A smooth, flat plate of length l = 6 m and width b = 4 m is placed in water with an upstream velocity of U = 0.5 m/s. Determine the boundary layer thickness and the wall shear stress at the center and the trailing edge of the plate. Assume a laminar boundary layer

Answer:

First we determine the tensile strength using the equation;

Tₓ (MPa) = 3.45 × HB

{ Tₓ is tensile strength, HB is Brinell hardness = 225 }

therefore

Tₓ = 3.45 × 225

Tₓ = 775 Mpa

From Conclusions, It is stated that in order to achieve a tensile strength of 775 MPa for a steel, the percentage of the cold work should be 10

When the percentage of cold work for steel is up to 10,the ductility is 16% EL.

And 16% EL is greater than 12% EL

Therefore, it is possible to cold work steel to a given minimum Brinell hardness of 225 and at the same time a ductility of at least 12% EL

Answer:

The answer is "conditionally unstable"

Explanation:

The conditional volatility is really a condition of uncertainty, which reflects on whether increasing air is polluted or not. It determines the rate of ambient delay, which has been between humid and dry adiabatic rates. In general, the environment is in an unilaterally unhealthy region.

Classification dependent on ELR:

Larger than 10 [tex]\frac{^{\circ}C}{1000}[/tex] m Around 10 and 6 [tex]\frac{^{\circ}C}{1000}[/tex] m or less 6 [tex]\frac{^{\circ}C}{1000}[/tex] m volatile implicitly unreliable Therefore ELR is implicitly unreliable 9 [tex]\frac{^{\circ}C}{1000}[/tex] m, that's why it is "conditionally unstable".

Answer:

The final volumetric flow rate will be "76.4 m³/s".

Explanation:

The given values are:

[tex]\dot{m_{1}}=10 \ kg/s[/tex]

[tex]\dot{m_{2}}=20 \ Kg/s[/tex]

[tex]T_{1}=293 \ K[/tex]

[tex]T_{2}=313 \ K[/tex]

[tex]P_{1}=P_{2}=P_{3}=10 \ bar[/tex]

As we know,

⇒  [tex]E_{in}=E_{out}[/tex]

[tex]\dot{m_{1}}h_{1}+\dot{m_{2}}h_{2}=\dot{m_{3}}h_{3}[/tex]

[tex]e_{1}\dot{v_{1}}h_{1}+e_{2}\dot{v_{2}}h_{2}=e_{3}\dot{v_{3}}h_{3}[/tex]

[tex]\frac{P_{1}}{RP_{1}}\dot{v_{1}} \ C_{p}T_{1}+ \frac{P_{2}}{RP_{2}}\dot{v_{2}} \ C_{p}T_{1}=\frac{P_{3}}{RP_{3}}\dot{v_{3}} \ C_{p}T_{3}[/tex]

⇒  [tex]\dot{v_{3}}=\dot{v_{1}}+\dot{v_{2}}[/tex]

         [tex]=\frac{\dot{m_{1}}}{e_{1}}+\frac{\dot{m_{2}}}{e_{2}}[/tex]

On substituting the values, we get

         [tex]=\frac{10}{10\times 10^5}\times 8314\times 293+\frac{20\times 8314\times 313}{10\times 10^5}[/tex]

         [tex]=76.4 \ m^3/s[/tex]

Answer:

Las declaraciones falsas incluyen

- El KOH es el agente reductor.

- La suma de electrones transferidos más el coeficiente mínimo de agua suman 16.

Todas las otras declaraciones son ciertas.

The false statements include

- The KOH is the reducing agent.

- The sum of transferred electrons plus the minimum coefficient of water add up to 16.

All the other statements are true.

Explanation:

Es evidente que esta es una reacción redox en presencia de medio básico. Entonces, equilibraremos esta reacción redox en pasos. I₂ + KOH → KIO₃ + KI + H₂O

Paso 1 Eliminar los iones espectadores; Estos son los iones que aparecen en ambos lados de la reacción. Es evidente que solo el ion de potasio (K⁺) es el ion espectador de esta reacción.

I₂ + OH⁻ → IO₃⁻ + I⁻ + H₂O

Paso 2

Separamos la reacción en las medias reacciones de oxidación y reductina. La oxidación es la pérdida de electrones que conduce a un aumento del número de oxidación del ion, mientras que la reducción es la ganancia de elecrones que conduce a una disminución en el número de oxidación del ion. También es evidente que es el gas de yodo el que se reduce y oxida para esta reacción.

El gas de yodo se reduce a I⁻ (el número de oxidación se reduce de 0 a -1) y el gas de yodo se oxida a IO₃⁻ (el número de oxidación de yodo aumenta de 0 en gas de yodo a +5 en IO₃⁻)

Reducción media reacción

I₂ → I⁻

Media reacción de oxidación

I₂ + OH⁻ → IO₃⁻ + H₂O

Paso 3

Equilibramos las medias reacciones y agregamos los respectivos electrones transferidos

Reducción media reacción

I₂ → 2I⁻

I₂ + 2e⁻ → 2I⁻

Media reacción de oxidación

I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O

I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O + 10e⁻

Paso 4

Balancee el número de electrones en las dos medias reacciones

[I₂ + 2e⁻ → 2I⁻] × 5

[I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O + 10e⁻] × 1

5I₂ + 10e⁻ → 10I⁻

I₂ + 12OH⁻ → 2IO₃⁻ + 6H₂O + 10e⁻

Paso 5

Agregue las dos medias reacciones y elimine cualquier especie que aparezca en ambos lados

