) Lexan               k = 0.19 W/m.K

           4) Masonite          k = 0.047 W/m.K

- Use the relation given above we can compute the thermal conductivity k with the given data:

                                 k = Q*L / (A*dT)

                                 k = (100 W * 7 m) / (8 m^2*10 C)

                                 k = 8.75 W/m.K

Answer:

Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L

Explanation:

The lagoon can be modelled as a Mixed flow reactor.

From the value of the decay constant (0.2/day), one can deduce that the decay reaction of the pollutant is a first order reaction.

The performance equation of a Mixed flow reactor is given from the material and component balance thus:

(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)

V = volume of the reactor (The lagoon) = ?

C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³

F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s

C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L

For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³

(-r) = kC (Since we know this decay process is a first order reaction)

This makes the performance equation to be:

(kVC₀/F₀) = (C₀ - C)/C

V = F₀(C₀ - C)/(kC₀C)

k = 0.2/day = 0.2/(24 × 3600s) = 2.31 × 10⁻⁶/s

V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)

V = 86580 m³

Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.

The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)

Hope this helps!

The volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".

The volume for the lagoon relation:

[tex]\to Q_1C_1 = Q_2C_2+KC_2V\\\\[/tex]

[tex]\to Q_1=0.10\\\\\to C_1=30\\\\\to Q_2=0.2\\\\\to C_2=10\\\\\to K=0.10[/tex]

Putting the value into the formula and calculating the volume:

[tex]\to (0.10 \times 30)=(0.2\times 10)+(0.10\times 10\times V \times (\frac{1}{24 \ hrs}) \times (\frac{1 \ hr}{3,600 \ sec})) \\\\\to (3)=(2)+(1\times V \times \frac{1}{86400}) \\\\\to 3-2=(1\times V \times \frac{1}{86400}) \\\\\to 1=(1\times V \times \frac{1}{86400}) \\\\\to 1 \times 86400 = V\\\\\to V= 8.64 \times 10^4 \ m^3\\\\[/tex]

Therefore, the calculated volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".

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The question is incomplete! Complete question along with answers and explanation is provided below.

Question:

A 15 Watt desk-type fluorescent lamp has an effective resistance of 200 ohms when operating (note: the 15 Watts is only associated with the lamp). It is in series with a ballast that has a resistance of 80 ohms and an inductance of .9H. The lamp and ballast are operated at 120V, 60Hz.

a) Draw the circuit

b) Calculate the power drawn by the lamp

c) Calculate the apparent power

d) Calculate the power factor

e) Calculate the reactive power

f) Calculate the size of the capacitor necessary to provide unity power factor correction

Explanation:

a) draw the circuit

Refer to the attached image.

As you can see in the attached drawing, it is a series circuit containing  two resistors and one inductor.

In a series circuit, current remains same throughout the circuit

The circuit is powered by an AC voltage source having voltage of 120 V and frequency 60 Hz.

The current flowing in the circuit can be found by ohm's law

 I = V/Z

where V is the voltage and Z is the total impedance of the circuit

 Z = R + XL

where  XL is the inductive reactance

XL = j2 π f L

XL = j2*π*60*0.9

XL = j339.29Ω

Total resistance is

R =200 + 80 = 280 Ω

Total impedance is

Z = 280 + j339.29 Ω

b) Calculate the power drawn by the lamp

First calculate the current

I = V/Z

I = 120/(280 + j339.29)

I = 0.272<-50.46° A  (complex notation)

P = I²R

P = (0.272)²200

P ≈ 15 W

Power drawn by the circuit

P=V*I*cos(50.46°)

P=20.77 W

c) Calculate the apparent power

A = VI*

A = 120*0.272<50.46°

A = 32.64<50.46° VA

d) Calculate the power factor

PF = cos(50.46)

PF = 0.63

e) Calculate the reactive power

Q = VIsin(50.46)

Q = 120*0.272<-50.46*sin(50.46)

Q = 25.13<-50.46  VAR

f) Calculate the size of the capacitor necessary to provide unity power factor correction

The required reactive compensation power is

Qc = P (tan(old) - tan(new))

Qc = 20.77 (tan(50.46) - tan(0))

Qc = 25.16 VAR

C = Qc/2πfV²

C = 25.16/2*π*60*120²

C = 4.63 uF

Hence adding a capacitor of 4.63 uF parallel to the load will improve the PF from 0.63 to 1.

Answer:

The Mach number of the throat for supersonic flow = M*  = 1

and the Mach number at exit = 2.44

For

a. Pa/Pt = 0.06, Me = 2.484 Supersonic flow

b. when Pa/Pt = 0.9725 Me = 0.1999 ≅ 0.2 or subsonic flow The mach number at the throat could also be determined given the temperature parameter

Explanation:

To solve the question we note that for a supersonic nozzle, the mach number at the throat = 1

Therefore M* = 1

[tex]\frac{A_{e} }{A^{*} } = 2.5[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(\gamma -1)M_{e} ^{2} }{\gamma +1} )^{\frac{\gamma +1}{2(\gamma -1)} }[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(0.4)M_{e} ^{2} }{2.4} )^{3 }[/tex] = [tex]\frac{1}{M_{e} } ({2+(0.4)M_{e} ^{2} })^{3 } = 34.56[/tex]

34.56Me = (2+(0.4)M²)³ expanding and collecting like terms we have

Possible solutions of  Me = 0.2395, 2.44, 0.90

Since flow is supersonic, Me = 2.44

a)

Solving for [tex]M_{e}[/tex] we have [tex]\frac{P_{a} }{P_{t} } =(1+\frac{\gamma -1}{2} M^{2} _{e} )^{\frac{-\gamma}{\gamma -1} }[/tex]

When Pa/Pt = 0.06 =[tex](1+\frac{1.4 -1}{2} M^{2} _{e} )^{\frac{-1.4}{1.4 -1} }[/tex] = [tex](1+0.2M^{2} _{e} )^{-3.5 }[/tex]

Solving, we get Me = 2.484 Supersonic flow

b)

When Pa/Pt = 0.9725,  Me = 0.1999 ≅ 0.2 or subsonic flow

The Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.

To find the Mach number at the throat and at the exit plane of a supersonic nozzle with an exit area 2.5 times the throat area for a steady, isentropic flow (gamma=1.4) discharging into an atmosphere with pressure Pa, we can use the following steps:

Calculate the critical pressure ratio:

pr_crit = (2 / (gamma + 1)) ** (gamma / (gamma - 1))

pr_crit = 0.5283

Calculate the Mach number at the throat:

Mt = sqrt((1 - pr_crit) / (gamma - 1))

Mt = 1.0859

Calculate the Mach number at the exit plane:

Me = sqrt((1 - (Pa / Pt)) / (gamma - 1))

Part a:

Pa/Pt = 0.06

Me = sqrt((1 - 0.06) / (1.4 - 1))

Me = 1.5329

Part b:

Pa/Pt = 0.9725

Me = sqrt((1 - 0.9725) / (1.4 - 1))

Me = 0.2622

Therefore, the Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.

For such more question on number

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Answer:

The question continues ; Determine the expressions for the velocity and acceleration of a point P as a function of time t, and determine the maximum velocity of point P during one cycle. Use the values r = 75mm and w = pie-rads/s

Explanation:

The diagram and the detailed step by step explanation is as shown in the attachment

In the Scotch-yoke mechanism, if the displacement is given by x = r sin(t), the velocity v = r cos(t) and acceleration a = -r sin(t). The negative sign in acceleration indicates that it is in the opposite direction to displacement.

The Scotch-yoke mechanism which converts rotary motion into reciprocating motion can be analyzed using principles of kinematics. If we have the displacement given by x = r sin(t), the velocity and acceleration can be derived from this displacement equation.

The velocity (v) is the time derivative of the displacement function, i.e., v = dx/dt = r cos(t).

The acceleration (a) is the time derivative of the velocity function, so a = dv/dt = -r sin(t).

The negative sign signifies that acceleration is in the opposite direction to displacement.

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Answer:

pressure is  825 lb/ft²

density is 1.20 × [tex]10^{-3}[/tex] slug/ft²

Explanation:

given data

p1 = 1800 lb/ft²

T1 = 500°

T2 = 400°

solution

we use here isentropic flow relation that is

[tex]\frac{P2}{P1} = (\frac{T2}{T1})^{\gamma / \gamma - 1 }[/tex]  

put here value we get pressure P2

P2 = 1800 ×  [tex](\frac{400}{500})^{3.5}[/tex]

P2 = 825 lb/ft²

and we know pressure is

pressure = [tex]\rho RT[/tex]

so for pressure 825 we get here  [tex]\rho[/tex]

825 = [tex]\rho[/tex] × 1716 × 400

[tex]\rho[/tex] = 1.20 × [tex]10^{-3}[/tex] slug/ft²

Answer:

Explanation:

