Consider a regenerative gas-turbine power plant with two stages of compression and two stages of expansion. The overall pressure ratio of the cycle is 9. The air enters each stage of the compressor at 300 K and each stage of the turbine at 1200 K. Accounting for the variation of specific heats with temperature, determine the minimum mass flow rate of air needed to develop a net power output of 105 MW.

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 tongue- and -groove joint

Explanation:

In this method, you first cut a a groove on one piece of the wood and then cut a matching tongue on the other piece. The cuts can be made using a dado head on a table saw. A glue is applied last in a line all around, thus it is vital to cut the tongue a bit smaller than the groove to allow gluing.

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:

12.84 mg/L

Explanation:

We are given;

Volume of lake; V = 1.1 x 10^(6) m³

decay coefficient; K = 0.10/day = 0.1/(24 × 60 × 60) /s = 0.00000115741 /s

Factory rate: Q_f = 4.3 m³/s

Factory concentration: C_f = 100 mg/L

Stream rate: Q_s = 34 m³/s

Stream Concentration: C_s = 2.3 mg/L

Now, to find the steady state concentration of pollutant in the lake, we will use the formula;

(Q_s•C_s) + (Q_f•C_f) = (Q_f + Q_s)C_L + (KV•C_L)

Where C_L is the steady state concentration of pollutant in the lake.

Thus, making C_L the subject, we have;

C_L = [(Q_s•C_s) + (Q_f•C_f)]/(Q_f + Q_s + K•V)

Plugging in the relevant values gives;

C_L = ((34 × 2.3) + (4.3 × 100))/(4.3 + 34 + (0.00000115741 × 1.1 × 10^(6)))

C_L = 12.84 mg/L

Answer:

the bar was stretched by [tex]\mathbf{\Delta L = 3.36 \ mm}[/tex]

the length of the after it was stretched is [tex]\mathbf{L_{new} = 603.36 \ mm}[/tex]

Explanation:

From the information given:

The strain gauge resistance R = 250 Ω

The gauge factor = 1

The original length L = 0.6 m = 600 mm

After the bar is being stretched under tension force;

the new resistance [tex]R_{new} = 251.42[/tex]

The gauge factor [tex]G = \dfrac{\Delta R/R}{\Delta L /L }[/tex]

where;

[tex]\Delta R = R_{new} - R[/tex]   and [tex]\Delta L = L_{new} - L[/tex]

ΔR = 251.4 - 250

ΔR = 1.4 Ω

[tex]\Delta L = L_{new} - L[/tex]

[tex]L_{new} = L + L (\dfrac{\Delta R/R}{G})[/tex]

[tex]L_{new} = 0.6 + 0.6 (\dfrac{\Delta 1.4/250}{1})[/tex]

[tex]L_{new} = 0.60336 \ m[/tex]

[tex]\mathbf{L_{new} = 603.36 \ mm}[/tex]

Thus, the length of the after it was stretched is [tex]\mathbf{L_{new} = 603.36 \ mm}[/tex]

Thus, the bar was stretched by [tex]\Delta L = L_{new} - L[/tex]

[tex]\Delta L = (603.36 - 600) \ mm[/tex]

[tex]\mathbf{\Delta L = 3.36 \ mm}[/tex]

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%  

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.

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Answer:

The lake can withdraw a maximum of [tex]1.464\times 10^{10}[/tex] cubic feet per year to provide water supply for the Triangle area.

Explanation:

The maximum amount of water that can be withdrawn from the lake is represented by the following formula:

[tex]V = V_{in}+V_{p}-V_{e}-V_{out}[/tex] (Eq. 1)

Where:

[tex]V[/tex] - Available amount of water for water supply in the Triangle area, measured in cubic feet per year.

[tex]V_{in}[/tex] - Inflow amount of water, measured in cubic feet per year.

[tex]V_{out}[/tex] - Amount of water released for the benefit of fish and downstream water users, measured in cubic feet per year.

[tex]V_{p}[/tex] - Amount of water due to precipitation, measured in cubic feet per year.

[tex]V_{e}[/tex] - Amount of evaporated water, measured in cubic feet per year.

