steam pressure

not enough information. Turbine efficiency, outlet pressure, steam quality, and steam flowrate.

Vicini is correct however you usually don't specify the steam flowrate as this is commonly a manipulated variable.

The 1st and 2nd Laws of thermodynamics are applied to solve the system.

With a few assumptions this problem can be simplified. Assume the turbine is adiabatic and reversible to remove considerations of heat loss and entropy changes. Now you're left with a simplified 1st and 2nd law

Ws,rev = ΔH = Hout,rev - Hin

ΔS = 0 --> Sout =Sin

Because steam is the fluid of choice, have a steam table handy to look up the enthaply and entropy at the inlet conditions. Steam is generally formed in a boiler and industrial size turbines require high pressure-superheated steam. My facility uses 4MPa steam at 450ºC in a turbine which expands the steam to 1 MPa, so I'll use these numbers. Note, outlet temperature doesn't need to be specified.

Hin = 3331 kJ/kg

Sin = 6.9386 kJ/kg-K

Because the change in entropy is zero, Sout equals Sin. Look up this entropy value at the outlet pressure spec and find its corresponding enthalpy. I found Hout,rev by linear interpolation to be 2950 kJ/kg.

So now Ws,rev = Hout - Hin = -381 kJ/kg. The negative is directional, meaning this energy is generated. However this is not the actual work produced since turbines are irreversible and some work is lost due to friction and other dissipated energy forms. That's why an efficiency factor comes into play. For the most part turbines can be assumed to be greater than 80% efficient. So now

Ws,act = Ws,rev * η = -381 * 0.80 = -305 kJ/kg

Finally the question is what steam flowrate (manipulated variable) will give a work output of 58kW (58kJ/s)?

58 kJ/s = 305 kJ/kg * m(dot)

Solving for m(dot) equals 0.19 kg/s steam or 684.6 kg/h

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The outlet temperature should be calculated to see what the quality of steam is (fraction of steam that is vapor). The actual work is used to back calculate the outlet enthalpy since the previous Hout was for the reversible process.

Hout,act = Ws,act + Hin = 3026 kJ/kg

The temperature can be found by looking up this enthalpy under the 1MPa steam table. The outlet temp is around 288 °C which is well above the saturation temperature (180 °C) at 1 MPa therefore no steam condenses.

This is important since liquid can damage turbines, so always stay above the dew point if possible!

-David