1.02kg of R-134a fills a 0.14m3 weighted piston cylinder device at a tempreture of -26.40C The container is now heated untill tempreture is 100.0C Determine the final volume of R134. Use data from the refrigrant table.
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1.02kg of R-134a fills a 0.14m3 weighted piston cylinder device at a tempreture of -26.40C The container is now heated untill tempreture is 100.0C Determine the final volume of R134. Use data from the refrigrant table.
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Asked by: Rajveer Prajapat
Created at: 2025-10-10 11:40:20
Asked by: Rajveer Prajapat
Created at: 2025-10-10 11:40:20
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# Final Volume of R-134a in a Piston-Cylinder Device
## Given Data
- **Mass of R-134a,** \( m = 1.02\, \text{kg} \)
- **Initial volume,** \( V_1 = .14\, \text{m}^3 \)
- **Initial temperature,** \( T_1 = -26.4^\circ\text{C} \)
- **Final temperature,** \( T_2 = 100.^\circ\text{C} \)
- **Device:** Weighted piston-cylinder (constant pressure)
## Step 1: Find Initial State
### 1. Calculate Initial Specific Volume
\[
v_1 = \frac{V_1}{m} = \frac{.14}{1.02} = .1373\, \text{m}^3/\text{kg}
\]
### 2. Use R-134a Property Tables
At \( T_1 = -26.4^\circ\text{C} \):
- From R-134a tables, the saturation pressure \( P_{\text{sat}} \approx 100\, \text{kPa} \).
- Saturated liquid specific volume \( v_f \approx .000757\, \text{m}^3/\text{kg} \).
- Saturated vapor specific volume \( v_g \approx .3666\, \text{m}^3/\text{kg} \).
Since \( v_1 \) is much greater than \( v_f \) and less than \( v_g \), the refrigerant is a **mixture** at -26.4°C.
### 3. Find Initial Pressure
Since the volume is fixed with a weighted piston, the pressure is constant. At \( T_1 \), the pressure is the saturation pressure at \( -26.4^\circ\text{C} \):
\[
P_1 = P_{\text{sat, at } -26.4^\circ\text{C}} = 100\, \text{kPa}
\]
## Step 2: Final State (After Heating to 100°C, at Constant Pressure)
- **Final temperature:** \( T_2 = 100.^\circ\text{C} \)
- **Pressure:** \( P_2 = P_1 = 100\, \text{kPa} \)
Check the phase at final state:
From R-134a tables, at \( P = 100\, \text{kPa} \):
- \( T_{\text{sat}} = -26.4^\circ\text{C} \)
- At \( T_2 = 100^\circ\text{C} \), the refrigerant is **superheated** at 100 kPa.
## Step 3: Find Final Specific Volume
From R-134a superheated tables:
- At \( P = 100\, \text{kPa} \), \( T = 100^\circ\text{C} \):
- \( v \approx .8845\, \text{m}^3/\text{kg} \)
## Step 4: Calculate Final Volume
\[
V_2 = m \cdot v_2 = 1.02 \times .8845 = .902\, \text{m}^3
\]
## **Summary Table**
| State | T (\( ^\circ\)C) | P (kPa) | v (\( \text{m}^3/\text{kg} \)) | V (\( \text{m}^3 \)) | Phase |
|-------|------------------|---------|-------------------------------|----------------------|-----------|
| 1 | -26.4 | 100 | .1373 | .14 | Mixture |
| 2 | 100. | 100 | .8845 | .902 | Superheated |
## **Final Answer**
\[
\boxed{V_2 = .90\, \text{m}^3}
\]
(Rounded to two decimal places)
---
**References:** R-134a refrigerant property tables (superheated and saturated).
**Note:** Actual values might vary slightly depending on the refrigerant property source used.
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