Waste-Heat Recovery Proposal That Fails an Energy Balance
A plant proposes to recover heat from 5.0 kg/s of clean exhaust gas at 320 °C to preheat 4.0 kg/s of process water from 20 °C to 60 °C in a gas-to-water heat exchanger. Use cp = 1.05 kJ/(kg·K) for the gas (approximately constant over this range) and cp = 4.18 kJ/(kg·K) for water. The exhaust is currently vented; there is no other use for it, and the gas leaves the exchanger to the stack. A vendor, competing for the same project, claims the identical gas stream can heat 4.0 kg/s of water to 150 °C, and offers a 15% lower price for a smaller exchanger to do it in.
1. Show that the 60 °C target is achievable and find the gas outlet temperature.
Read the worked answer
Heat required by the water:
$$\dot{Q} = \dot{m}c_p\Delta T = 4.0(4.18)(60-20) = 668.8 \text{ kW}$$
Gas-side capacity rate: $\dot{m}c_p = 5.0(1.05) = 5.25$ kW/K
Gas temperature drop:
$$\Delta T_{gas} = \frac{668.8}{5.25} = 127.4 \text{ K} \Rightarrow T_{out} = 320 - 127.4 = 192.6,°\text{C}$$
The gas still leaves at 192.6 °C, far above the water inlet of 20 °C, so there is no temperature-crossing violation and the duty is thermodynamically achievable. Energy balance closes: 5.25 × 127.4 = 668.8 kW in = 668.8 kW out.
2. Size the exchanger (find UA and, with U = 0.045 kW/(m²·K), the area) for the 60 °C case in counterflow, and comment on which side controls.
Read the worked answer
Terminal differences (counterflow):
- Hot end: 320 − 60 = 260 K
- Cold end: 192.6 − 20 = 172.6 K
$$LMTD = \frac{260 - 172.6}{\ln(260/172.6)} = \frac{87.4}{\ln 1.506} = \frac{87.4}{0.4098} = 213.3 \text{ K}$$
$$UA = \frac{\dot{Q}}{LMTD} = \frac{668.8}{213.3} = 3.14 \text{ kW/K}$$
$$A = \frac{3.14}{0.045} = 69.7 \text{ m}^2$$
With U ≈ 45 W/(m²·K), the overall coefficient is dominated by the gas-side film coefficient (gas-side coefficients are typically 10–20× lower than water-side). That has two consequences: the area is large for the duty, and any improvement should come from finned or extended gas-side surface, not from making the water side larger. Note also that a smaller exchanger than ~70 m² will not deliver the duty at this U - the vendor's "smaller unit, lower price" claim must be checked against exactly this arithmetic.
3. Evaluate the vendor's claim of heating the same water to 150 °C from the same gas stream.
Read the worked answer
First-law ceiling. The absolute maximum heat recoverable is obtained by cooling the gas all the way to the water inlet temperature, 20 °C:
$$\dot{Q}_{max} = 5.0(1.05)(320 - 20) = 1575 \text{ kW}$$
Heat demanded by the claim:
$$\dot{Q}_{claim} = 4.0(4.18)(150 - 20) = 2174 \text{ kW}$$
The claim requires 38% more energy than exists in the stream - it violates the first law even before considering that cooling the gas to 20 °C would require an infinite area (zero pinch).
What is actually achievable: even with zero approach temperature at the cold end, the water rise is limited to $1575/(4.0 \times 4.18) = 94.2$ K, i.e. 114 °C, and a real exchanger with a 20–30 K pinch would reach roughly 85–90 °C.
Conclusion: the vendor's proposal is not a design option - it is either a unit/rate misunderstanding (e.g., a different, larger gas flow) or a misrepresentation. Do not sign a performance guarantee against it. Also flag the pressure issue: 150 °C water in a non-pressurized circuit would flash; the claim may have been produced by someone who did not check saturation conditions.