For these problems, assume air behaves as an ideal gas with R = 0.287 k J k g K. A compressor operates at steady state and takes in air from ambient 0 kPa, gage and 300 K. The outlet pressure is 50 kPa, gage and 400 K. Determine: the mass flow rate if the inlet area is 10 cm2 and the inlet pressure is -2 kPa, gage. the minimum outlet temperature that is possible for this compressor. the isentropic efficiency of the compressor, assuming no heat loss. if there is a heat loss of 30 kJ/kg, the work required to run the compressor, and the new isentropic efficiency. A turbine receives air at 50 kPa, gage and 800 K. It discharges to 0 kPa, gage, and the outlet temperature is measured as 500 K. The mass flow rate is the same as in the previous problem. Determine: the maximum work the turbine can do under these conditions. the heat loss, if the turbine works isentropically.

Answer:

To answer this question we assumed that the area units and the thickness units are given in inches.

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

[tex] V = A*t [/tex]

Where:

A: is the surface area = 160

t: is the thickness = 0.002

Assuming that the units given above are in inches we proceed to calculate the volume:

[tex]V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}[/tex]    

Now, using the density we can find the mass:

[tex] m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g [/tex]

Finally, with the Avogadros number ([tex]N_{A}[/tex]) and with the atomic mass (A) we can find the number of atoms (N):

[tex] N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms [/tex]    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

Answer:

F=265.39N

Explanation:

In order to solve this problem, we must first draw a diagram of what the problem looks like (see attached picture).

So, in order to find the torque we must use the torque formula:

[tex]\tau=rxF[/tex]

or

[tex]\tau=rFsin\theta[/tex]

In this case we can either use 125° or 55°, the answer should be the same. So when solving the formula for F we get:

[tex]F=\frac{\tau}{r sin\theta}[/tex]

so we can now substitute values:

[tex]F=\frac{50Nm}{(23x10^{-2}m)sin 125^{o}}[/tex]

so

F=265.39N

Answer:

F = 0.0022N

Explanation:

Given:

Surface area (A) = 4,000mm² = 0.004m²

Viscosity = µ = 0.55 N.s/m²

u = (5y-0.5y²) mm/s

Assume y = 4

Computation:

F/A = µ(du/dy)

F = µA(du/dy)

F = µA[(d/dy)(5y-0.5y²)]

F = (0.55)(0.004)[(5-1(4))]

F = 0.0022N

Answer:

COP(heat pump) = 2.66

COP(Theoretical maximum) = 14.65

Explanation:

Given:

Q(h) = 200 KW

W = 75 KW

Temperature (T1) = 293 K

Temperature (T2) = 273 K

Find:

COP(heat pump)

COP(Theoretical maximum)

Computation:

COP(heat pump) = Q(h) / W

COP(heat pump) = 200 / 75

COP(heat pump) = 2.66

COP(Theoretical maximum) = T1 / (T1 - T2)

COP(Theoretical maximum) = 293 / (293 - 273)

COP(Theoretical maximum) = 293 / 20

COP(Theoretical maximum) = 14.65

Answer:

The function is [tex]-\sin x[/tex] and the constant of integration is [tex]C = - 1[/tex].

Explanation:

The resultant expression is equal to the sum of a constant multiplied by the integral of a given function and an integration constant. That is:

[tex]a = k\cdot \int\limits {f(x)} \, dx + C[/tex]

Where:

[tex]k[/tex] - Constant, dimensionless.

[tex]C[/tex] - Integration constant, dimensionless.

By comparing terms, [tex]k = 2[/tex], [tex]C = -1[/tex] and [tex]\int {f(x)} \, dx = \cos x[/tex]. Then, [tex]f(x)[/tex] is determined by deriving the cosine function:

[tex]f(x) = \frac{d}{dx} (\cos x)[/tex]

[tex]f(x) = -\sin x[/tex]

The function is [tex]-\sin x[/tex] and the constant of integration is [tex]C = - 1[/tex].

Answer:

(b) Given the Weibull parameters of example 11-3, the factor by which the catalog rating must be increased if the reliability is to be increased from 0.9 to 0.99.

Equation 11-1: F*L^(1/3) = Constant

Weibull parameters of example 11-3: xo = 0.02 (theta-xo) = 4.439 b = 1.483

Explanation:

(a)The Catalog rating(C)

   Bearing life:[tex]L_1 = L , L_2 = 2L[/tex]

   Catalog rating: [tex]C_1 = C , C_2 = ? ,[/tex]

From given equation bearing life equation,

[tex]F\times\frac{1}{3} (L_1) = C_1 ...(1) \\\\ F\times\frac{1}{3} (L_2) =C_2...(2)[/tex]

we Dividing eqn (2) with (1)

[tex]\frac{C_2}{C_1} =\frac{1}{3} (\frac{L_2}{L_1})\\\\ C_2 = C*(\frac{2L}{L})\frac{1}{3} \\\\ C_2 = 1.26 C[/tex]

The Catalog rating increased by factor of 1.26

(b) Reliability Increase from 0.9 to 0.99

[tex]R_1 = 0.9 , R_2 = 0.99[/tex]

