Subtopic: Calculation of power, torque, fuel consumption, heat load and power losses
Question Statement:
A 4-cylinder, 4-stroke tractor diesel engine running at $N = 1800\text{ rpm}$ consumes fuel at a rate of $\dot{m}_f = 7.20\text{ kg/h}$. The density of the diesel fuel is $\rho_f = 840\text{ kg/m}^3$. The volume of fuel injected per cylinder per power stroke in $\text{mm}^3$ is ________ (round off to 1 decimal place).
Answer Key:
39.7
Step-by-Step Solution & Explanation
```
Fuel Injection Quantity Calculation:
Cylinders: k = 4
Speed: N = 1800 rpm
Power strokes per min per cyl: N/2 = 900
Total power strokes/min: 4 * 900 = 3600
Fuel rate: 7.20 kg/h
Density: 840 kg/m^3
```
**Step-by-step Calculation:**
1. Total fuel mass injected per hour:
$\dot{m}_f = 7.20\text{ kg/h} = \frac{7.20}{60} = 0.120\text{ kg/min}$
2. Total fuel volume injected per minute:
$V_{\text{total}} = \frac{0.120\text{ kg/min}}{840\text{ kg/m}^3} = 1.42857 \times 10^{-4}\text{ m}^3/\text{min} = 142857\text{ mm}^3/\text{min}$
3. Total number of injection strokes across all 4 cylinders per minute ($n_{\text{inj}}$):
$n_{\text{inj}} = k \times \frac{N}{2} = 4 \times \frac{1800}{2} = 3600\text{ injections/min}$
4. Fuel volume injected per cylinder per cycle ($V_{\text{stroke}}$):
$V_{\text{stroke}} = \frac{142857\text{ mm}^3/\text{min}}{3600\text{ injections/min}} = 39.6825\text{ mm}^3 \approx 39.7\text{ mm}^3$