Give step-by-step solution with explanation and final answer:Problem #2 [35 Points] Polytropic Processes
A piston-cylinder undergoes an adiabatic expansion followed by a constant
pressure compressio

Question

Give step-by-step solution with explanation and final answer:Problem #2 [35 Points] Polytropic Processes
A piston-cylinder undergoes an adiabatic expansion followed by a constant
pressure compression process. The adiabatic expansion can be modeled as a
polytropic process PV® = constant with a volume exponent n = 1.4. At state 1 the
pressure of the system is P1 = 100 psi, the volume is V1 = 10.86 ft* and the internal
energy U1 = 2311 BTU. The absolute magnitude of the work input during the
compression process is 2W3 = 75.2 BTU and the absolute magnitude of the heat
transfer output for this process is 2Q3 = 687.4 BTU. The final system pressure is
P3 = 60 psi. Neglecting kinetic and potential energy effects sketch the processes on a
P-v diagram and determine final system volume V3 in ft’ and the final internal
energy Us in BTU.
Given: [3 Points] Find: [2 Points]
Svstem Energy Diagram 1-52: [4 Points] Assume:
w=/Pdy,
w= [Pve— Pwvil{ll - n]
w= Biyuln [veivi]
AU=Q-W =mg -w)
1 Btu = 778 ftlbr
+ 112 = 144 in?
Sistem Energy Dingram 2-23: [4 Points]

VIPSolutions · Accepted Answer

Let's break down and solve the problem step by step:

### **Given Data**
- **State 1:**  
  $ P_1 = 100 $ psi  
  $ V_1 = 10.86 $ ft³  
  $ U_1 = 2311 $ BTU

- **Polytropic Expansion (1→2):**  
  $ PV^n = \text{constant} $  
  $ n = 1.4 $

- **After Expansion, Compression (2→3):**  
  **Constant Pressure Compression:**  
  $ P_3 = 60 $ psi

- **Work and Heat Data:**  
  $ |W_{2\rightarrow3}| = 75.2 $ BTU (compression, so work is on the system, **W is negative**)  
  $ |Q_{2\rightarrow3}| = 687.4 $ BTU (**heat is out of the system, so Q is negative**)

- **Find:**  
  Final system volume $ V_3 $ (in ft³)  
  Final internal energy $ U_3 $ (in BTU)

---

## **Step 1: Find State 2 Properties**

### **A. Polytropic Expansion (1→2)**
- **Equation:**  
  $ P_1 V_1^n = P_2 V_2^n $

But we don't have $ P_2 $ or $ V_2 $ yet.  
However, after this process, the pressure at state 3 will be $ P_3 = 60 $ psi.

### **B. The Constant Pressure Compression (2→3):**
- $ P_2 = P_3 = 60 $ psi

---

## **Step 2: Calculate Volume at State 2**

From polytropic process $ P_1 V_1^n = P_2 V_2^n $:

$$
V_2 = \left( \frac{P_1}{P_2} \right)^{1/n} V_1
$$

- $ P_1 = 100 $ psi
- $ P_2 = 60 $ psi
- $ n = 1.4 $
- $ V_1 = 10.86 $ ft³

$$
V_2 = \left( \frac{100}{60} \right)^{1/1.4} \times 10.86
$$

Calculate $ \frac{100}{60} = 1.6667 $

$$
1.6667^{1/1.4} = e^{\frac{1}{1.4} \ln(1.6667)}
$$
$$
\ln(1.6667) \approx 0.5108
$$
$$
\frac{1}{1.4} \approx 0.7143
$$
$$
0.7143 \times 0.5108 \approx 0.3654
$$
$$
e^{0.3654} \approx 1.441
$$

$$
V_2 = 1.441 \times 10.86 \approx 15.66 \text{ ft}^3
$$

---

## **Step 3: Find Work Done (1→2), $ W_{1\rightarrow2} $**

For a polytropic process:
$$
W = \frac{P_2 V_2 - P_1 V_1}{1 - n}
$$

Convert psi·ft³ to BTU:
- $ 1 \text{ BTU} = 778 \text{ ft} \cdot \text{lbf} $
- $ 1 \text{ psi} \cdot \text{ft}^3 = 144 \text{ lbf/ft}^2 \cdot \text{ft}^3 = 144 \text{ ft} \cdot \text{lbf} $