Now calculating life adjustment factor for both value of reliability from Weibull parametres

[tex]a_1 = x_o + (\theta - x_o){ ln(\frac{1}{R_1} ) }^{\frac{1}{b}}[/tex]

[tex]= 0.02 + 4.439{ ln(\frac{1}{0.9} ) }^{\frac{1}{1.483}} \\\\ = 0.02 + 4.439( 0.1044 )^{0.67}\\\\a_1 = 0.9968[/tex]

Similarly

[tex]a_2 = x_o + (\theta - x_o){ ln(\frac{1}{R_2} ) }^{\frac{1}{b} }\\\\ = 0.02 + 4.439{ ln(1/0.99) }^{\frac{1}{1.483} }\\\\ = 0.02 + 4.439( 0.0099 )^{0.67}\\\\a_2 = 0.2215[/tex]

Now calculating bearing life for each value

[tex]L_1 = a_1 * LL_1 = 0.9968LL_2 = a_2 * LL_2 = 0.2215L[/tex]

Now using given ball bearing life equation and dividing each other similar to previous problem

[tex]\frac{C_2}{C_1} = (\frac{L_2}{L_1} )^{\frac{1}{3} }\\\\ C_2 = C* (\frac{0.2215L }{0.9968L} )^{1/3}\\\\ C_2 = 0.61 C[/tex]

Catalog rating increased by factor of 0.61

Complete Question:

A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.

Answer:

[tex]T_{min} =[/tex] 26 mins 40 secs

Explanation:

Reduction in depth, Δd = 20 mm

Depth of cut, [tex]d_c = 0.5 mm[/tex]

Number of passes necessary for this reduction, [tex]n = \frac{\triangle d}{d_c}[/tex]

n = 20/0.5

n = 40 passes

Tool width, w = 5 mm

Width of metal plate, W = 200 mm

For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times

Speed of tool, v = 100 mm/s

[tex]Time/pass = \frac{40*400}{400} \\Time/pass = 40 sec[/tex]

minimum time required to reduce the depth of the plate by 20 mm:

[tex]T_{min} =[/tex] number of passes * Time/pass

[tex]T_{min} =[/tex] n * Time/pass

[tex]T_{min} =[/tex] 40 * 40

[tex]T_{min} =[/tex]  1600 = 26 mins 40 secs

Answer:

the minimum time required to reduce the depth of the plate by 20 mm is 26 minutes 40 seconds

Explanation:

From the given information;

Assuming the tool moves 100 mm/sec

The number of passes required to reduce the depth from 30 mm to 20 mm can be calculated as:

Number of passes = [tex]\dfrac{30-20}{0.5}[/tex]

Number of passes = 20

We know that the width of the tool is 5 mm; therefore, to reduce the depth per pass; the tool have to travel 20 times

However; the time per passes is;

Time/pass = [tex]\dfrac{20*L}{velocity \ of \ the \ feed}[/tex]

where;

length L = 400mm

velocity of the feed is assumed as 100

Time/pass  [tex]=\dfrac{20*400}{100}[/tex]

Time/pass = 80 sec

Thus; the minimum time required to reduce the depth of the plate by 20 mm can be estimated as:

[tex]T_{min} = Time/pass *number of passes[/tex]

[tex]T_{min} = 20*80[/tex]

[tex]T_{min} = 1600 \ sec[/tex]

[tex]T_{min}[/tex] = 26 minutes 40 seconds

Answer:

[F cos θ - k(m2g - Fsin θ) + m1g]/(m1 + m2)

Explanation:

Since F cos θ - k(m2g - Fsin θ) - m1(a + g) = m2a

Expanding the bracket containing a, we have

F cos θ - k(m2g - Fsin θ) - m1a + m1g = m2a

Collecting the terms in a to the right-hand-side of the equation, we have

F cos θ - k(m2g - Fsin θ) + m1g =  m1a + m2a

Factorizing a out, we have

F cos θ - k(m2g - Fsin θ) + m1g = (m1 + m2)a

Dividing both sides by (m1 + m2), we have

[F cos θ - k(m2g - Fsin θ) + m1g]/(m1 + m2) = a

So, a = [F cos θ - k(m2g - Fsin θ) + m1g]/(m1 + m2)

Mathematical modeling aids in technological design by simulating how proposed system might behave. The correct option is 2.

Mathematical modelling describes a real world problem in mathematical terms or in the form of equations. This makes an engineer to discover new features about the problem and designer to alter his design for better function and output.

Mathematical models allow engineers and designers to understand how the proposed model and actual prototype will be produced.

Thus, the correct option is 2.

Learn more about mathematical modelling

https://brainly.com/question/731147

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Answer:

D = 1311.94 lb/ft

Explanation:

We are given the velocity as;

V = 55 mph.

