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MariettaO [177]
3 years ago
15

Why is there an Engineering Process?

Engineering
2 answers:
Inga [223]3 years ago
8 0
There is a engineering process to plan your work out better!
Studentka2010 [4]3 years ago
7 0

Answer:

well if there wasnt a process tony starks suit wouldn’t work properly, and it would probably fall apart. it is to make sure everything works exactly as they planned, our civilization has gotten this far because we created rules on how to build stuff.

Explanation:

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If extension done in a stork position, during backward bending in a standing position of a special test for lumbar spine, produc
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The answer is 7 because I just took the test!
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2 years ago
Estimate the endurance strength, Se, of a 37.5-mm- diameter rod of AISI 1040 steel having a machined finish and heat-treated to
7nadin3 [17]

Answer:

endurance length is 236.64 MPa

Explanation:

data given:

d = 37.5 mm

Sut = 760MPa

endurance limit is

Se = 0.5 Sut

   = 0.5*760 = 380 MPa

surface factor is

Ka = a*Sut^b

where

Sut is ultimate strength

for AISI 1040 STEEL

a = 4.51, b = -0.265

Ka = 4.51*380^{-0.265}

Ka = 0.93

size factor is given as

Kb =1.29 d^{-0.17}

Kb = 0.669

Se = Sut *Ka*Kb

    = 380*0.669*0.93

Se = 236.64 MPa

therefore endurance length is 236.64 MPa

4 0
3 years ago
You don't have to notify employees that a lockout/tagout is about to begin?
Sergeu [11.5K]

Answer:

When the imposter is sus : O

Explanation:

3 0
3 years ago
Type the correct answer in the box. Spell all words correctly.
Otrada [13]

Answer:

solar engineering field

5 0
3 years ago
Read 2 more answers
Air enters the compressor of an air-standard Brayton cycle with a volumetric flow rate of 60 m3/s at 0.8 bar, 280 K. The compres
natima [27]

Answer:

a) The Net power developed in this air-standard Brayton cycle is 43.8MW

b) The rate of heat addition in the combustor is 84.2MW

c) The thermal efficiency of the cycle is 52%

Explanation:

To solve this cycle we need to determinate the enthalpy of each work point of it. If we consider the cycle starts in 1, the air is compressed until 2, is heated until 3 and go throw the turbine until 4.

Considering this:

h_{i} =T_{i}C_{pair}=T_{i}1.005\frac{KJ}{Kg K}

\mu_{comp}=\frac{h_{2S}-h_{1}}{h_{2}-h_{1}}

\mu_{comp}=\frac{h_{3}-h_{4}}{h_{3}-h_{4S}}

G_{m} =\frac{PMG_{v}}{TR} =59.73\frac{Kg}{s}

Now we can calculate the enthalpy of each work point:

h₁=281.4KJ/Kg

h₂=695.41KJ/Kg

h₃=2105KJ/Kg

h₄=957.14KJ/Kg

The net power developed:

P_{net}=P_{Tur}-P_{Comp}=G_{m}((h_{3}-h_{4})-(h_{2}-h_{1}))

The rate of heat:

Q=G_{m}(h_{3}-h_{2})

The thermal efficiency:

\mu_{ther}=\frac{P_{net}}{Q}

3 0
3 years ago
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