Question 2 of 4: Engine Indicated Power, Displacement and Compression Ratio

Problem 2: Engine Indicated Power, Displacement and Compression Ratio (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.

Find. (a) Indicated power $\text{IP}$ (kW). (b) Piston (swept) displacement per cylinder and total. (c) Compression ratio.

Approach. Use the printed formula directly for indicated power (piston area $A$ in m$^2$, stroke $S$ kept in mm, speed $N$ in rpm, pressure $P$ in kPa — the "$\times2$" folds in the four-stroke fact that only one revolution in two is a power stroke); get displacement from swept volume $V_d=\tfrac{\pi}{4}B^2S$; then compression ratio $CR=(V_d+V_c)/V_c$.

1. (a) Piston area and indicated power. $$A = \frac{\pi}{4}B^2 = \frac{\pi}{4}(0.109\ \text{m})^2 = 9.331\times10^{-3}\ \text{m}^2$$ $$\text{IP} = \frac{P\,S\,A\,N\,n}{60{,}000\times2} = \frac{(1050)(115)(9.331\times10^{-3})(3000)(3)}{120{,}000} = \boxed{84.5\ \text{kW}}$$
2. (b) Swept (piston) displacement. Per cylinder, using $B,S$ in cm for a direct-litre result: $$V_{d,\text{cyl}} = \frac{\pi}{4}(10.9\ \text{cm})^2(11.5\ \text{cm}) = 1073\ \text{cm}^3 = 1.073\ \text{L}$$ $$V_{d,\text{total}} = n\,V_{d,\text{cyl}} = 3(1.073) = \boxed{3.22\ \text{L}}$$
3. (c) Compression ratio (per cylinder). $$CR = \frac{V_{d,\text{cyl}}+V_c}{V_c} = \frac{1.073+0.063124}{0.063124} = \boxed{18.0}$$ Discussion: $CR=18$ is typical of a compression-ignition (diesel) engine, which usually runs at roughly 14–22, and is well above the practical knock limit of a spark-ignition engine (about 8–12), so this is a diesel unit. The mean effective pressure of 1050 kPa is within the normal range for both engine types, so the compression ratio, not the pressure, is what identifies the engine type. The swept volume is about 17 times the clearance volume, so the charge is compressed to about 1/18 of its bottom-dead-centre volume. That compression gives the air temperature needed for the injected fuel to self-ignite.