5I₂ + 10e⁻ + I₂ + 12OH⁻ → 10I⁻ + 2IO₃⁻ + 6H₂O + 10e⁻

Entonces, eliminamos los 10 electrones que fueron transferidos en la reacción balanceada

6I₂ + 12OH⁻ → 10I⁻ + 2IO₃⁻ + 6H₂O

Paso 6

Reintroducimos la especie eliminada desde el principio (el ion potasio)

6I₂ + 12KOH → 10KI + 2KIO₃ + 6H₂O

Los coeficientes mínimos son entonces

3I₂ + 6KOH → 5KI + KIO₃ + 3H₂O

Luego verificamos cada una de las declaraciones proporcionadas para elegir las falsas.

- La suma de los coeficientes mínimos del agua y el agente reductor es 6.

El gas yodo es el agente reductor y oxidante. Coeficiente mínimo de agua y gas de yodo = 3 + 3 = 6 Esta afirmación es cierta.

- El KI es la forma reductora KI resulta de la semirreacción de reducción.

Por lo tanto, es la forma reducida del gas de yodo. Esta afirmación es cierta. - El KOH es el agente reductor. KOH no es el agente reductor. Esta afirmación es falsa.

- La suma de los electrones transferidos más el coeficiente mínimo de agua suman 16.

Electrones transferidos = 10

Coeficiente mínimo de agua = 3

Suma = 13 y no 16.

Esta afirmación es falsa.

- La proporción del agente oxidante y el agente reductor es 1.

Dado que el gas yodo es el agente reductor y oxidante, la proporción de estos dos es verdaderamente 1. Esta afirmación es cierta.

¡¡¡Espero que esto ayude!!!

Answer:

The voltage input to the inverting terminal is 60μV

Explanation:

Given;

open-loop gain, A = 150,000

output voltage, V₀ = 15 V

voltage at the inverting input, [tex]V_n[/tex] = −40 μV = 40 x 10⁻⁶ V

The relationship between output voltage and voltage at the inverting input is given as;

[tex]V_o = A(V_p -V_n)\\\\15 = 150,000(V_p -(-40*10^{-6}))\\\\15 = 150,000 (V_p +40*10^{-6}) \\\\V_p +40*10^{-6} = \frac{15}{150,000} \\\\V_p + 40*10^{-6} = 1 *10^{-4}\\\\V_p + 40*10^{-6} = 100 *10^{-6}\\\\V_p = 100 *10^{-6} - 40*10^{-6}\\\\V_p = 60 *10^{-6}\\\\V_p = 60 \ \mu V[/tex]

Therefore, the voltage input to the inverting terminal is 60μV

Answer:

Explanation:

GIVEN DATA

external load applied (p) = 85 kips

bolt stiffness ( Kb ) = 3(10^6) Ibf / in

Member stiffness (Km) = 12(10^6) Ibf / in

Diameter of bolts ( d ) = 1/2 in - 13 UNC grade 8

Number of bolts = 6

assumptions

for unified screw threads UNC and UNF

tensile stress area ( A ) = 0.1419 in^2

SAE specifications for steel bolts for grade 8

we have

Minimum proff strength ( Sp) = 120 kpsi

Minimum tensile strength (St) = 150 Kpsi

Load Bolt (p) = external load / number of bolts = 85 / 6 = 14.17 kips

Given the following values

Fi = 75%* Sp*At = (0.75*120*0.1419 ) = 12.771 kip

Preload stress

αi = 0.75Sp = 0.75 * 120 = 90 kpsi

stiffness constant

C = [tex]\frac{Kb}{Kb + Km}[/tex]  = [tex]\frac{3}{3+2}[/tex] = 0.2

A) yielding factor of safety

nP = [tex]\frac{sPAt}{Cp + Fi}[/tex] = [tex]\frac{120* 0.1419}{0.2*14.17 + 12.771}[/tex]

nP = 77.028 / 15.605 = 4.94 ≈ 4.9

B) Determine the overload factor safety

[tex]nL = \frac{SpAt - Fi}{CP}[/tex] = ( 120 * 0.1419) - 12.771 / 0.2 * 14.17

= 17.028 - 12.771 / 2.834

= 1.50

Answer:

1. Loss of income.

2. Rental costs.

3. Utility bills.

4. Loss of rent.

5. Storage costs.

Explanation:

Liquidated damages can be defined as pre-determined damages or clauses that are highlighted or indicated at the time of entering into a contract between a contractor and a client which is mainly based on evaluation of the actual loss the client may incur should the contractor fail to meet the agreed completion date.

Generally, liquidated damages are meant to be fair rather than being a penalty or punitive to the defaulter. It is usually calculated on a daily basis for the loss.

When entering into a contract with another, liquidated damages are intended to represent anticipated losses to the owner based upon circumstances existing at the time the contract was made.

Listed below are five (5) types of potential losses to the owner that would qualify for determination of such potential losses;

1. Loss of income.

2. Rental costs.

3. Utility bills.

4. Loss of rent.

5. Storage costs.

Answer:

  I = 6.364∠3.8° A

Explanation:

You can use KCL or KVL to write node or mesh equations for the voltages and currents in the network. Those require a matrix equation solver capable of working with complex numbers. Some calculators can do that. However, we're going to try a different approach here.

