Question 2 of 8: Heat Recovery Steam Generator

Question 2: Heat Recovery Steam Generator (10 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 once-through counterflow heat-recovery steam generator raising superheated steam at a single pressure from gas-turbine exhaust.

Find. The temperature profile of the two streams, the enthalpies at the key states, the steam production, the pinch-point temperature difference, its design significance, and the physical arrangement of the heating surfaces.

Approach. Take an overall energy balance on the whole generator to get the steam flow, then a partial balance on the superheater-plus-evaporator alone to locate the gas temperature at the point where the water first reaches saturation.

1. Part (a) — establish the shape of the profile. The gas is a single-phase stream of constant specific heat, so its temperature falls linearly with the heat surrendered. The water changes phase at constant pressure, so its line has three parts: sensible heating of liquid (economiser), evaporation at the saturation temperature (evaporator), and superheating (superheater). Numbering as on the figure: 1 gas in, 2 gas at the pinch, 3 gas out, 4 feedwater in, 5 saturated liquid, 6 saturated vapour, 7 steam out.
2. Part (b) — enthalpies at the key points. At 1.40 MPa the saturation temperature is 195.0 °C. The feedwater at 30 °C is compressed liquid, whose enthalpy is that of saturated liquid at 30 °C plus the small flow-work term $v(p-p_{sat}) \approx 1.4$ kJ/kg: $$h_4 = 127\ \text{kJ/kg},\quad h_5 = h_f = 830\ \text{kJ/kg},\quad h_6 = h_g = 2789\ \text{kJ/kg},\quad h_7 = 3562\ \text{kJ/kg}.$$ The latent heat at this pressure is $h_{fg} = 1959$ kJ/kg, which is the flat part of the cold-stream line.
3. Part (c) — overall energy balance gives the steam flow. All the heat given up by the gas is taken by the water: $$M_{gas}\,c_{p,g}\,(T_{g,in}-T_{g,out}) = M_{steam}\,(h_7-h_4).$$ The gas side releases $125 \times 1.148 \times (560-130) = 61\,705$ kW, and each kilogram of steam absorbs $3562 - 127 = 3435$ kJ, so $$\boxed{M_{steam} = \frac{61\,705}{3435} = 17.96\ \text{kg/s}}$$ which is consistent with the 20 kg/s assumed in Question 1 once the small heat loss and blowdown a real unit carries are allowed for.
4. Part (d) — locate the pinch. The pinch sits where the water first reaches saturation, because upstream of it the cold line is flat while the gas line keeps falling. Balance the superheater and evaporator together: $$M_{steam}\,(h_7-h_5) = M_{gas}\,c_{p,g}\,(T_{g,in}-T_{g,pinch}).$$ The duty above the pinch is $17.96 \times (3562-830) = 49\,079$ kW, and the gas capacity rate is $125 \times 1.148 = 143.5$ kW/K, so the gas cools by $49\,079/143.5 = 342.0$ K and reaches $$T_{g,pinch} = 560 - 342.0 = 218.0\ \text{°C}.$$ The water at that station is saturated liquid at 195.0 °C, so $$\boxed{\Delta T_{pinch} = 218.0 - 195.0 = 23.0\ \text{K}}$$
5. Confirm the profile closes. The economiser must then take the gas from 218.0 °C to the stated outlet. Its duty is $17.96 \times (830-127) = 12\,626$ kW, which cools the gas a further $12\,626/143.5 = 88.0$ K to exactly 130 °C. The profile is self-consistent and no temperature crossover occurs anywhere along it.
6. Part (e) — why the pinch governs the design. The pinch is the tightest approach anywhere in the exchanger, so it is the point that limits how much heat can be recovered and it sets the surface area required. Choosing it is a direct trade of capital against fuel. A small pinch — say 10 K — lets the evaporator start further down the gas path, so more steam is raised and the stack temperature falls, but the driving temperature difference at that station is small and the area needed to move the duty rises steeply, roughly as the reciprocal of the approach. A large pinch — say 40 K — makes the unit small and cheap but pushes heat up the stack unrecovered. Typical single-pressure HRSG practice lands at 15–25 K, which is where this unit sits. A second consequence matters operationally: because the pinch fixes the ratio of gas heat above and below the saturation temperature, it also fixes the steam flow, so the pinch chosen at design determines the bottoming-cycle output and cannot be improved later without adding surface.
7. Part (f) — the physical arrangement. The three surfaces are stacked in the gas path in the reverse order of the water flow, so that the hottest gas meets the hottest steam. Gas entering the top of the casing passes first over the superheater (7), then over the evaporator tubes (5–6), and finally over the economiser (4) before leaving at 130 °C. The steam drum sits outside the gas path at the top, connected to the evaporator bank by downcomers and risers; it separates the water and steam produced by natural circulation and delivers dry saturated steam to the superheater. Feedwater enters the economiser at the cold end and rises through the unit against the gas.