First of all, let's convert it to ft/s

V = 55 × (5280/3600) ft/s

V = 80.67 ft/s

The equation for the aerodynamic drag force on the car is given as;

D = C_d•½ρ•V²•A

Where;

C_d is drag coefficient = 0.21

ρ is density of air

V is velocity = 80.67 ft/s

A is area 55 ft²

Now, from tables, the density of air under S.T.P condition is 1.225 kg/m³. Converting to lb/ft³ gives; 0.0624 lb/ft³

Plugging in the relevant values, we have;

D = 0.21•½•0.0624•80.67²•30

D = 1311.94 lb/ft

Question:Technician A say's that The most two-stroke engines have a pressure type lubrication system. Technician be says that four stroke engines do not require the mixing of oil with gasoline . Which of them is correct ?

Answer: Technician  B is   correct

Explanation: Two types of engines exist , the two stroke (example, used in chainsaws)  is a type of engine that uses two strokes--a compression stroke and a return stroke to produce power in a crankshaft combustion cycle and the  four stroke engines(eg lawnmowers) which  uses four strokes,  2-strokes during  compression and exhaustion accompanied by 2 return strokes for each of the initial process to produce power in a combustion cycle.

While a 2 stroke system engine, requires mixing of oil and fuel to the crankshaft before  forcing  the mixture into the cylinder and do not require a pressurized system.  The 4 stroke system uses a splash and pressurized system where oil is not mixed with gasoline but drawn from the sump and  directed to  the main moving  parts of  crankshaft through its channels.

We can therefore say that Technician A is wrong while Technician B is  correct

Answer:

0.31

126.23 kg/s

Explanation:

Given:-

- Fluid: Water

- Turbine: P3 = 8MPa , P4 = 10 KPa , nt = 85%

- Pump: Isentropic

- Net cycle-work output, Wnet = 100 MW

Find:-

- The thermal efficiency of the cycle

- The mass flow rate of steam

Solution:-

- The best way to deal with questions related to power cycles is to determine the process and write down the requisite properties of the fluid at each state.

First process: Isentropic compression by pump

       P1 = P4 = 10 KPa ( condenser and pump inlet is usually equal )

      h1 = h-P1 = 191.81 KJ/kg ( saturated liquid assumption )

       s1 = s-P1 = 0.6492 KJ/kg.K

       v1 = v-P1 = 0.001010 m^3 / kg

       P2 = P3 = 8 MPa( Boiler pressure - Turbine inlet )

       s2 = s1 = 0.6492 KJ/kg.K   .... ( compressed liquid )

- To determine the ( h2 ) at state point 2 : Pump exit. We need to determine the wok-done by pump on the water ( Wp ). So from work-done principle we have:

                           [tex]w_p = v_1*( P_2 - P_1 )\\\\w_p = 0.001010*( 8000 - 10 )\\\\w_p = 8.0699 \frac{KJ}{kg}[/tex]

- From the following relation we can determine ( h2 ) as follows:

                          h2 = h1 + wp

                          h2 = 191.81 + 8.0699

                          h2 = 199.88 KJ/kg

Second Process: Boiler supplies heat to the fluid and vaporize

- We have already evaluated the inlet fluid properties to the boiler ( pump exit property ).

- To determine the exit property of the fluid when the fluid is vaporized to steam in boiler ( super-heated phase ).

              P3 = 8 MPa

              T3 = ?  ( assume fluid exist in the saturated vapor phase )

              h3 = hg-P3 = 2758.7 KJ/kg

              s3 = sg-P3 = 5.7450 KJ/kg.K

- The amount of heat supplied by the boiler per kg of fluid to the water stream. ( qs ) is determined using the state points 2 and 3 as follows:

                          [tex]q_s = h_3 - h_2\\\\q_s = 2758.7 -199.88\\\\q_s = 2558.82 \frac{KJ}{kg}[/tex]

Third Process: The expansion ( actual case ). Turbine isentropic efficiency ( nt ).

- The saturated vapor steam is expanded by the turbine to the condenser pressure. The turbine inlet pressure conditions are similar to the boiler conditions.

- Under the isentropic conditions the steam exits the turbine at the following conditions:

             P4 = 10 KPa

             s4 = s3 = 5.7450 KJ/kg.K ... ( liquid - vapor mixture phase )

- Compute the quality of the mixture at condenser inlet by the following relation:

                           [tex]x = \frac{s_4 - s_f}{s_f_g} \\\\x = \frac{5.745- 0.6492}{7.4996} \\\\x = 0.67947[/tex]

- Determine the isentropic ( h4s ) at this state as follows:

                          [tex]h_4_s = h_f + x*h_f_g\\\\h_4_s = 191.81 + 0.67947*2392.1\\\\h_4_s = 1817.170187 \frac{KJ}{kg}[/tex]        

- Since, we know that the turbine is not 100% isentropic. We will use the working efficiency and determine the actual ( h4 ) at the condenser inlet state:

                         [tex]h4 = h_3 - n_t*(h_3 - h_4_s ) \\\\h4 = 2758.7 - 0.85*(2758.7 - 181.170187 ) \\\\h4 = 1958.39965 \frac{KJ}{kg} \\[/tex]

- We can now compute the work-produced ( wt ) due to the expansion of steam in turbine.

                        [tex]w_t = h_3 - h_4\\\\w_t = 2758.7-1958.39965\\\\w_t = 800.30034 \frac{KJ}{kg}[/tex]

- The net power out-put from the plant is derived from the net work produced by the compression and expansion process in pump and turbine, respectively.

