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KonstantinChe [14]
3 years ago
15

I am sending an email to my teacher. Is this mature enough?

Engineering
2 answers:
nirvana33 [79]3 years ago
7 0
I dislike how you wrote the 0’s. But honestly Teachers don’t really care as long as they can read it. So sure send it I guess
xenn [34]3 years ago
6 0
Take out the 0’s but other then that yes.
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A 600 MW power plant has an efficiency of 36 percent with 15
ololo11 [35]

Answer:

401.3 kg/s

Explanation:

The power plant has an efficiency of 36%. This means 64% of the heat form the source (q1) will become waste heat. Of the waste heat, 85% will be taken away by water (qw).

qw = 0.85 * q2

q2 = 0.64 * q1

p = 0.36 * q1

q1 = p /0.36

q2 = 0.64/0.36 * p

qw = 0.85 *0.64/0.36 * p

qw = 0.85 *0.64/0.36 * 600 = 907 MW

In evaporation water becomes vapor absorbing heat without going to the boiling point (similar to how sweating takes heat from the human body)

The latent heat for the vaporization of water is:

SLH = 2.26 MJ/kg

So, to dissipate 907 MW

G = qw * SLH = 907 / 2.26 = 401.3 kg/s

8 0
3 years ago
Read 2 more answers
What is the difference between an arch and a dome?
bonufazy [111]
This is an arch, its basically a half circle attach to a rectangle, you could also think of it as an upside down U. A dome is a Sphere with the inside hollowed out.

1 difference is a dome is a 3 dimensional shape while an arch is normally not. Or that a dome is the complete shape with a arch act as it’s diameter.

4 0
2 years ago
Regression analysis is a statistical procedure for developing a mathematical equation that describes how _____. Group of answer
Oduvanchick [21]

Answer:

one dependent and on or more independent variables are related.

4 0
3 years ago
Flank wear data were collected in a series of turning tests using a coated carbide tool on hardened alloy steel at a feed of 0.3
Paladinen [302]

Answer:

A) n =  0.6143, c ≈ 640m/min

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

Explanation:

using the given data

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

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

is attached below

calculate for the constant N from the second plot

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

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

= - N = \frac{In(V2) - In(V1)}{In(T2)-ln(T1)}

therefore  - N = \frac{5.043 - 4.605}{2.302 -3.015}

                       = - 0.6143

THEREFORE  ( N ) = 0.6143

Determine for the constant C from the second plot as well

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

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

hence   C ≈ 640m/min

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

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

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

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

tool life = T = 20.4 min and 10 min

also constant  n and c are obtained from the previous plot

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

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

also using the second values of  v and T

taylor tool life equation = 155 * 10 = C

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

Equate equation 2 and equation 3 and solve simultaneously

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

To find N

take natural log of both sides of the equation

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

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

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

= 0.7129 n = 0.438

therefore n = 0.6143

To find C

substitute 0.6143 for n in equation 2

C = (100)(20.4) ^ 0.6143

C = 637.53 m/min

Attached are the two plots for solution A

7 0
3 years ago
What is the thermal energy contained within a cubic kilometre of granite (in barrels of oil equivalent) for every one degree cha
AnnZ [28]

Answer:

348643.34 barrels of oil

Explanation:

Given;

Volume of the granite = 1 km³ = 1000³ m³

Specific heat of granite, C = 790 J/kg.°C

Density of granite = 2700 kg/m³

Energy in 1 barrel of oil = 6.118 × 10⁹ J

For every 1° change in temperature, ΔT = 1°C

Now,

The thermal energy stored is given as;

Thermal energy = mCΔT

where, m is the mass

the mass of 1 km³ of granite = Density × Volume

or

the mass of 1 km³ of granite = 2700 × 1000³ = 27 × 10¹¹ Kg

therefore,

Thermal energy = 27 × 10¹¹ × 790 × 1

or

Thermal energy = 21330 × 10¹¹ J

hence,

the thermal energy in terms of barrels of oil

= Total thermal energy / Energy stored in 1 barrel of oil

= 21330 × 10¹¹ J / ( 6.118 × 10⁹ J per barrel )

= 348643.34 barrels of oil

4 0
3 years ago
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