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kicyunya [14]
3 years ago
14

Why do engineers play a variety of roles in the engineering process? They are part of a team. They need to act as both scientist

s and engineers when working on a project. They have to brainstorm, build, and design all of their projects alone. They have to be cyclical in their process.
Engineering
2 answers:
Sidana [21]3 years ago
8 0

Answer:

Explanation:

They showed that many people worked on the project and demonstrate how hard they worked.

Kamila [148]3 years ago
4 0

Answer: They showed that many people worked on the project and demonstrate how hard they worked.

Explanation:

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Holmes owns two suits: one black and one tweed. He always wears either a tweed suit or sandals. Whenever he wears his tweed suit
vazorg [7]

Answer:

He wore his black suit, another color of shirt (not purple) and shoes

Explanation:

Holmes owns two suits: one black and one tweed.

Whenever he wears his tweed suit and a purple shirt, he chooses not to wear a tie and whenever he wears sandals, he always wears a purple shirt.

So, if he wore a bow tie yesterday, it means he wore his black suit, another color of shirt (not purple) and shoes because the shirt color is not purple

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3 years ago
True or false It is legal to pass in Florida when approaching within 100 feet of or traversing any railroad crossing grade croea
Iteru [2.4K]
That is a false statement.
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3 years ago
What would happen if an exposed film was accidentally placed in the fixer before being placed in the developer
Eduardwww [97]
Do you still want this answered
5 0
3 years ago
A road has a crest curve, where the PVI station is a 71 35. The road transitions from a 2.1% grade to a -3.4% grade. The highest
sveticcg [70]

Answer:

Stat PVC = Stat(82+98.5)

Stat PVT = Stat(59+71.5)

Explanation

PVI = 71 + 35

Let G1 = Grade 1; G2 = Grade 2

G1 = +2.1% ; G2 = -3.4%

Highest point of curve at station = 74 + 10

General equation of a curve:

y = ax^{2} +bx+c\\dy/dx=2ax+b\\

At highest point of the curve dy/dx=o

2ax+b=0\\x=-b/2a\\x=G1L/(G2-G1)\\x=L/2 +(stat 74+10)-(stat 71+35)\\x=L/2 + 275

-G1L/(G2-G1) = (L/2 + 275)/100\\L = -2327 ft\\Station PVC = Stat(71+35)+(-2327/2)\\\\Stat PVC = 7135-1163.5\\Stat PVC = Stat(82+98.5)\\

Station PVT

Station PVT = Stat PVI + (L/2)\\Station PVT = Stat(71+35)+(-2327/2)\\Station PVT = 7135-1163.5\\Stat PVT = Stat(59+71.5)

3 0
3 years ago
The outer surface of an engine is situated in a place where oil leakage can occur. When leaked oil comes in contact with ahot su
VladimirAG [237]

Answer:

TBC thickness of 4 mm is insufficient to prevent fire hazard

Explanation:

Given:

- Temperature of hot-fluid inner surface T_i = 333°C

- The convection coefficient hot-fluid h_i = 7 W/m^2K

- The thermal conductivity of engine cover k_1 = 14 W/mK

- The thickness of engine cover L_1 = 0.01 m

- The thermal conductivity of TBC layer k_2 = 1.1 W/mK  ... (Typing error)

- The thickness of TBC layer L_2 = 0.004 m

- Temperature of ambient air outer surface T_o = 69°C

- The convection coefficient ambient air h_o = 7 W/m^2K

Find:

Would a TBC layer of 4 mm thickness be sufficient to keep the engine cover surface below autoignition temperature of 200°C to prevent fire hazard?

Solution:

- We will use thermal circuit analogy for the 1-D problem and steady state conduction with no heat generation in the cover or TBC layer.

 The temperature at each medium interface and the Thermal resistance for each medium is given in the attachment schematic and circuit analogy.

 - We will calculate the total heat flux for the entire system q:

                       q = ( T_i - T_o ) / R_total

- R_total is the equivalent thermal resistance of the entire circuit. Since all resistances are in series we have:

                       R_total = 1 / h_i + L_1 / k_1 + L_2 / k_2 + 1 / h_o

- Plug in the values and compute:

                       R_total = 1 / 7 + 0.01 / 14 + 0.004 / 1.1 + 1 / 7

                       R_total = 0.2900649351 T-m^2 / W

- Calculate the Total heat flux q:

                       q = ( 333 - 69 ) / 0.2900649351

                       q = 910.141 W / m^2

- Just like the total current in a circuit remains same, the total heat flux remains same. We will use the total heat flux q to calculate the temperature of outer engine surface T_2 as follows:

                      q = ( T_i - T_2 ) / R_i2

Where,

                      R_i2 = 1 / h_i + L_1 / k_1

                      R_i2 = 1 / 7 + 0.01 / 14 = 0.14357 T-m^2 / W

Hence,

                      ( T_i - T_2 ) = q*R_i2

                        T_2 = T_i - q*R_i2

Plug the values in:

                        T_2 = 333 - 910.141*0.14357

                        T_2 = 202.33°C

- The outer surface of the engine cover has a temperature above T_ignition = 200°C. Hence, the TBC thickness of 4 mm is insufficient to prevent fire hazard

3 0
3 years ago
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