22-Mec-A1 Applied Thermodynamics and Heat Transfer · May 2018
Question 2 of 8: Ideal regenerative steam cycle with a closed feedwater heater
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Examination 16-Mec-A1, 3 hours, open book. Eight questions of equal value: Part A — Thermodynamics (Q1–Q4) and Part B — Heat Transfer (Q5–Q8). A complete paper is any five questions (three from one part and two from the other).
Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed., McGraw-Hill) — first-law cycle analysis, wet-region steam properties, the ideal regenerative Rankine cycle with a closed feedwater heater, the air-standard Diesel cycle, and isentropic-efficiency compressor analysis; Çengel & Ghajar, Heat and Mass Transfer (6th ed.) and Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — steady conduction with internal generation in a composite cylinder, internal-flow convection at constant wall temperature, combined convection–radiation surface balances, and the LMTD method for counterflow heat exchangers. Water/steam properties are IAPWS-consistent; air is treated as an ideal gas with constant specific heats.
Question 2 — Ideal regenerative steam cycle with a closed feedwater heater (Part A, equal value)
Given. Ideal (isentropic turbine and pump) regenerative Rankine cycle with a single closed feedwater heater (FWH); properties from IAPWS steam tables.
Quantity
Value
Turbine inlet, state 1
4.0 MPa, 325 °C
Extraction pressure, state 2
0.7 MPa
Condenser pressure, states 3/4
15.0 kPa
Feedwater exit from FWH
at $T_\text{sat}(0.7\,\text{MPa})=165.0$ °C
FWH drain
saturated liquid at 0.7 MPa, trapped to condenser
Find. (a) the $T$–$s$ layout; (b) the extraction fraction $y$, the net work per unit mass entering the turbine, and the thermal efficiency.
Figure 2 — $T$–$s$ diagram. Steam expands isentropically 1→2→3; part of the flow ($y$) is bled at 0.7 MPa (state 2) to the closed FWH, the remainder expands to the condenser (state 3). Condensate is pumped (4→5) and warmed in the FWH to 165 °C (state 6) before the boiler returns it to state 1. The bled steam condenses in the FWH and drains through a trap to the condenser.
Approach. Read the isentropic turbine states from $s_1$, apply an energy balance on the closed FWH to get $y$, then combine the (reduced-flow) turbine work, pump work and boiler heat.
Turbine inlet and isentropic expansion. At 4 MPa, 325 °C: $h_1=3029.5\ \text{kJ/kg}$, $s_1=6.5843\ \text{kJ/kg}\cdot\text{K}$. Expanding at constant entropy:
$$h_2=h(0.7\,\text{MPa},s_1)=2663.1\ \text{kJ/kg},\qquad h_3=h(15\,\text{kPa},s_1)=2098.6\ \text{kJ/kg}\;(x_3=0.789)$$
Condenser exit and pump. Saturated liquid at 15 kPa: $h_4=225.9\ \text{kJ/kg}$, $v_4=0.001014\ \text{m}^3/\text{kg}$. The pump raises it to 4 MPa:
$$w_p=v_4(P_\text{boiler}-P_\text{cond})=0.001014(4000-15)=4.04\ \text{kJ/kg}\;\Rightarrow\;h_5=230.0\ \text{kJ/kg}$$
Closed-FWH energy balance for the extraction fraction $y$. The bled steam ($h_2$) condenses to saturated liquid at 0.7 MPa ($h_7=697.0$), giving up heat to the full feed stream, which leaves at 165 °C and 4 MPa ($h_6=698.9\ \text{kJ/kg}$, compressed liquid):
$$y\,(h_2-h_7)=(h_6-h_5)\;\Rightarrow\;y=\frac{698.9-230.0}{2663.1-697.0}=\boxed{0.239}$$
Extraction fraction $y=0.239$
Net work per unit mass entering the turbine. The full flow drives the HP portion 1→2, and only $(1-y)$ continues 2→3:
$$w_t=(h_1-h_2)+(1-y)(h_2-h_3)=(3029.5-2663.1)+(0.761)(2663.1-2098.6)=796.2\ \text{kJ/kg}$$
$$w_\text{net}=w_t-w_p=796.2-4.0=792.2\ \text{kJ/kg}$$
Boiler heat and thermal efficiency. Heat is added from the FWH exit (state 6) to the turbine inlet:
$$q_\text{in}=h_1-h_6=3029.5-698.9=2330.6\ \text{kJ/kg}$$
$$\eta_\text{th}=\frac{w_\text{net}}{q_\text{in}}=\frac{792.2}{2330.6}=\boxed{0.340}$$
$w_\text{net}=792$ kJ/kg, $\;\eta_\text{th}=34.0\%$
Check — table interpolation sensitivity.
The turbine exit enthalpies and the FWH balance rest on interpolated steam-table values; a neighbouring-row read shifts $h_2$ and $h_3$ by $\pm0.5$–1 kJ/kg, moving $y$ by $\pm0.001$ and $\eta_\text{th}$ by about $\pm0.1$ percentage point. The compressed-liquid enthalpy $h_6$ is approximated as $h_f(165\,{}^\circ\text{C})$ plus the small pressure correction; using $h_f(165\,{}^\circ\text{C})$ alone changes $\eta_\text{th}$ by well under 0.1 point.
