Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:Problem: A four-cylinder, four-stroke, 2.5 L gasoline engine operates on an Otto cycle with a
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Question

Act like a helpful tutor and exlain me :
Give step-by-step solution with explanation and final answer:Problem: A four-cylinder, four-stroke, 2.5 L gasoline engine operates on an Otto cycle with a
compression ratio of 7. The air is at 300 kPa and 40 °C at the beginning of the compression
process, and the maximum pressure during the cycle is 8 MPa. The air/fuel ratio (the amount
of air divided by the amount of fuel intake) is 10. The compression and expansion processes
may be modeled as polytropic (PV = constant) with k = 1.344. Assuming the properties of the
working fluid (specific heat, etc.) are constant and are equal to the values for air at 400 K,
determine:
a. the temperature at the end of the expansion process,
b. the net work output,
c. the thermal efficiency, and
d. the mean effective pressure Stage [7 [Ppa v (u®/kg)
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d J 3
Sr 4 . &
4 o Fill-up this table
N Low o Show the calculation steps as shown in PDF.
f Co o Upload in the blackboard
) vr o Due within 48 hours
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VIPSolutions · Accepted Answer

Let's break down and solve the Otto cycle problem step-by-step. **We'll first outline the process, list assumptions, and then solve each part.**

---

## **Given Data**
- Engine: 4-cylinder, four-stroke, 2.5 L gasoline
- Compression ratio ($ r $): **7**
- Initial pressure ($ P_1 $): **300 kPa**
- Initial temperature ($ T_1 $): **40°C = 313.15 K**
- Maximum pressure ($ P_3 $): **8 MPa = 800 kPa**
- Air/fuel ratio: **10**
- Polytropic exponent ($ k $): **1.344**
- Assume air properties at **400 K** (constant cp, cv, R, etc.)

---

## **Stages in Otto Cycle**

| State | Description              |
|-------|--------------------------|
| 1     | Start of compression     |
| 2     | End of compression       |
| 3     | End of heat addition     |
| 4     | End of expansion         |

---

## **Step 1: Fill the Table - State Properties**

### **1.1. Calculate Specific Volume at State 1 ($ v_1 $)**

- Displacement = 2.5L = $ .0025 $ m³ (total for 4 cylinders)
- For one cycle (since it's four-stroke), only half the displacement is filled per revolution:
  - **Volume per cycle = $ .0025 / 2 = .00125 $ m³**
- Assume 1 kg of air for calculations.

Use the ideal gas law:
$$
v_1 = \frac{RT_1}{P_1}
$$

Where $ R = .287 $ kJ/kg·K (for air).

$$
v_1 = \frac{.287 \cdot 313.15}{300} = .299 \, \text{m}^3/\text{kg}
$$

---

### **1.2. State 2 (End of Compression)**

$$
\text{Compression ratio } r = \frac{v_1}{v_2} = 7 \implies v_2 = \frac{v_1}{7}
$$
$$
v_2 = \frac{.299}{7} = .0427 \, \text{m}^3/\text{kg}
$$

For **Polytropic Compression** ($ PV^k = \text{const} $):

$$
\frac{T_2}{T_1} = r^{k-1}
$$
$$
T_2 = T_1 \cdot r^{k-1} = 313.15 \cdot 7^{1.344-1} = 313.15 \cdot 7^{.344}
$$
$$
7^{.344} \approx 1.870
$$
$$
T_2 = 313.15 \times 1.870 = 585 \, \text{K}
$$

$$
\frac{P_2}{P_1} = r^k = 7^{1.344} \approx 12.8
$$
$$
P_2 = 300 \times 12.8 = 384 \, \text{kPa}
$$

---

### **1.3. State 3 (End of Heat Addition, at Maximum Pressure)**

$$
P_3 = 800 \, \text{kPa}
$$
$$
v_3 = v_2 \quad (\text{constant volume process})
$$
$$
\frac{T_3}{T_2} = \frac{P_3}{P_2}
$$
$$
T_3 = T_2 \cdot \frac{P_3}{P_2} = 585 \times \frac{800}{384} = 585 \times 2.083 = 1218 \, \text{K}
$$

---

### **1.4. State 4 (End of Expansion)**

$$
v_4 = v_1 \quad (\text{constant volume})
$$

**Polytropic Expansion**:
$$
\frac{T_4}{T_3} = \left(\frac{v_3}{v_4}\right)^{k-1}
$$
But $ \frac{v_4}{v_3} = r \implies \frac{v_3}{v_4} = \frac{1}{r} $
$$
T_4 = T_3 \cdot \left(\frac{1}{r}\right)^{k-1} = 1218 \cdot 7^{-.344} = 1218 \cdot .535 = 652 \, \text{K}
$$

$$
P_4 = P_3 \cdot \left(\frac{v_3}{v_4}\right)^{k} = 800 \times (1/7)^{1.344} = 800 \times .0781 = 625 \, \text{kPa}
$$

---

### **Summary Table**

| Stage | $ T $ (K) | $ P $ (kPa) | $ v $ (m³/kg) |
|-------|-------------|---------------|-----------------|
| 1     | 313         | 300           | .299           |
| 2     | 585         | 384          | .0427          |
| 3     | 1218        | 800          | .0427          |
| 4     | 652         | 625           | .299           |

---

## **Step 2: Answer the Questions**

### **a. Temperature at the end of expansion ($ T_4 $)**
**Answer:** $ T_4 = 652 \, \text{K} $

---

### **b. Net Work Output per Cycle (per kg air)**

$$
w_{net} = q_{in} - q_{out}
$$

- For air (ideal gas), $ q = c_v (T_3 - T_2), \, q_{out} = c_v (T_4 - T_1) $
- **At 400 K**, $ c_v \approx .718 \, \text{kJ/kg·K} $

$$
q_{in} = c_v (T_3 - T_2) = .718 (1218 - 585) = .718 \times 633 = 454.6 \, \text{kJ/kg}
$$
$$
q_{out} = c_v (T_4 - T_1) = .718 (652 - 313) = .718 \times 339 = 243.6 \, \text{kJ/kg}
$$
$$
w_{net} = 454.6 - 243.6 = 211 \, \text{kJ/kg}
$$

---

### **c. Thermal Efficiency**

$$
\eta = 1 - \frac{q_{out}}{q_{in}} = 1 - \frac{243.6}{454.6} = .4639 = 46.4\%
$$

---

### **d. Mean Effective Pressure (MEP)**

$$
MEP = \frac{w_{net}}{v_{max} - v_{min}}
$$
$$
v_{max} = v_1 = .299 \, \text{m}^3/\text{kg}, \quad v_{min} = v_2 = .0427 \, \text{m}^3/\text{kg}
$$
$$
MEP = \frac{211}{.299 - .0427} = \frac{211}{.2563} = 823 \, \text{kPa}
$$

---

## **Summary Table of Results**

| Property                   | Value            |
|----------------------------|------------------|
| $ T_4 $                  | **652 K**        |
| Net work ($ w_{net} $)   | **211 kJ/kg**    |
| Thermal efficiency ($ \eta $) | **46.4%**    |
| Mean effective pressure    | **823 kPa**      |

---

### **Let me know if you need any clarifications or calculations for the total engine output (using 2.5 L, cycles per second, etc.)!**

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