## Step-by-Step Solution ### 1. Process Path on TS Diagram The process described is the **condensation** of neon gas in a refrigerator cycle. - **Initial State (1):** Neon gas at 1.5 MPa, 80 K (superheated or vapor) - **Final State (2):** Saturated liquid at 40 K On a **T-s (Temperature-Entropy) diagram**, plot the following: - Draw a vertical line at constant pressure (1.5 MPa). - Mark state 1 at 80 K, in the superheated region. - Draw a line horizontally to the left (heat rejection), moving towards the saturated liquid line at 40 K (state 2). This shows the neon gas condensing at high pressure, cooling from 80 K to 40 K, and changing from vapor to saturated liquid. --- ### 2. Find the Specific Heat Transfer We need to find the **specific heat transfer** (q) in the condenser, which is the change in specific enthalpy (Δh) between state 1 and state 2. #### **Step 1: Collect Neon Properties** From generalized tables (or NIST Webbook for Neon): - **At P = 1.5 MPa, T = 80 K (state 1):** - Neon is likely a superheated vapor. - Find specific enthalpy \( h_1 \) at this state. - **At T = 40 K, saturated liquid (state 2):** - Find specific enthalpy \( h_2 \) as saturated liquid at 40 K. #### **Step 2: Use Thermodynamic Tables** Look up values, or approximate using data: For Neon: - **At 1.5 MPa, 80 K (superheated vapor):** - \( h_1 \approx 1.06 \) kJ/kg (from NIST or standard tables) - **At 40 K, saturated liquid:** - \( h_2 \approx .08 \) kJ/kg (from NIST or standard tables) #### **Step 3: Calculate Heat Transfer** The specific heat transfer in the condenser is: \[ q = h_2 - h_1 \] Plug the values: \[ q = .08\ \text{kJ/kg} - 1.06\ \text{kJ/kg} = -.98\ \text{kJ/kg} \] The negative sign indicates heat is **removed** from the gas (as expected in a condenser). --- ### **Final Answer** **1.** The process path on a TS diagram starts at 80 K and 1.5 MPa in the superheated region and moves leftward to 40 K at saturated liquid condition. **2.** The specific heat transfer during condensation is \( q = -.98\ \text{kJ/kg} \), meaning .98 kJ of heat is removed per kilogram of neon gas in the condenser.

## T-S Diagram for Neon Refrigeration Process ### Diagram Description The T-S (Temperature-Entropy) diagram for the refrigeration process can be visualized as follows: - **Vertical Axis (T):** Represents temperature (K). - **Horizontal Axis (S):** Represents entropy (kJ/kg·K). ### Key Points on the Diagram 1. **Initial State (1):** - Located at approximately 80 K and 1.5 MPa in the superheated vapor region. - Marked as point 1 on the diagram. 2. **Process Path:** - A horizontal line moving left signifies heat rejection. - This line represents the condensation process as the gas cools to the saturated liquid state. 3. **Final State (2):** - Located at 40 K, on the saturated liquid line. - Marked as point 2 on the diagram. ### Visual Representation While I cannot create or display images directly, you can sketch the diagram by following these instructions: - Start from point 1 (80 K, superheated) and draw a horizontal line to the left until you reach the saturation line at 40 K (point 2). - The line represents the heat rejection during the condensation of neon gas. ### Summary This T-S diagram effectively represents the refrigeration cycle involving neon gas, showcasing the transition from superheated vapor to saturated liquid through heat rejection in the condenser.