Pseudocodes are used as a prototype of an actual program.
The error in the pseudocode is that, the while loop in the pseudocode will run endlessly.
From the pseudocode, the first line is:
<em>Declare Boolean finished = false</em>
The while loop is created to keep running as long as <em>finished = false.</em>
So, for the while loop to end, the finished variable must be updated to true.
This action is not implemented in the pseudocode.
Hence, the error in the pseudocode is that, the while loop is an endless loop
Read more about pseudocodes at:
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Answer with Explanation:
There are various factors that needed to be taken into account while deciding the factor of safety some of which are summarized below as:
1) Importance of the structure: When we design any structure different structures have different importance in our society. Take an example of hospital, in case a natural disaster struck's a place the hospital should be the designed to withstand the disaster as it's role in the crisis management following a disaster is well understood. Thus while designing it we need it to have a higher factor of safety against failure when compared to a local building.
2) Errors involved in estimation of strength of materials: when we design any component of any machine or a structure we need to have an exact idea of the behavior of the material and know the value of the strength of the material. But many materials that we use in structure such as concrete in buildings have a very complex behavior and we cannot estimate the strength of the concrete absolutely, thus we tend to decrease the strength of the concrete more if errors involved in the estimation of strength are more to give much safety to the structure.
3) Variability of the loads that may act on the structure: If the loads that act on the structure are highly variable such as earthquake loads amd dynamic loads then we tend to increase the factor of safety while estimating the loads on the structure while designing it.
4) Economic consideration: If our project has abundant funds then we can choose a higher factor of safety while designing the project.
Answer: a 8143.71 kJ/kg
b 393.15 K
Explanation:
This system is an isobaric process in which there is no change in pressure a quasistatic process where a pressure distribution exists
a since no change in pressure =0 the system does work thus
FOR HELIUM properties in standard thermodynamic chart
cv = 3.1 kJ/kgK
M = Molar mass = 4 kg/kmol
R = Universal gas constant = 8.314 kJ/kg K
cp ≈ cv +R /M = 3.1 + 8.314 /4 = 5.1785 kJ/kgK
Cp = cp * M = 5.1785 kJ/kgK * 4 kg/kmol = 20.714 kJ/kgkmol
T = 120 °C to Kelvin = 120 + 273.15k = 393.15 K
W =n Cp ΔT = 1 kmol * 20.714 kJ/kg kmol* 393.15 K = 8143.71 kJ/kg
b convert T °C = T K thus 120 + 273.15 K = 393.15 K
P₁/T₁ = P₂/T₂
200 kPa/ 393.15 K = 200 kPa/T₂
T₂ = 200 kPa * 393.15 K/ 200 kPa = 393.15 K or 120 k
Answer:
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Explanation:
