Question

Throttling process is a/an___________process
A) reversible and constant enthalpy
B) irreversible and constant enthalpy
C) reversible and isothermal
D) none of these

Answer 1

Throttling process is a process used for the reduction of the pressure of a system.
The value of friction cannot be zero for real life process or practical process, its only zero for ideal process which is theoretically possible.

Complete step by step answer:
So in the question we asked to comment on the throttling process i.e. which type of thermodynamic process is the throttling process.
Before that let’s discuss the throttling process uses and its real life examples.
Throttling process is a thermodynamic process which is used to reduce or decrease the pressure of a fluid. In this process a liquid having high pressure is converted into low pressure fluid.
So now know why throttling process is used, and lets now move on to the daily examples its practical applications in our daily life:
Flow through a partially open valve-one of the best examples for this method of reduction of pressure is the capillary present in the refrigerants and window ACs.
Next the best example is the knob of deodorants, there the pressure is reduced by the method of flow through a small opening.
Now let’s discuss the options given and through which we can trace the characteristics of the throttling process.
So in option (A) ,it is given throttling process is a reversible and constant enthalpy, but we know that the real life process cannot be reversible ,since there will be friction acting between the surface and the molecules in the fluid and also between the molecules, friction can’t be zero. So the process can’t be reversible.
So the option is not correct.
In option (B) it is given as the process is irreversible and constant enthalpy.as we have explained above real life processes are irreversible, so that part is correct. Now let's move to the second part, the process is isenthalpic in nature i.e. the process proceeding with constant enthalpy.
So if we consider the steady flow equation, which is,
$steady\,flow\,energy={{h}_{1}}+\dfrac{{{c}_{1}}^{2}}{2}+{{g}_{{{z}_{1}}}}+Q={{h}_{2}}+\dfrac{{{c}_{2}}^{2}}{2}+{{g}_{{{z}_{2}}}}+W$
The heat transfer and work done in this process is having the value zero and the value for change in kinetic and potential energy is also zero.
Hence the equation becomes,$$$$
$steady\,flow\,energy={{h}_{1}}={{h}_{2}}=H$
The initial and final enthalpy of the process is equal hence we can conclude that the option given is correct.
The process is irreversible and with constant enthalpy.
In option (C), it is given as, the process is reversible which is incorrect.

Hence we now know that the correct option for the question is option (B).

Note: Characteristics of throttling process:
-No work is done
-There is no heat transfer to the system from the surroundings to the system.
-It is an isenthalpic process and the process is irreversible.
For an ideal gas, the enthalpy is the function of temperature only since the internal energy of the ideal gas only depends on the temperature, i.e. U= temperature.