Calculate the maximum load that a 7075 series aluminum alloy bar (with a T6 temper heat treatment) can support without permanent
1 answer:
Answer:
The maximum load the bar can withstand = 35.43 KN
Explanation:
Ultimate tensile strength of the given aluminium bar = 540 M pa
Cross section area of the bar = = 65.61
We know that the ultimate strength of the bar is calculated from
= 540 × 65.61
= 35.43 KN
Therefore the maximum load the bar can withstand = 35.43 KN
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Answer:
Assumption:
1. The kinetic and potential energy changes are negligible
2. The cylinder is well insulated and thus heat transfer is negligible.
3. The thermal energy stored in the cylinder itself is negligible.
4. The process is stated to be reversible
Analysis:
a. This is reversible adiabatic(i.e isentropic) process and thus
From the refrigerant table A11-A13
sat vapor
m=
b.) We take the content of the cylinder as the sysytem.
This is a closed system since no mass leaves or enters.
Hence, the energy balance for adiabatic closed system can be expressed as:
ΔE
ΔU
)
workdone during the isentropic process
=5.8491(246.82-219.9)
=5.8491(26.91)
=157.3993
=157.4kJ
Answer:
Consider A is square matrix of order 4 x 4 generated using magic function. Augmented matrix can be generated using:
Aug=[A eye(size(A))]
Above command is tested in MATLAB command window and is attached in figure below
