A 5.3 kg block initially at rest is pulled to the right along a frictionless, horizontal surface by a constant horizontal force
2 answers:
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Answer:
a) ΔEC=-23.4kW
b)W=12106.2kW
c)
Explanation:
A)
The kinetic energy is defined as:
(vel is the velocity, to differentiate with v, specific volume).
The kinetic energy change will be: Δ ()=
Δ ()=
Where 1 and 2 subscripts mean initial and final state respectively.
Δ()=
This amount is negative because the steam is losing that energy.
B)
Consider the energy balance, with a neglective height difference: The energy that enters to the turbine (which is in the steam) is the same that goes out (which is in the steam and in the work done).
We already know the last quantity: =
Δ (
)=23400W
For the steam enthalpies, review the steam tables (I attach the ones that I used); according to that,
The exit state is a liquid-vapor mixture, so its enthalpy is:
Finally, the work can be obtained:
C) For the area, consider the equation of mass flow:
where
is the density, and A the area. The density is the inverse of the specific volume, so
The specific volume of the inlet steam can be read also from the steam tables, and its value is: , so: