Turning operations that require heavy material removal typically use what setting on the
engine lathe?

Answer:

A

Explanation:

Answer:

A) Engineers must clearly determine exactly what needs to be solved or designed.

Explanation:

Answer:

D = 0.060732 in

Explanation:

given data

sp. wt. = 500 lb/ft³

diameter = 0.036 in

solution

we get here maximum diameter of rod that is express as

D = [tex]\sqrt{\frac{8 \sigma }{\pi y}}[/tex]   ......................1

here [tex]\sigma[/tex] surface tension of water at 60⁰f  = 5.03 × [tex]10^{-3}[/tex]  lb/ft and y = 500 lb/ft³

so put here value and we will get

D = [tex]\sqrt{\frac{8 \times 5.03 \times 10^{-3} }{\pi \times 500}}[/tex]

D = 0.005061 ft

D = 0.060732 in

Answer:

its a gun

Explanation:

The Accuracy International AWM (Arctic Warfare Magnum or AI-Arctic Warfare Magnum) is a bolt-action sniper rifle manufactured by Accuracy International designed for magnum rifle cartridges. ...

Effective firing range: 1,100 m (1,203 yd) (.300 ...

Manufacturer: Accuracy International

In service: 1996–present

Answer:

$8.89

Explanation:

$8.80/a day for heat and $0.09kwh for electricity

Answer:

b. The universe may continue to expand.

Explanation:

The most supported theory about the future of the universe is that, it may continue to expand up to such point that it will become too cold to allow any forms of life to pull through. This is popularly known as the "Big Freeze."

Such  acceleration of expansion is caused by the so-called, "dark energy." This also explains why the acceleration of expansion is not constant. This thereby causes the distance between galaxies to grow farther apart.

Answer:

b

Explanation:

edge 2021

Answer:

A) 22.22 kg

B) 0.1364

Explanation:

When we perform a degree of freedom (DOF) analysis on the bypass subsystem, we will have 2 unknown masses which are; (m1, m2).l with degree of freedom as 1.

Secondly, when we perform a degree of freedom analysis on the overall system, we will have 2 unknown masses namely (m3, m5) with degree of freedom as 0.

Thirdly, when we perform a degree of freedom analysis on the evaporator, we will have 3 unknown masses namely (m1, m3, m4) with Degree of freedom as 1.

Lastky,when we perform a degree of freedom analysis on mixing point, we will have 3 unknown masses namely, (m2, m4, m5) with degree of freedom as 1.

We are given;

Fresh orange = 10.0 wt% solids

Frozen juice concentrate = 45.0wt% solids

Mixture produced = 67.0 wt% solids

Final concentrate = 45.0 wt% solids

Now, making use of the overall solids balance, we have;

0.1 × 100 = 0.45(m5)

m5 = (0.1 × 100)/0.45

m5 = 22.22 kg

Making use of overall mass balance, we have;

100 kg = m3 + m5

m3 = 100 - 22.22

m3 = 77.78 kg

Now, making use of the Mixing point mass balance, we have;

m4 + m2 = m5

So; m4 + m2 = 22.22 - - - (eq 1)

Also, making use of the Mixing point solids balance, we have;

0.67m4 + 0.1m2 = 0.45m5

0.67m4 + 0.1m2 = 9.999 - - - (eq 2)

Solving eq 1 and eq 2 simultaneously, we have;

m2 = 13.64 kg

m4 = 8.58 kg

Final concentrate is m5 = 22.22 kg

Now, fraction = m2/100 = 13.64/100 = 0.1364

Answer: (d−4) 2

Explanation: Factoring  d2-8d+16

The first term is,  d2  its coefficient is  1 .

The middle term is,  -8d  its coefficient is  -8 .

The last term, "the constant", is  +16

Step-1 : Multiply the coefficient of the first term by the constant   1 • 16 = 16

Step-2 : Find two factors of  16  whose sum equals the coefficient of the middle term, which is   -8 .

     -16    +    -1    =    -17

     -8    +    -2    =    -10

     -4    +    -4    =    -8    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -4  and  -4

                    d2 - 4d - 4d - 16

Step-4 : Add up the first 2 terms, pulling out like factors :

                   d • (d-4)

             Add up the last 2 terms, pulling out common factors :

                   4 • (d-4)

Step-5 : Add up the four terms of step 4 :

                   (d-4)  •  (d-4)

            Which is the desired factorization

  Multiply  (d-4)  by  (d-4)

The rule says : To multiply exponential expressions which have the same base, add up their exponents.

