A gas pressure difference is applied to the legs of a U-tube manometer filled with a liquid with specific gravity of 1.7. The ma
2 answers:
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Answer:
the elongation of the metal alloy is 21.998 mm
Explanation:
Given the data in the question;
K = σT/ (εT)ⁿ
given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,
strain-hardening exponent n = 0.22
we substitute
K = 345 /
K = 815.8165 Mpa
next, we determine the true strain
(εT) = (σT/ K)^1/n
given that σT = 412 MPa
we substitute
(εT) = (412 / 815.8165 )^(1/0.22)
(εT) = 0.04481 mm
Now, we calculate the instantaneous length
=
given that = 480 mm
we substitute
=
×
= 501.998 mm
Now we find the elongation;
Elongation =
we substitute
Elongation = 501.998 mm - 480 mm
Elongation = 21.998 mm
Therefore, the elongation of the metal alloy is 21.998 mm
Explanation:
A.
H = Aeσ^4
Using the stefan Boltzmann law
When we differentiate
dH/dT = 4AeσT³
dH/dT = 4(0.15)(0.9)(5.67)(10^-8)(650)³
= 8.4085
Exact error = 8.4085x20
= 168.17
H(650) = 0.15(0.9)(5.67)(10^-8)(650)⁴
= 1366.376watts
B.
Verifying values
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(670)⁴
= 1542.468
H(T+ΔT) = 0.15(0.9)(5.67)(10^-8)(630)⁴
= 1205.8104
Error = 1542.468-1205.8104/2
= 168.329
ΔT = 40
H(T+ΔT) = 0.15(0.9)(5.67)(10)^-8(690)⁴
= 1735.05
H(T-ΔT) = 0.15(0.9)(5.67)(10^-8)(610)⁴
= 1735.05-1059.83/2
= 675.22/2
= 337.61
