The key to solving
this problem is to recognize that any process that is both adiabatic
and reversible is ISENTROPIC.This means that S2 = S1 and this allows you to
fix state 2 and evaluate U2.Use U2 in the 1st Law to evaluate W.
Process is internally reversible.
Changes in kinetic and potential energies are negligible.
work is the only form of work that crosses the system boundary.
/ Data / Solve:
Begin by applying the 1st Law to the process, assuming changes in kinetic and potential energies are negligible:
The process is adiabatic so Eqn 1 can be simplified to :
Use the NIST Webbook to obtain properties
for state 1, saturated vapor at -10oC :
Because the process is
both reversible and adiabatic, it is isentropic.
At this point we know
values of two intensive variables for state 2, so we can use the NIST Webbook to determine the value
of any other property.In this case, we need U2.First we need to
determine the phases
that exist at state 2.
At P2 :
Because S2 > Ssat vap at P2, we can conclude that state 2 is a superheated vapor.We could have reached
the same conclusion after careful consideration of a TS Diagram.
We can get the
following data from the superheated Ammonia Tables or from the NIST Webbook :
Interpolation yields :
Now, we can plug
values back into Eqn 2 :
None of the
assumptions made in this problem solution can be verified.