Given

Take off speed [tex]v=162\ mph\approx 237.6\ ft/s[/tex]

distance traveled in runway [tex]d=5000 ft[/tex]

using motion of equation

[tex]v^2-u^2=2as[/tex]

where v=final velocity

u=initial velocity

a=acceleration

s=displacement

[tex](237.6)^2=2\times a\times 5000[/tex]

[tex]a=5.64\ ft/s^2[/tex]

Acceleration after take off [tex]a_2=3\ ft/s^2[/tex]

time taken to reach [tex]237.6 ft/s[/tex]

[tex]v=u+at[/tex]

[tex]237.6=0+5.64\times t[/tex]

[tex]t=42.127\ s [/tex]

after take off it take [tex]t_2[/tex] time to reach [tex]220 mph\approx 322.67[/tex]

[tex]322.67=237.6+3\times t_2[/tex]

[tex]t_2=28.35\ s[/tex]

total time taken [tex]t_0=t+t_1[/tex]

[tex]t_0=70.48\ s[/tex]

To determine the minimum cross-sectional area of the steel cables, calculate the allowable stress and use it along with the provided load. The result is a minimum area of 50 mm² for each cable.

The question involves calculating the minimum cross-sectional area of each steel cable designed to support a load with a safety factor, given the yield stress of the material. First, determine the allowable stress by dividing the yield stress by the factor of safety. In this case, the allowable stress is 200 MPa (400 MPa / 2.0). To find the minimum cross-sectional area (A), use the formula A = P / σ, where P = 10 kN = 10,000 N (the load) and σ (sigma) is the allowable stress in N/m². Convert 200 MPa to N/m² to get 200,000 N/m². Therefore, the minimum cross-sectional area required for each cable is 50 mm² (10,000 N / 200,000 N/m²).

Answer:

The fluid level difference in the manometer arm = 22.56 ft.

Explanation:

Assumption: The fluid in the manometer is incompressible, that is, its density is constant.

The fluid level difference between the two arms of the manometer gives the gage pressure of the air in the tank.

And P(gage) = ρgh

ρ = density of the manometer fluid = 60 lbm/ft³

g = acceleration due to gravity = 32.2 ft/s²

ρg = 60 × 32.2 = 1932 lbm/ft²s²

ρg = 1932 lbm/ft²s² × 1lbf.s²/32.2lbm.ft = 60 lbf/ft³

h = fluid level difference between the two arms of the manometer = ?

P(gage) = 9.4 psig = 9.4 × 144 = 1353.6 lbf/ft²

1353.6 = ρg × h = 60 lbf/ft³ × h

h = 1353.6/60 = 22.56 ft

A diagrammatic representation of this setup is presented in the attached image.

Hope this helps!

Answer:

6.30 miles/hour

Explanation:

Newton's second law applies here. In simple terms:

[tex]F = ma[/tex]

where F = Force (Thrust) in N

           a = acceleration (m/s²)

The acceleration can be given by rearranging the  formula to give:

[tex]a = \frac{F}{N}[/tex]

  = [tex]\frac{(686.68*10^{6} )}{24811505120150.2656} \\= 0.0000277 m/s\\= 6.03 miles/hr[/tex]

Answer:

a. Calories required = 622710 calories

b. Energy = 1000 times much energy

Calories = 622710000 calories

Explanation:

Given:

h = Depth of pot = 20cm = 0.2m

Diameter of pot = 20cm = 0.2m

r = ½ *diameter = ½ * 0.2

r = 0.1m

Density = 1000kg/m³

Water temperature = 1°C

a.

First, we calculate the volume of the water(pot)

V = Volume = πr²h

V = 22/7 * 0.1² * 0.2

V = 0.044/7

V = 0.00629m³

M = Mass of water = Volume * Density

M = 0.00629m³ * 1000kg/m³

M = 6.29kg

M = 6.29 * 1000 grams

M = 6290g

The water is at 1°C, so it needs to gain 99°C to reach boiling point

So, Calories = 99 * 6290

Calories required = 622710 calories

b.

If we consider that D for the pot is 20 cm, approximately how much more energy is needed to heat a hot tub with D = 2 m? How many calories is that?

Depth of pot = 20cm

Depth of pot = 0.2m

Depth of hot tube = 2m

Energy is directly proportional to D³

Since the depth of hot the is 10 times greater than that of the pot

It'll require 10³ much more energy

Energy = 10³

Energy = 1000 times much energy

Calories required = 622710 * 1000

Calories = 622710000 calories