Question 3 of 6: Steam Plant Turbomachinery — expansion line and feed pump efficiency

Question 3: Steam Plant Turbomachinery — expansion line and feed pump efficiency (15 marks)

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. A throttled, reheated turbine train and a feed pump test, both to be solved from the bound steam tables.

Find. Part I: the enthalpy at every point of the expansion line, the temperature and moisture at the HP inlet, the HP exit and the LP exit, and the total turbine power. Part II: the internal efficiency of the feed pump from the two pressures and the two temperatures alone.

Approach. Throttling is isenthalpic, so state 2 sits at the same enthalpy as state 1 but at the lower pressure; each expansion is then found by dropping vertically at constant entropy to the exhaust pressure and taking 80 % of that ideal enthalpy drop. Power is flow times actual enthalpy drop, machine by machine.

1. Part (a), step 1 — state 1 and the throttle to state 2. Dry saturated steam at 6 MPa has, from the saturation table, $T_{sat}=275.6^\circ\text{C}$ and$$h_1=h_g=h_f+h_{fg}=1213.4+1571.0=2784.4\ \text{kJ/kg}$$A control valve does no work and is adiabatic, so $h_2=h_1=2784.4$ kJ/kg at 3 MPa. At 3 MPa the saturated vapour enthalpy is 2803.2 kJ/kg, which is higher than $h_2$, so the throttled steam is very slightly wet rather than superheated:$$x_2=\dfrac{h_2-h_f}{h_{fg}}=\dfrac{2784.4-1008.4}{1795.7}=0.9890$$$$s_2=s_f+x_2 s_{fg}=2.6457+0.9890\times 3.5412=6.148\ \text{kJ/kg K}$$This is worth pausing over: $h_g$ for steam peaks near 3 MPa, so throttling from 6 MPa down to 3 MPa lands just below the saturated vapour line rather than above it. On the Mollier chart the move is a horizontal line to the right that finishes marginally inside the wet region.
2. Part (a), step 2 — the high pressure expansion, states 3s and 3. Dropping at constant entropy from state 2 to 0.6 MPa:$$x_{3s}=\dfrac{s_2-s_f}{s_{fg}}=\dfrac{6.148-1.9312}{4.8288}=0.8732\qquad h_{3s}=670.6+0.8732\times 2086.3=2492.4\ \text{kJ/kg}$$The internal efficiency then converts the ideal drop into the real one:$$\Delta h_{HP}=\eta_i\,(h_2-h_{3s})=0.80\times (2784.4-2492.4)=\boxed{233.6\ \text{kJ/kg}}$$$$h_3=2784.4-233.6=\boxed{2550.8\ \text{kJ/kg}}$$
3. Part (a), step 3 — reheat to state 4. At 0.6 MPa the saturation temperature is 158.9 °C, so 255 °C is superheated by about 96 K. Interpolating the superheat table between 250 and 300 °C:$$h_4=2957.2+\tfrac{5}{50}(3061.6-2957.2)=2967.6\ \text{kJ/kg}\qquad s_4=7.1816+\tfrac{5}{50}(7.3724-7.1816)=7.201\ \text{kJ/kg K}$$
4. Part (a), step 4 — the low pressure expansion, states 5s and 5. Dropping at constant entropy from state 4 to 0.004 MPa:$$x_{5s}=\dfrac{7.201-0.4226}{8.0520}=0.8418\qquad h_{5s}=121.5+0.8418\times 2432.9=2169.4\ \text{kJ/kg}$$$$\Delta h_{LP}=0.80\times (2967.6-2169.4)=\boxed{638.6\ \text{kJ/kg}}\qquad h_5=2967.6-638.6=\boxed{2329.1\ \text{kJ/kg}}$$
5. Part (b) — terminal temperatures and moisture. Each of the three points asked for lies in the wet region, so the temperature is the saturation temperature at that pressure and the moisture is $1-x$:$$\text{HP inlet (3 MPa, after the throttle): } T=233.9^\circ\text{C},\quad x=0.9890,\quad \text{moisture}=\boxed{1.1\ \%}$$$$\text{HP exit (0.6 MPa): } x_3=\dfrac{2550.8-670.6}{2086.3}=0.9012,\quad T=158.9^\circ\text{C},\quad \text{moisture}=\boxed{9.9\ \%}$$$$\text{LP exit (0.004 MPa): } x_5=\dfrac{2329.1-121.5}{2432.9}=0.9074,\quad T=29.0^\circ\text{C},\quad \text{moisture}=\boxed{9.3\ \%}$$For completeness the steam before the throttle is dry saturated at 6 MPa, 275.6 °C with zero moisture. Both exhausts sit just inside the 10 % erosion limit that governs the last stages of a saturated-steam turbine, which is why the reheat is fitted at all.
6. Part (c) — power output of the whole turbine. Each machine earns its own flow times its own actual enthalpy drop:$$P_{HP}=600\times 233.6=140\,133\ \text{kW}\qquad P_{LP}=500\times 638.6=319\,277\ \text{kW}$$$$P_{total}=\boxed{459\,410\ \text{kW}\approx 459.4\ \text{MW}}$$The 100 kg/s that does not reach the LP turbine is bled between the machines for feedwater heating and moisture separation, which the question implies by quoting two different flows.
7. Part II, step 1 — what an internal efficiency means for a pump. Efficiency is the ideal work divided by the actual work, both per kilogram, so the unmeasurable flow cancels completely and the test needs only the two pressures and the two temperatures:$$\eta_P=\dfrac{w_{ideal}}{w_{actual}}=\dfrac{h_{2s}-h_1}{h_2-h_1}$$
8. Part II, step 2 — actual work from the compressed liquid table. At 2.5 MPa and 210.0 °C the state is barely above saturation ($p_{sat}=1.91$ MPa) so $h_1=897.8$ kJ/kg. The delivery state must be read from the compressed liquid table at 20 MPa, interpolating between 200 and 220 °C:$$h_2=860.4+\tfrac{14.3}{20}\,(949.3-860.4)=923.9\ \text{kJ/kg}$$$$w_{actual}=h_2-h_1=923.9-897.8=\boxed{26.1\ \text{kJ/kg}}$$
9. Part II, step 3 — ideal work and the efficiency. For a liquid the reversible steady-flow work is $\int v\,dp$, evaluated at the mean specific volume over the pressure rise ($v=0.0011721$ at inlet and 0.0011605 m³/kg at outlet):$$w_{ideal}=\bar v\,(p_2-p_1)=0.0011663\times (20\,000-2500)=20.41\ \text{kJ/kg}$$$$\eta_P=\dfrac{20.41}{26.1}=\boxed{0.78\ (78\ \%)}$$which is the value expected of a large barrel-type boiler feed pump. The corresponding isentropic temperature rise is only 3.1 K against the 4.3 K measured, and the 1.2 K difference is the frictional heating — which is exactly why the test works.