                       [tex]W_n_e_t = flow(m) * ( w_t - w_p )\\\\flow ( m ) = \frac{W_n_e_t}{w_t - w_p} \\\\flow ( m ) = \frac{100000}{800.30034-8.0699} \\\\flow ( m ) = 126.23 \frac{kg}{s}[/tex]

Answer: The mass flow rate of the steam would be 126.23 kg/s

- The thermal efficiency of the cycle ( nth ) is defined as the ratio of net work produced by the cycle ( Wnet ) and the heat supplied by the boiler to the water ( Qs ):

                        [tex]n_t_h = \frac{W_n_e_t}{flow(m)*q_s} \\\\n_t_h = \frac{100000}{126.23*2558.82} \\\\n_t_h = 0.31[/tex]

Answer: The thermal efficiency of the cycle is 0.31

Answer:

height ≈ 60.60 m

Explanation:

The surveyor is trying to find the height of the hill . He takes a sight on the top of the hill and finds the angle of elevation is 40°. The distance from the hill where he measured the angle of elevation of 40° is not known.

Now he moves 150 m on level ground directly away from the hill and take a second sight from this point and measures the angle of elevation as 22°. This illustration forms a right angle triangle. The opposite side of the triangle is the height of the hill. The adjacent side of the triangle which is 150 m is the distance on level ground directly away from the hill.

Using tangential ratio,

tan 22° = opposite/adjacent

tan 22° = h/150

h = 150 × tan 22°

h = 150 × 0.40402622583

h = 60.6039338753

height ≈ 60.60 m

Answer:

amplitudes : 1 , 0.05, 0.05

frequencies : 50/[tex]\pi[/tex],   105/[tex]2\pi[/tex],  95/2[tex]\pi[/tex]

phases : [tex]\pi /2 , \pi /2 , \pi /2[/tex]

Explanation:

signal  s(t) = ( 1 + 0.1 cos 5t )cos 100t

signal s(t) = cos100t + 0.1cos100tcos5t . using the identity for cosacosb

         s(t) = cos100t + [tex]\frac{0.1}{2}[/tex] [cos(100+5)t + cos (100-5)t]

          s(t) = cos 100t + 0.05cos ( 100+5)t + 0.05cos (100-5)t

               =  cos100t + 0.05cos(105)t + 0.05cos 95t

             = cos 2 [tex](\frac{50}{\pi } )t + 0.05cos2 (\frac{105}{2\pi } )t + 0.05cos2 (\frac{95}{2\pi } )t[/tex] [ ∵cos (∅) = sin(/2 +∅ ]

= sin ( 2 [tex](\frac{50}{\pi } ) t[/tex]  + /2 ) + 0.05sin ( 2 [tex](\frac{105}{2\pi } ) t + /2 )[/tex] + 0.05sin ( 2 [tex](\frac{95}{2\pi } )t + /2[/tex] )

attached is the remaining part of the solution

Explanation:

Inputs and Outputs:

There are 3 inputs = I₁, I₂, and S

There are 2 outputs = O₁ and O₂

The given problem is solved in three major steps:

Step 1: Construct the Truth Table

Step 2: Obtain the logic equations using Karnaugh map

Step 3: Draw the logic circuit

Step 1: Construct the Truth Table

The given logic is

When S = 0 then O₁ = I₁ and O₂ = I₂

When S = 1 then O₁ = I₂ and O₂ = I₁

I₁     |     I₂     |    S    |    O₁    |    O₂

0     |     0     |    0    |    0    |     0

0     |     0     |    1     |    0    |     0

0     |     1      |    0    |    0    |     1

0     |     1      |    1     |    1     |     0

1      |     0     |    0    |    1     |     0

1      |     0     |    1     |    0    |     1

1      |     1      |    0    |    1     |     1

1      |     1      |    1     |    1     |     1

Step 2: Obtain the logic equations using Karnaugh map

Please refer to the attached diagram where Karnaugh map is set up.

The minimal SOP representation for output O₁

[tex]$ O_1 = I_1 \bar{S} + I_2 S $[/tex]

The minimal SOP representation for output O₂

[tex]$ O_2 = I_2 \bar{S} + I_1 S $[/tex]

Step 3: Draw the logic circuit

Please refer to the attached diagram where the circuit has been drawn.

Answer: find the answer in the explanation.

Explanation:

Some of the methods of science are hypothesis, observations and experiments.

Scientific research and findings cut across many areas and field of life. Field like engineering, astronomy and medicine e.t.c

Using models in research is another great method in which researchers and scientists can master a process of a system and predict how it works.

Wilbur and Orville Wright wouldn't have died if the first airplane built by these two wonderful brothers was first researched by using models and simulations before embarking on a test.

Modelling in science involves calculations by using many mathematical methods and equations and also by using many laws of Physics. In this present age, many of these are computerised.

In space science, it will save lives, time and money to first model and simulate if any new thing is discovered in astronomy rather than people going to the space by risking their lives.

Also in medicine, drugs and many medical equipment cannot be tested by using people. This may lead to chaos and loss of lives.

