Bc that’s the way it’s posta be
Answer:
it is true
Explanation:
I don’t know why but maybe it’s because they just want you to pay for more than you goes?
Answer:
Please check explanation for answer
Explanation:
Here, we are concerned with stating the advantages and disadvantages of using a 6 tube passes instead of a 2 tube passes of the same diameter:
<u>Advantages</u>
* By using a 6 tube passes diameter, we are increasing the surface area of the heat transfer surface
* As a result of increasing the heat transfer surface area, the rate of heat transfer automatically increases too
Thus, from the above, we can conclude that the heat transfer rate of a 6 tube passes is higher than that of a 2 tube passes of the same diameter.
<u>Disadvantages</u>
* They are larger in size and in weight when compared to a 2 tube passes of the same diameter and therefore does not find use in applications where space conservation is quite necessary.
* They are more expensive than the 2 tube passes of the same diameter and thus are primarily undesirable in terms of manufacturing costs
Answer:
i) 796.18 N/mm^2
ii) 1111.11 N/mm^2
Explanation:
Initial diameter ( D ) = 12 mm
Gage Length = 50 mm
maximum load ( P ) = 90 KN
Fractures at = 70 KN
minimum diameter at fracture = 10mm
<u>Calculate the engineering stress at Maximum load and the True fracture stress</u>
<em>i) Engineering stress at maximum load = P/ A </em>
= P /
= 90 * 10^3 / ( 3.14 * 12^2 ) / 4
= 90,000 / 113.04 = 796.18 N/mm^2
<em>ii) True Fracture stress = P/A </em>
= 90 * 10^3 / ( 3.24 * 10^2) / 4
= 90000 / 81 = 1111.11 N/mm^2
Answer:
E= 15 GPa.
Explanation:
Given that
Length ,L = 0.5 m
Tensile stress ,σ = 10.2 MPa
Elongation ,ΔL = 0.34 mm
lets take young modulus = E
We know that strain ε given as
![\varepsilon =\dfrac{\Delta L}{L}](https://tex.z-dn.net/?f=%5Cvarepsilon%20%3D%5Cdfrac%7B%5CDelta%20L%7D%7BL%7D)
![\varepsilon =\dfrac{0.34}{0.5\times 1000}](https://tex.z-dn.net/?f=%5Cvarepsilon%20%3D%5Cdfrac%7B0.34%7D%7B0.5%5Ctimes%201000%7D)
![\varepsilon =0.00068](https://tex.z-dn.net/?f=%5Cvarepsilon%20%3D0.00068)
We know that
![\sigma = \varepsilon E\\\\E=\dfrac{10.2}{0.00068}\\E= 15000\ MPa\\E=15\ GPa](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Cvarepsilon%20%20E%5C%5C%5C%5CE%3D%5Cdfrac%7B10.2%7D%7B0.00068%7D%5C%5CE%3D%0915000%5C%20MPa%5C%5CE%3D15%5C%20GPa)
Therefore the young's modulus will be 15 GPa.