NivaarExam PrepOfficial exam papers ↗

22-Mec-A1 Applied Thermodynamics and Heat Transfer · Undated paper

Question 3 of 8: Two-stage steam turbine with reheat

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Examination 16-Mec-A1 Applied Thermodynamics and Heat Transfer — 3 hours, open book. Two parts: Part A (Thermodynamics, Q1–Q4) and Part B (Heat Transfer, Q5–Q8). The rubric asks for five questions (three from one part, two from the other), each of equal value; all eight are solved.

Reference texts: Çengel & Boles, Thermodynamics: An Engineering Approach (9th ed.); Çengel & Ghajar, Heat and Mass Transfer (6th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Moran & Shapiro, Fundamentals of Engineering Thermodynamics. Property data: saturated ammonia and R-134a tables printed with the paper.

Question 3 — Two-stage steam turbine with reheat (Part A, equal value)

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.

StateConditionEnthalpy / entropy
1 — HP inlet7.5 MPa, 550 °C$h_1=3526.7$ kJ/kg, $s_1=6.914$
2 — HP exit (isentropic)0.5 MPa, $s_2=s_1$$h_2=2788.7$ kJ/kg
3 — LP inlet (reheated)0.5 MPa, 350 °C$h_3=3168.1$ kJ/kg, $s_3=7.635$
4s — LP isentropic exit10 kPa, $s=s_3$$h_{4s}=2419.8$ kJ/kg
4 — LP actual exit10 kPa, $x_4=0.95$$h_4=2464.3$ kJ/kg

Find. stage works $w_{HP},w_{LP}$ and $\dot W$; $\eta_{LP}$ and $\eta_\text{overall}$; net $\Delta s$; the $T$–$s$ sketch.

Flow schematic HP Reheat LP 1 2 3 4 → condenser T–s diagram sT sat. dome 1 2 3 4
HP expansion 1→2 (isentropic, vertical), constant-pressure reheat 2→3, LP expansion 3→4 (irreversible, entropy rises to $x_4=0.95$).

Approach. Fix each state from the steam tables, take stage work as the enthalpy drop, compare actual to isentropic drops for the efficiencies, and sum the entropy rises of the two expansions for the net $\Delta s$.

  1. Stage works and total power. Each adiabatic stage delivers its enthalpy drop: $$w_{HP}=h_1-h_2=3526.7-2788.7=738.0\ \text{kJ/kg},\quad w_{LP}=h_3-h_4=3168.1-2464.3=703.8\ \text{kJ/kg}$$ $$\dot W=\dot m\,(w_{HP}+w_{LP})=5.0(738.0+703.8)=\boxed{7.21\ \text{MW}}$$
  2. Isentropic efficiencies. The LP stage compares the actual drop to the isentropic drop to $h_{4s}=2419.8$ kJ/kg: $$\eta_{LP}=\frac{h_3-h_4}{h_3-h_{4s}}=\frac{703.8}{748.3}=\boxed{0.941}$$ For the whole machine (HP taken ideal), $\eta_\text{overall}=\dfrac{w_{HP}+w_{LP}}{(h_1-h_2)+(h_3-h_{4s})}=\dfrac{1441.8}{1486.0}=\boxed{0.970}$.
  3. Net entropy change per kg. The HP leg is isentropic ($\Delta s_{1\to2}=0$); the reheat adds entropy externally (heat) and the LP leg generates entropy internally. Across the turbine control volume the fluid entropy change is $$\Delta s=(s_2-s_1)+(s_4-s_3)=0+(7.774-7.635)=\boxed{+0.139\ \text{kJ/kg}\cdot\text{K}}$$ with $s_4=s_{f}+x_4 s_{fg}$ at 10 kPa. The positive value is the irreversibility of the LP expansion.
QuantityResult
HP / LP specific work738.0 / 703.8 kJ/kg
Total power $\dot W$7.21 MW
LP isentropic efficiency0.941
Overall isentropic efficiency0.970
Net entropy change+0.139 kJ/kg·K (generated in LP)
Check — dropped "fraction returned for reheating" sub-part.
The 0.30-confidence page 3 also carries the phrase "the fraction of the steam ... returned to the boiler for reheating." In a straight reheat cycle (no extraction) all of the HP exhaust passes through the reheater, so that fraction is unity; the overlaid figures show no bleed line, so the reconstruction treats the machine as a simple two-stage reheat turbine. State values; enthalpy differences are used throughout, so the datum is irrelevant.