Compute the volume percent of graphite, VGr, in a 3.1 wt% C cast iron, assuming that all the carbon exists as the graphite phase. Assume densities of 7.9 and 2.3 g/cm3 for ferrite and graphite, respectively.

Answer:

false

Explanation:

the changing of a prisoner sentence or another penalty to another less severe

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:

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:

1211 N.

Explanation:

Okay, we are given the following data or parameters or information in the question above;

=> "aluminum (E = 70 GPa) tube of length 8-m.

=> "The aluminum is used as a simply supported column carrying a 1.2 kN axial load"

The axial load to provide a factor of safety of 2 against buckling =[ (22/7)^2 × 224348 × 1000 × 70] ÷ (8 × 1000)^2 × 2.

The axial load to provide a factor of safety of 2 against buckling = 1211 N.

Answer:

False

Explanation:

Because

The effectiveness (ϵ) of a heat exchanger is defined as the ratio of the actual heat transfer to the maximum possible heat transfer.

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]

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:

Wizkid

Explanation:

wizkid is the best musician in Nigeria

Explanation:

I have no clue??????????

Answer:

1) 287760.4 Hp

2) 18410899.5 kPa

Explanation:

The parameters given are;

p₁ = 14.7 lbf/in² = 101325.9 Pa

v₁ = 0.0196 ft³ = 0.00055501 m³

T₁ = 80°F = 299.8167 K

k = 1.4

Assumptions;

1) Air standard conditions are appropriate

2) There are negligible potential and kinetic energy changes

3) The air behaves as an ideal gas and has constant specific heat capacities of temperature and pressure

1) Process 1 to 2

Isentropic compression

T₂/T₁ = (v₁/v₂)^(1.4 - 1) = 10^0.4

p₂/p₁ = (v₁/v₂)^(1.4)

p₂ = p₁×10^0.4 =  101325.9*10^0.4 = 254519.153 Pa

T₂ = 299.8167*10^0.4 = 753.106 K

p₃ = 1080 lbf/in² = 7,446,338 Pa

Stage 2 to 3 is a constant volume process

p₃/T₃ = p₂/T₂

7,446,338/T₃ =   254519.153/753.106

T₃ = 7,446,338/(254519.153/753.106) = 22033.24 K

T₃/T₄ = (v₁/v₂)^(1.4 - 1) = 10^0.4

T₄ = 22033.24/(10^0.4) = 8771.59 K

The heat supplied, Q₁ = cv(T₃ - T₂) = 0.718*(22033.24 -753.106) = 15279.14 kJ

The heat rejected = cv(T₄ - T₁) = 0.718*(8771.59 - 299.8167) = 6082.73 kJ

W(net) = The heat supplied - The heat rejected = (15279.14 - 6082.73) = 9196.41 kJ

The power = W(net) × RPM/2*1/60 = 9196.41 * 2800/2*1/60 = 214582.9 kW

The power by the engine = 214582.9 kW = 287760.4 Hp

2) The mean effective pressure, MEP  = W(net)/(v₁ - v₂)

v₁ = 0.00055501 m³

v₁/v₂ = 10

v₂ = v₁/10 = 0.00055501/10 = 0.000055501

MEP  = 9196.41/(0.00055501 -  0.000055501) = 18410899.5 kPa

Answer:

The power of force F is 115.2 W

Explanation:

Use following formula

Power  = F x V

[tex]F_{H}[/tex] = F cos0

[tex]F_{H}[/tex] = (30) x 4/5

[tex]F_{H}[/tex] = 24N

Now Calculate V using following formula

V = [tex]V_{0}[/tex] + at

[tex]V_{0}[/tex] = 0

a = [tex]F_{H}[/tex] / m

a = 24N / 20 kg

a = 1.2m / [tex]S^{2}[/tex]

no place value in the formula of V

V = 0 + (1.2)(4)

V = 4.8 m/s

So,

Power = [tex]F_{H}[/tex] x V

Power = 24 x 4.8

Power = 115.2 W

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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Answer:

point B where [tex]Q_B = 101.25 \ in^3[/tex]  has the largest Q value at section a–a

Explanation:

The missing diagram that is suppose to be attached to this question can be found in the attached file below.

So from the given information ;we are to determine the  point that  has the largest Q value at section a–a

In order to do that; we will work hand in hand with the image attached below.

From the image attached ; we will realize that there are 8 blocks aligned on top on another in the R.H.S of the image with the total of 12 in; meaning that each block contains 1.5 in each.

We also have block partitioned into different point segments . i,e A,B,C, D

For point A ;

Let Q be the moment of the Area A;

SO ; [tex]Q_A = Area \times y_1[/tex]

where ;

[tex]y_1 = (6 - \dfrac{1.5}{2})[/tex]

[tex]y_1 = (6- 0.75)[/tex]

[tex]y_1 = 5.25 \ in[/tex]

[tex]Q_A =(L \times B) \times y_1[/tex]

[tex]Q_A =(6 \times 1.5) \times 5.25[/tex]

[tex]Q_A =47.25 \ in^3[/tex]

For point B ;

Let Q be the moment of the Area B;

SO ; [tex]Q_B = Area \times y_2[/tex]

where ;

[tex]y_2 = (6 - \dfrac{1.5 \times 3}{2})[/tex]

[tex]y_2= (6 - \dfrac{4.5}{2}})[/tex]

[tex]y_2 = (6 -2.25})[/tex]

[tex]y_2 = 3.75 \ in[/tex]

[tex]Q_B =(L \times B) \times y_1[/tex]

