Engine performance and testing of engines

Let $p_m$ be the mean effective pressure (Pa), $L$ the stroke (m), $A$ the piston area (m$^2$), $N$ the speed (rev/s), $k = 1$ for two-stroke and $k = \tfrac12$ for four-stroke, $n$ the number of cylinders.

Notes:

• $\dot{m}_f$ is the fuel mass flow (kg/s) and $CV$ the calorific value in kJ/kg.
• Brake mean effective pressure: $bmep = BP/(LAkNn)$.
• A typical SI engine has $\eta_m$ = 80-90%, $\eta_{bth}$ = 25-30%, bsfc = 0.27-0.35 kg/kWh; a diesel engine has $\eta_{bth}$ = 30-40%, bsfc = 0.2-0.25 kg/kWh.

Given: $D = 0.08$ m, $L = 0.1$ m, $n = 4$, $N = 3000$ rpm, $p_{mi} = 9 \times 10^5$ Pa, $W = 24 \times 9.81 = 235.4$ N, $R = 0.5$ m, $\dot{m}_f = 11.2$ kg/h, $CV = 44\,000$ kJ/kg.

Indicated power

For a four-stroke engine the number of power strokes per cylinder per second is $N/2 = 25$.

$$\begin{aligned} A &= \frac{\pi}{4}(0.08)^2 = 5.027 \times 10^{-3}\ \text{m}^2 \\ IP &= \frac{p_{mi} L A \cdot (N/2) \cdot n}{1000} = \frac{9 \times 10^5 \times 0.1 \times 5.027 \times 10^{-3} \times 25 \times 4}{1000} = 45.24\ \text{kW} \end{aligned}$$

Brake power

$$\begin{aligned} T &= W R = 235.44 \times 0.5 = 117.7\ \text{N m} \\ BP &= \frac{2\pi N T}{60\,000} = \frac{2\pi \times 3000 \times 117.7}{60\,000} = 36.98\ \text{kW} \end{aligned}$$

Friction power and mechanical efficiency

$$\begin{aligned} FP &= IP - BP = 45.24 - 36.98 = 8.26\ \text{kW} \\ \eta_m &= \frac{BP}{IP} = \frac{36.98}{45.24} = 0.8175 \end{aligned}$$

Thermal efficiencies

$$\begin{aligned} \text{Fuel energy} &= \frac{11.2}{3600} \times 44\,000 = 136.9\ \text{kW} \\ \eta_{bth} &= \frac{36.98}{136.9} = 0.270 \\ \eta_{ith} &= \frac{45.24}{136.9} = 0.330 \end{aligned}$$

Check: $\eta_m \eta_{ith} = 0.8175 \times 0.330 = 0.270$.

Specific fuel consumption

$$bsfc = \frac{11.2}{36.98} = 0.303\ \text{kg/kWh}, \qquad isfc = \frac{11.2}{45.24} = 0.248\ \text{kg/kWh}$$

Answer: IP = 45.2 kW, BP = 37.0 kW, FP = 8.26 kW, $\eta_m$ = 81.8%, $\eta_{ith}$ = 33.0%, $\eta_{bth}$ = 27.0%, bsfc = 0.303 kg/kWh, isfc = 0.248 kg/kWh.

Morse test

It is used for multi-cylinder engines. The engine is run at a fixed speed and load, and the brake power $B$ is measured with all cylinders firing. Then one cylinder is cut out (by shorting the spark plug in an SI engine, or cutting the fuel in a CI engine) and the load is reduced so that the speed returns to the original value. The new brake power $B_1$ is measured. The other cylinders keep running, and the friction and pumping power of the dead cylinder is still the same, so

$$IP_1 = B - B_1$$

is the indicated power of the cut-out cylinder. The process is repeated for each cylinder and $IP = \sum IP_i$. Then $FP = IP - B$ and $\eta_m = B/IP$. It gives a result without any indicator, but it assumes that friction does not change when a cylinder is cut out.

Calculation

$$\begin{aligned} IP &= 17.4 + 18.0 + 16.9 + 17.7 = 70.0\ \text{kW} \\ FP &= 70.0 - 62.5 = 7.5\ \text{kW} \\ \eta_m &= \frac{62.5}{70.0} = 0.893 \end{aligned}$$

Answer: cylinder indicated powers 17.4, 18.0, 16.9 and 17.7 kW; IP = 70.0 kW; FP = 7.5 kW; $\eta_m$ = 89.3%.

