Let's break down the problem step by step: **Given:** - Initial temperature, \( T_1 = 15^\circ C \) - Throttling from \( P_1 = 5\, \text{atm} \) to \( P_2 = 1\, \text{atm} \) - Final temperature, \( T_2 = ? \) - Volume expansivity \( \beta = 1.5 \times 10^{-4} \, \text{K}^{-1} \) (assuming typo in question, it should be \(10^{-4}\)) - Surroundings temperature \( T_0 = 20^\circ C = 293.15\, \text{K} \) - Water is a liquid (incompressible assumption) - Use steam tables as needed We are to determine: 1. The temperature change of the water after throttling (\( \Delta T \)) 2. The lost work per kg (\( W_{lost} \)) during the process --- ## **Step 1: Throttling Process Overview** **Throttling process:** - Isenthalpic (\( h_1 = h_2 \)) - Involves a significant pressure drop - No heat or work transfer (adiabatic, no shaft work) --- ## **Step 2: Isenthalpic Relation for Liquids** For a small pressure change in a liquid, the enthalpy change can be written as: \[ dh = c_p dT + \left[ v - T\left(\frac{\partial v}{\partial T}\right)_P \right]dP \] But for **incompressible liquids**, \( v \) is nearly constant, and \( (\partial v/\partial T)_P = v\beta \): \[ dh = c_p dT + \left[ v - T v \beta \right] dP \] For **throttling** (\( dh = 0 \)): \[ 0 = c_p dT + \left[ v - T v \beta \right] dP \] \[ c_p dT = - \left[ v - T v \beta \right] dP \] \[ dT = -\frac{v}{c_p} [1 - T \beta] dP \] Integrate from \( P_1 \) to \( P_2 \): \[ \Delta T = -\frac{v}{c_p} [1 - T \beta] (P_2 - P_1) \] --- ## **Step 3: Gather Data** At \( 15^\circ C \): - \( v \) (specific volume of liquid water) ≈ \( 1.001 \times 10^{-3} \, \text{m}^3/\text{kg} \) - \( c_p \) (specific heat at constant pressure for liquid water) ≈ \( 4.18\, \text{kJ}/\text{kg}\cdot\text{K} = 4180\, \text{J}/\text{kg}\cdot\text{K} \) - \( \beta = 1.5 \times 10^{-4} \, \text{K}^{-1} \) - \( T = 288.15 \, \text{K} \) - Pressure drop: \( P_2 - P_1 = 1\,\text{atm} - 5\,\text{atm} = -4\, \text{atm} \) - \( 1\, \text{atm} = 101,325\, \text{Pa} \) So, \[ P_2 - P_1 = -4 \times 101325 = -405300\, \text{Pa} \] --- ## **Step 4: Calculate the Temperature Change** \[ \Delta T = -\frac{v}{c_p} [1 - T\beta] (P_2 - P_1) \] Calculate \( 1 - T\beta \): \[ T\beta = 288.15 \times 1.5 \times 10^{-4} = 0.0432 \] \[ 1 - T\beta = 1 - 0.0432 = 0.9568 \] Plug in numbers: \[ \Delta T = -\frac{1.001 \times 10^{-3}}{4180} \times 0.9568 \times (-405300) \] First, compute the numerator of the fraction: \[ 1.001 \times 10^{-3} \times 0.9568 = 0.0009578 \] \[ \Delta T = -\frac{0.0009578}{4180} \times (-405300) \] \[ \frac{0.0009578}{4180} = 2.292 \times 10^{-7} \] \[ \Delta T = -2.292 \times 10^{-7} \times (-405300) = 0.0928\, \text{K} \] So the temperature **increases** by about **0.093°C**. --- ## **Step 5: Lost Work (Irreversibility) per Kilogram** Lost work (irreversibility) per kg is given by: \[ w_{lost} = T_0 \Delta s_{gen} \] where \( \Delta s_{gen} \) is the entropy generated per kg. For a throttling process with liquid water (assuming negligible temperature change): \[ \Delta s = c_p \ln\frac{T_2}{T_1} - R \ln\frac{P_2}{P_1} \] But for liquids, the entropy change is approximated as: \[ \Delta s = c_p \ln\frac{T_2}{T_1} - v \frac{(P_2-P_1)}{T_1} \] Since \( \Delta T \) is very small, \( \ln\frac{T_2}{T_1} \approx \frac{\Delta T}{T_1} \). So, \[ \Delta s \approx c_p \frac{\Delta T}{T_1} - v \frac{(P_2-P_1)}{T_1} \] Plug in values: \( T_1 = 288.15\, \text{K} \), \( c_p = 4180\, \text{J/kg K} \), \( \Delta T = 0.093\, \text{K} \), \( v = 1.001 \times 10^{-3} \, \text{m}^3/\text{kg} \), \( P_2 - P_1 = -405300\, \text{Pa} \) \[ c_p \frac{\Delta T}{T_1} = 4180 \times \frac{0.093}{288.15} = 4180 \times 0.000323 = 1.35\, \text{J/kg K} \] \[ v \frac{(P_2-P_1)}{T_1} = 1.001 \times 10^{-3} \times \frac{-405300}{288.15} \] \[ = 1.001 \times 10^{-3} \times (-1,407.15) = -1.409 \, \text{J/kg K} \] So, \[ \Delta s \approx 1.35 - (-1.409) = 2.76\, \text{J/kg K} \] Now, \[ w_{lost} = T_0 \Delta s_{gen} = 293.15 \times 2.76 = 808.1\, \text{J/kg} \] --- ## **Step 6: Final Answers and Assumptions** ### **Summary** **1. Temperature change of the water:** \[ \Delta T = +0.093^\circ C \] **2. Lost work per kilogram:** \[ w_{lost} = 808\, \text{J/kg} \] ### **Assumptions:** - Water is incompressible and liquid throughout the process. - The temperature rise is small, so properties (v, c_p, β) are taken at the initial state. - The process is adiabatic and at steady state, with no shaft work. - The expansion is isenthalpic (throttling process). - Surroundings temperature is constant at 20°C. - Changes in kinetic and potential energy are negligible. --- ## **Final Answers** **Temperature change of the water after throttling:** \[ \boxed{+0.093^\circ \text{C} \text{ (increase)}} \] **Lost work (irreversibility) per kilogram of water:** \[ \boxed{808\, \text{J/kg}} \] If you have any follow-up or need more details on any step, just let me know!