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Question

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VIPSolutions · Accepted Answer

Let's break down and solve the **Diesel cycle** problem step by step.

---

# 1. **Problem Recap**

- **Compression ratio ($r$)**: 20  
- **Cutoff ratio ($r_c$)**: 1.3  
- **Initial pressure ($P_1$)**: 100 kPa  
- **Initial temperature ($T_1$)**: 60°C = 333.15 K  
- **Output power ($\dot{W}_{\text{net, out}}$)**: 200 kW  
- **Working fluid:** Air (ideal gas, use constant specific heats at room temp)

**Tasks:**
a. Find maximum temperature of the air  
b. Find mass flow rate of air ($\dot{m}$)  
c. Find rate of heat addition ($\dot{Q}_{\text{in}}$)  
d. Find thermal efficiency ($\eta_{th}$) using given equations  
e. Sketch P-v diagram

---

# 2. **Diesel Cycle Background and Useful Equations**

- **Compression ratio ($r$):** $ r = \frac{V_1}{V_2} $
- **Cutoff ratio ($r_c$):** $ r_c = \frac{V_3}{V_2} $
- **Ideal gas law:** $ PV = nRT $ or $ PV = mRT $
- **Diesel cycle efficiency:**  
  $$
  \eta_{th} = 1 - \frac{1}{r^{k-1}} \cdot \frac{r_c^k - 1}{k(r_c - 1)}
  $$
  where $ k $ (air) $\approx 1.4 $

- **Enthalpy for air:** $ h = c_p T $ (assume $ c_p \approx 1.005 $ kJ/kg·K)

---

# 3. **Step-by-Step Solution**

## **a) Maximum Temperature ($T_3$)**

### **Step 1: State 1 (Initial State)**
- $ P_1 = 100 $ kPa
- $ T_1 = 333.15 $ K

### **Step 2: State 2 (End of Compression)**
- Isentropic compression:
  $$
  \frac{T_2}{T_1} = r^{k-1}
  $$
  $$
  T_2 = T_1 \cdot r^{k-1}
  $$
  $$
  T_2 = 333.15 \times 20^{.4}
  $$
  Evaluate $ 20^{.4} $:

$$
  20^{.4} = e^{.4 \ln 20} \approx e^{.4 \times 2.9957} \approx e^{1.1983} \approx 3.314
  $$
  $$
  T_2 = 333.15 \times 3.314 \approx 1103.6\ \text{K}
  $$

### **Step 3: State 3 (End of Heat Addition)**
- At constant pressure: $ \frac{T_3}{T_2} = r_c $
  $$
  T_3 = T_2 \times r_c = 1103.6 \times 1.3 \approx 1434.7\ \text{K}
  $$

**Final Answer (a):**  
$$
\boxed{T_3 = 1435\ \text{K}\ (\text{rounded})}
$$

---

## **b) Mass Flow Rate ($\dot{m}$)**

Given: Net work output $ \dot{W}_{\text{net, out}} = 200 $ kW

- Work per cycle for Diesel cycle (per kg):
  $$
  w_{\text{net, out}} = \eta_{th} \cdot q_{in}
  $$

### **First, Find Efficiency ($\eta_{th}$)**

$$
\eta_{th} = 1 - \frac{1}{r^{k-1}} \cdot \frac{r_c^k - 1}{k(r_c - 1)}
$$

Calculate:
- $ r^{k-1} = 20^{.4} \approx 3.314 $
- $ r_c^k = 1.3^{1.4} \approx e^{1.4 \ln 1.3} \approx e^{1.4 \times .2624} \approx e^{.3674} \approx 1.444 $
- $ r_c - 1 = .3 $

Plug in:

$$
\eta_{th} = 1 - \frac{1}{3.314} \cdot \frac{1.444 - 1}{1.4 \times .3}
$$
$$
= 1 - .3017 \times \frac{.444}{.42}
$$
$$
= 1 - .3017 \times 1.057
$$
$$
= 1 - .319
$$
$$
= .681
$$

$$
\eta_{th} \approx \boxed{.68}
$$

---

### **Find Heat Added ($q_{in}$, per kg):**

$$
q_{in} = c_p (T_3 - T_2)
$$
$$
= 1.005 \times (1435 - 1104) = 1.005 \times 331 = 332.66\ \text{kJ/kg}
$$

### **Find Specific Work Output ($w_{\text{net, out}}$):**

$$
w_{\text{net, out}} = \eta_{th} \cdot q_{in} = .68 \times 332.66 = 226.2\ \text{kJ/kg}
$$

### **Mass Flow Rate:**

$$
\dot{m} = \frac{\dot{W}_{\text{net, out}}}{w_{\text{net, out}}}
$$
$$
= \frac{200,000\ \text{W}}{226,200\ \text{J/kg}}
= .884\ \text{kg/s}
$$

**Final Answer (b):**  
$$
\boxed{\dot{m} = .88\ \text{kg/s}}
$$

---

## **c) Rate of Heat Addition ($\dot{Q}_{\text{in}}$)**

$$
\dot{Q}_{\text{in}} = \dot{m} \cdot q_{in}
$$
$$
= .884 \times 332.66 = 294\ \text{kW}
$$

**Final Answer (c):**  
$$
\boxed{\dot{Q}_{\text{in}} = 294\ \text{kW}}
$$

---

## **d) Thermal Efficiency ($\eta_{th}$)**

Already calculated above:

$$
\boxed{\eta_{th} = .68}
$$

---

## **e) Sketch P-v Diagram**

A Diesel cycle P-v diagram consists of:
- 1-2: Isentropic compression  
- 2-3: Constant pressure heat addition  
- 3-4: Isentropic expansion  
- 4-1: Constant volume heat rejection

```
        3
     /----\
    /      \
  2         4
   \       /
    \_____/
      1
```

**Key Points:**
- 1-2: Steeply rising curve (compression)
- 2-3: Horizontal (constant pressure)
- 3-4: Downward (expansion)
- 4-1: Vertical (constant volume)

---

# **Summary Table**

| Part | Answer                   |
|------|--------------------------|
| a    | $T_3 = 1435$ K         |
| b    | $\dot{m} = .88$ kg/s  |
| c    | $\dot{Q}_{\text{in}} = 294$ kW |
| d    | $\eta_{th} = .68$     |

---

If you need the actual diagram, let me know or draw it by hand based on the description!

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