NivaarExam PrepOfficial exam papers ↗

25-Nav-A1 Fundamentals of Naval Architecture · May-98-Mar-A1 2017

Question 2 of 8: Air-Standard Dual Cycle

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

Notes on this paper

National Examinations, May 2017 — 98-Mar-A1 Applied Thermodynamics and Heat Transfer, 3 hours, open book (Part A: Thermodynamics, Part B: Heat Transfer; 5 of 8 questions required, all 8 answered below for full study coverage).

Reference texts: Cengel & Boles, Thermodynamics: An Engineering Approach; Sonntag, Borgnakke & Van Wylen, Fundamentals of Thermodynamics; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

It is solved as the thermodynamics/heat-transfer exam it actually is.

Question 2: Air-Standard Dual Cycle (20 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. Cold air-standard dual cycle, states 1 (BDC, start compression) → 2 (end isentropic compression) → 3 (end constant-volume heat addition) → 4 (end constant-pressure heat addition) → 5 (end isentropic expansion, $v_5=v_1$) → 1 (constant-volume heat rejection).

Given data
QuantitySymbolValue
Compression ratio$r_v$16.5
Initial pressure, temperature$P_1,T_1$101.325 kPa, 25°C (298.15 K)
Temp. after const-vol. combustion$T_3$1170°C (1443.15 K)
Temp. after const-P combustion$T_4$1595°C (1868.15 K)
Air properties$k,c_p,c_v$1.4, 1.005, 0.718 kJ/kg·K
Entropy sTemperature T1 (BDC)2 (end compr.)3 (end CV heat add)4 (end CP heat add)5 (end expansion)1→2 isentropic compression · 2→3 const-vol heat add · 3→4 const-P heat add · 4→5 isentropic expansion · 5→1 const-vol heat reject
Air-standard dual cycle on the T–s plane (schematic, cold-air-standard).

Find. Cut-off ratio $r_c$, constant-volume pressure ratio $r_p$, heat added and rejected per unit mass, and thermal efficiency.

Approach. Walk the cycle process-by-process with the isentropic and ideal-gas relations to fix every state, then form $q_{in}$, $q_{out}$ and $\eta_{th}=1-q_{out}/q_{in}$.

  1. State 2 (end of isentropic compression 1→2). $$T_2=T_1r_v^{k-1}=298.15\times16.5^{0.4}=915.0\text{ K}$$
  2. Constant-volume pressure ratio (2→3, $v$ fixed so $r_p=P_3/P_2=T_3/T_2$). $$\boxed{r_p=\dfrac{T_3}{T_2}=\dfrac{1443.15}{915.0}=1.577}$$
  3. Cut-off ratio (3→4, $P$ fixed so $r_c=v_4/v_3=T_4/T_3$). $$\boxed{r_c=\dfrac{T_4}{T_3}=\dfrac{1868.15}{1443.15}=1.294}$$
  4. Heat added. Constant-volume leg $q_{23}=c_v(T_3-T_2)=0.718(1443.15-915.0)=379.2\text{ kJ/kg}$; constant-pressure leg $q_{34}=c_p(T_4-T_3)=1.005(1868.15-1443.15)=427.1\text{ kJ/kg}$. $$\boxed{q_{in}=q_{23}+q_{34}=806.3\text{ kJ/kg}}$$
  5. State 5 (isentropic expansion 4→5 to $v_5=v_1$). $v_4/v_1=r_c/r_v=1.294/16.5=0.07845$, so the expansion volume ratio is $v_5/v_4=1/0.07845=12.75$, and $$T_5=T_4(v_4/v_5)^{k-1}=1868.15\times(1/12.75)^{0.4}=674.9\text{ K}$$
  6. Heat rejected and thermal efficiency. $q_{51}=c_v(T_5-T_1)=0.718(674.9-298.15)=270.5\text{ kJ/kg}$. $$\boxed{\eta_{th}=1-\dfrac{q_{51}}{q_{in}}=1-\dfrac{270.5}{806.3}=0.6645=66.5\%}$$
Question 2 — final results
QuantityValue
Constant-volume pressure ratio$r_p = 1.577$
Cut-off ratio$r_c = 1.294$
Heat added$q_{in} = 806.3\text{ kJ/kg}$
Heat rejected$q_{out} = 270.5\text{ kJ/kg}$
Thermal efficiency$\eta_{th} = 66.5\%$