Starting from upper left (j4) and working clockwise around the outside, label the impedances Z1 .. Z4. Label the horizontal branch across the middle Z5. We're going to transform the Δ of Z1, Z2, Z5 into a Y of ZA, ZB, ZC that will facilitate computing the effective impedance of the bridge to the source voltage.

The Δ-Y transformation is symmetrical. The numerator of the equivalent impedance connected to each node is the product of the values currently connected to that node; the denominator is the sum of the values around the loop of the Δ.

So, If we transform the Δ of Z1, Z2, Z5 to a Y of ZA, ZB. ZC with ZA connected where Z1 and Z2 are now connected, ZB connected to Z4, and ZC connected to Z3, the network becomes a series-parallel network with an effective impedance of ...

  Z = ZA + ((ZB +Z4) ║ (ZC +Z3))

__

For starters, we have ...

  ZA = Z1·Z2/(Z1 +Z2 +Z5) = (j4)(-j3)/(j4-j3+8+j5) = 12/(8+j6) = 0.96-j0.72

  ZB = (j4)(8+j5)/(8+j6) = (20+j32)/(8+j6) = 0.32+j3.76

  ZC = (-j3)(8+j5)/(8+j6) = (15-j24)/(8+j6) = -0.24-j2.82

So, the left branch of the parallel combination is ...

  ZB +Z4 = (0.32+j3.76) +(5-j2) = 5.32+j1.76

And the right branch is ...

  ZC +Z3 = (-0.24-j2.82) +10 = 9.76-j2.82

Then the series-parallel combination we want is ...

  ZA + (ZB+Z4)(ZC+Z3)/(ZB+Z4+ZC+Z3) ≈ 4.703671 -j0.3126067

That is, the impedance of the bridge circuit to the source voltage is about ...

  4.7140478∠-3.802°

Dividing the source voltage by this impedance gives the source current, ...

  I = (30∠0°)/(4.7140478∠-3.802°)

  I ≈ 6.363958∠3.802° . . . amperes

Answer:

the width of the turning roadway = 15 ft

Explanation:

Given that:

A ramp from an expressway with a design speed(u) =  30 mi/h connects with a local road

Using 0.08 for superelevation(e)

The minimum radius of the curve on the road can be determined by using the expression:

[tex]R = \dfrac{u^2}{15(e+f_s)}[/tex]

where;

R= radius

[tex]f_s[/tex] = coefficient of friction

From the tables of coefficient of friction for a design speed at 30 mi/h ;

[tex]f_s[/tex] = 0.20

So;

[tex]R = \dfrac{30^2}{15(0.08+0.20)}[/tex]

[tex]R = \dfrac{900}{15(0.28)}[/tex]

[tex]R = \dfrac{900}{4.2}[/tex]

R = 214.29 ft

R ≅ 215 ft

However; given that :

The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.

From the tables of "Design widths of pavement for turning roads"

For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation

Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.

As such in Case 1 operation that falls under traffic condition B  in accordance with the Design widths of pavement for turning roads;

If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft

Hence; the width of the turning roadway = 15 ft

Answer:

Ductile

Explanation:

So, from the question, we have the following information or parameters or data which is going to help us in solving this particular problem or question;

=> " impact testing a sample = -100oC shows that the fracture surface is very DULL AND FIBROUS"

TAKE NOTE: DULL AND FIBROUS.

IMPACT TESTING is used by engineers in the configuration of a sample or object.

In order to determine whether a specimen is ductile or brittle, it can be shown from its appearance for instance;

A DUCTILE SAMPLE will be DULL AND FIBROUS thus, our answer!

But a brittle sample will have a crystal shape.

Answer:

Check Explanation.

Explanation:

ENGINEERING ECONOMY:

In a simple way, Engineering Economy simply refers to the study of Economics which is related to engineers that is the study of Economic decisions by people in the engineering field. The study of Engineering Economy is very important because Engineering is a major manufacturing part in every country's economy.

With the study of Economics by Engineering that is Engineering Economy, engineers can make rational decisions after seeing alternatives.

The foundation of Engineering Economy in terms of seven basic principles:

(A). Creation of Alternatives: there will always be a problem and every problem had one or more solutions. When a problem has been seen as a problem alternative solutions come in.

(B). Differences in the Alternatives : this part is when engineers makes the best decision(choice) among alternates.

(C). Your viewpoint should be consistent: consistency is power. In order to make decisions in Engineering works or projects, viewpoint should be consistent.

(D). Develop Common Performance Measures: in order to make sure that the project is perfected there should be common performance measures.

(E). Considering Relevant Criteria: relevant Criteria will be met before the best choice is decided

(F). Risk making: Engineering projects should not be put under risk and thus is why this principle is very important.

(G). Decision retargeting: go back to the alternatives and recheck your choices.

Answer:

[tex]\frac{e'_z}{e_z} = 0.87142[/tex]

Explanation:

Given:-

- The diameter of the cylinder, d = 50 mm.

- The compressive load, F = 80 KN.

Solution:-

- We will form a 3-dimensional coordinate system. The z-direction is along the axial load, and x-y plane is categorized by lateral direction.

- Next we will write down principal strains ( εx, εy, εz ) in all three directions in terms of corresponding stresses ( σx, σy, σz ). The stress-strain relationships will be used for anisotropic material with poisson ratio ( ν ).