In our case, the common base is  (d-4)  and the exponents are :

         1 , as  (d-4)  is the same number as  (d-4)1

and   1 , as  (d-4)  is the same number as  (d-4)1

The product is therefore,  (d-4)(1+1) = (d-4)2

Please mark me brainlyest

Answer:

a. [tex]Total\ Cost = \$2667[/tex]

b. [tex]Total\ Cost = \$667[/tex]

Explanation:

(a)

Given

[tex]Distance = 10000\ miles[/tex]

[tex]Rate = 15\ miles/gallon[/tex]

[tex]Price = \$4.00/gallon[/tex]

Required

Determine the total amount spent in a year

First, we need to determine the number of gallons used in a year;

[tex]Total\ Gallons = Distance/Rate[/tex]

[tex]Total\ Gallons = 10000miles /\frac{15miles}{gallon}[/tex]

[tex]Total\ Gallons = 10000miles * \frac{1\ gallon}{15\ miles}[/tex]

[tex]Total\ Gallons = 10000 * \frac{1\ gallon}{15}[/tex]

[tex]Total\ Gallons = \frac{10000\ gallons}{15}[/tex]

[tex]Total\ Gallons = \frac{10000}{15}\ gallons[/tex]

Next, is to determine the total cost:

[tex]Total\ Cost = Total\ Gallons * Price[/tex]

[tex]Total\ Cost = \frac{10000}{15}\ gallon * \frac{\$4}{gallon}[/tex]

[tex]Total\ Cost = \frac{10000}{15} * \$4[/tex]

[tex]Total\ Cost = \frac{10000 * \$4}{15}[/tex]

[tex]Total\ Cost = \frac{\$40000}{15}[/tex]

[tex]Total\ Cost = $2666.66666667[/tex]

[tex]Total\ Cost = \$2667[/tex] (approximated)

(b)

Given

[tex]Rate = 60\ miles/gallon[/tex]

Required

Determine the total amount spent in a year

First, we need to determine the number of gallons used in a year;

[tex]Total\ Gallons = Distance/Rate[/tex]

[tex]Total\ Gallons = 10000miles /\frac{60\ miles}{gallon}[/tex]

[tex]Total\ Gallons = 10000miles * \frac{1\ gallon}{60\ miles}[/tex]

[tex]Total\ Gallons = 10000 * \frac{1\ gallon}{60}[/tex]

[tex]Total\ Gallons = \frac{10000\ gallons}{60}[/tex]

[tex]Total\ Gallons = \frac{10000}{60}\ gallons[/tex]

Next, is to determine the total cost:

[tex]Total\ Cost = Total\ Gallons * Price[/tex]

[tex]Total\ Cost = \frac{10000}{60}\ gallon * \frac{\$4}{gallon}[/tex]

[tex]Total\ Cost = \frac{10000}{60} * \$4[/tex]

[tex]Total\ Cost = \frac{10000 * \$4}{60}[/tex]

[tex]Total\ Cost = \frac{\$40000}{60}[/tex]

[tex]Total\ Cost = \$666.666666667[/tex]

[tex]Total\ Cost = \$667[/tex] (approximated)

Answer:

American Welding Society filler metal specification A5. 5-96

Explanation:

You prepare low alloy electrodes by adding the correct alloying elements to the electrode coating. This specification covers the following;

Answer:

A

Explanation:

Answer:

Yes, it was caused by an insider or outsider.

Explanation:

Yes, I believe that this event was caused by an insider or outsider. This is because, if a USB removable flash drive is attached to all the office computers by an insider and now an outsider sends a mail to the company’s General group mail; if the mail has a link that contains some infected worms and virus; once anybody clicks on the link in the mail then, all the computers will be affected by the virus.

Therefore, this event may be caused by the insider who clicked the infected link or by the outsider sent the infected link in the mail.

