Find the heat flow from the composite wall as shown in figure. Assume one dimensional flow KA=150 W/m°C , KB=25 W/m°C, KC=60 W/m°C , KD=60 W/m°C

Answer:

Please check explanation for answer

Explanation:

Here, we are concerned with stating the advantages and disadvantages  of using a 6 tube passes instead of a 2 tube passes of the same diameter:

Advantages

* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface

* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too

            Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.

Disadvantages

* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.

* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of  manufacturing costs

Answer:

the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s

Explanation:

Given that:

Pressure of the feed water = 7000 kPa

Temperature of the closed feedwater heater = 260 ° C

Pressure of of the turbine = 6000 kPa

Temperature of the turbine = 325 ° C

The  objective is to calculate the amount of bleed steam required to heat 1 kg of feedwater in this unit.

From the table A-4 of saturated water temperature table at temperature  260° C at state 1 ;

Enthalpies:

[tex]h_1 = h_f = 1134.8 \ kJ/kg[/tex]

From table A-6 superheated water at state 3 ; the value of the enthalpy relating to the pressure of the turbine at 6000 kPa and temperature of 325° C  is obtained by the interpolating the temperature between 300 ° C and 350 ° C

At 300° C; enthalpy = 2885.6 kJ/kg

At 325° C. enthalpy = 3043.9 kJ/kg

Thus;

[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-{h_{300^0}}}{{h_{350^0}}- {h_{300^0}}}[/tex]

[tex]\dfrac{325-300}{350-300}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]

[tex]\dfrac{25}{50}=\dfrac{h_{325^0}-2885.6}{3043.9-2885.6 }}[/tex]

[tex]h_{325^0} = 2885.6 + \dfrac{25}{50}({3043.9-2885.6 )[/tex]

[tex]h_{325^0} = 2885.6 + 0.5({3043.9-2885.6 )[/tex]

[tex]h_{325^0} =2964.75 \ kJ/kg[/tex]

At pressure  of 7000 kPa at state 6; we obtain the enthalpies corresponding to the pressure at table A-5 of the saturated water pressure tables.

[tex]h_6 = h_f = 1267.5 \ kJ/kg[/tex]

From state 4 ;we obtain the specific volume corresponding to the pressure of 6000 kPa at table A-5 of the saturated water pressure tables.

[tex]v_4 = v_f = 0.001319\ m^3 /kg[/tex]

However; the specific work pump can be determined by using the formula;

[tex]W_p = v_4 (P_5-P_4)[/tex]

where;

[tex]P_4[/tex] = pressure at state 4

[tex]P_5[/tex] = pressure at state 5

[tex]W_p = 0.001319 (7000-6000)[/tex]

[tex]W_p = 0.001319 (1000)[/tex]

[tex]W_p =1.319 \ kJ/kg[/tex]

Using the energy balance equation of the closed feedwater heater to calculate the amount of bleed steam required to heat 1 kg of feed water ; we have:

[tex]E_{in} = E_{out} \\ \\ m_1h_1 +m_3h_3 + m_3W_p = (m_1+m_3)h_6[/tex]

where;

[tex]m_1 = 1 \ kg[/tex]

Replacing our other value as derived above into the energy balance equation ; we have:

[tex]1 \times 1134.8 +m_3 \times 2964.75 + m_3 \times 1.319 = (1+m_3)\times 1267.5[/tex]

[tex]1134.8 + 2966.069 \ m_3 = 1267.5 + 1267.5m_3[/tex]

Collect like terms

[tex]2966.069 \ m_3- 1267.5m_3 = 1267.5-1134.8[/tex]

[tex]1698.569 \ m_3 =132.7[/tex]

[tex]\ m_3 = \dfrac{132.7}{1698.569}[/tex]

[tex]\mathbf{ m_3 = 0.078 \ kg/s}[/tex]

Hence; the amount of bleed steam required to heat 1 kg of feedwater in this unit is 0.078 kg/s

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]

Learn more about the normality and the molarity of a solution here:

https://brainly.com/question/6532653

https://brainly.com/question/14112872

Answer:

Explanation:

coarse aggregate (ca) = 65%,   SG = 2.701

Fine aggregate = 35%,    SG = 2.625

A) Bulk specific gravity of aggregate

   = [tex]\frac{65*2.701 + 35*2.625}{100} = 2.674[/tex]

