A four-cylinder, four-stroke internal combustion engine operates at 2800 RPM. The processes within each cylinder are modeled as an air-standard Otto cycle with a pressure of 14.7 lbf/in. 2 , a temperature of 80 8 F, and a volume of 0.0196 ft 3 at the beginning of compression. The compression ratio is 10, and maximum pressure in the cycle is 1080 lbf/in.Write possible Assumptions no less than three assumptionsDetermine, using a cold air-standard analysis with k 5 1.4, the power developed by the engine, in horsepower, and the mean effective pressure, in lbf/in.

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]

Learn more about Air compressors:

brainly.com/question/15181914

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:

a. [tex]\eta _{th}[/tex] = 77.65%

b. bwr = 6.5%

c. 3538.986 kW

d. -163.169 kJ

Explanation:

a. The given property  are;

P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa

p₄/p₁ = 10

P₂/P₁ = p₄/p₃ = √10

p₂ = 100·√10

[tex]T_{2s}[/tex] = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K

T₂ = T₁ + ([tex]T_{2s}[/tex] - T₁)/[tex]\eta _c[/tex] = 300 + (416.85 - 300)/0.8 = 446.0625 K

p₄ = 10×p₁ = 10×100 = 1000 kPa

p₄/p₃ = √10 =

p₃ = 100·√10

T₃ = 300 K

T₃/[tex]T_{4s}[/tex] = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)

[tex]T_{4s}[/tex] = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K

T₄ = T₃ + ([tex]T_{4s}[/tex] - T₃)/[tex]\eta _c[/tex] = 300 + (215.905- 300)/0.8 = 194.881 K

The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28

T₄ = 446.0625 K

T₆ = 1400 K

[tex]T_{7s}[/tex]/T₆ = (1/√10)^(0.4/1.4)

[tex]T_{7s}[/tex] = 1400×(1/√10)^(0.4/1.4)  = 1007.6 K

T₇ = T₆ - [tex]\eta _t[/tex](T₆ - [tex]T_{7s}[/tex]) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K

T₈ = 1400 K

T₉ = 1086.08 K

T₅ = T₄ + [tex]\epsilon _{regen}[/tex](T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K

[tex]\eta _{th}[/tex] =(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))

(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765

[tex]\eta _{th}[/tex] = 77.65%

b. Back work ratio, bwr = [tex]bwr = \dfrac{w_{c,in}}{w_{t,out}}[/tex]

((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))

40.9435/627.84 = 6.5%

c. [tex]w_{net, out} = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)][/tex]

Power developed is given by the relation;

[tex]\dot m \cdot w_{net, out}[/tex]

[tex]\dot m \cdot w_{net, out}[/tex]= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW

d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)

-163.169 kJ

Answer:

A) n =  0.6143, c ≈ 640m/min

B) n = 0.6143 , c = 637.53m/min

Explanation:

using the given data

A) A plot of flank wear as a function of time and also A plot for tool when

Flank wear is 0.75 and cutting edge speed is 100m/min, Time of cutting edge is said to be 20.4 min  also for cutting edge speed of 155m/min , time for cutting edge is 10 min

is attached below

calculate for the constant N from the second plot

note : the slope will be negative because cutting speed decreases as time of cutting increase

V1 = 100m/min , V2 = 155m/min,  T1 = 20.4 min, T2 = 10 min

= - N = [tex]\frac{In(V2) - In(V1)}{In(T2)-ln(T1)}[/tex]

therefore  - N = [tex]\frac{5.043 - 4.605}{2.302 -3.015}[/tex]

                       = - 0.6143

THEREFORE  ( N ) = 0.6143

Determine for the constant C from the second plot as well

note : C is the intercept on the cutting speed axis in 1 min tool life

connecting the two points with a line and extend it to touch the cutting speed axis and measure the value at that point

hence   C ≈ 640m/min

B) Calculate the values of  N and C in the Taylor equation solving simultaneous equations

using the above cutting speed and time of cutting values we can find the constant N via Taylor tool life equation

Taylor tool life equation = vT = C ------------- equation 1

cutting speed = v = 100m/min and 155m/min

tool life = T = 20.4 min and 10 min

also constant  n and c are obtained from the previous plot

back to taylor tool life equation = 100 * 20.4 = C

therefore C = (100)(20.4)^n  ---------------- equation 2

also using the second values of  v and T

taylor tool life equation = 155 * 10 = C

therefore C = ( 155 )(10)^n ----------------- equation 3

Equate equation 2 and equation 3 and solve simultaneously

(100)(20.4)^n = (155)(10)^n

To find N

take natural log of both sides of the equation

= In ((100)(20.4)^n) = In((155)(10)^n)

= In (100) + nIn(20.4) = In(155) + nIn(10)^n

= n(3.0155) - n (2.3026) = 5.043 - 4.605

= 0.7129 n = 0.438

therefore n = 0.6143

To find C

substitute 0.6143 for n in equation 2

C = (100)(20.4) ^ 0.6143

C = 637.53 m/min

Attached are the two plots for solution A

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:

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:

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:

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: 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:

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

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:

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:

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:

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:

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:

Explanation:

La vaca

El pato

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 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:

The Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.

