clean the tubes and fins with a high-pressure jet of air or mechanical scrubbing
ensure that the condenser fans are operating properly
Answer:
The strength coefficient is
and the strain-hardening exponent is 
Explanation:
Given the true strain is 0.12 at 250 MPa stress.
Also, at 350 MPa the strain is 0.26.
We need to find
and the
.

We will plug the values in the formula.

We will solve these equation.
plug this value in 

Taking a natural log both sides we get.

Now, we will find value of 

So, the strength coefficient is
and the strain-hardening exponent is
.
Answer:
the answer is below
Explanation:
a) The conductivity of graphite (σ) is calculated using the formula:

where f = frequency = 100 MHz, δ = skin depth = 0.16 mm = 0.00016 m, μ = 0.0000012
Substituting:

b) f = 1 GHz = 10⁹ Hz.

Answer:
The preferred steel type for W-shapes is structural steel and the its preferred ASTM designation is ASTM A992.
Explanation:
The ASTM A992 is a structural steel and it's the most available for w-shapes; besides its availabilty, its ductility improvements makes it the preferred choice; other common designations for this shapes are ASTM A572 Grade 50,0r ASTM A36, but this designations aren't as available as ASTM A992, and it has to be confirmed prior to their specification.
This question is incomplete, the complete question is;
(Laminar flow) A fluid flows through two horizontal pipes of equal length which are connected together to form a pipe of length 2
. The flow is laminar and fully developed. The pressure drop for the first pipe is 1.657 times greater than it is for the second pipe. If the diameter of the first pipe is D, determine the diameter of the second pipe.
D₃ = _____D.
{ the tolerance is +/-3% }
Answer:
the diameter of the second pipe D₃ is 1.13D
Explanation:
Given the data in the question;
Length = 2
pressure drop in the first pipe is 1.657 times greater than it is for the second pipe.
Now, we know that for Laminar Flow;
V' = πD⁴ΔP / 128μL
where V'₁ = V'₂ and ΔP₁₋₂ = 1.657 ΔP₂₋₃
Hence,
V'₁ = πD⁴ΔP₁₋₂ / 128μL = V'₃ = πD₃⁴ΔP₂₋₃ / 128μL
so
D₃ = D
ΔP₁₋₂ / ΔP₂₋₃ 
we substitute
D₃ = D
1.657 
D₃ = D( 1.134568 )
D₃ = 1.13D
Therefore, the diameter of the second pipe D₃ is 1.13D