Heat balance sheet

It is an account of the fuel energy supplied to an engine and how it is distributed: brake work, coolant, exhaust gases, and the unaccounted losses (radiation, friction heat to oil, incomplete combustion). It shows where the energy goes and how to improve the engine, and is prepared for a test at a stated load and speed, usually on a per minute or per second (kW) basis.

Heat supplied

$$Q_s = \frac{14}{3600} \times 42\,000 = 163.3\ \text{kW}$$

Heat distribution

$$\begin{aligned} \text{Brake work} &= 50\ \text{kW} \\ Q_w &= \frac{50}{60} \times 4.187 \times 14 = 48.85\ \text{kW} \\ \dot{m}_{exh} &= \dot{m}_f + \dot{m}_a = \frac{14 \times (1 + 22)}{3600} = 0.0894\ \text{kg/s} \\ Q_{exh} &= 0.0894 \times 1.05 \times (380 - 27) = 33.15\ \text{kW} \\ Q_{unacc} &= 163.33 - 50 - 48.85 - 33.15 = 31.33\ \text{kW} \end{aligned}$$

The exhaust heat is the sensible heat above the ambient temperature.

Balance sheet

Answer: brake work 30.6%, cooling water 29.9%, exhaust 20.3%, unaccounted 19.2% of the 163.3 kW supplied.

Air density and orifice head

$$\begin{aligned} \rho_a &= \frac{p}{RT} = \frac{101\,325}{287 \times 300} = 1.177\ \text{kg/m}^3 \\ H_a &= h_w \frac{\rho_w}{\rho_a} = 0.100 \times \frac{1000}{1.177} = 84.97\ \text{m of air} \end{aligned}$$

Air flow through the orifice

$$\begin{aligned} A_o &= \frac{\pi}{4}(0.048)^2 = 1.810 \times 10^{-3}\ \text{m}^2 \\ \dot{V}_a &= C_d A_o \sqrt{2 g H_a} = 0.62 \times 1.810 \times 10^{-3} \times \sqrt{2 \times 9.81 \times 84.97} \\ &= 0.62 \times 1.810 \times 10^{-3} \times 40.83 = 0.04581\ \text{m}^3/\text{s} \end{aligned}$$

Air mass flow $= 0.04581 \times 1.177 = 0.05391$ kg/s $= 194.1$ kg/h.

Swept volume rate

$$\begin{aligned} V_s &= \frac{\pi}{4}(0.09)^2(0.11) = 6.998 \times 10^{-4}\ \text{m}^3 \\ \dot{V}_s &= 4 \times 6.998 \times 10^{-4} \times \frac{2500}{2 \times 60} = 0.05832\ \text{m}^3/\text{s} \end{aligned}$$

Volumetric efficiency and A/F

$$\begin{aligned} \eta_v &= \frac{0.04581}{0.05832} = 0.7855 \\ A/F &= \frac{194.1}{9.5} = 20.4 \end{aligned}$$

An air box is used between the engine and the orifice to damp the pulsation of the intake, so that the orifice reading is steady.

Answer: $\eta_v = 78.6\%$ and $A/F = 20.4$.

Methods of finding friction power

1. Willans line: plot fuel consumption against brake power at constant speed and extrapolate back to zero fuel; the intercept on the negative BP axis is the friction power. Suitable for CI engines, where the line is straight to about 70% load.
2. Morse test: for multi-cylinder engines; cylinders cut out one by one (details in another answer).
3. Motoring test: the engine is driven by an electric motor (fuel off) at the test speed; the power taken equals FP. Gives hot FP if done soon after a run.
4. Retardation test: the engine is run at no load at rated speed; the ignition (or fuel) is cut and the speed fall is timed, once with the engine alone and once with a known extra load. The deceleration times give FP.
5. From indicator diagram: $FP = IP - BP$ with IP found by an engine indicator.

Willans line from the data

The values lie on a straight line.

 fuel
 kg/h |             . 40
  8.7 |         . 30
  6.6 |     . 20
  4.5 | . 10
  2.4 |.   (BP = 0)
  0.0 +---+--------------- BP
    -11.4 0

$$\text{Slope} = \frac{10.8 - 4.5}{40 - 10} = 0.21\ \text{kg/h per kW}$$

Fuel at zero BP $= 4.5 - 0.21 \times 10 = 2.4$ kg/h. The line meets zero fuel when

$$BP = -\frac{2.4}{0.21} = -11.43\ \text{kW}$$

so $FP = 11.43$ kW (assumed constant at the speed).