So, to convert psi·ft³ to BTU: divide by 778 and multiply by 144.

$$
W = \frac{P_2 V_2 - P_1 V_1}{1 - n}
$$

Plug in numbers:

$$
P_2 V_2 = 60 \times 15.66 = 939.6 \text{ psi·ft}^3
$$
$$
P_1 V_1 = 100 \times 10.86 = 1086 \text{ psi·ft}^3
$$
$$
W_{1\rightarrow2} = \frac{939.6 - 1086}{1 - 1.4} = \frac{-146.4}{-0.4} = 366 \text{ psi·ft}^3
$$

Convert to BTU:
$$
366 \times 144 = 52704 \text{ ft} \cdot \text{lbf}
$$
$$
\frac{52704}{778} \approx 67.77 \text{ BTU}
$$

---

## **Step 4: Find Internal Energy at State 2, $ U_2 $**

For adiabatic (no heat):
$$
\Delta U_{1\rightarrow2} = Q_{1\rightarrow2} - W_{1\rightarrow2}
$$
$$
Q_{1\rightarrow2} = 0
$$
$$
\Delta U_{1\rightarrow2} = -W_{1\rightarrow2}
$$
$$
U_2 = U_1 + \Delta U_{1\rightarrow2}
$$
$$
U_2 = 2311 + (-67.77) = 2243.23 \text{ BTU}
$$

---

## **Step 5: Find Volume at State 3, $ V_3 $**

For constant pressure:
$$
W_{2\rightarrow3} = P (V_3 - V_2)
$$

Given: **Work is done on the system, so sign is negative**
$$
W_{2\rightarrow3} = P_3 (V_3 - V_2)
$$
Given $ |W_{2\rightarrow3}| = 75.2 $ BTU (**work input, so should be negative: $ W_{2\rightarrow3} = -75.2 $ BTU**)

Convert -75.2 BTU to $ \text{psi} \cdot \text{ft}^3 $:

$$
-75.2 \text{ BTU} \times 778 = -58465.6 \text{ ft} \cdot \text{lbf}
$$
$$
\frac{-58465.6}{144} = -406 \text{ psi} \cdot \text{ft}^3
$$

So,
$$
P_3 (V_3 - V_2) = -406
$$
$$
60 (V_3 - 15.66) = -406
$$
$$
V_3 - 15.66 = \frac{-406}{60} = -6.767
$$
$$
V_3 = 15.66 - 6.767 = 8.89 \text{ ft}^3
$$

---

## **Step 6: Find Internal Energy at State 3, $ U_3 $**

$$
\Delta U_{2\rightarrow3} = Q_{2\rightarrow3} - W_{2\rightarrow3}
$$
Given:
- $ Q_{2\rightarrow3} = -687.4 $ BTU (heat out)
- $ W_{2\rightarrow3} = -75.2 $ BTU (work input, so negative)

$$
\Delta U_{2\rightarrow3} = (-687.4) - (-75.2) = -687.4 + 75.2 = -612.2 \text{ BTU}
$$
$$
U_3 = U_2 + \Delta U_{2\rightarrow3} = 2243.23 - 612.2 = 1631.03 \text{ BTU}
$$

---

## **Final Answers**

### **Final System Volume $ V_3 $:**
$$
\boxed{8.89 \ \text{ft}^3}
$$

### **Final Internal Energy $ U_3 $:**
$$
\boxed{1631 \ \text{BTU}}
$$

---

### **Summary Table**

| State | P (psi) | V (ft³) | U (BTU)  |
|-------|---------|---------|----------|
| 1     | 100     | 10.86   | 2311     |
| 2     | 60      | 15.66   | 2243     |
| 3     | 60      | 8.89    | 1631     |

---

**Let me know if you want the process plotted on a P-V diagram!**

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