Then, we can expand this expression as follows:

[tex]V = f_{in}\cdot \Delta t+h_{p}\cdot A_{l}-h_{e}\cdot A_{l}-f_{out}\cdot \Delta t[/tex]

[tex]V = (f_{in}-f_{out})\cdot \Delta t +(h_{p}-h_{e})\cdot A_{l}[/tex] (Eq. 2)

Where:

[tex]f_{in}[/tex] - Average watershed inflow, measured in cubic feet per second.

[tex]f_{out}[/tex] - Average flow to be released, measured in cubic feet per second.

[tex]\Delta t[/tex] - Yearly time, measured in seconds per year.

[tex]h_{p}[/tex] - Change in lake height due to precipitation, measured in feet per year.

[tex]h_{e}[/tex] - Change in lake height due to evaporation, measured in feet per year.

[tex]A_{l}[/tex] - Surface area of the lake, measured in square feet.

If we know that [tex]f_{in} = 900\,\frac{ft^{3}}{s}[/tex], [tex]f_{out} = 300\,\frac{ft^{3}}{s}[/tex], [tex]\Delta t = 31,536,000\,\frac{second}{yr}[/tex], [tex]h_{p} = 32\,\frac{in}{yr}[/tex], [tex]h_{e} = 55\,\frac{in}{yr}[/tex] and [tex]A_{l} = 47,000\,acres[/tex], the available amount of water for supply purposes in the Triangle area is:

[tex]V = \left(900\,\frac{ft^{2}}{s}-300\,\frac{ft^{3}}{s} \right)\cdot \left(31,536,000\,\frac{s}{yr} \right) +\left(32\,\frac{in}{yr}-55\,\frac{in}{yr} \right)\cdot \left(\frac{1}{12}\,\frac{ft}{in}\right)\cdot (47000\,acres)\cdot \left(43560\,\frac{ft^{2}}{acre} \right)[/tex][tex]V = 1.464\times 10^{10}\,\frac{ft^{3}}{yr}[/tex]

The lake can withdraw a maximum of [tex]1.464\times 10^{10}[/tex] cubic feet per year to provide water supply for the Triangle area.

Answer:

a) the tension in the cord T is 1200 N

b)

reaction at C along x-axis Cy is 400 N

reaction at C along y-axis Cy is 1200 N

reaction at C along z-axis Cz is 0 N

reaction at D along X-axis Dx is 1600 N

reaction at D along y-axis Dy is -480 N

Explanation:

a)

to find the tension in the cord, we say;

∑ Mz = 0

720(200) - T(120) = 0

144000 - 120T = 0

T = 144000/120

T = 1200 N

the tension in the cord T is 1200 N

b)

the reactions at C and D. Assume that the bearing at D does not exert any axial thrust.

we make use of the moment law of equilibrium at point D about y-axis.

∑ MDy = 0

Cx(120) - T(40) = 0

we substitute value of T

Cx(120) - 1200(40) = 0

Cx 120 = 48000

Cx = 48000/120

Cx = 400 N

reaction at C along x-axis Cy is 400 N

​  

Next we also​ apply the moment law of equilibrium at point D about x-axis.

∑MD z = 0

-Cy(120) + 720(80 + 120) = 0

-Cy(120) = - 144000

Cy = -144000 / - 120

Cy = 1200 N

reaction at C along y-axis Cy is 1200 N

we also​ apply the force law of equilibrium along z direction

∑Fz = 0

Cz = 0 N

reaction at C along z-axis Cz is 0 N

we also​ apply the force law of equilibrium along x direction

∑Fx = 0

Cx + Dx + T = 0

Substitute 400N for Cx and 1200 N for Dx

so

400 + Dx + 1200 = 0

Dx = 1600 N

therefore reaction at D along X-axis Dx is 1600 N

we also​ apply the force law of equilibrium along y direction

∑ Fy = 0

Cy + Dy - 720 = 0

we substitute Cy = 1200 N

so

1200 + Dy - 720 = 0

Dy = - 480 N

therefore reaction at D along y-axis Dy is -480 N

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.