Mechanical efficiency

$$\begin{aligned} \eta_m(30\ \text{kW}) &= \frac{30}{30 + 11.43} = 0.724 \\ \eta_m(40\ \text{kW}) &= \frac{40}{40 + 11.43} = 0.778 \end{aligned}$$

Answer: FP = 11.4 kW; $\eta_m$ = 72.4% at 30 kW and 77.8% at 40 kW.

Measurement of brake power

Brake power is obtained from the shaft torque $T$ and speed $N$ by a dynamometer, which absorbs the engine power and measures the torque:

$$BP = \frac{2\pi N T}{60\,000}\ \text{kW}$$

($T$ in N m, $N$ in rpm.) Dynamometers are of two kinds: absorption types (Prony, rope, hydraulic, eddy current), which convert the power to heat, and transmission types (torsion) for power transmitted without absorbing.

Rope brake dynamometer

        flywheel with water-cooled rim
       ______
      /  O   \   rope turns around, ends carry:
      \______/   S (spring balance)   W (dead load)

Several turns of rope are wound around the flywheel drum of the engine. One end carries a dead weight $W$ and the other a spring balance reading $S$. Wood blocks keep the rope in place. Net load $= W - S$ and torque is $T = (W - S)(R + r_{rope})$, where $R$ is the drum radius. Cooling water is circulated in the drum. It is simple and cheap and suits small engines, but heats up, and the friction is not steady.

Hydraulic (water brake) dynamometer

A rotor with vanes (on the engine shaft) turns in a casing full of water. The vanes churn the water against fixed casing pockets, and the casing, free to rotate, tends to turn with the rotor. The torque is balanced by a weight or load cell at an arm, and the water supply is adjusted to vary the absorbing load. Heat is taken away by the flowing water. It is compact and good for high-power engines, easy to control and stable.

Quantities measured in a performance test

• Speed (tachometer) and torque (dynamometer) for the brake power.
• Fuel consumption (burette and stopwatch or flow meter) for sfc.
• Air consumption (air box with orifice, or hot-wire meter) for A/F and volumetric efficiency.
• Temperatures of the cooling water inlet/outlet, exhaust gas, oil and ambient; and the cooling water flow, for the heat balance.
• Exhaust gas composition (gas analyser), and indicated diagram if IP is needed.

A test may be a constant-speed test (Willans line, load variation), a variable-speed test at full throttle (performance curves) or a heat balance test.

Fuel energy

$$Q_f = \dot{m}_f \times CV = \frac{6.2}{3600} \times 44\,000 = 75.78\ \text{kW}$$

Efficiencies

$$\begin{aligned} \eta_m &= \frac{BP}{IP} = \frac{18}{22} = 0.818 \\ \eta_{ith} &= \frac{IP}{Q_f} = \frac{22}{75.78} = 0.290 \\ \eta_{bth} &= \frac{BP}{Q_f} = \frac{18}{75.78} = 0.2375 \end{aligned}$$

Air-standard efficiency and relative efficiency

$$\begin{aligned} \eta_{Otto} &= 1 - \frac{1}{r^{\gamma-1}} = 1 - 7^{-0.4} = 0.541 \\ \eta_{rel} &= \frac{\eta_{ith}}{\eta_{Otto}} = \frac{0.290}{0.541} = 0.537 \end{aligned}$$

The relative efficiency of 53.7% shows how well the real engine approaches the ideal cycle: the rest is lost to the real-gas properties, heat loss, finite burning time and incomplete combustion. Typical values are 50-60% for SI engines.

Specific fuel consumption

$$bsfc = \frac{6.2}{18} = 0.344\ \text{kg/kWh}, \qquad isfc = \frac{6.2}{22} = 0.282\ \text{kg/kWh}$$

Answer: $\eta_m$ = 81.8%, $\eta_{ith}$ = 29.0%, $\eta_{bth}$ = 23.8%, $\eta_{rel}$ = 53.7%, bsfc = 0.344 kg/kWh, isfc = 0.282 kg/kWh.