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.


1 Answer 1


Since the mass flow into the turbine is equal to the mass flow out, you can factor it out of the right-hand side of the energy equation you provided, then divide both sides by it. The actual power W divided by the mass flow rate m_dot is the specific work (w). If working in SI units, power has units of watts, or joules/second. Power divided by mass flow rate in kg/s reduces as follows: J/s / kg/s=J/kg. These units are energy per unit mass, which are the same units as specific enthalpy (h). The concept of "specific" variables appears in several ways in thermodynamics, such as specific work, specific enthalpy, and specific volume.


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