# Assumption of specific enthalpy

Given an adiabat turbine with state 1 fluid data as:

$$h_1 = a$$ $$p_1 = b$$ $$t_1 = c$$ and state 2 data fluid data as:

$$h_2 = unknown$$ $$p_2 = e << b$$ $$t_2 = f < c$$ $$h_2' = g$$ $$h_2'' = h$$

With kinetic and potential energy small enough to be ignored, a steam content of $$x_2 = 0.965$$

What is the technical work $$w_{t,12}$$

With the equation:

$$h_2 = h_2' + x_2(h_2''-h_2')$$

The specific enthalpy of $$h_2$$ can be easily solved. Now taking the first Thermo D. law, and using an energy balance, the technical work can be solved via:

$$W_{t,12} = \dot{m_{in}}(h_1 + \frac{c_{in}^2}{2}) - \dot{m_{out}}(h_2 + \frac{c_{out}^2}{2})$$

Whereby $$\dot{m_{in}} = \dot{m_{out}} = \dot{m}$$

Atleast, this is what I had thought though I am missing $$c_{in} \text{ and } c_{out}$$, when checking the solution, it is suggest the specific work is solved "simply"

$$w_{t,12} = h_1 - h_2$$

Whereby the mass velocity and fluid velocity are unneeded.

My assumption in this solution, is that the author assumes $$c_{in} \text{ and } c_{out}$$ are equivalent thus equal zero in the equation, but I fail to see how $$W_{t,12}$$ can be divided by the mass velocity $$\dot{m}$$.

What am I missing compared to my assumed solution and the authors, where should I have understood that $$\dot{m}$$ is not required.