Shaft Work of a Steam Turbine: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2014, Q1

Question 1: Shaft Work of a Steam Turbine (Part A — 20%)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Superheated steam expands through the turbine; the stream is pure H$_2$O at 105 mol/s (mass flow $\dot m = 105\times0.018015 = 1.892$ kg/s). Heat loss is specified per unit mass, so the whole balance is written per kilogram.

Quantity	Inlet (1)	Outlet (2)
Pressure	1.0 MPa	0.1 MPa
Temperature	800 K (526.85 °C)	400 K (126.85 °C)
Velocity $u$	50 m/s	80 m/s
Enthalpy $h$ (steam tables)	3538.0 kJ/kg	2729.9 kJ/kg

Find. The shaft work $W_s$ delivered by the turbine (per kg and as total power).

Figure 1 — steam enters the turbine at state 1 (1.0 MPa, 800 K, 50 m/s), expands and leaves at state 2 (0.1 MPa, 400 K, 80 m/s); the turbine loses Q = −1.16×10⁴ J/kg to the surroundings and delivers shaft work Wₛ.

Approach

Apply the steady-state open-system (first-law) energy balance including the kinetic-energy change, look up the two steam enthalpies from the superheated tables, and solve for the shaft work.

Write the steady-flow energy balance per unit mass. Neglecting potential energy, $$\Delta h + \tfrac{1}{2}\Delta u^{2} = Q - W_s,$$ where $Q$ is heat added to the fluid and $W_s$ is shaft work done by the fluid, both per kg.
Enthalpy change from the steam tables. At 1.0 MPa, 800 K, $h_1 = 3538.0$ kJ/kg (interpolated between 500 °C and 600 °C); at 0.1 MPa, 400 K, $h_2 = 2729.9$ kJ/kg (interpolated between 100 °C and 150 °C). Hence $$\Delta h = 2729.9 - 3538.0 = -808.1\ \mathrm{kJ/kg}.$$
Kinetic-energy change. $$\tfrac{1}{2}\Delta u^2 = \tfrac{1}{2}\left(80^2-50^2\right) = 1950\ \mathrm{J/kg} = 1.95\ \mathrm{kJ/kg}.$$ The kinetic term is tiny (0.24 % of $\Delta h$) but is retained because the velocities are given.
Solve for the shaft work. With $Q=-11.60$ kJ/kg, $$W_s = Q - \Delta h - \tfrac{1}{2}\Delta u^2 = -11.60 -(-808.1) - 1.95 = 794.5\ \mathrm{kJ/kg}.$$
Shaft work $W_s \approx 794$ kJ per kg of steam (positive ⇒ work delivered by the turbine).
Total power. Multiply by the mass flow: $$\dot W_s = \dot m\,W_s = (1.892\ \mathrm{kg/s})(794.5\ \mathrm{kJ/kg}) = 1503\ \mathrm{kW}.$$
Turbine power $\dot W_s \approx 1.50\ \mathrm{MW}$.

Quantity	Value
Enthalpy change $\Delta h$	−808.1 kJ/kg
Kinetic-energy change	+1.95 kJ/kg
Shaft work (specific)	794 kJ/kg
Turbine power	1.50 MW

Concept & Theory

Topic. Steady-flow energy balance (first law for an open system) on a turbine.

Key relations. $\Delta h + \tfrac12\Delta u^2 + g\Delta z = Q - W_s$; enthalpy from superheated-steam tables; $\dot m = \dot n M$.

Why this works. Enthalpy already bundles internal energy and flow work ($h=u+Pv$), so for a control volume the useful shaft work is whatever is left of the enthalpy drop after heat loss and the small kinetic change. Because the drop is 808 kJ/kg while the heat loss is only 12 kJ/kg, the turbine is very nearly adiabatic and the work tracks the enthalpy fall.

Common pitfall. Sign errors on $Q$ and $W_s$: here $Q$ is negative (heat leaves) and $W_s$ comes out positive (work leaves). Also forgetting to convert $\tfrac12\Delta u^2$ from J/kg to kJ/kg before adding.

Source. Smith, Van Ness & Abbott, Introduction to Chemical Engineering Thermodynamics, ch. 2 (energy balances for open systems); steam tables, Cengel & Boles appendix.

Quantity	Value
Enthalpy change \(\Delta h\)	−808.1 kJ/kg
Kinetic-energy change	+1.95 kJ/kg
Shaft work (specific)	794 kJ/kg
Turbine power	1.50 MW

Question 1 of 8: Shaft Work of a Steam Turbine

Question 1: Shaft Work of a Steam Turbine (Part A — 20%)

Approach