So research using models in astronomy, medicine, engineering and many aspects of science and technology can indeed save lives.

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

Answer:

A. 336.3 ft.

B. 279.83 ft.

Explanation:

 The formula for Stopping Site Distance SSD is given s follows;

[tex]SSD=\dfrac{2 \times R \times \theta \times \pi}{180}[/tex]

[tex]R = \dfrac{M}{1 - cos\theta}[/tex]

R = 5730/D = 5730/10 = 573 ft

[tex]\theta = \dfrac{SSD \times D}{200}[/tex]

D = Degree of curvature

θ = (SSD×D)/20 = (28.65×SSD)/573

Where:

[tex]\dfrac{M}{R} = 1 - cos \left(\dfrac{28.65 \cdot SSD}{R} \right)[/tex]

D = 10°

M is the distance of the obstruction from the highway = 24.5 ft

[tex]\dfrac{24.5}{573} = 1 - cos \left(\dfrac{28.65 \cdot SSD}{573} \right)[/tex]

[tex]cos \left(\dfrac{28.65 \cdot SSD}{573} \right) = 1 - \dfrac{24.5}{573} =\dfrac{1097}{1146}[/tex]

[tex]cos^{-1} \left(\dfrac{1097}{1146}\right) = \dfrac{28.65 \cdot S}{573} =16.815[/tex]

573×16.815= 28.65×SSD

SSD = 9635.13/28.65 = 336.3 ft

B. When M = 17, we have

[tex]\dfrac{17}{573} = 1 - cos \left(\dfrac{28.65 \cdot SSD}{573} \right)[/tex]

[tex]cos \left(\dfrac{28.65 \cdot SSD}{573} \right) = 1 - \dfrac{17}{573} = \dfrac{556}{573}[/tex]

[tex]cos^{-1} \left( \dfrac{556}{573}\right) = \dfrac{28.65 \cdot SSD}{573} = 13.99 \approx 14[/tex]

SSD = 573×14/28.65 = 279.83 ft.

Answer:

A.) Time = 13.75 seconds

B.) Total distance = 339 m

C.) V = 11.18 m/s

D.) V = 10.2 m/s

Explanation: Given that the automobile travels along a straight road at 15.65 m/s through a 11.18 m/s speed zone.

Then,

Initial velocity U of the motorist = 15.65m/s

acceleration a = - 3.05 m/s^2

Initial velocity u of the police man = 11.18 m/s

Acceleration a = 1.96 m/s^2

The police will overtake at distance S as the motorist decelerate and come to rest.

Where V = 0 and a = negative

While the police accelerate.

Using 2nd equation of motion for the motorist and the police

S = ut + 1/2at^2

Since the distance S covered will be the same, so

15.65t - 1/2×3.05t^2 = 11.18t +1/2×1.96t^2

Solve for t by collecting the like terms

15.56t - 1.525t^2 = 11.18t + 0.98t^2

15.56t - 11.18t = 0.98t^2 + 1.525t^2

4.38t = 2.505t^2

t = 4.38/2.505

t = 1.75 seconds approximately

But the motorist noticed the police car in his rear view mirror 12 s after the police car started the pursuit.

Therefore, the total time required for the police car to overtake the automobile will be:

12 + 1.75 = 13.75 seconds

B.) Using the same formula

S = ut + 1/2at^2

Where S = total distance travelled

Substitutes t into the formula

S = 11.18(13.75) + 1/2 × 1.96 (13.75)^2

S = 153.725 + 185.28

S = 339 m approximately

C.) The speed of the police car at the time it overtakes the automobile will be constant = 11.18 m/s

D.) Using first equation of motion

V = U - at

Since the motorist is decelerating

V = 15.65 - 3.05 × 1.75

V = 15.65 - 5.338

V = 10.22 m/s

Therefore, the speed of the automobile at the time it was overtaken by the police car is 10.2 m/ s approximately

The given question is incomplete. The complete question is as follows.

Steam is contained in a closed rigid container with a volume of 1 m3. Initially, the pressure and temperature of the steam are 10 bar and 500°C, respectively. The temperature drops as a result of heat transfer to the surroundings. Determine

(a) the temperature at which condensation first occurs, in [tex]^{o}C[/tex],

(b) the fraction of the total mass that has condensed when the pressure reaches 0.5 bar.

(c) What is the volume, in [tex]m^{3}[/tex], occupied by saturated liquid at the final state?

Explanation:

Using the property tables

  [tex]T_{1} = 500^{o}C[/tex],    [tex]P_{1}[/tex] = 10 bar

  [tex]v_{1} = 0.354 m^{3}/kg[/tex]

(a) During the process, specific volume remains constant.

  [tex]v_{g} = v_{1} = 0.354 m^{3}/kg[/tex]

  T = [tex](150 - 160)^{o}C[/tex]

Using inter-polation we get,

      T = [tex]154.71^{o}C[/tex]

The temperature at which condensation first occurs is [tex]154.71^{o}C[/tex].

(b) When the system will reach at state 3 according to the table at 0.5 bar then

  [tex]v_{f} = 1.030 \times 10^{-3} m^{3}/kg[/tex]

  [tex]v_{g} = 3.24 m^{3} kg[/tex]

Let us assume "x" be the gravity if stream

   [tex]v_{1} = v_{f} + x_{3}(v_{g} - v_{f})[/tex]

   [tex]x_{3} = \frac{v_{1} - v_{f}}{v_{g} - v_{f}}[/tex]

               = [tex]\frac{0.3540 - 0.00103}{3.240 - 0.00103}[/tex]

               = 0.109

At state 3, the fraction of total mass condensed is as follows.

  [tex](1 - x_{5})[/tex] = 1 -  0.109

                = 0.891

The fraction of the total mass that has condensed when the pressure reaches 0.5 bar is 0.891.

(c) Hence, total mass of the system is calculated as follows.

     m = [tex]\frac{v}{v_{1}}[/tex]

         = [tex]\frac{1}{0.354}[/tex]

         = 2.825 kg

Therefore, at final state the total volume occupied by saturated liquid is as follows.

     [tex]v_{ws} = m \times v_{f}[/tex]

                 = [tex]2.825 \times 0.00103[/tex]

                 = [tex]2.9 \times 10^{-3} m^{3}[/tex]

The volume occupied by saturated liquid at the final state is [tex]2.9 \times 10^{-3} m^{3}[/tex].

Answer:

a. The magnitude of the line source voltage is

Vs = 4160 V

b. Total real and reactive power loss in the line is

Ploss = 12 kW

Qloss = j81 kvar

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

Ss = 540.046 + j476.95 kVA

Ps = 540.046 kW

Qs = j476.95 kvar

Explanation:

a. The magnitude of the line voltage at the source end of the line.

The voltage at the source end of the line is given by

Vs = Vload + (Total current×Zline)

Complex power of first load:

S₁ = 560.1 < cos⁻¹(0.707)

S₁ = 560.1 < 45° kVA

Complex power of second load:

S₂ = P₂×1 (unity power factor)

S₂ = 132×1

S₂ = 132 kVA

S₂ = 132 < cos⁻¹(1)

S₂ = 132 < 0° kVA

Total Complex power of load is

S = S₁ + S₂

S = 560.1 < 45° + 132 < 0°

S = 660 < 36.87° kVA

Total current is

I = S*/(3×Vload)   ( * represents conjugate)

The phase voltage of load is

Vload = 3810.5/√3

Vload = 2200 V

I = 660 < -36.87°/(3×2200)

I = 100 < -36.87° A

The phase source voltage is

Vs = Vload + (Total current×Zline)

Vs = 2200 + (100 < -36.87°)×(0.4 + j2.7)

Vs = 2401.7 < 4.58° V

The magnitude of the line source voltage is

Vs = 2401.7×√3

Vs = 4160 V

b. Total real and reactive power loss in the line.

The 3-phase real power loss is given by

Ploss = 3×R×I²

Where R is the resistance of the line.

Ploss = 3×0.4×100²

Ploss = 12000 W

Ploss = 12 kW

The 3-phase reactive power loss is given by

Qloss = 3×X×I²

Where X is the reactance of the line.

Qloss = 3×j2.7×100²

Qloss = j81000 var

Qloss = j81 kvar

Sloss = Ploss + Qloss

Sloss = 12 + j81 kVA

c. Real power and reactive power supplied at the sending end of the line

The complex power at sending end of the line is

Ss = 3×Vs×I*

Ss = 3×(2401.7 < 4.58)×(100 < 36.87°)

Ss = 540.046 + j476.95 kVA

So the sending end real power is

Ps = 540.046 kW

So the sending end reactive power is

Qs = j476.95 kvar

Answer:

11.87 ft

Explanation:

Volumetric flow rate of the turbine Q = 1.7 million gal/min

Q = 1.7 x [tex]10^{6}[/tex] gal/min

1 gal = 0.00378541 m^3

1 min = 60 sec

1.7 x [tex]10^{6}[/tex] gal/min = (1.7 x [tex]10^{6}[/tex] x 0.00378541)/60 = 107.25 m^3

average velocity of flow through turbine = 34 ft/s

1 ft = 0.3048 m

34 ft/s = 34 x 0.3048 = 10.36 m/s

According to continuity equation, Q = AV

where Q = volumetric flow rate

A = Area of conduit

V = velocity of flow through turbine

A = Q/V = 107.25/10.36 = 10.35 m^2

Area of conduit = [tex]\pi r ^{2}[/tex]

radius r = [tex]\sqrt{\frac{area}{\pi } }[/tex] = [tex]\sqrt{\frac{10.35}{3.142} }[/tex] = 1.81 m

diameter = 2 x radius = 2 x 1.81 = 3.62 m

diameter = 3.62 m = 3.62 x 3.28084 = 11.87 ft

Answer:

In consolidated AB Co 300 shares.

Explanation:

Consolidation is a process in which two different organizations are united. In this question A Co and B Co are consolidated and a new Co names AB Co is formed. The shares of both the companies will be combined and their total share capital will be increased.

