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3 years ago

Q6: What the differences between CV and CC welding machine?

Engineering

1 answer:

zhenek [66]3 years ago

6 0

Answer:

A CC power source will maintain current at a relatively constant level, while a CV power source will maintain voltage at a relatively constant level.

Explanation:

A CC power source will maintain current at a relatively constant level, regardless of fairly large changes in voltage, while a CV power source will maintain voltage at a relatively constant level, regardless of fairly large changes in current.

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3 years ago

A thin aluminum sheet is placed between two very large parallel plates that are maintained at uniform temperatures T1 = 900 K, T

Maru [420]

The net radiation heat transfer between the two plates per unit surface area of the plates with shield and without shied are respectively; 2282.76 W/m² and 9766.75 W/m²

<h3>How to find the net radiation heat transfer?</h3>

We are given;

Temperature 1; T₁

Temperature 2; T₂

Temperature 3; T₃

Emissivity 1; ε₁ = 0.3

Emissivity 2; ε₂ = 0.7

Emissivity 3; ε₃ = 0.2

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with shield is;

Q'₁₂ = σ(T₁⁴ - T₂⁴)]/[((1/ε₁) + (1/ε₂) - 1) + ((1/ε₃,₁) + (1/ε₃,₂) - 1)]

Q'₁₂ = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[((1/0.3) + (1/0.7) - 1) + ((1/0.15) + (1/0.15) - 1)]

Q'₁₂,shield = 2282.76 W/m²

The net rate of radiation heat transfer with a thin aluminum shield per unit area of the plates with no shield is;

Q'₁₂,no shield = σ(T₁⁴ - T₂⁴)]/((1/ε₁) + (1/ε₂) - 1))

Q'₁₂,no shield = 5.67 * 10⁻⁸(900⁴ - 300⁴)/[(1/0.3) + (1/0.7) - 1)]

Q'₁₂,no shield = 9766.75 W/m²

Then the ratio of radiation heat transfer for the two cases becomes;

Q'₁₂,shield/Q'₁₂,no shield = 2282.76/9766.75 = 0.2337 or 4/17

Read more about Net Radiation Heat Transfer at; brainly.com/question/14148915

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2 years ago

Linear Time Invariant Systems For each of the systems below an input x(t) and the output y(t) are plotted. Determine whether eac

STALIN [3.7K]

Answer:

The system can be described by a convolution

Explanation:

Thinking process:

If we consider a discrete input to a linear time-invariant system, then the system will be periodic with respect to the period, say N. This therefore, means that the output must also be periodic. The proof is as follows:

The LTI system can be written for the system where:

y (n+N) = ∑ $h(k)x(n + N - k)$

= ∑ $h(k)x(n-k)\\= y(n)$

From the proof, it turns out that y(y + N) = y(n) for any value of n, then the output will be the periodic with the period N.

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3 years ago