Explanation:

This is what we called HALF WAVE RECTIFIER in which the Peak value of the output voltage is less or lower than that of the peak value of the input voltage by 0.6V reason been that the voltage is tend to drop across the diode.

Therefore this is the formula for Half wave rectifier

Vrms = Vm/2 and Vdc

= Vm/π:

Where,

Vrms = rms value of input

Vdc = Average value of input

Vm = peak value of output

Hence, half wave rectifier is a rectifier which allows one half-cycle of an AC voltage waveform to pass which inturn block the other half-cycle which is why this type of rectifiers are often been used to help convert AC voltage to a DC voltage, because they only require a single diode to inorder to construct.

Answer:

a. 0.4544 N

b. [tex]5.112 \times 10^{-5 M}[/tex]

Explanation:

For computing the normality and molarity of the acid solution first we need to do the following calculations

The balanced reaction

[tex]H_2SO_4 + 2NaOH = Na_2SO_4 + 2H_2O[/tex]

[tex]NaOH\ Mass = Normality \times equivalent\ weight \times\ volume[/tex]

[tex]= 0.3200 \times 40 g \times 21.30 mL \times 1L/1000mL[/tex]

= 0.27264 g

[tex]NaOH\ mass = \frac{mass}{molecular\ weight}[/tex]

[tex]= \frac{0.27264\ g}{40g/mol}[/tex]

= 0.006816 mol

Now

Moles of [tex]H_2SO_4[/tex] needed  is

[tex]= \frac{0.006816}{2}[/tex]

= 0.003408 mol

[tex]Mass\ of\ H_2SO_4 = moles \times molecular\ weight[/tex]

[tex]= 0.003408\ mol \times 98g/mol[/tex]

= 0.333984 g

Now based on the above calculation

a. Normality of acid is

[tex]= \frac{acid\ mass}{equivalent\ weight \times volume}[/tex]

[tex]= \frac{0.333984 g}{49 \times 0.015}[/tex]

= 0.4544 N

b. And, the acid solution molarity is

[tex]= \frac{moles}{Volume}[/tex]

[tex]= \frac{0.003408 mol}{15\ mL \times 1L/1000\ mL}[/tex]

= 0.00005112

=[tex]5.112 \times 10^{-5 M}[/tex]

We simply applied the above formulas

The volume of the 0.3200 M, NaOH required to neutralize the H₂SO₄, is

21.30 mL, which gives the following acid solution approximate values;

1) Normality of the acid solution is 0.4544 N

2) The molarity of the acid is 0.2272

Molarity of the NaOH = 0.3200 M

Volume of NaOH required = 21.30 mL

1) The normality of the acid solution is found as follows;

The chemical reaction is presented as follows;

H₂SO₄(aq) + 2NaOH (aq) → Na₂SO₄ (aq) + H₂O

Number of moles of NaOH in the reaction is found as follows;

[tex]n = \dfrac{21.30}{1,000} \times 0.3200 \, M = \mathbf{0.006816 \, M}[/tex]

Therefore;

The number of moles of H₂SO₄ = 0.006816 M ÷ 2 = 0.003408 M

[tex]Normality = \mathbf{ \dfrac{Mass \ of \, Acid \ in \ reaction}{Equivalent \ mass \times Volume \ of \ soltute}}[/tex]

Which gives;

[tex]Normality = \dfrac{ 98 \times 0.003408 }{49 \times 0.015} = \mathbf{0.4544}[/tex]

2) The molarity is found as follows;

[tex]Molarity = \dfrac{0.003408 \, moles}{0.015 \, L} = \mathbf{0.2272 \, M}[/tex]

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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: 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:

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:

(a) The potential difference between any two points is zero.

Explanation:

A conservative field is;

i. a vector field that is the gradient of some function. Electrostatic field is the gradient of scalar potential, hence it is conservative.

ii. a vector field where the integral along every closed path is zero. This means that the work done in a closed cycle is zero. For an electrostatic field, the charge along closed path inside the field is zero. Hence, electrostatic field is conservative.

iii. a vector field if curl of its potential(vector product of the del operator and the potential) is zero. The curl of electrostatic field is identically zero everywhere.

iv. a vector field whose circulation is zero along any path.

v. a vector field whose potential difference between two points is independent of the path taken. The potential difference between any two points is not necessarily zero.

