Answer:
The answer is "+9.05 kw"
Step-by-step 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+ (v^2_1)/(z)+ gz_1 \}+Q= m \{h_2+ (v^2_2)/(2)+ gz_2\}+w\\\\\ change \ energy\\\\0.1[1.005 * 800]-1= 0.01[1.005* 700]+w\\\\w= +9.05 \ kw\\\\](https://img.qammunity.org/2021/formulas/engineering/college/hkabuy66d0um54e7inuu2js9zlgzf0317u.png)
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.