                          εx = - [ σx - ν( σy + σz ) ] / E

                          εy = - [ σy - ν( σx + σz ) ] / E

                          εz = - [ σz - ν( σy + σx ) ] / E

- First we will investigate the "no-restraint" case. That is cylinder to expand in lateral direction as usual and contract in compressive load direction. The stresses in the x-y plane are zero because there is " no-restraint" and the lateral expansion occurs only due to compressive load in axial direction. So σy= σx = 0, the 3-D stress - strain relationships can be simplified to:

                          εx =  [ ν*σz ] / E

                          εy = [ ν*σz ] / E

                          εz = - [ σz ] / E   .... Eq 1

- The "restraint" case is a bit tricky in the sense, that first: There is a restriction in the lateral expansion. Second: The restriction is partial in nature, such, that lateral expansion is not completely restrained but reduced to half.

- We will use the strains ( simplified expressions ) evaluated in " no-restraint case " and half them. So the new lateral strains ( εx', εy' ) would be:

                         εx' = - [ σx' - ν( σy' + σz ) ] / E = 0.5*εx

                         εx' = - [ σx' - ν( σy' + σz ) ] / E =  [ ν*σz ] / 2E

                         εy' = - [ σy' - ν( σx' + σz ) ] / E = 0.5*εy

                         εx' = - [ σy' - ν( σx' + σz ) ] / E =  [ ν*σz ] / 2E

- Now, we need to visualize the "enclosure". We see that the entire x-y plane and family of planes parallel to ( z = 0 - plane ) are enclosed by the well-fitted casing. However, the axial direction is free! So, in other words the reduction in lateral expansion has to be compensated by the axial direction. And that compensatory effect is governed by induced compressive stresses ( σx', σy' ) by the fitting on the cylinderical surface.

- We will use the relationhsips developed above and determine the induced compressive stresses ( σx', σy' ).

Note:  σx' = σy', The cylinder is radially enclosed around the entire surface.

Therefore,

                        - [ σx' - ν( σx'+ σz ) ] =  [ ν*σz ] / 2

                          σx' ( 1 - v ) = [ ν*σz ] / 2

                          σx' = σy' = [ ν*σz ] / [ 2*( 1 - v ) ]

- Now use the induced stresses in ( x-y ) plane and determine the new axial strain ( εz' ):

                           εz' = - [ σz - ν( σy' + σx' ) ] / E

                           εz' = - { σz - [ ν^2*σz ] / [ 1 - v ] } / E

                          εz' = - σz*{ 1 - [ ν^2 ] / [ 1 - v ] } / E  ... Eq2

- Now take the ratio of the axial strains determined in the second case ( Eq2 ) to the first case ( Eq1 ) as follows:

                            [tex]\frac{e'_z}{e_z} = \frac{- \frac{s_z}{E} * [ 1 - \frac{v^2}{1 - v} ] }{-\frac{s_z}{E}} \\\\\frac{e'_z}{e_z} = [ 1 - \frac{v^2}{1 - v} ] = [ 1 - \frac{0.3^2}{1 - 0.3} ] \\\\\frac{e'_z}{e_z} = 0.87142[/tex]... Answer

Answer:

a) The planar defect that exists is twin boundary defect.

b) The planar defect that exists is the stacking fault.

Explanation:      

I am using bold and underline instead of a vertical line.

a. A B C A B C B A C B A

In this stacking sequence, the planar defect that occurs is twin boundary defect because the stacking sequence at one side of the bold and underlined part of the sequence is the mirror image or reflection of the stacking sequence on the other side. This shows twinning. Hence it is the twin boundary inter facial defect.

b. A B C A B C  B C A B C

In this stacking sequence the planar defect that occurs is which occurs is stacking fault defect. This underlined region is HCP like sequence. Here BC is the extra plane hence resulting in the stacking fault defect. The fcc stacking sequence with no defects should be A B C A B C A B C A B C. So in the above stacking sequence we can see that A is missing in the sequence. Instead BC is the defect or extra plane. So this disordering of the sequence results in stacking fault defect.

Answer: The largest torque that can be applied at A is the smaller of both

Tᵇc and Tᵃᵇ

which is (Tᵃᵇ = 530.14376 Nm)

Explanation:

First we take a look at Torque formula

T = M × Shear stress

T = π/2 × (d/2)³ × shear stress

{ where M = π/2 × (d/2)³ = π/2 × d³/8

For Shaft AB

d = 30mm = 30 × 10⁻³m

shaft stress = 100Mpa = 100 × 10⁶ N/m²

now Torque at A due to AB

Tᵃᵇ = π/2 × (d/2)³ × shear stress

Tᵃᵇ = π/2 × (30 × 10⁻³)³/2³ × 100 × 10⁶

Tᵃᵇ =  π/2 × (0.000027 / 8) × 100 × 10⁶

Tᵃᵇ = 530.14376 Nm

For Shaft BC

the value for M is changed

M = π/2 × { (d₂/2)⁴ - (d₁/2)⁴}

d₁ = 30mm = 30 × 10⁻³m

d₂ = 50mm = 50 × 10⁻³m

M = π/2 × { (50 × 10⁻³/2)⁴ - (30 × 10⁻³/2)⁴}

M = 5.34 × 10⁻⁷

Torque formula is also changed which is

T = M × share stress / (d₂/2)

shear stress = 60Mpa = 60 × 10⁶ N/m²

so Torque A due to BC is

Tᵇc =  (5.34 × 10⁻⁷ × 60 × 10⁶) / ( 50 × 10⁻³ / 2

Tᵇc  = 1281.6 Nm

Therefore the largest torque that can be applied at A is the smaller of both

Tᵇc and Tᵃᵇ

which is (Tᵃᵇ = 530.14376 Nm)

