Complete Question
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of this material elongate when a true stress of 411 MPa (59610 psi) is applied if the original length is 470 mm (18.50 in.)?Assume a value of 0.22 for the strain-hardening exponent, n.
Answer:
The elongation is 
Explanation:
In order to gain a good understanding of this solution let define some terms
True Stress
A true stress can be defined as the quotient obtained when instantaneous applied load is divided by instantaneous cross-sectional area of a material it can be denoted as
.
True Strain
A true strain can be defined as the value obtained when the natural logarithm quotient of instantaneous gauge length divided by original gauge length of a material is being bend out of shape by a uni-axial force. it can be denoted as
.
The mathematical relation between stress to strain on the plastic region of deformation is

Where K is a constant
n is known as the strain hardening exponent
This constant K can be obtained as follows

No substituting
from the question we have


Making
the subject from the equation above




From the definition we mentioned instantaneous length and this can be obtained mathematically as follows

Where
is the instantaneous length
is the original length



We can also obtain the elongated length mathematically as follows



Answer:
1700kJ/h.K
944.4kJ/h.R
944.4kJ/h.°F
Explanation:
Conversions for different temperature units are below:
1K = 1°C + 273K
1R = T(K) * 1.8
= (1°C + 273) * 1.8
1°F = (1°C * 1.8) + 32
Q/delta T = 1700kJ/h.°C
T (K) = 1700kJ/h.°C
= 1700kJ/K
T (R) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8R
= 944.4kJ/h.R
T (°F) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8°F
= 944.4kJ/h.°F
Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C
= 2°C
(273+5) - (273+3)
= 2 K
Its C .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........
Answer:
The work of the cycle.
Explanation:
The area enclosed by the cycle of the Pressure-Volume diagram of a Carnot engine represents the net work performed by the cycle.
The expansions yield work, and this is represented by the area under the curve all the way to the p=0 line. But the compressions consume work (or add negative work) and this is substracted fro the total work. Therefore the areas under the compressions are eliminated and you are left with only the enclosed area.