[tex]Q_B=(6 \times 4.5) \times 3.75[/tex]

[tex]Q_B = 101.25 \ in^3[/tex]

For point C ;

Let Q be the moment of the Area C;

SO ; [tex]Q_C = Area \times y_3[/tex]

where ;

[tex]y_3 = (6 - \dfrac{1.5 \times 2}{2})[/tex]

[tex]y_3 = (6 - 1.5})[/tex]

[tex]y_3= 4.5 \ in[/tex]

[tex]Q_C =(L \times B) \times y_1[/tex]

[tex]Q_C =(6 \times 3) \times 4.5[/tex]

[tex]Q_C=81 \ in^3[/tex]

For point D ;

Let Q be the moment of the Area D;

SO ; [tex]Q_D = Area \times y_4[/tex]

since there is no area about point D

Area = 0

[tex]Q_D =0 \times y_4[/tex]

[tex]Q_D = 0[/tex]

Thus; from the foregoing ; point B where [tex]Q_B = 101.25 \ in^3[/tex]  has the largest Q value at section a–a

Complete question is;

A hot plate with a temperature of 60 °C will be cooled by adding 50 triangular profile needle blades (k = 204 W/m.K) with a length of 54 mm and diameter 10 mm. According to the ambient temperature 20 °C and the heat transfer coefficient on the surface 20 W/m².K. Calculate,

a-) Wing efficiency

b-) Total heat transfer rate (W) from the wings,

c-) Calculate the effectiveness of a wing.

Answer:

A) Efficiency = 96.05 %

B) Total heat transfer rate = 166.68 W/m

C) Wing Effectiveness = 10.42

Explanation:

Please find attached explanation for all the answers given.

Answer:

Entropy change is same for both systems because entropy is a state function i.e., it doesn't depend on the path by which the system arrived at it's present state (500 kPa, 600 K) from initial state (300 kPa, 350 K) .

Explanation:

Answer:

c) 1.75 g/cm³

Explanation:

Given that

Radii of the A ion, r(c) = 0.137 nm

Radii of the X ion, r(a) = 0.241 nm

Atomic weight of the A ion, A(c) = 22.7 g/mol

Atomic weight of the X ion, A(a) = 91.4 g/mol

Avogadro's number, N = 6.02*10^23 per mol

Solution is attached below

Answer:

Option A - fail/ not fail

Explanation:

For this given problem, if the yield strength is now 45 ksi, using Distortion Energy Theory the material will _fail______ and using the Maximum Shear Stress Theory the material will ___not fail_______

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:

c. stochastic.

Explanation:

A stochastic model is a tool in statistics, used to estimate the probability distributions of intended outcomes by the allowance of random variation in any number of the inputs over time. For a stochastic model, Inputs to a quantitative model are uncertain, and the value of the output from a stochastic model cannot be easily determined, even if the value of the input that can be determined is known. The distributions of the resulting outcomes of a stochastic model is usually due to the large number of simulations involved, and it is widely used as a statistical tool in the life sciences.

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 temperature will be greater than 25°C

Explanation:

In an adiabatic process, heat is not transferred to or from the boundary of the system. The gain or loss of internal heat energy is solely from the work done on the system, or work done by the system. The work done on the system by the environment adds heat to the system, and work done by the system on its environment takes away heat from the system.

mathematically

Change in the internal energy of a system ΔU = ΔQ + ΔW

in an adiabatic process, ΔQ = 0

therefore

ΔU = ΔW

where ΔQ is the change in heat into the system

ΔW is the work done by or done on the system

when work is done on the system, it is conventionally negative, and vice versa.

also W = pΔv

where p is the pressure, and

Δv = change in volume of the system.

In this case, work is done on the gas by compressing it from an initial volume to the new volume of the cylinder. The result is that the temperature of the gas will rise above the initial temperature of 25°C

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:

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:

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:

115.2 W

Explanation:

The computation is shown below:

As we know that

Power = F . v

[tex]F_H = F cos \theta[/tex]

[tex]F_H = 30 \frac{4}{5}[/tex]

[tex]F_H = 24N[/tex]

Now we solve for V

[tex]V = V_0 + at[/tex]            a = 24N ÷ 20Kg

But V_0 = 0          a = 1.2 m/s^2

F_H = ma             V = 0 + (1.2) (4)

a = F_H ÷ m        V = 4.8 m/s

Therefore

Power = F_Hv

= (24) (4.8)

= 115.2 W

By applying the above formuals we can get the power

Answer:

degree of hotness of coldness of a substance

Answer:

Explanation:

La vaca

El pato

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:

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.

Find out more on Personal  protective equipment at: https://brainly.com/question/13720623

When scaffolds are now being construct or deconstruct, a competent person must supervise the work and train everybody who'll be assisting, and the further discussion can be defined as follows:

Therefore, the final answer is "Option B".

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Answer:

Hello your question is incomplete here is the complete question and the attached is the required diagram and solution

Determine the equivalent state of stress on an element at the point which represents the principal stresses and the maximum in-plane shear stress and the associated average normal stress. Also, for each case, determine the corresponding orientation of the element with respect to the element shown in the figure below. Suppose that ?x = 70 MPa . (Figure 1) . a)  Determine the orientation of principal planes of stress.

Answer : The orientation of principal planes of stress are :  -11.6⁰ and 78.4⁰

Explanation:

Given data from the diagram

[tex]t_{xy}[/tex] =  - 15 MPA

α[tex]_{x}[/tex] = 70 MPA

The Calculation of  the principal stresses and the maximum in-plane shear stress are attached below