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:
[tex]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\\\\[/tex]
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.