Answer:

α = kt

ω = [tex]\frac{kt^2}{2}[/tex]

θ = [tex]\frac{kt^3}{6}[/tex]  

Explanation:

given data

Initial velocity ω = 0

time t = 10 s

Number of revolutions = 20

solution

we will take here first α = kt     .....................1

and as α = [tex]\frac{d\omega}{dt}[/tex]

so that

[tex]\frac{d\omega}{dt}[/tex] = kt      ..................2

now we will integrate it then we will get

∫dω  = [tex]\int_{0}^{t} kt\ dt[/tex]   .......................3

so

ω = [tex]\frac{kt^2}{2}[/tex]

and

ω = [tex]\frac{d\theta}{dt}[/tex]     ..............4

so that

[tex]\frac{d\theta}{dt}[/tex]  = [tex]\frac{kt^2}{2}[/tex]

now we will integrate it then we will get

∫dθ = [tex]\int_{0}^{t}\frac{kt^2}{2} \ dt[/tex]      ...............5

solve it and we get

θ = [tex]\frac{kt^3}{6}[/tex]  

Answer:

(a) 48.2 %

(b) 0.4137

(c) 2385.9 kW

Explanation:

The given values are:

Initial pressure,

p₁ = 100 kPa

Initial temperature,

T₁ = 300 K

Mass,

M = 6 kg/s

Pressure ration,

r = 10

Inlent temperature,

T₃ = 1400 K

Specific heat ratio,

k = 1.4

At T₁ and p₁,

⇒  [tex]c_{p}=1.005 \ KJ/Kg.K[/tex]

Process 1-2 in isentropic compression, we get

⇒  [tex]\frac{T_{2}}{T_{1}}=(\frac{p_{2}}{p_{1}})^{\frac{k-1}{k}}[/tex]

    [tex]T_{2}=(\frac{p_{2}}{p_{1}})^{\frac{k-1}{k}}. T_{1}[/tex]

On putting the estimated values, we get

         [tex]=(10)^{\frac{1.4-1}{1.4}}(300)[/tex]

         [tex]=579.2 \ K[/tex]

Process 3-4,

⇒  [tex]\frac{T_{4}}{T_{3}}=(\frac{p_{4}}{p_{3}})^{\frac{k-1}{k}}[/tex]

    [tex]T_{4}=(\frac{1}{10})^{\frac{1.4-1}{1.4}}(1400)[/tex]

         [tex]=725.13 \ K[/tex]

(a)...

The thermal efficiency will be:

⇒  [tex]\eta =\frac{\dot{W_{t}}-\dot{W_{e}}}{\dot{Q_{in}}}[/tex]

    [tex]\eta=1-\frac{\dot{Q_{out}}}{\dot{Q_{in}}}[/tex]

⇒  [tex]\dot{Q_{in}}=\dot{m}(h_{1}-h_{2})[/tex]

           [tex]=\dot{mc_{p}}(T_{3}-T_{2})[/tex]

           [tex]=6\times 1005\times (1400-579.2)[/tex]

           [tex]=4949.4 \ kJ/s[/tex]

⇒  [tex]\dot{Q_{out}}=\dot{m}(h_{4}-h_{1})[/tex]

             [tex]=6\times 1.005\times (725.13-300)[/tex]

             [tex]=2563.5 \ KJ/S[/tex]

As we know,

⇒  [tex]\eta=1-\frac{\dot{Q_{out}}}{\dot{Q_{in}}}[/tex]

On putting the values, we get

       [tex]=1-\frac{2563.5}{4949.4}[/tex]

       [tex]=0.482 \ i.e., \ 48.2 \ Percent[/tex]

(b)...

Back work ratio will be:

⇒  [tex]bwr=\frac{\dot{W_{e}}}{\dot{W_{t}}}[/tex]

Now,

⇒  [tex]\dot{W_{e}}=\dot{mc_{p}}(T_{2}-T_{1})[/tex]

On putting values, we get

          [tex]=6\times 1.005\times (579.2-300)[/tex]

          [tex]=1683.6 \ kJ/s[/tex]

⇒  [tex]\dot{W_{t}}=\dot{mc_{p}}(T_{3}-T_{4})[/tex]

          [tex]=6\times 1.005\times (1400-725.13)[/tex]

          [tex]=4069.5 \ kJ/s[/tex]

So that,

⇒  [tex]bwr=\frac{1683.6}{4069.5}=0.4137[/tex]

(c)...

