Question 4 of 9: Exhaust-Gas Heat Recovery Steam Generator (HRSG)
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exam 04-BS-10, Thermodynamics — May 2017. 3 hours, Closed-Book Exam (approved
calculator and one double-sided 8.5x11-inch aid sheet permitted; property tables and charts supplied
in an appendix, interpolation not required). Part A: answer 2 of Questions 1-3 (20 marks each). Part
B: answer 4 of Questions 4-9 (15 marks each), for a 100-mark paper. Only the first two Part-A and
first four Part-B questions as they appear in the answer book are marked. All nine questions (Part A
complete, Part B complete) are solved below for completeness.
Reference texts: Cengel & Boles, Thermodynamics: An Engineering
Approach, 8th ed.; Moran, Shapiro, Boettner & Bailey, Fundamentals of Engineering
Thermodynamics, 8th ed. All state properties (water/steam, R-134a, air, N₂, CO₂,
O₂, moist air) were computed from high-accuracy equations of state in place of printed property-table interpolation; every boxed numeric
result.
Check: no separate dead-state $T_0$ is stated; the problem's own "ambient
temperature of 20°C" (the feedwater inlet condition) is taken as $T_0=20\ ^\circ$C for the
exergy calculation, the standard convention for this Cengel/Moran problem archetype.
Given. Exhaust gas (treated as ideal-gas air): $T_{g,in}=400\ ^\circ$C,
$P=150$ kPa, $\dot m_g=0.8$ kg/s, $T_{g,out}=350\ ^\circ$C. Water enters at $T_{w,in}=20\ ^\circ$C
and leaves as saturated vapor at $T_{sat}=200\ ^\circ$C ($P_{sat}=1554.9$ kPa). $T_0=20\ ^\circ$C.
Insulated (adiabatic to the surroundings) heat exchanger, steady flow, no work.
Fig. Q4 — Counter-flow gas-to-water heat
exchanger: hot exhaust gas stream (top→bottom) transfers heat to the feedwater/steam stream
(left→right), sized to produce saturated steam at 200°C.
Approach
Use the gas-side energy balance to find the total duty transferred, then divide by the
water-side specific enthalpy rise (feedwater at 20°C up to saturated vapor at 200°C, both
evaluated at the boiler pressure) to get the steam production rate. Exergy destruction is the sum of
the two streams' entropy-rate changes, scaled by $T_0$; second-law efficiency compares the exergy
gained by the water to the exergy given up by the gas.
Gas-side energy balance. Using ideal-gas variable-specific-heat air properties,
$h_{g,in}=810.94$ kJ/kg (673.15 K), $h_{g,out}=757.81$ kJ/kg (623.15 K):
$$\dot Q=\dot m_g(h_{g,in}-h_{g,out})=0.8\times(810.94-757.81)=42.50\ \text{kW}.$$
Water-side states at the boiler pressure. $P_{sat}(200\ ^\circ\text{C})=1554.9$
kPa. Saturated vapor: $h_{g,steam}=2792.01$ kJ/kg, $s_{g,steam}=6.4302$ kJ/kg·K. Compressed
liquid feedwater at 20°C, 1554.9 kPa: $h_{w,in}=85.37$ kJ/kg, $s_{w,in}=0.29616$ kJ/kg·K.
Rate of steam production (part a).
$$\dot m_{steam}=\frac{\dot m_g(h_{g,in}-h_{g,out})}{h_{g,steam}-h_{w,in}}=\frac{42.50}{2792.01-85.37}
=\boxed{0.01570\ \text{kg/s}}.$$
Entropy generation and exergy destruction (part b). Gas cools at constant
pressure ($\Delta s_{gas}=s^\circ(T_{out})-s^\circ(T_{in})$, negative); water heats to saturated vapor
(positive):
$$\dot S_{gen}=\dot m_g\big[s^\circ(T_{g,out})-s^\circ(T_{g,in})\big]+\dot m_{steam}(s_{g,steam}-s_{w,in})
=0.8\times(-0.0820)+0.01570\times6.1340$$
$$=-0.0656+0.0963=0.0307\ \text{kW/K}.$$
$$\dot X_{dest}=T_0\dot S_{gen}=293.15\times0.0307=\boxed{9.01\ \text{kW}}.$$
Second-law efficiency (part c). Exergy given up by the gas and gained by the
water:
$$\dot X_{gas,drop}=\dot m_g\big[(h_{g,in}-h_{g,out})-T_0(s^\circ_{g,in}-s^\circ_{g,out})\big]=23.27\ \text{kW}.$$
$$\dot X_{water,gain}=\dot m_{steam}\big[(h_{g,steam}-h_{w,in})-T_0(s_{g,steam}-s_{w,in})\big]=14.27\ \text{kW}.$$
$$\eta_{II}=\frac{\dot X_{water,gain}}{\dot X_{gas,drop}}=\frac{14.27}{23.27}=\boxed{0.6130\ (61.3\%)}
\quad(\text{check: }23.27-14.27=8.99\approx\dot X_{dest}\ \checkmark).$$