Answer:

a) 34 mm

b) 39 mm

c) 93.16%

Explanation:

power transmitted P = 28 kW 28000 W

angular speed N = 440 rpm

angular speed in rad/s Ω = 2[tex]\pi[/tex]N/60

Ω = (2 x 3.142 x 440)/60 = 46.08 rad/s

allowable shear stress τ = 80 MPa = 80 x [tex]10^{6}[/tex] Pa

torque T = P/Ω = 28000/46.08 = 607.64 N-m

a)  for the minimum diameter of a solid shaft, we use the equation

τ[tex]d^{3}[/tex]= [tex]\frac{16T}{ \pi}[/tex]

80 x [tex]10^{6}[/tex] x [tex]d^{3}[/tex] =  [tex]\frac{16*607.64}{3.142}[/tex] = 3094.28

[tex]d^{3}[/tex] = 3094.28/(80 x [tex]10^{6}[/tex]) = 0.0000386785

d = [tex]\sqrt[3]{0.0000386785}[/tex] ≅ 0.034 m = 34 mm

b) For a hollow shaft with outside diameter D = 40 mm = 0.04 m

we use the equation,

T = [tex]\frac{16}{\pi }[/tex] x τ x [tex]\frac{D^{4} - d^{4}}{D^{4} }[/tex]

where d is the internal diameter of the pipe

607.64 =  [tex]\frac{16}{3.142}[/tex] x 80 x [tex]10^{6}[/tex] x  [tex]\frac{0.04^{4} - d^{4}}{0.04^{4} }[/tex]

3.82 x [tex]10^{-12}[/tex] = [tex]0.04^{4} - d^{4}[/tex]

[tex]d^{4}[/tex] = [tex]\sqrt[4]{2.56*10^{-6} }[/tex]

d = 0.039 m = 39 mm

c) we assume weight is proportional to cross-sectional area

for solid shaft,

area = [tex]\pi r^{2}[/tex]

r = diameter/2 = 34/2 = 17 mm

area = 3.142 x [tex]17^{2}[/tex] = 907.92 mm^2

for hollow shaft, radius is also gotten as before

external area =  [tex]\pi r^{2}[/tex] = 3.142 x [tex]20^{2}[/tex] = 1256.64 mm^2

internal diameter =  [tex]\pi r^{2}[/tex]  = 3.142 x [tex]19.5^{2}[/tex] = 1194.59 mm^2

true area of hollow shaft = external area minus internal area

area = 1256.64 - 1194.59 = 62.05 mm^2

material weight saved is proportional to 907.92 - 62.05 = 845.87 mm^2

percentage weight saved is proportional to  845.87/907.92 x 100%

= 93.15%

Answer:

Explanation:

The value of a will be zero as it is provided that the particle is on the x-axis.

Calculate the velocity of particles along x-axis.

[tex]{\bf{V}} = 2{x^2}t{\bf{\hat i}} + [4y(t - 1) + 2{x^2}t]{\bf{\hat j}}{\rm{ m/s}}[/tex]

Substitute 0 for y.

[tex]\begin{array}{c}\\{\bf{V}} = 2{x^2}t{\bf{\hat i}} + \left( {4\left( 0 \right)\left( {t - 1} \right) + 2{x^2}t} \right){\bf{\hat j}}{\rm{ m/s}}\\\\ = 2{x^2}t{\bf{\hat i}} + 2{x^2}t{\bf{\hat j}}{\rm{ m/s}}\\\end{array}[/tex]

Here,

[tex]A = 2{x^2}t \ \ and\ \ B = 2{x^2}t[/tex]

Calculate the magnitude of vector V .

[tex].\left| {\bf{V}} \right| = \sqrt {{A^2} + {B^2}}[/tex]

Substitute

[tex]2{x^2}t \ \ for\ A\ and\ 2{x^2}t \ \ for \ B.[/tex]

[tex]\begin{array}{c}\\\left| {\bf{V}} \right| = \sqrt {{{\left( {2{x^2}t} \right)}^2} + {{\left( {2{x^2}t} \right)}^2}} \\\\ = \left( {2\sqrt 2 } \right){x^2}t\\\end{array}[/tex]

The velocity of the fluid particles on the x-axis is [tex]\left( {2\sqrt 2 } \right){x^2}t{\rm{ m/s}}[/tex]

Calculate the direction of flow.

[tex]\theta = {\tan ^{ - 1}}\left( {\frac{B}{A}} )[/tex]

Here, θ is the flow from positive x-axis in a counterclockwise direction.

Substitute [tex]2{x^2}t[/tex] as A and [tex]2{x^2}t[/tex] as B.

[tex]\begin{array}{c}\\\theta = {\tan ^{ - 1}}\left( {\frac{{2{x^2}t}}{{2{x^2}t}}} \right)\\\\ = {\tan ^{ - 1}}\left( 1 \right)\\\\ = 45^\circ \\\end{array}[/tex]

The direction of flow is [tex]45^\circ[/tex] from the positive x-axis.

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

[tex]E = G \times H[/tex]

Where G represents complex rated power and H is the inertia constant of turbo-generator.