Net power is equivalent to,

⇒  [tex]\dot{W}_{eyele}=\dot{W_{t}}-\dot{W_{e}}[/tex]

On substituting the values, we get

               [tex]= 4069.5-1683.6[/tex]

               [tex]=2385.9 \ kW[/tex]

Following are the solution to the  given points:

Given :  

Initial pressure [tex]p_1 = 100\ kPa \\\\[/tex]

Initial temperature [tex]T_1 = 300\ K \\\\[/tex]

Mass flow rate of air [tex]m= 6\ \frac{kg}{s}\\\\[/tex]  

Compressor pressure ratio [tex]r =10\\\\[/tex]

Turbine inlet temperature [tex]T_3 = 1400\ K\\\\[/tex]

Specific heat ratio [tex]k=1.4\\\\[/tex]

Temperature [tex]\ T_1 = 300\ K[/tex]

pressure [tex]p_1 = 100\ kPa\\\\[/tex]

[tex]\to c_p=1.005\ \frac{kJ}{kg\cdot K}\\\\[/tex]

Process 1-2 is isen tropic compression  

[tex]\to \frac{T_2}{T_1}=(\frac{P_2}{P_1})^{\frac{k-1}{k}} \\\\[/tex]

[tex]\to T_2=(\frac{P_2}{P_1})^{\frac{k-1}{k}} \ T_1 \\\\[/tex]

         [tex]=(10)^{\frac{1.4-1}{1.4}} (300)\\\\ =(10)^{\frac{0.4}{1.4}} (300) \\\\[/tex]

[tex]\to T_2 = 579.2\ K \\\\[/tex]

Process 3-4 is isen tropic expansion  

[tex]\to \frac{T_4}{T_3}=(\frac{P_4}{P_3})^{\frac{k-1}{k}}\\\\ \to T_4=(\frac{1}{10})^{\frac{1.4-1}{1.4}} (1400)\\\\\to T_4= 725.13\ K \\\\[/tex]

For point a:

The thermal efficiency of the cycle:

[tex]\to \eta = \frac{W_i-W_e}{Q_{in}} \\\\\to \eta = \frac{Q_{in}- Q_{out}}{Q_{in}}\\\\\to \eta =1 - \frac{Q_{out}}{Q_{in}} \\\\\to Q_{in}= m(h_3-h_1) = mc_p (T_4-T_1) =(6)(1.005)(725.13-300) = 2563 \ \frac{kJ}{S}\\\\\to \eta =1- \frac{Q_{out}}{Q_{in}}\\\\[/tex]

       [tex]=1-\frac{2563.5}{4949.4}\\\\ = 0.482\\\\[/tex]

 [tex]\eta = 48.2\%\\\\[/tex]

  For point b:  

The back work ratio  

[tex]\to bwr =\frac{W_e}{W_t}[/tex]

Now

[tex]\to W_e =mc_p (T_2 -T_1)[/tex]

          [tex]=(6) (1.005)(579.2 -300)\\\\ =1683.6 \ \frac{kJ}{S}\\\\[/tex]

[tex]\to W_t=mc_p(T_3-T_4)[/tex]

         [tex]=(6)(1.005)(1400 - 725.13)\\\\ = 4069.5 \frac{KJ}{s}[/tex]

[tex]\to bwr =\frac{W_s}{W_t}= \frac{1683.6}{4069.5}=0.4137[/tex]

   For point c:

The net power developed is equal to

 [tex]\to W_{cycle} = W_t-W_e \\\\[/tex]

                [tex]= ( 4069.5-1683.6)\\\\ = 2385.9 \ kW\\[/tex]

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Answer:

Transverse shear stress formula

Explanation:

Transverse shear stress also known as the beam shear, is the shear stress due to bending of a beam.

Generally, when a beam is made to undergo a non-uniform bending, both bending moment (I) and a shear force (V) acts on its cross section or width (t).

Transverse shear stress formula is used to determine the shear stress at point P over the section supporting a downward shear force in the -y direction.

Mathematically, the transverse shear stress is given by the formula below;

[tex]T' = \frac{VQ}{It}[/tex]

Also note, T' is pronounced as tau.

Where;

V is the total shear force with the unit, Newton (N).

I is the Moment of Inertia of the entire cross sectional area with the unit, meters square (m²).

t is the thickness or width of cross sectional area of the material perpendicular to the shear with the unit centimeters (cm).

Q is the statical moment of area.

Mathematically, Q is given by the formula;

[tex]Q = y'P^{*} = ∑y'P^{*}[/tex]

Where [tex]P^{*}[/tex] is the section supporting a downward shear force in the y' direction.

Answer:

Nup = 8

Explanation:

u(r) = C1 and T(r) = T(s) = C2 [ 1 - r/ro)^2 ]

The temperature and velocity particles at a particular axial location is Uniform and parabolic

The Nusset Number Nup at this location can be obtained by first determining/calculating for the convection coefficient = [tex]h = \frac{q^I s}{TsTn}[/tex]

Therefore the mean temperature is calculated as follows

attached is the detailed solution and free body diagram

It is motor oil, as oil is used to reduce friction

Answer:

Use cases are known to be a set of instruction  or processes between a User/Actor with the system to produce a desired input.

With the aid of a diagram, the set of use cases that are carried out in this ATM are given below:

Insert PIN

(1)Perform required transaction

(2)Withdrawal

(3)Deposit

(4)Transfer

(5)Change PIN

(6)Exit

Note: Kindly find an attached diagram of the Use case as part of the solution to process carried out at the ATM

Sources: The diagram of the Use case for ATM was researched and taken from Quizlet.

Explanation:

Solution

Use cases are normally a set of instruction  or processes between a User/Actor with the system to produce a desired input.

A use case diagram or image is a graphical representation of all the use case or processes that connects or interact with the system

The use case diagram is a part of Unified Modelling Language also called the UML.

The set of use cases that are carried out in this ATM use case diagram to know the requirements of the ATM is shown below:

Now both the customer/client and Bank are seen as Actors.

