Question

A piston-cylinder assembly having 250 mm diameter contains 0.01 kg of water vapor at 1 MPa and 200°C. The specific volume of the vapor is 0.020602 m\(^3\)/kg. The system expands as per the relation \( p v^n = {constant} \), where \( p \) is pressure and \( v \) is the specific volume. The expansion of water vapor displaces the piston by 50 mm. If the final pressure is 0.35 MPa, the value of the exponent \( n \) is ........ (rounded off to two decimal places).

Hint:

For a polytropic process, use the relationship \( p v^n = {constant} \) to find the exponent \( n \) by applying the initial and final pressures and volumes.

Answer 1

Correct Answer: 1.3

We are given:
- Initial volume \( V_1 = 0.01 \, {kg} \times 0.020602 \, {m}^3/{kg} = 0.00020602 \, {m}^3 \),
- Initial pressure \( p_1 = 1 \, {MPa} \),
- Final pressure \( p_2 = 0.35 \, {MPa} \),
- The displacement of the piston is 50 mm.
The relation between pressure and specific volume is given as: \[ p v^n = {constant} \] where \( n \) is the polytropic index. Step 1: We first calculate the initial volume \( V_1 \) using the formula: \[ V_1 = m \times v_1 = 0.01 \, {kg} \times 0.020602 \, {m}^3/{kg} = 0.00020602 \, {m}^3 \] Step 2: The diameter of the piston is given as \( d = 250 \, {mm} = 0.25 \, {m} \). The cross-sectional area of the piston is: \[ A = \frac{\pi d^2}{4} = \frac{\pi (0.25)^2}{4} = 0.0491 \, {m}^2 \] The volume displaced by the piston is: \[ \Delta V = A \times \Delta h = 0.0491 \, {m}^2 \times 0.05 \, {m} = 0.002455 \, {m}^3 \] Step 3: The final volume \( V_2 \) is the sum of the initial volume and the volume displaced: \[ V_2 = V_1 + \Delta V = 0.00020602 + 0.002455 = 0.00266102 \, {m}^3 \] Step 4: The relation \( p v^n = {constant} \) implies: \[ p_1 V_1^n = p_2 V_2^n \] Substitute the known values: \[ 1000 \times (0.00020602)^n = 350 \times (0.00266102)^n \] \[ \frac{1000}{350} = \left( \frac{0.00266102}{0.00020602} \right)^n \] \[ 2.8571 = \left( 12.9 \right)^n \] Step 5: Now, solve for \( n \) by taking the logarithm of both sides: \[ \log(2.8571) = n \log(12.9) \] \[ n = \frac{\log(2.8571)}{\log(12.9)} = \frac{0.456} {1.110} = 1.30 \] Step 6: Therefore, the value of the exponent \( n \) is approximately 1.30, which is within the given range.