Answer:

1) 13 mg/liters

2)  72.22 mg/lit

3)  The Alkanity in raw water is not sufficient enough and the kinds of chemicals needed to enhance its acidity are

Explanation:

1) plot of turbidity versus dose and optimal dose of Alum ( mg/L )

Optimal dose of Alum = 13 mg/liters from the graph attached below

2) Theoretical amount of alkalinity consumed at the optimal dose can be calculated as follows

Alkanity is due to [tex]HCO^-_{3}[/tex]

given optimal dose of Alum = 13 mg/liters for question 1

I mole of alum = 2 moles of AL(OH)3

666 grams of alum = 2*27 = 54 grams of AL(OH)3

hence 1 mole of [tex]AL^{+3}[/tex]  = (13/54 ) mMole / lit

The  moles of HCO3 = 6 * [tex]\frac{13}{54}[/tex]   because 1 mole of Alum reacts with 6 moles of HC03

[HCO3] as CaCO3 = 6 * (13/54) * 50

                               = 72.22 mg/lit (theoretical amount of alkalinity consumed)

3) The Alkanity in raw water is not sufficient enough and the kinds of chemicals needed to enhance its acidity are

Answer:

Explanation:

Given that:

At state 1:

Pressure P₁ = 20 bar

Volume V₁ = 0.03 [tex]\mathbf{m^{3}}[/tex]

From the tables at saturated vapour;

Temperature T₁ = 212.4⁰ C  ; [tex]v_1 = vg_1[/tex] = 0.0996 [tex]\mathbf{m^{3}}[/tex] / kg

The mass inside the cylinder is m = 0.3 kg, which is constant.

The specific internal energy u₁ = ug₁ = 2599.2 kJ/kg

At state 2:

Temperature T₂ = 200⁰ C

Since the 1 - 2 occurs in an isochoric process v₂ = v₁ = 0.099 [tex]\mathbf{m^{3}}[/tex] / kg

From temperature T₂ = 200⁰ C

[tex]v_f_2 = 0.0016 \ m^3/kg[/tex]  

[tex]vg_2 = 0.127 \ m^3/kg[/tex]  

Since  [tex]vf_2 < v_2<vg_2[/tex] , the saturated pressure at state 2 i.e. P₂ = 15.5 bar

Mixture quality [tex]x_2 = \dfrac{v_2-vf_2}{vg_2 -vf_2}[/tex]

[tex]x_2 = \dfrac{(0.099-0.0016)m^3/kg}{(0.127 -0.0016) m^3/kg}[/tex]

[tex]x_2 = \dfrac{(0.0974)m^3/kg}{(0.1254) m^3/kg}[/tex]

[tex]\mathsf{x_2 =0.78}[/tex]

At temperature T₂, the specific internal energy [tex]u_f_2 = 850.6 \ kJ/kg[/tex] , also [tex]ug_2 = 2594.3 \ kJ/kg[/tex]

Thus,

[tex]u_2 = uf_2 + x_2 (ug_2 -uf_2)[/tex]

[tex]u_2 =850.6 +0.78 (2594.3 -850.6)[/tex]

[tex]u_2 =850.6 +1360.086[/tex]

[tex]u_2 =2210.686 \ kJ/kg[/tex]

At state 3:

Temperature [tex]T_3=T_2 = 200 ^0 C ,[/tex]

[tex]V_3 = 2V_1 = 0.06 \ m^3[/tex]

Specific volume [tex]v_3 = 0.2 \ m^3/kg[/tex]

Thus; [tex]vg_3 =vg_2 = 0.127 \ m^3/kg[/tex] ,

SInce [tex]v_3 > vg_3[/tex], therefore, the phase is in a superheated vapour state.

From the tables of superheated vapour tables; at [tex]v_3 = 0.2 \ m^3/kg[/tex] and T₃ = 200⁰ C

The pressure = 10 bar and v =0.206 [tex]\ m^3/kg[/tex]

The specific internal energy [tex]u_3[/tex] at the pressure of 10 bar = 2622.3 kJ/kg

The changes in the specific internal energy is:

[tex]u_2-u_1[/tex]

= (2210.686 - 2599.2) kJ/kg

= -388.514 kJ/kg

≅ - 389 kJ/kg

[tex]u_3-u_2[/tex]

= (2622.3 - 2210.686)  kJ/kg

= 411.614 kJ/kg

≅ 410 kJ/kg  

We can see the correct sketches of the T-v plot showing the diagrammatic expression in the image attached below.

Answer:

gdyc ddxtfvytg4dgtfxdwcftcd3rcby

If the crankshaft end play is out of specifications, check for:

Thrust Bearings

Main Bearings

Crankshaft

Crankshaft Thrust Washers

Engine Block

To understand the crankshaft when it is out of specifications, check for:

Thrust Bearings: Check the condition of the thrust bearings, which are located at the front and/or rear of the engine block. Excessive wear or damage to the thrust bearings can contribute to increased crankshaft end play.