Actors are the ones or people that interface with the system.

The ATM is often used to withdraw money and it also have some requirement.

Some of the use-cases that could serve as a basis for understanding the requirements for an ATM system are;

Customer (actor)  often uses bank ATM to know or see the Balances of his/her bank accounts. They use it also to Deposit Funds and Withdraw Cash.

They can also use it to Transfer Funds (use cases). ATM Technician are known to give a form of Maintenance and Repairs. All these used cases is one that involve Bank as the actor  when it is liked to customer transactions or to the ATM servicing.

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Answer:

A. Yes

B. Yes

Explanation:

We want to evaluate the validity of the given assertions.

1. The first statement is true

The sine rule stipulates that the ratio of a side and the sine of the angle facing the side is a constant for all sides of the triangle.

Hence, to use it, it’s either we have two sides and an angle and we are tasked with calculating the value of the non given side

Or

We have two angles and a side and we want to calculate the value of the side provided we have the angle facing this side in question.

For notation purposes;

We can express the it for a triangle having three sides a, b, c and angles A,B, C with each lower case letter being the side that faces its corresponding big letter angles

a/Sin A = b/Sin B = c/Sin C

2. The cosine rule looks like the Pythagoras’s theorem in notation but has a subtraction extension that multiplies two times the product of the other two sides and the cosine of the angle facing the side we want to calculate

So let’s say we want to calculate the side a in a triangle of sides a, b , c and we have the angle facing the side A

That would be;

a^2 = b^2 + c^2 -2bcCosA

So yes, the cosine rule can be used for the scenario above

Answer:

a) the mass flow rate of the steam is  [tex]\mathbf{m_1 =6.92 \ kg/s}[/tex]

b) the exit velocity of the steam  is [tex]\mathbf{V_2 = 562.7 \ m/s}[/tex]

c) the exit area of the nozzle is  [tex]A_2[/tex] = 0.0015435 m²

Explanation:

Given that:

A steam with 5 MPa and 400° C enters a nozzle steadily

So;

Inlet:

[tex]P_1 =[/tex] 5 MPa

[tex]T_1[/tex] = 400° C

Velocity V = 80 m/s

Exit:

[tex]P_2 =[/tex] 2 MPa

[tex]T_2[/tex] = 300° C

From the properties of steam tables  at [tex]P_1 =[/tex] 5 MPa and [tex]T_1[/tex] = 400° C we obtain the following properties for enthalpy h and the speed v

[tex]h_1 = 3196.7 \ kJ/kg \\ \\ v_1 = 0.057838 \ m^3/kg[/tex]

From the properties of steam tables  at [tex]P_2 =[/tex] 2 MPa and [tex]T_1[/tex] = 300° C we obtain the following properties for enthalpy h and the speed v

[tex]h_2 = 3024.2 \ kJ/kg \\ \\ v_2= 0.12551 \ m^3/kg[/tex]

Inlet Area of the nozzle = 50 cm²

Heat lost Q = 120 kJ/s

We are to determine the following:

a) the mass flow rate of the steam.

From the system in a steady flow state;

[tex]m_1=m_2=m_3[/tex]

Thus

[tex]m_1 =\dfrac{V_1 \times A_1}{v_1}[/tex]

[tex]m_1 =\dfrac{80 \ m/s \times 50 \times 10 ^{-4} \ m^2}{0.057838 \ m^3/kg}[/tex]

[tex]m_1 =\dfrac{0.4 }{0.057838 }[/tex]

[tex]\mathbf{m_1 =6.92 \ kg/s}[/tex]

b) the exit velocity of the steam.

Using Energy Balance equation:

[tex]\Delta E _{system} = E_{in}-E_{out}[/tex]

In a steady flow process;

[tex]\Delta E _{system} = 0[/tex]

[tex]E_{in} = E_{out}[/tex]

[tex]m(h_1 + \dfrac{V_1^2}{2})[/tex] [tex]= Q_{out} + m (h_2 + \dfrac{V_2^2}{2})[/tex]

[tex]- Q_{out} = m (h_2 - h_1 + \dfrac{V_2^2-V^2_1}{2})[/tex]

[tex]- 120 kJ/s = 6.92 \ kg/s (3024.2 -3196.7 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]

[tex]- 120 kJ/s = 6.92 \ kg/s (-172.5 + \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]

[tex]- 120 kJ/s = (-1193.7 \ kg/s + 6.92\ kg/s ( \dfrac{V_2^2- 80 m/s^2}{2}) \times (\dfrac{1 \ kJ/kg}{1000 \ m^2/s^2})[/tex]

[tex]V_2^2 = 316631.29 \ m/s[/tex]

[tex]V_2 = \sqrt{316631.29 \ m/s[/tex]

[tex]\mathbf{V_2 = 562.7 \ m/s}[/tex]

c) the exit area of the nozzle.

The exit of the nozzle can be determined by using the expression:

[tex]m = \dfrac{V_2A_2}{v_2}[/tex]

making [tex]A_2[/tex] the subject of the formula ; we have:

[tex]A_2 = \dfrac{ m \times v_2}{V_2}[/tex]

[tex]A_2 = \dfrac{ 6.92 \times 0.12551}{562.7}[/tex]

[tex]A_2[/tex] = 0.0015435 m²

Answer:

The type of ICS/SCADA i choose is the Paint processing plant.