Other examples of conservative fields are;

i. gravitational field.

ii. magnetic field.

When we say that electrostatic field is conservative, we do not mean that the potential difference between any two points is zero.

A conservative field refers to a form of force between the Earth and another mass whose work is determined only by the final displacement of the object acted upon.

What we mean by saying an electrostatic field is conservative includes:

Hence, when we say that electrostatic field is conservative, we do not mean that the potential difference between any two points is zero.

Therefore, the Option A is correct.

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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:

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:

d= 2.80inch

Explanation:

Given:

Axial force= 30kip

d= 1inch

CHECK THE ATTACHMENT FOR DETAILED EXPLANATION

A) The average normal stress in the concrete and in each bar are; σ_st = 15.52 kpi ; σ_con = 2.25 kpi

B) The required diameter of each bar so that 60% of the axial force is carried by concrete is; 0.94 inches

We are told that;

Column has eight A992 steel reinforcing bars.

Column is subjected to an axial force of 200 kip.

A) Diameter of each bar is 1 inch.

Using equations of equilibrium, we have;

∑fy = 0;

8P_st + P_con = 200      ------(eq 1)

Using compatibility concept, we know from the image attached that;

δ_st = δ_con

where δ_st is change in length of steel and δ_con is change in length of concrete.

Thus;

δ_st = (P_st * L)/(A_st * E_st)

where;

P_st is tensile force of steel

L is length of steel = 3 ft = 36 inches

A_st is area of steel = π/4 * 1² = 0.7854 in²

E_st is young's modulus of steel = 29000 ksi

Similarly;

δ_con = (P_con * L)/(A_con * E_con)

where;

P_con is tensile force of concrete

L is length of concrete = 3 ft = 36 inches

E_con is young's modulus of concrete = 4200 ksi

A_con is area of concrete with diameter of 8 inches = (π/4 * 8²) - 6(π/4 * 1²) = 45.5531 in²

Thus;

From δ_st = δ_con;

(P_st * 36)/(0.7854 * 29000) = (P_con * 36)/(45.5531 * 4200)

Solving this gives;

P_st = 0.119P_con    -----(eq 2)

Put 0.119P_con for P_st in eq 1 to get;

8(0.119P_con) + P_con = 200  

1.952P_con = 200

P_con = 102.459 kip

Thus; P_st = 12.193 kip

Thus, average normal stress is;

Steel; σ_st = P_st/A_st

σ_st = 12.193/0.7854

σ_st = 15.52 kpi

Concrete; σ_con = P_con/A_con

σ_con = 102.459/45.5531

σ_con = 2.25 kpi

B) Since 60% of the axial force is carried by the concrete. Then it means that 40% will be carried by the steel.

Thus;

P_con = 60% * 200 = 120 kip

P_st = 40% * 200 = 80 kip

Using compatibility again;

δ_st = δ_con

Thus;

(P_st * L)/(A_st * E_st) = (P_con * L)/(A_con * E_con)

6(π/4 * d²)) = (80 * ((π/4 * 8²) - 6(π/4 * d²)) * 4200)/(120 * 29000)

⇒ 4.712d² = 0.09655(50.2655 - 4.712d²)

⇒ 4.712d²/0.09655 = 50.2655 - 4.712d²

⇒ 48.8037d² = 50.2655 - 4.712d²

Solving this gives;

d = 0.94 inches

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Answer:

The height of the oil on the side of the bar when the soap is floating in only the oil is 1.236 cm

Explanation:

The water level on the bar soap = 1.1 m mark

Therefore, the proportion of the bar soap that is under the water is given by the relation;

Volume of bar soap = LW1.7

Volume under water = LW1.1

Volume floating = LW0.6

The relative density of the bar soap = Density of bar soap/(Density of water)

= m/LW1.7/(m/LW1.1) = 1.1/1.7

Given that the oil density = 890 kg/m³

Relative density of the oil to water = Density of the oil/(Density of water)

Relative density of the oil to water = 890/1000 = 0.89

Therefore, relative density of the bar soap to the relative density of the oil = (1.1/1.7)/0.89

Relative density of the bar soap to the oil = (1.1/0.89/1.7) = 1.236/1.7

Given that the relative density of the bar soap to the oil = Density of bar soap/(Density of oil) = m/LW1.7/(m/LWX) = X/1.7 = 1.236/1.7

Where:

X  = The height of the oil on the side of the bar when the soap is floating in only the oil

Therefore;

X = 1.236 cm.

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