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
extension lines,sketches,leader lines,dimensions describes all illustrations created by freehand.
Answer:Counter,
0.799,
1.921
Explanation:
Given data




Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger
Equating Heat exchange
![m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]](https://tex.z-dn.net/?f=m_hc_%7Bph%7D%5Cleft%20%5B%20T_%7Bh_i%7D-T_%7Bh_o%7D%5Cright%20%5D%3Dm_cc_%7Bpc%7D%5Cleft%20%5B%20T_%7Bc_o%7D-T_%7Bc_i%7D%5Cright%20%5D)
=
we can see that heat capacity of hot fluid is minimum
Also from energy balance

=


NTU=1.921





Answer:
375 KPa
Explanation:
From the question given above, the following data were obtained:
Initial pressure (P₁) = 125 KPa
Initial temperature (T₁) = 300 K
Final temperature (T₂) = 900 K
Final pressure (P₂) =?
The new (i.e final) pressure of the gas can be obtained as follow:
P₁/T₁ = P₂/T₂
125 / 300 = P₂ / 900
Cross multiply
300 × P₂ = 125 × 900
300 × P₂ = 112500
Divide both side by 300
P₂ = 112500 / 300
P₂ = 375 KPa
Thus, the new pressure of the gas is 375 KPa
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