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Sladkaya [172]
3 years ago
12

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lskjdfghlskjfdhglkjkfhgldkjhgsldhglsjghlkjghlkjdfhglkdfjfjghldsfjghldjghdffgjhlsffghsldffghdfkj
Engineering
1 answer:
vodka [1.7K]3 years ago
5 0

Answer:

(⌐■-■) (⌐■-■) (⌐■-■) (⌐■-■)

You might be interested in
Is a diesel truck less expensive to drive than a gas truck?
MArishka [77]

Answer:

Typically, diesel trucks cost more than those with gas engines, especially when you're first buying them, as diesel is usually featured as an add-on for gas-powered cars. Diesel add-ons can cost over $5,000 for midsize trucks and around $10,000 for heavy-duty trucks.

Explanation:

Make me brain pls

3 0
2 years ago
1. A pipeline constructed of carbon steel failed after 3 years of operation. On examination it was found that the wall thickness
jek_recluse [69]

Answer:

check the explanation

Explanation:

1.

Thickness Loss = t =\frac{t_{o}-t_{i}}{2} = \frac{114.3-102.3}{2} = 2mm

t_{f} = \frac{1}{2}*6 = 3mm

Hence Rate of Corrosion = 6*\frac{1-0.5}{3} = 1mm/year = 0.03 inches per year

2.

As the expected future life is 7 years,

40 carbon steel pipe has to be replaced every 3 years as given in the question,

Cost per unit length is the sum of material cost and installation cost.

Cost of 40 carbon steel = (5 dollars + 16.5 dollars) * 3 = 64.5 dollars

For 80 carbon steel pipe, first calculate the thickness loss,

\frac{114.3-97.2}{2} = 8.55mm

The critical thickness is given to be 3mm, Hence change in thickness is 8.55-3 = 5.5mm

This 80 carbon steel pipe has to be replaced one more time

Hence, Cost per unit length is the sum of material cost and installation cost.

Cost of 80 carbon steel = (8.3 dollars + 16.5 dollars) * 2 = 49.6 dollars

The best is of stainless steel which does not undergo corrosion at all and thus it needs to be replaced only once throughout the plant operation. Its cost = 24.8 dollars + 16.5 dollars = 41.3 dollars

Hence, stainless steel is the recommended pipe to be used.

3 0
3 years ago
The stagnation chamber of a wind tunnel is connected to a high-pressure airbottle farm which is outside the laboratory building.
Natasha2012 [34]

This question is not complete, the complete question is;

The stagnation chamber of a wind tunnel is connected to a high-pressure air bottle farm which is outside the laboratory building. The two are connected by a long pipe of 4-in inside diameter. If the static pressure ratio between the bottle farm and the stagnation chamber is 10, and the bottle-farm static pressure is 100 atm, how long can the pipe be without choking? Assume adiabatic, subsonic, one-dimensional flow with a friction coefficient of 0.005

Answer:

the length of the pipe is 11583 in or 965.25 ft

Explanation:

Given the data in the question;

Static pressure ratio; p1/p2 = 10

friction coefficient f = 0.005

diameter of pipe, D =4 inch

first we obtain the value from FANN0 FLOW TABLE for pressure ratio of ( p1/p2 = 10 )so

4fL_{max} / D = 57.915

we substitute

(4×0.005×L_{max}) / 4  = 57.915

0.005L_{max} = 57.915

L_{max} = 57.915 / 0.005

L_{max}  = 11583 in

Therefore, the length of the pipe is 11583 in or 965.25 ft

6 0
3 years ago
Determine the reactor volume (assume a CSTR activated sludge aerobic reactor at steady state) required to treat 5 MGD of domesti
12345 [234]

Answer:

1.0MG

Explanation:

to solve this problem we use this formula

S₀-S/t = ksx --- (1)

the values have been given as

concentration = S₀ = 250mg

effluent concentration = S= 10mg

value of K = 0.04L/day

x = 3000 mg

when we put these values into this equation,

250-10/t = 0.04x10x3000

240/t = 1200

we cross multiply from this stage

240 = 1200t

t = 240/1200

t = 0.2

remember the question says that 5MGD is required to be treated

so the volume would be

v = 0.2x5

= 1.0 MG

4 0
2 years ago
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phas
jarptica [38.1K]

Answer:

The heat input from the combustion phase is 2000 watts.

Explanation:

The energy efficiency of the heat engine (\eta), no unit, is defined by this formula:

\eta = \frac{\dot W}{\dot Q} (1)

Where:

\dot Q - Heat input, in watts.

\dot W - Power output, in watts.

If we know that \dot W = 600\,W and \eta = 0.3, then the heat input from the combustion phase is:

\eta = \frac{\dot W}{\dot Q}

\dot Q = \frac{\dot W}{\eta}

\dot Q = \frac{600\,W}{0.3}

\dot Q = 2000\,W

The heat input from the combustion phase is 2000 watts.

8 0
3 years ago
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