Main Bearings: Inspect the main bearings, which support the crankshaft within the engine block. Worn or damaged main bearings can cause excessive movement of the crankshaft.

Crankshaft: Check the crankshaft itself for any signs of damage, such as scoring or bending. A damaged crankshaft may not sit properly within the bearings, leading to increased end play.

Crankshaft Thrust Washers: Some engines use thrust washers to control the end play of the crankshaft. Inspect these washers for wear, damage, or improper installation.

Engine Block: Check the engine block for any signs of damage or distortion that may affect the alignment and support of the crankshaft.

To learn more about Crankshaft, refer:

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Answer: At least I’m not the only one looking for answers lol

Explanation:

This question is incomplete, the complete question is;

Steam heated at constant pressure in a steam generator enters the first stage of a supercritical reheat cycle at 28 MPa, 520°C. Steam exiting the first-stage turbine at 6 MPa is reheated at constant pressure to 500°C. Each turbine stage has an isentropic efficiency of 78% while the pump has an isentropic efficiency of 82%. Saturated liquid exits the condenser that operates at constant pressure of 6 kPa.

Determine the quality of the steam exiting the second stage of the turbine and the thermal efficiency.

Answer:

- the quality of the steam exiting the second stage of the turbine is 0.9329  

- the thermal efficiency is 36.05%  

Explanation:

get the properties of steam at pressure p1 = 28 MPa and temperature T2 = 520°C .

Specific enthalpy h1= 3192.3 kJ/kg

Specific entropy s1 = 5.9566 kJ/kg.K  

Process 1 to 2s is isentropic expansion process in the turbine

S1 = S2s

get the enthalpy at state 2s at pressure p2 = 6 MPa and S2s = 5.9566 kJ/kg.K

h2s = 2822.2 kJ/kg

get the enthalpy at state 2 using isentropic turbine efficiency of the turbine. nT1 = (h1 - h2) / (h1 - h2s)

0.78 = (3192.3 - h2) / (3192.3 - 2822.2)

h2 = 2903.6 kJ/kg

get the enthalpy at state 3 at pressure p2 = p3 = 6 MPa and T3 = 500°C

h3 = 3422.2 kJ/kg

s3 = 6.8803 kJ/kg.K

Process 3 to 4s is isentropic expansion process in the turbine

S3 = S4s

get the enthalpy at state 4s at pressure p4s = p4 = 6 kPa and S4s = 6.8803 kJ/kg.K

h4s = 2118.8 kJ/kg

get the enthalpy at state 4 using isentropic turbine efficiency of the turbine. nT2 = (h3 - h4) / (h3 - h4s)

0.78 = (3422.2 - h4) / ( 3422.2 - 2118.8 )

h4 = 2405.5 kJ/kg

get the properties at pressure, p5 = 6 kPa

h5 = hf

= 151.53 kJ/kg

v5 = Vf  

= 0.0010064 m³/kg  

get the enthalpy at state 6 using isentropic pump efficiency of the turbine, at

p6 = p1 = 28 MPa

np = v5( p6 - p5) / (h6 - h5)

0.82 =  ((0.0010064)( 28000 - 6)) / (h6 - 151.53)

h6 = 185.89 kJ/kg  

Now to find the quality of the steam at the exit of the second stage of the turbine

At stat4, p4 = 6kPa  

h4f = 151.53 kJ/kg

h4fg = 2415.9 kJ/kg  

h4 = h4f + x4h4fg

2405.5 = 151.53 + (x4 (2415.9))

x4 = 0.9329  

the quality of the steam exiting the second stage of the turbine is 0.9329  

Also to find the efficiency of the power plant, we use the following equation;

n = Wnet / Qin  

= (Wt1 + Wt2 - Wp) / (Q61 + Q23)

=  [(h1 - h2) + (h3 - h4) - (h6 - h5)] / [(h1 - h6) + (h3 - h2)]

[(3192.3 - 2903.6) + (3422.2 - 2405.5) - (185.89 - 151.53)] / [(3192.3 - 185.89) + (3422.2 - 2903.6)]

= 0.3605

n = 36.05%  

therefore the thermal efficiency is 36.05%  

Technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.

Hybrid vehicle are defined as a powered by a combustion engine and/or a number of electric motors that draw power from batteries. A gas-powered car simply has a traditional gas engine, but a hybrid car also features an electric motor.

