Answer:
b) false
Explanation:
We know that Otto cycle is the ideal cycle for all petrol working engine.In Otto cycle all process are consider is ideal ,means there is no any ir-reversibility in the processes.
It consist four processes
1-2:Reversible adiabatic compression
2-3:Constant volume heat addition
3-4:Reversible adiabatic expansion
3-4:Constant volume heat rejection
Along with above 4 processes intake and exhaust processes are parallel to each other.From the P-v diagram we can see that all processes.
But actually in general we are not showing intake and exhaust line then it did not mean that in Otto cycle did not have intake and exhaust processes.
Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by


78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
Answer:
The answer is "+9.05 kw"
Explanation:
In the given question some information is missing which can be given in the following attachment.
The solution to this question can be defined as follows:
let assume that flow is from 1 to 2 then
Q= 1kw
m=0.1 kg/s
From the steady flow energy equation is:
![m\{n_1+ \frac{v^2_1}{z}+ gz_1 \}+Q= m \{h_2+ \frac{v^2_2}{2}+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 \times 800]-1= 0.01[1.005\times 700]+w\\\\w= +9.05 \ kw\\\\](https://tex.z-dn.net/?f=m%5C%7Bn_1%2B%20%5Cfrac%7Bv%5E2_1%7D%7Bz%7D%2B%20gz_1%20%5C%7D%2BQ%3D%20m%20%5C%7Bh_2%2B%20%5Cfrac%7Bv%5E2_2%7D%7B2%7D%2B%20gz_2%5C%7D%2Bw%5C%5C%5C%5C%5C%20change%20%5C%20energy%5C%5C%5C%5C0.1%5B1.005%20%5Ctimes%20800%5D-1%3D%200.01%5B1.005%5Ctimes%20700%5D%2Bw%5C%5C%5C%5Cw%3D%20%2B9.05%20%5C%20kw%5C%5C%5C%5C)
If the sign of the work performed is positive, it means the work is done on the surrounding so, that the expected direction of the flow is right.