I will consider both the the time and event based as we need to mix various colors at different time interval and different quantities.

The PLC is used to process different tasks based on the commands assigned to it. In a paint processing plant, when a command from a computer is given to PLC for processing that tasks at that time whatever is the quantity is considered to mix it is carried out by PLC.

Explanation:

Solution

From the given question, i will select the Paint processing plant.

Here, the Supervisory Control and Data Acquisition System (SCADA) refers to a  control system which uses computer network to control and manage the various processes from a single computer.

Since we consider the paint processing system for this we make use of both the time and event based as we need to mix various colors at different time interval and different quantities.

The programming Logic Controller (PLC): This is used to process different inputs based on the commands assigned to it.

In paint processing plant, when a command from a computer is assigned to PLC for processing that function at that time whatever is the quantity is required to mix it is carried out by PLC.

So, PLC is very useful device which is also the main processing device which carries out tasks assigned by the SCADA.

Answer and Explanation:

Let A denote its switch first after that we will assume B which denotes the next switch and then we will assume C stand for both the bulb. we assume 0 mean turn off while 1 mean turn on, too. The light is off, as both switches are in the same place. This may be illustrated with the below table of truth:

A                    B                       C (output)

0                    0                        0

0                    1                          1

1                     0                         1

1                     1                          0

The logic circuit is shown below

C = A'B + AB'

If the switches are in multiple places the bulb outcome will be on on the other hand if another switches are all in the same place, the result of the bulb will be off. This gate is XOR. The gate is shown in the diagram adjoining below.

Answer: read your vehicle owner's manual for any special directions or warnings.

Answer:

B. read your vehicle owner's manual for any special directions or warnings.

Explanation:

Answer:

Moore's Law

Explanation:

An observation that the number of transistors in a dense integrated circuit doubles about every two years (24 months), was made by Gordon E. Moore, the co-founder of Intel, and this observation became Moore's Law in 1965.

Therefore, Moore's Law is based on the observation that the rate of increase in transistor density on microchips had increased steadily, roughly doubling every 18 to 24 months.

Answer:

Working volttage

Explanation:

SMT electrolytic capacitors are marked with working voltage. The value of these capacitors is measured in micro farads. It is a surface mount capacitor which is used for high volume manufacturers. They are small lead less and are widely used. They are placed on modern circuit boards.

Answer:

Temperature to which the shaft must be cooled, [tex]\theta_2 = -180.61 ^0C[/tex]

Explanation:

Diameter of the shaft at room temperature, d₁ = 40 mm

Room temperature, θ₁ = 21°C

Coefficient of thermal expansion, [tex]\alpha = 24.8 * 10^{-6} / ^0 C[/tex]

The shaft is reduced in size by 0.20 mm:

Δd = - 0.20 mm

The temperature to which the shaft must be cooled, θ₂ = ?

The coefficient of thermal expansion is given by the equation:

[tex]\alpha = \frac{\triangle d}{d_1 * \triangle \theta}\\\\24.8 * 10^{-6} = \frac{-0.20}{40 * \triangle \theta}\\\\\triangle \theta = \frac{-0.20 }{24.8 * 10^{-6} * 40} \\\\\triangle \theta = - 201.61 ^0 C\\\triangle \theta = \theta_2 - \theta_1\\\\- 201.61 = \theta_2 - 21\\\\\theta_2 = -201.61 + 21\\\\\theta_2 = -180.61 ^0C[/tex]

Answer:

Option C = internal energy stays the same.

Explanation:

The internal energy will remain the same or unchanged because this question has to do with a concept in physics or classical chemistry (in thermodynamics) known as Free expansion.

So, the internal energy will be equals to the multiplication of the change in temperature, the heat capacity (keeping volume constant) and the number of moles. And in free expansion the internal energy is ZERO/UNCHANGED.

Where, the internal energy, ∆U = 0 =quantity of heat, q - work,w.

The amount of heat,q = Work,w.

In the concept of free expansion the only thing that changes is the volume.

Answer:

F(F) = 15037 lb

F(E) = 24481.5 lb

F(D) =  24481.5 lb

Explanation:

(The diagram of the figure and Free Body Diagram is attached)

W(A) = 50,000 lb

W(B) = 8000 lb

W(C) = 6000 lb

F(F) + F(E) + F(D) - W(A) - W(B) - W(C) = 0

F(F) + F(E) + F(D) = W(A) + W(B) + W(C)

F(F) + F(E) + F(D) = 50000 + 8000 + 6000

F(F) + F(E) + F(D) = 64000 lb

∑M(o) = M(F) + M(E) + M(D) + M(A) + M(B) + M(C)

Where

M(F) = 27i × F(F)k = -27F(F)j

M(E) = 14j × F(E)k = 14F(E)i

M(D) = -14j × F(D)k = -14F(D)i

M(A) = 7i × -50000k = 350,000j

M(B) = (4i - 6j) × -8000k = 48000i + 32000j

M(C) = (4i + 8j) × -6000k = -48000i + 24000j

∑M(i) = 14F(E) - 14F(D) = 0

F(E) = F(D)

∑M(j) = -27F(F) + 350,000 + 32,000 + 24,000 = 0

27F(F) = 406,000

F(F) = 15037 lb

F(E) = F(D)

F(F) + 2F(E) = 64000

2F(E) = 64000 - 15037

2F(E) = 48963

F(E) = 24481.5 lb

F(D) =  24481.5 lb