One important advantage of hybrid cars is their capacity to reduce the size of the main engine, which improves fuel efficiency. Many hybrid vehicles employ electric motors to accelerate slowly at first until they reach higher speeds. They then use gasoline-powered engines to increase fuel efficiency.

Thus, technician A says that the unitized structure of a hybrid vehicle is considerably different when compared to the same conventional model is right.

To learn more about hybrid vehicle, refer to the link below:

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Answer:

m = 228 gm

ΔQ = 4.35 Btu

P₂ = 43.2 lbf/in² = 2.98 x 10⁵ Pa

Explanation:

FOR MASS OF OXYGEN:

We will use ideal gas equation for initial conditions:

P₁V₁ = nRT₁

P₁V₁ = mRT₁/M

where,

P₁ = Initial Pressure = (16.8 lbf/in²)(6894.76 Pa/1 lbf/in²) = 1.15 x 10⁵ Pa

V₁ = Initial Volume = (5 ft³)(0.028316 m³/1 ft³) = 0.1415 m³

m = mass of oxygen = ?

R = Universal Gas Constant = 8.314 J/mol.k

T₁ = Initial Temperature = 35°F = 274.67 k

M = Molecular Mass of Oxygen = 32 gm

Therefore,

(1.15 x 10⁵ Pa)(0.1415 m³) = m(8.314 J/mol.k)(274.67 k)/(32 gm)

m = (1.15 x 10⁵ Pa)(0.1415 m³)(32 gm)/(8.314 J/mol.k)(274.67 k)

m = 228 gm

FOR AMOUNT OF HEAT TRANDFER:

Using First Law of Thermodynamics:

ΔQ = ΔU +W

Since, this is a constant volume process. Therefore, the work done in this process must be equal to zero:

ΔQ = ΔU + 0

ΔQ = ΔU

and,

ΔU = m Cv ΔT

Therefore,

ΔQ = m Cv ΔT

where,

ΔQ = Heat Transfer Amount = ?

m = mass of oxygen = 228 g = 0.228 kg

Cv = Molar Specific Heat of Oxygen at Constant Volume = 0.659 KJ/kg.k

ΔT = Change in temperature = 305.22 k - 274.67 k = 30.55 k

Therefore,

ΔQ = (0.228 kg)(0.659 KJ/kg.k)(30.55 k)

ΔQ = (4.6 KJ)(0.9478 Btu/1 KJ)

ΔQ = 4.35 Btu

FOR FINAL PRESSURE IN TANK:

Using equation of state:

P₁V₁/T₁ = P₂V₂/T₂

Here, the volume will be constant due to rigid tank. Hence,

V₁ = V₂ = V

Therefore,

P₁V/T₁ = P₂V/T₂

P₁/T₁ = P₂/T₂

(16.8 lbf/in²)/(35°F) = P₂/(90°F)

P₂ = (16.8 lbf/in²)(90°F)/(35°F)

P₂ = 43.2 lbf/in² = 2.98 x 10⁵ Pa

Answer:

A) V_o = 0.7543 V

B) W = 328.42 nm

C) X_n = 298.56 nm and X_p = 29.86 nm

D) Q_j = 47.767 × 10^(-15)

Explanation:

We are given;

Acceptor concentration; N_A = 10^(17) /cm³

donor concentration; N_D = 10^(16) /cm³

ni = 1.5 × 10^(10) /cm3

A) The formula for the built - in - voltage at the junction is given by;

V_o = V_T(In (N_A × N_D/ni²))

Where V_T is thermal voltage at room temperature with a value from online sources as 25.9 mV

Thus;

V_o = 25.9(In (10^(17) × 10^(16))/(1.5 × 10^(10))²

V_o = 0.7543 V

B) Now, formula for the width of the depletion region (W)is given as;

W = √(2ε_s/q[(1/N_A) + (1/N_D)]V_o)

Where;

ε_s is the permittivity in the semiconductor with a constant value of 1.04 × 10^(-12) F/cm

q is the electron charge = 1.6 × 10^(-19) C

Thus;

W = √(2 × 1.04 × 10^(-12)/(1.6 × 10^(-19)) [(1/10^(17)) + (1/10^(16)]0.7543)

W = 32.842 × 10^(-6) cm

Converting to m gives;

W = 328.42 nm

C) Formula for extent of the n region is;

X_n = W × (N_A/(N_A + N_D))