B) Wm = 1257.9 g { weight in air }

    Ww = 740 g { submerged weight }

   therefore Bulk specific gravity of compacted specimen

   = [tex]\frac{Wm}{Wm-Ww}[/tex]  =  [tex]\frac{1257.9}{1257.9 - 740 }[/tex]  =  2.428

   Theoretical specific gravity = 2.511

Percent stone

= 100 - asphalt content - Vv

= 100 - 5 - 3.305 = 91.695%

c) percent of void

= [tex]\frac{9.511-2.428}{2.511} * 100[/tex]    Vv = 3.305%

d) let effective specific gravity in stone

     = [tex]\frac{91.695*unstone+ 5 *1.030 }{96.695} = 2.511[/tex]

    = Instone = 2.592 effective specific gravity of stone

e) Vv = 3.305%

f ) volume filled with asphalt (Vb) = [tex]\frac{\frac{Wb}{lnb} }{\frac{Wm}{Inm} } * 100[/tex]

           Vb = [tex]\frac{5 * 2.428}{1.030 * 100} * 100[/tex]

          Vb = 11.786 %

Volume of mineral aggregate = Vb + Vv

              VMA = 11.786 + 3.305 = 15.091

g) percent void filled with alphalt

     = Vb / VMA * 100

    VMA = 11.786 + 3.305 = 15.091

   percent void filled with alphalt

     = Vb / VMA * 100 = (11.786 / 15.091) * 100 = 78.1%

Answer:

B. restricting the dislocation motion

Explanation:

Solid solution strengthening is a type of alloying that is carried out by the addition of the atoms of the element used for the alloying to the crystallized lattice structure of the base metal, which the metal that would be strengthened. The purpose of this act is to increase the strength of metals. It actually works by impeding or restricting the motion in the crystal lattice structure of metals thus making them more difficult to deform.

The solute atoms used for strengthening could be interstitial or substitutional. The interstitial solute atoms work by moving in between the space in the atoms of the base metal while the substitutional solute atoms make a replacement with the solvent atoms in the base metal.

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:

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:

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:

  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:

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 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:

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:

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:

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 answer is given below

Explanation:

A 60 Hz three-phase, three-wire overhead line has solid cylindrical conductors  arranged in the form of an equilateral triangle with 4 ft conductor spacing. Conductor  diameter is 0.5 in.

Given that:

The spacing between the conductors (D) = 4 ft

1 ft = 0.3048 m

D = 4 ft = 4 × 0.3048 m = 1.2192 m

The conductor diameter = 0.5 in

Radius of conductor (r) = 0.5/2 = 0.25 in = 0.00635 m

Frequency (f) = 60 Hz

The capacitance-to-neutral is given by:

[tex]C_n=\frac{2\pi \epsilon_0}{ln(\frac{D}{r} )} =\frac{2\pi *8.854*10^{-12}}{ln(1.2192/0.00635)}=1.058*10^{-11}\ F/m[/tex]

The admittance-to-neutral is given by:

[tex]Y_n=j2\pi fC_n=j*2\pi *60*1.058*10^{-11}*\frac{1000\ m}{1\ km}=j3.989*10^{-6}\ S/km[/tex]

Answer:

Explanation:

La vaca

El pato

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:

effectiveness of the heat exchanger will not be double when the length of the heat exchanger is doubled.

Because effectiveness depends on NTU and not necessarily the length of the heat exchanger

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:

The correct answer would be a) Engineering Controls.

Explanation:

If the controls are handled correctly, you can reduce and eliminate hazards so no one gets hurt. Engineering controls are absolutely necessary to prevent hazards.

Hope this helped! :)

Personal  protective equipment (PPE) is appropriate for controlling hazards

PPE are used for exposure to hazards when safe work practices and other forms of administrative controls cannot provide sufficient additional protection, a supplementary method of control is the use of protective clothing or equipment. PPE may also be appropriate for controlling hazards  while engineering and work practice controls are being installed.

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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

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

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:

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:

The volume percentage of graphite is 10.197 per cent.

