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inna [77]
3 years ago
11

A gas metal arc welder is also known as a _____ welder. A) TIGB) GTAWC) GMAWD) Resistance spot

Engineering
1 answer:
nignag [31]3 years ago
7 0

Answer:

GMAW

Explanation:

It's literally the initials of that type of welding

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Soil is to be excavated from a borrow pit to construct a 100-m-long trapezoidal levee that is 3 meters high, 5 meters wide at th
shepuryov [24]

Answer:

The answer is 2217.32m³

Explanation:

Given that:

V₂  = (15 + 5/2) * 3 * 100 = 3000 m³

r₁ = 21.7kN/m3

w (water) = 11.4%.

e₂ (void ratio) = 0.47

V₁ = ?

Now,

rd₁ =  r₁/1 + w₁ = 21.7/1 +(11.4/100)

= 19.47 kN/m3

Thus,

rd₂ = r₁/1 + e₂ = 21.7/1+ 0.47 = 14.76 kN/m3

Then,

V₁/V₂ = rd₁/ rd₂

V₁ = 3000 * 14.76/19.97

V₁ = 2217.32m³

Therefore, volume that should be excavated from the borrow pit is 2217.32m³

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3 years ago
Which employer is the least likely for careers in Energy?
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Answer: Self-Employed,

Explanation: Edge 2020.

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A power cycle receives QH by heat transfer from a hot reservoir at TH = 1200 K and rejects energy QC by heat transfer to a cold
PIT_PIT [208]

Answer:

a) Irreversible, b) Reversible, c) Irreversible, d) Impossible.

Explanation:

Maximum theoretical efficiency for a power cycle (\eta_{r}), no unit, is modelled after the Carnot Cycle, which represents a reversible thermodynamic process:

\eta_{r} = \left(1-\frac{T_{C}}{T_{H}} \right)\times 100\,\% (1)

Where:

T_{C} - Temperature of the cold reservoir, in Kelvin.

T_{H} - Temperature of the hot reservoir, in Kelvin.

The maximum theoretical efficiency associated with this power cycle is: (T_{C} = 400\,K, T_{H} = 1200\,K)

\eta_{r} = \left(1-\frac{400\,K}{1200\,K} \right)\times 100\,\%

\eta_{r} = 66.667\,\%

In exchange, real efficiency for a power cycle (\eta), no unit, is defined by this expression:

\eta = \left(1-\frac{Q_{C}}{Q_{H}}\right) \times 100\,\% = \left(\frac{W_{C}}{Q_{H}} \right)\times 100\,\% = \left(\frac{W_{C}}{Q_{C} + W_{C}} \right)\times 100\,\% (2)

Where:

Q_{C} - Heat released to cold reservoir, in kilojoules.

Q_{H} - Heat gained from hot reservoir, in kilojoules.

W_{C} - Power generated within power cycle, in kilojoules.

A power cycle operates irreversibly for \eta < \eta_{r}, reversibily for \eta = \eta_{r} and it is impossible for \eta > \eta_{r}.

Now we proceed to solve for each case:

a) Q_{H} = 900\,kJ, W_{C} = 450\,kJ

\eta = \left(\frac{450\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 50\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

b) Q_{H} = 900\,kJ, Q_{C} = 300\,kJ

\eta = \left(1-\frac{300\,kJ}{900\,kJ} \right)\times 100\,\%

\eta = 66.667\,\%

Since \eta = \eta_{r}, the power cycle operates reversibly.

c) W_{C} = 600\,kJ, Q_{C} = 400\,kJ

\eta = \left(\frac{600\,kJ}{600\,kJ + 400\,kJ} \right)\times 100\,\%

\eta = 60\,\%

Since \eta < \eta_{r}, the power cycle operates irreversibly.

d) Since \eta >  \eta_{r}, the power cycle is impossible.

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