X_n = 32.842 × 10^(-6) × (10^(17)/(10^(17) + 10^(16))

X_n = 29.856 × 10^(-6) cm

Converting to m gives;

X_n = 298.56 nm

Formula for extent of the p region is;

X_p = W - X_n

X_p = 328.42 nm - 298.56 nm

X_p = 29.86 nm

D) The charge stored on either side of the junction is given by the formula;

Q_j = Aq[(N_A × N_D)/(N_A + N_D)]W

Where A is junction Area and we are given junction area as 10 µm² = 100 × 10^(-8) cm

Thus;

Q_j = (100 × 10^(-8) × 1.6 × 10^(-19))[(10^(17) × 10^(16))/(10^(17) + 10^(16)] × 32.842 × 10^(-6)

Q_j = 47.767 × 10^(-15)

Answer: the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Explanation:

from the T-S diagram, we get the overall pressure ratio of the cycle is 9

Calculate the pressure ratio in each stage of compression and expansion. P1/P2 = P4/P3  = √9 = 3

P5/P6 = P7/P8  = √9 =3  

get the properties of air from, "TABLE A-17 Ideal-gas properties of air", in the text book.

At temperature T1 =300K

Specific enthalpy of air h1 = 300.19 kJ/kg

Relative pressure pr1 = 1.3860  

At temperature T5 = 1200 K

Specific enthalpy h5 = 1277.79 kJ/kg

Relative pressure pr5 = 238  

Calculate the relative pressure at state 2

Pr2 = (P2/P1) Pr5

Pr2 =3 x 1.3860 = 4.158  

get the two values of relative pressure between which the relative pressure at state 2 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure pr = 4.153

The corresponding specific enthalpy h = 411.12 kJ/kg  

Relative pressure pr = 4.522

The corresponding specific enthalpy h = 421.26 kJ/kg  

Find the specific enthalpy of state 2 by the method of interpolation

(h2 - 411.12) / ( 421.26 - 411.12) =  

(4.158 - 4.153) / (4.522 - 4.153 )

h2 - 411.12 = (421.26 - 411.12) ((4.158 - 4.153) / (4.522 - 4.153))  

h2 - 411.12 = 0.137

h2 = 411.257kJ/kg  

Calculate the relative pressure at state 6.

Pr6 = (P6/P5) Pr5

Pr6 = 1/3 x 238 = 79.33  

Obtain the two values of relative pressure between which the relative pressure at state 6 lies and take the corresponding values of specific enthalpy from, "TABLE A-17 Ideal-gas properties of air", in the text book.  

Relative pressure Pr = 75.29

The corresponding specific enthalpy h = 932.93 kJ/kg  

Relative pressure pr = 82.05

The corresponding specific enthalpy h = 955.38 kJ/kg  

Find the specific enthalpy of state 6 by the method of interpolation.

(h6 - 932.93) / ( 955.38 - 932.93) =  

(79.33 - 75.29) / ( 82.05 - 75.29 )

(h6 - 932.93) = ( 955.38 - 932.93) ((79.33 - 75.29) / ( 82.05 - 75.29 )

h6 - 932.93 = 13.427

h6 = 946.357 kJ/kg

Calculate the total work input of the first and second stage compressors

(Wcomp)in = 2(h2 - h1 ) = 2( 411.257 - 300.19 )

= 222.134 kJ/kg  

Calculate the total work output of the first and second stage turbines.

(Wturb)out = 2(h5 - h6) = 2( 1277.79 - 946.357 )

= 662.866 kJ/kg  

Calculate the net work done

Wnet = (Wturb)out  - (Wcomp)in

= 662.866 - 222.134

= 440.732 kJ/kg  

Calculate the minimum mass flow rate of air required to generate a power output of 105 MW

W = m × Wnet

(105 x 10³) kW = m(440.732 kJ/kg)

m = (105 x 10³) / 440.732

m = 238.2 kg/s

therefore the minimum mass flow rate of air required to generate a power output of 105 MW is 238.2 kg/s

Answer:

- the final equilibrium temperature in the cylinder is  85.67 °C  

- Also the answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats.  