Explanation:

The volume percent of graphite is the ratio of the volume occupied by the graphite phase to the volume occupied by the graphite and ferrite phases. The weight percent in the cast iron is 3.2 wt% (graphite) and 96.8 wt% (ferrite). The volume percentage of graphite is:

[tex]\%V_{gr} = \frac{V_{gr}}{V_{gr}+V_{fe}} \times 100\,\%[/tex]

Where:

[tex]V_{gr}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.

[tex]V_{fe}[/tex] - Volume occupied by the graphite phase, measured in cubic centimeters.

The expression is expanded by using the definition of density and subsequently simplified:

[tex]\%V_{gr} = \frac{\frac{m_{gr}}{\rho_{gr}} }{\frac{m_{gr}}{\rho_{gr}}+\frac{m_{fe}}{\rho_{fe}}}\times 100\,\%[/tex]

Where:

[tex]m_{fe}[/tex], [tex]m_{gr}[/tex] - Masses of the ferrite and graphite phases, measured in grams.

[tex]\rho_{fe}, \rho_{gr}[/tex] - Densities of the ferrite and graphite phases, measured in grams per cubic centimeter.

[tex]\%V_{gr} = \frac{1}{1+\frac{\frac{m_{fe}}{\rho_{fe}} }{\frac{m_{gr}}{\rho_{gr}} } }\times 100\,\%[/tex]

[tex]\%V_{gr} = \frac{1}{1 + \left(\frac{\rho_{gr}}{\rho_{fe}} \right)\cdot\left(\frac{m_{fe}}{m_{gr}} \right)} \times 100\,\%[/tex]

If [tex]\rho_{gr} = 2.3\,\frac{g}{cm^{3}}[/tex], [tex]\rho_{fe} = 7.9\,\frac{g}{cm^{3}}[/tex], [tex]m_{gr} = 3.2\,g[/tex] and [tex]m_{fe} = 96.8\,g[/tex], the volume percentage of graphite is:

[tex]\%V_{gr} = \frac{1}{1+\left(\frac{2.3\,\frac{g}{cm^{3}} }{7.9\,\frac{g}{cm^{3}} } \right)\cdot \left(\frac{96.8\,g}{3.2\,g} \right)} \times 100\,\%[/tex]

[tex]\%V_{gr} = 10.197\,\%V[/tex]

The volume percentage of graphite is 10.197 per cent.

Following are the solution to the given points:

[tex]\to C_{Gr} = 100\\\\ \to C_{\alpha}= 0[/tex]From [tex]Fe-F_{\frac{e}{3}} c[/tex] diagram.  

[tex]\to W_{\alpha} =\frac{C_{Gr}-C_{o}}{C_{Gr}-C_{\alpha}}[/tex]

           [tex]= \frac{100-3.6}{100-0} \\\\= \frac{100-3.6}{100} \\\\= \frac{96.4}{100} \\\\=0.964[/tex]

Calculating the weight fraction of graphite:  

[tex]\to W_{Gr}=\frac{C_0 - c_d}{C_{Gr} -c_d}[/tex]

            [tex]= \frac{3.6-0}{100-0} \\\\ = \frac{3.6}{100} \\\\= 0.036[/tex]

Calculating the volume percent of graphite:

[tex]\to V_{Gr}=\frac{\frac{W_{Gr}}{P_{Gr}}}{\frac{w_{\alpha}}{P_{\alpha}}+ \frac{W_{Gr}}{P_{Gr}}}[/tex]

           [tex]=\frac{\frac{0.036}{2.3}}{\frac{0.964}{7.9}+\frac{0.036}{2.3}}\\\\=0.11368 \times 100\%\\\\=11.368\%[/tex]

Therefore, the final answer is "0.964, 0.036, and 11.368%"

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It is to be noted that a hypertonic solution have the capacity to make cells to shrink.

In a hypertonic solution, the concentration of solutes (e.g., salts, sugars) outside the cell is higher than inside the cell.

As a result, water moves out of the cell through osmosis, trying to equalize the concentration, causing the cell to lose water and shrink.

This process is commonly observed in biology when examining the effect of different solutions on cells, such as in red blood cells or plant cells.

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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:

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:

Explanation:

For a job of applications engineer which require excellent technical skills, commitment  to working , ability to deal well and confidently with customers a willingness to travel and very intelligent and well-balanced personality.

The ten questions you should ask Maria to determine if she is qualified for the job are :

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.