[tex]E = 100 \times 8 \\\\E = 800 \: MJ[/tex]

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

[tex]$ P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2} $[/tex]

Where M is given by

[tex]$ M = \frac{E}{180 \times f} $[/tex]

[tex]$ M = \frac{800}{180 \times 50} $[/tex]

[tex]M = 0.0889 \: MJ \cdot s/ elec \: \: deg[/tex]

So, the rotor acceleration is

[tex]$ P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2} $[/tex]

[tex]$ 30 = 0.0889 \frac{d^2 \delta}{dt^2} $[/tex]

[tex]$ \frac{d^2 \delta}{dt^2} = \frac{30}{0.0889} $[/tex]

[tex]$ \frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2 $[/tex]

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

[tex]$ \Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2 $[/tex]

Where t is given by

[tex]1 \: cycle = 1/f = 1/50 \\\\10 \: cycles = 10/50 = 0.2 \\\\t = 0.2 \: sec[/tex]

So,

[tex]$ \Delta \delta = \frac{1}{2} \cdot 337.46 \cdot (0.2)^2 $[/tex]

[tex]$ \Delta \delta = 6.75 \: elec \: deg[/tex]

The change in torque in rpm/s is given by

[tex]$ \Delta \delta = \frac{337.46 \cdot 60}{2 \cdot 360\circ } $[/tex]

[tex]$ \Delta \delta =28.12 \: \: rpm/s $[/tex]

The rotor speed in revolutions per minute at the end of this period (10 cycles) is given by

[tex]$ Rotor \: speed = \frac{120 \cdot f}{P} + (\Delta \delta)\cdot t $[/tex]

Where P is the number of poles of the turbo-generator.

[tex]$ Rotor \: speed = \frac{120 \cdot 50}{4} + (28.12)\cdot 0.2 $[/tex]

[tex]$ Rotor \: speed = 1500 + 5.62 $[/tex]

[tex]$ Rotor \: speed = 1505.62 \:\: rpm[/tex]

Answer: b. STP

Explanation:

Twisted Pair Cables are best used for network cabling but are usually prone to EMI (Electromagnetic Interference) which affects the electrical circuit negatively.

The best way to negate this effect is to use Shielding which will help the cable continue to function normally. This is where the Shielded Twisted Pair (STP) cable comes in.

As the name implies, it comes with a shield and that shield is made out of metal which can enable it conduct the Electromagnetic Interference to the ground. The Shielding however makes it more expensive and in need of more care during installation.

Answer:

a.) I = 7.8 × 10^-4 A

b.) V(20) = 9.3 × 10^-43 V

Explanation:

Given that the

R1 = 20 kΩ,

R2 = 12 kΩ,

C = 10 µ F, and

ε = 25 V.

R1 and R2 are in series with each other.

Let us first find the equivalent resistance R

R = R1 + R2

R = 20 + 12 = 32 kΩ

At t = 0, V = 25v

From ohms law, V = IR

Make current I the subject of formula

I = V/R

I = 25/32 × 10^3

I = 7.8 × 10^-4 A

b.) The voltage across R1 after a long time can be achieved by using the formula

V(t) = Voe^- (t/RC)

V(t) = 25e^- t/20000 × 10×10^-6

V(t) = 25e^- t/0.2

After a very long time. Let assume t = 20s. Then

V(20) = 25e^- 20/0.2

V(20) = 25e^-100

V(20) = 25 × 3.72 × 10^-44

V(20) = 9.3 × 10^-43 V

Answer: Non of them.

Explanation:

There are many throttle position sensors voltage check. Computer reference voltage to TP sensor is about 5V. Voltage drop should be less than 0.5 v while TP sensor output to computer is about 0.65v.

When a vehicle is being tested using a scan tool, the only the throttle position (TP) sensor that should read between 0 and 1 volt is the base voltage reading. The reading according to specifications is around 0.05v.

When checking for proper voltage for the opening and closing of the throttle, the voltage rises from 1 volt to a maximum of 5 volts.

Ammeter or multimeter can be used to check for proper voltage and current of the shift solenoids.

We can therefore conclude that both technicians A and B are incorrect. This is, non of them is correct.

Answer:

a. The buoyant force on the chamber is 220029.6 N

b. The tension in the cable is 7369.6 N

Explanation:

The diameter of the sphere cannot be in millimeter (mm), if the chamber must occupy a big mass as 21700kg

Given;

diameter of the sphere, d = 3.50 m

radius of the sphere, r = 1.75 mm = 1.75 m

mass of the chamber, m = 21700 kg

density of water, ρ = 1000 kg/m³

(a)

Buoyant force is the weight of water displaced, which is calculated as;

Fb = ρvg

where;

v is the volume of sphere, calculated as;

[tex]V = \frac{4}{3} \pi r^3\\\\V = \frac{4}{3} \pi (1.75)^3\\\\V = 22.452 \ m^3[/tex]

Fb = 1000 x 22.452 x 9.8

Fb = 220029.6 N

(b)

The tension in the cable will be calculated as;

T = Fb - mg

T = 220029.6 N - (21700 x 9.8)

T =  220029.6 N - 212660 N

T = 7369.6 N