Explanation:

Given That;

Compartment Nitrogen:

Volume V1 = 1 m³, Pressure P1 = 500 kPa , Temperature T = 120°C

Compartment of helium:

Volume V1H = 1 m³, Pressure P1H = 500 kPa,  Temperature T1H = 40°C  

From the ideal specific heat of gases

-Nitrogen

The gas k (constant) and the k (constant) volume specific heats are;

R = 0.2968 kJ

/kg.K

Cv = 0.743 kJ/kg.K

also From the ideal specific heat of gases

-helium  

R = 2.0769 kJ /kg.K

Cv = 3.1156 kJ/kg.K  

we know that

PV = mRT

Mass of the nitrogen  

mN2 = (P1V1 /RT1)_N2

mN2 = (500)(1) / (0.2968)(393)

= 4.29kg

mass of helium

mHe = (P1HVIH /RT1H)_He

mHe = (P1V1 /RT1)_N2

mHe  = (500)(1) / (2.0769)(313)

= 0.769kg

Taking the whole contents of the cylinder,

the 1st law relation can be expressed as;  

Ein - Eout = ΔEsystem

0 = ΔU = (ΔU)_N2 + (ΔU)_He

0 = [mcV(T2 -T1)]_N2 +  [mcV(T2 - Ti)]He  

Since T2 =T_F

0 = [mcV(T_F -T1)]_N2 +  [mcV(T_F - Ti)]He  

we Substitute  

(4.29)(0.743)(T_F - 120) + (0.769)(3.1156)(T_F - 40) = 0

T_F = 85.67 °C  

therefore the final equilibrium temperature in the cylinder is  85.67 °C  

Also the answer would be the same if the piston were not free to move since it would effect only pressure and not the specific heats.  

Answer:Itaka

Explanation:

Answer:

Tora!

Explanation:

It was an expression that illustrated the complete surprise of the attack.

Answer: A. steer around a hazard

Explanation: say there was a vehicle next to you on your left and there was a pile of rocks it would prevent you from steer from going straight but instead have you steer around the rocks and cross into another lane that’s what it said in the lesson

Driving next to another vehicle can sometimes take away the option to A. steer around a hazard

This refers to the rules and regulations that guide road users and pedestrians to ensure their safety.

Hence, we can see that when driving on the road next to another vehicle, the option to steer around a hazard is taken away, and hence, it is not advisable to do this.

Read more about traffic rules here:

https://brainly.com/question/1071840

#SPJ2

Answer:

Written in C++

#include<iostream>

#include<cmath>

using namespace std;

int main()

{

float degreeC, degreeF;

cout<<"Degree Fahrenheit: ";

cin>>degreeF;

degreeC = 5 * (degreeF - 32)/9;

cout<<"Degree Celsius: "<<degreeC<<" C";  

return 0;

}

Explanation:

The question requests that input should be in degree Fahrenheit

Declare all necessary variables

float degreeC, degreeF;

Prompt user for input in degrees Fahrenheit as stated in the question

cout<<"Degree Fahrenheit: ";

Get User Input

cin>>degreeF;

Convert degree Fahrenheit to Celsius

degreeC = 5 * (degreeF - 32)/9;

Display output

cout<<"Degree Celsius: "<<degreeC<<" C";  

Answer:

Explanation:

r the open-loop plant:

P(S) = 10/(s2 + 2s + 5) and a controller: k(s +1) /(s) = s(s +2) structured in a unity feedback control architecture: = w r u y C(s) P(s) (f) (numerical) Suppose the maximum control effort the actuator can provide is Umar = 0.52. Run a separate iteration, this time beginning with k = 0.01, and incrementing by 0.01 to determine the largest value of k such that the actuator does not exceed its limit. Simulate for 30 seconds during each iteration. With this k, display a 1 x 2 subplot showing the control command u(t) and the output y(t). On the u(t) plot, show the maximum value Umax being less than 0.52. Also display the k value that yields this response. Note: the loop will probably take a long time to run since it's running the simulation each time.

(a) (numerical) Next, run the simulation for 60 seconds, this time introducing a unit step distur- bance w(t) beginning at 30 seconds. With the value of k found in (f), plot the output y(t) and the control command u(t). Is the system capable of rejecting a constant disturbance? E(S) (h) (by hand) As a follow-up to

(b), compute the transfer function T(s) in terms of k, and W(s) note the number of zeros at the origin. Keep this number in mind as we move into the next topic, system TYPE & steady-state error. It can tell us definitively what types of disturbances we can reject and what types of references we can track.

Answer:

Your answer is: Evaporator, the condensed water drip comes from something called the evaporator. When warm air touches the cold coil ( or the cold pipe) the water vapor condensed while the cold coil absorbs.

Explanation:

Hope this helped : )