Sure! Let’s solve the problem step-by-step. --- # **Step 1: Understanding the Problem** - **Initial state:** - Mass, \( m_1 = 1.2 \) kg - Pressure, \( P_1 = 700 \) kPa - Temperature, \( T_1 = 200^\circ C = 473.15 \) K - **Final state:** - Pressure drops to \( P_2 = 600 \) kPa (because piston starts to move at this pressure) - Volume decreases to 80% of initial (\( V_2 = 0.8 V_1 \)) - Heat lost \( Q = -40 \) kJ (heat leaves the gas, so it's negative) - **Exit temperature** is average of initial and final: \( T_{\text{exit}} = \frac{T_1 + T_2}{2} \) - **Air properties:** - \( C_p = 1.02 \) kJ/kg·K, \( C_v = 0.733 \) kJ/kg·K, \( R = C_p - C_v = 0.287 \) kJ/kg·K --- # **Step 2: Initial Volume Calculation** Use the ideal gas law for initial state: \[ P_1 V_1 = m_1 R T_1 \] \[ V_1 = \frac{m_1 R T_1}{P_1} = \frac{1.2 \times 0.287 \times 473.15}{700} \] \[ V_1 = \frac{162.972}{700} = 0.233 \text{ m}^3 \] \[ V_2 = 0.8 V_1 = 0.8 \times 0.233 = 0.186 \text{ m}^3 \] --- # **Step 3: Energy (First Law) Analysis** **Control mass:** The air remaining in the cylinder. **Control volume:** The cylinder. First Law for unsteady flow with mass leaving (neglecting kinetic and potential energy): \[ Q - W = (m_2 u_2 - m_1 u_1) + m_e h_e \] - \( Q \) = Heat lost (negative) - \( W \) = Boundary work (to be found) - \( m_1, m_2 \) = Initial and final mass in cylinder - \( u_1, u_2 \) = Initial and final specific internal energy - \( m_e \) = Mass that escaped - \( h_e \) = Enthalpy of escaping air Where: - \( u = C_v T \) - \( h = C_p T \) --- ## **Let’s Find \( m_2 \) in Terms of \( T_2 \):** From ideal gas law at final state: \[ P_2 V_2 = m_2 R T_2 \] \[ m_2 = \frac{P_2 V_2}{R T_2} = \frac{600 \times 0.186}{0.287 \times T_2} = \frac{111.6}{0.287 T_2} \] --- ## **Let’s Find \( m_e \) in Terms of \( T_2 \):** \[ m_e = m_1 - m_2 \] --- ## **Exit Temperature:** \[ T_e = \frac{T_1 + T_2}{2} \] --- # **Step 4: Apply the First Law** Plug in all expressions: \[ Q - W = [m_2 u_2 - m_1 u_1] + m_e h_e \] \[ Q - W = [m_2 (C_v T_2) - m_1 (C_v T_1)] + (m_1 - m_2) [C_p T_e] \] \[ -40 - W = [m_2 C_v T_2 - m_1 C_v T_1] + (m_1 - m_2) C_p \left(\frac{T_1 + T_2}{2}\right) \] --- # **Step 5: Work Done (\(W\))** Work is done only when piston moves (i.e., when pressure is 600 kPa). \[ W = P_2 (V_1 - V_2) = 600 \times (0.233 - 0.186) = 600 \times 0.047 = 28.2 \text{ kJ} \] --- # **Step 6: Plug Numbers and Solve for \( T_2 \)** Let’s substitute all values (all energy units in kJ): - \( C_v = 0.733 \) kJ/kg·K - \( C_p = 1.02 \) kJ/kg·K - \( m_1 = 1.2 \) - \( T_1 = 473.15 \) K - \( Q = -40 \) kJ - \( W = 28.2 \) kJ - \( m_2 = \frac{111.6}{0.287 T_2} \) - \( m_e = 1.2 - m_2 \) - \( T_e = \frac{T_1 + T_2}{2} \) Plug into the First Law: \[ -40 - 28.2 = [m_2 (0.733) T_2 - 1.2 (0.733) \times 473.15] + (1.2 - m_2) \cdot 1.02 \cdot \left(\frac{473.15 + T_2}{2}\right) \] Calculate \( 1.2 \times 0.733 \times 473.15 = 415.38 \) kJ So: \[ -68.2 = [0.733 m_2 T_2 - 415.38] + (1.2 - m_2) \cdot 1.02 \cdot \frac{473.15 + T_2}{2} \] Let’s write \( m_2 = \frac{111.6}{0.287 T_2} \): Now expand and collect terms (let’s denote \( x = T_2 \) for simplicity): \[ -68.2 = 0.733 m_2 x - 415.38 + 1.02 \cdot \frac{1.2 - m_2}{2} (473.15 + x) \] \[ -68.2 + 415.38 = 0.733 m_2 x + 0.61(1.2 - m_2)(473.15 + x) \] \[ 347.18 = 0.733 m_2 x + 0.61 \times 1.2 \times (473.15 + x) - 0.61 m_2 (473.15 + x) \] \[ 347.18 = 0.733 m_2 x + 0.732(473.15 + x) - 0.61 m_2 (473.15 + x) \] Expand: \[ 347.18 = 0.733 m_2 x + 346.11 + 0.732 x - 0.61 m_2 \times 473.15 - 0.61 m_2 x \] Combine like terms: \[ 347.18 - 346.11 = [0.733 m_2 x - 0.61 m_2 x] + 0.732 x - 0.61 m_2 \times 473.15 \] \[ 1.07 = (0.123 m_2 x) + 0.732 x - 0.61 \times 473.15 \times m_2 \] Now plug \( m_2 = \frac{111.6}{0.287 x} \): - \( 0.123 m_2 x = 0.123 \times \frac{111.6}{0.287 x} \times x = 0.123 \times \frac{111.6}{0.287} \) - \( 0.61 \times 473.15 \times m_2 = 0.61 \times 473.15 \times \frac{111.6}{0.287 x} \) Calculate constants: - \( 0.123 \times \frac{111.6}{0.287} = 47.86 \) - \( 0.61 \times 473.15 = 288.62 \) - \( 288.62 \times \frac{111.6}{0.287} = 112,188 \) So, \[ 1.07 = 47.86 + 0.732 x - \frac{112,188}{x} \] So, \[ 1.07 - 47.86 = 0.732 x - \frac{112,188}{x} \] \[ -46.79 = 0.732 x - \frac{112,188}{x} \] \[ 0.732 x - \frac{112,188}{x} + 46.79 = 0 \] --- # **Step 7: Solve for \( T_2 \) Numerically** Let’s solve this equation: \[ 0.732 x + 46.79 = \frac{112,188}{x} \] \[ 0.732 x^2 + 46.79 x = 112,188 \] \[ 0.732 x^2 + 46.79 x - 112,188 = 0 \] Use quadratic formula: \[ x = \frac{-46.79 \pm \sqrt{(46.79)^2 - 4 \times 0.732 \times (-112,188)}}{2 \times 0.732} \] Calculate discriminant: \[ (46.79)^2 = 2,190.71 \] \[ 4 \times 0.732 \times 112,188 = 328,372 \] \[ 2,190.71 + 328,372 = 330,563 \] \[ \sqrt{330,563} = 575.01 \] So, \[ x = \frac{-46.79 + 575.01}{1.464} = \frac{528.22}{1.464} = 360.8 \text{ K} \] (negative root will be negative, so ignore) --- ## **Final Answers** ### **a) Final temperature, \( T_2 \):** \[ \boxed{T_2 = 361 \text{ K} \approx 88^\circ C} \] --- ### **b) Mass that escaped** \[ m_2 = \frac{111.6}{0.287 \times 360.8} = \frac{111.6}{103.34} = 1.08 \text{ kg} \] \[ m_e = m_1 - m_2 = 1.2 - 1.08 = 0.12 \text{ kg} \] --- ### **c) Work Done** Previously calculated: \[ W = 28.2 \text{ kJ} \] --- ### **d) Sketch the Process on \( P-v \) and \( T-v \) Diagrams** #### **P-v diagram** - **State 1:** \( P_1 = 700 \) kPa, \( v_1 = V_1/m_1 = 0.233/1.2 = 0.194 \) m³/kg - **State 2:** \( P_2 = 600 \) kPa, \( v_2 = V_2/m_2 = 0.186/1.08 = 0.172 \) m³/kg Draw a horizontal line at \( P=700 \) kPa as volume decreases with constant pressure, then a drop to 600 kPa at constant pressure. #### **T-v diagram** - **State 1:** \( T_1 = 473.15 \) K, \( v_1 = 0.194 \) m³/kg - **State 2:** \( T_2 = 360.8 \) K, \( v_2 = 0.172 \) m³/kg Draw a line connecting these two points. --- ## **Summary Table** | | Initial | Final | |---------|--------------------|--------------------| | P (kPa) | 700 | 600 | | T (K) | 473 | 361 | | m (kg) | 1.2 | 1.08 | | V (m³) | 0.233 | 0.186 | | v (m³/kg)| 0.194 | 0.172 | --- ## **Diagrams** ### **P-v Diagram** ``` | P | o (1)-------------------o (2) | 700 kPa 600 kPa +---------------------------------------> v 0.194 0.172 (m³/kg) ``` ### **T-v Diagram** ``` | T | o (1) | \ | o (2) +---------------------> v 0.194 0.172 (m³/kg) ``` --- ## **Final Answers** a) **Final temperature:** \( \boxed{361\, \text{K}} \) b) **Mass escaped:** \( \boxed{0.12\, \text{kg}} \)