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Answer:

The answer is below

Explanation:

a) To solve this problem, we are going to use the smith chart. After entering the value of Zo = 200 ohm and ZL = (50 − j25) Ω, this give us [tex]z_L\ and\ y_L[/tex], the intersection between [tex]y_L[/tex] and the SWR line gives:

[tex]y(d)=1.026-j1.54\\\\We\ are\ using\ the\ imaginary\ part\ to calculate\ the\ capacitance,hence:\\\\wC=1.54*Y_o\\\\C=\frac{1.54}{Z_o*w}=\frac{1.54}{200*2\pi*800*10^6}=1.53*10^{-12}\\ \\C=1.53*10^{-12}F[/tex]

b) Also, we get:

[tex]y(d)=1.0-j1.52\\\\We\ are\ using\ the\ imaginary\ part\ to\ calculate\ the\ inductance,hence:\\\\1/wL=1.52*Y_o\\\\L=\frac{Z_o}{1.52*w}=\frac{200}{1.52*2\pi*800*10^6}=2.6*10^{-8}\\ \\L=2.6*10^{-8}H[/tex]

Answer:

drugs

Explanation:

Answer: A. Makes a commodity available for sale or exchange

Explanation: hope it helps ^w^

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:

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:

$8.89

Explanation:

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

Answer:

Scalability

Explanation:

just took a test and it was right.

Answer:

Humans have been on Earth for a very short amount of time.

Explanation:

edgeunity 2021

Answer:

Humans have been on Earth for a very short amount of time<3

D.

Explanation:

Answer:

The time taken will be "1 hour 51 min". The further explanation is given below.

Explanation:

The given values are:

Number of required layers:

= [tex]\frac{38}{0.25}[/tex]

= [tex]152 \ layers[/tex]

Diameter (d):

= 1.25 mm

Velocity (v):

= 40 mm/s

Now,

The area of one layer will be:

= [tex]38\times 38 \ mm^2[/tex]

= [tex]1444 \ mm^2[/tex]

The area covered every \second will be:

= [tex]d\times v[/tex]

= [tex]1.25\times 40[/tex]

= [tex]50 \ mm^2[/tex]

The time required to deposit one layer will be:

= [tex]\frac{1444}{50}[/tex]

= [tex]28.88 \ sec[/tex]

The time required for one layer will be:

= [tex]15 \ sec[/tex]

∴ Total times required for one layer will be:

= [tex]15+28.88[/tex]

= [tex]43.88 \ sec[/tex]

So,

Number of layers = 152

Therefore,

Total time will be:

= [tex]152\times 43.88[/tex]

= [tex]6669.76 \ sec[/tex]

= [tex]1 \ hour \ 51 \ min[/tex]

Given values are:

Now,

Area of one layer will be:

= [tex]38\times 38[/tex]

= [tex]1444 \ mm^2[/tex]

Area covered per second:

= [tex]d\times v[/tex]

= [tex]1.25\times 40[/tex]

= [tex]50 \ mm^2[/tex]

= [tex]\frac{1444}{50}[/tex]

= [tex]28.88 \ sec[/tex]

= [tex]15 \ sec[/tex]  

∴ Total time for 1 layer will be:

= [tex]15+28.88[/tex]

= [tex]43.88 \ sec[/tex]

hence,

The total time required:

= [tex]152\times 43.88[/tex]

= [tex]6669.76 \ sec[/tex]

= [tex]1 \ hour \ 51 \ minutes[/tex]

Thus the above response is right.

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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 : )

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:

The water of the saturated clayed soil is 66.67 %.

Explanation:

Given;

mass of saturated clayed soil, [tex]M_s[/tex] = 600 g

mass of dry soil sample, [tex]M_d[/tex] = 200 g

mass of water content, [tex]M_w[/tex] = [tex]M_s[/tex] - [tex]M_d[/tex] = 600 g - 200 g = 400 g

The water content is determined as;

[tex]M_w(\%) = \frac{M_s - M_d}{M_s} *100\%\\\\M_w(\%) = \frac{600-200}{600} *100 \% \\\\M_w(\%) = 66.67 \%[/tex]

Therefore, the water of the saturated clayed soil is 66.67 %.

Answer:

I think it is the 2,3,5 and 1 ones

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:

it wqas red

Explanation:

red