# Step-by-Step Solution ## **Given Data** - **Inlet pressure**: \( P_1 = 10\,\text{kPa} \) - **Inlet state**: Saturated liquid **Exit pressure**: \( P2 = 5\,\text{MPa} = 500\text{k} \) - **entropic efficiency pump**: \(eta_{\text{pump}} =.90) - **Ambient temperature**: \( T_ = 25^\circ \text{C} = 298.15\,\text{K} \) ## **1. Find Properties at Inlet (\(P_1\), saturated liquid)** From steam tables at \(P_1 = 10\,\text{kPa}\): - **Saturated liquid enthalpy** (\(h_1\)): \( h_f@10\,\text{kPa} \approx 191.81\,\text{kJ/kg} \) - **Saturated liquid entropy** (\(s_1\)): \( s_f@10\,\text{kPa} \approx .6492\,\text{kJ/kgK} \) - **Specific volume** (\(v_1\)): \( v_f@10\,\text{kPa} \approx .001010\,\text{m}^3/\text{kg} \) ## **2. Isentropic Work and Exit Enthalpy** For a pump, \[ w_{\text{pump, isen}} = v_1 (P_2 - P_1) \] \[ w_{\text{pump, isen}} = .001010 \times (500 - 10)\,\text{kPa} \] Convert kPa to kJ/m³: \(1\,\text{kPa} = 1\,\text{kN/m}^2 = 1\,\text{kJ/m}^3\) \[ w_{\text{pump, isen}} = .001010 \times 499 = 5.04\,\text{kJ/kg} \] So, \[ h_{2s} = h_1 + w_{\text{pump, isen}} = 191.81 + 5.04 = 196.85\,\text{kJ/kg} \] ## **3. Actual Pump Work and Actual Exit Enthalpy** The isentropic efficiency of a pump is: \[ \eta_{\text{pump}} = \frac{h_{2s} - h_1}{h_2 - h_1} \] \[ .90 = \frac{196.85 - 191.81}{h_2 - 191.81} \] \[ h_2 - 191.81 = \frac{5.04}{.90} = 5.60 \] \[ h_2 = 191.81 + 5.60 = 197.41\,\text{kJ/kg} \] ## **4. Entropy at Exit (\(s_2\))** For a liquid, entropy change can be estimated by: \[ \Delta s \approx v \frac{\Delta P}{T} \] But more accurately, use: \[ s_2 = s_1 + \frac{h_2 - h_1}{T_{\text{avg}}} \] However, for a liquid, \(s_2 \approx s_1\) (pump work is a small fraction), but let's use the isentropic process for \(s_{2s}\): \[ s_{2s} = s_1 \] But for the actual process, use: \[ \Delta s = \frac{h_2 - h_1}{T_{\text{avg}}} \] Let’s estimate \(T_{\text{avg}}\). For water, the temperature rise across the pump is small. At 10 kPa, \(T_{sat} \approx 45.8^\circ C\). At 5 MPa, the temperature is still close to the initial value for liquid water. So, \(T_{\text{avg}} \approx 45.8^\circ C = 318.95\,\text{K}\). \[ \Delta s = \frac{197.41 - 191.81}{318.95} = \frac{5.60}{318.95} = .0176\,\text{kJ/kgK} \] \[ s_2 = s_1 + .0176 = .6492 + .0176 = .6668\,\text{kJ/kgK} \] ## **5. Exergy Destroyed in the Pump** Exergy destroyed is due to irreversibility: \[ \dot{E}_{\text{destroyed}} = T_ \Delta s_{\text{gen}} \] \[ \Delta s_{\text{gen}} = s_2 - s_1 \] \[ E_{\text{destroyed}} = T_ (s_2 - s_1) \] \[ = 298.15 \times (.6668 - .6492) \] \[ = 298.15 \times .0176 = 5.25\,\text{kJ/kg} \] --- ## **Final Answer** ### **Exergy destroyed in the pump:** \[ \boxed{5.25\,\text{kJ/kg}} \] --- **Summary Table** | Quantity | Value | |------------------|-----------------------| | \( h_1 \) | 191.81 kJ/kg | | \( h_2 \) | 197.41 kJ/kg | | \( s_1 \) | .6492 kJ/kgK | | \( s_2 \) | .6668 kJ/kgK | | \( E_{\text{destroyed}} \) | 5.25 kJ/kg |

# Step-by-Step Solution for Exergy Destroyed in Pump ## **Given Data** - **Inlet pressure**: \( P_1 = 10\,\text{kPa} \) - **Inlet state**: Saturated liquid - **Exit pressure**: \( P_2 = 5\,\text{MPa} = 5000\,\text{kPa} \) - **Isentropic efficiency of the pump**: \( \eta_{\text{pump}} = 0.90 \) - **Ambient temperature**: \( T_0 = 25^\circ \text{C} = 298.15\,\text{K} \) ## **1. Find Properties at Inlet (\(P_1\), saturated liquid)** Using steam tables at \(P_1 = 10\,\text{kPa}\): - **Saturated liquid enthalpy** (\(h_1\)): \( h_f@10\,\text{kPa} \approx 191.81\,\text{kJ/kg} \) - **Saturated liquid entropy** (\(s_1\)): \( s_f@10\,\text{kPa} \approx 0.6492\,\text{kJ/kgK} \) - **Specific volume** (\(v_1\)): \( v_f@10\,\text{kPa} \approx 0.001010\,\text{m}^3/\text{kg} \) ## **2. Isentropic Work and Exit Enthalpy** Calculate isentropic work done by the pump: \[ w_{\text{pump, isen}} = v_1 (P_2 - P_1) \] Convert pressures to kPa: \[ w_{\text{pump, isen}} = 0.001010 \times (5000 - 10)\,\text{kPa} = 0.001010 \times 4980 = 5.02\,\text{kJ/kg} \] Now, calculate the isentropic exit enthalpy: \[ h_{2s} = h_1 + w_{\text{pump, isen}} = 191.81 + 5.02 = 196.83\,\text{kJ/kg} \] ## **3. Actual Pump Work and Actual Exit Enthalpy** Using the isentropic efficiency: \[ \eta_{\text{pump}} = \frac{h_{2s} - h_1}{h_2 - h_1} \] Rearranging for \(h_2\): \[ 0.90 = \frac{196.83 - 191.81}{h_2 - 191.81} \] \[ h_2 - 191.81 = \frac{5.02}{0.90} = 5.58 \] \[ h_2 = 191.81 + 5.58 = 197.39\,\text{kJ/kg} \] ## **4. Entropy at Exit (\(s_2\))** For the actual process: \[ s_2 = s_1 + \frac{h_2 - h_1}{T_{\text{avg}}} \] Estimate \(T_{\text{avg}}\) around \(45.8^\circ C\): \[ T_{\text{avg}} \approx 318.95\,\text{K} \] Calculate the change in entropy: \[ \Delta s = \frac{197.39 - 191.81}{318.95} = \frac{5.58}{318.95} \approx 0.0175\,\text{kJ/kgK} \] Now, find \(s_2\): \[ s_2 = s_1 + \Delta s = 0.6492 + 0.0175 = 0.6667\,\text{kJ/kgK} \] ## **5. Exergy Destroyed in the Pump** Exergy destroyed due to irreversibility: \[ \Delta s_{\text{gen}} = s_2 - s_1 \] \[ E_{\text{destroyed}} = T_0 (s_2 - s_1) = 298.15 \times (0.6667 - 0.6492) \] Calculating: \[ = 298.15 \times 0.0175 \approx 5.21\,\text{kJ/kg} \] --- ## **Final Answer** ### **Exergy destroyed in the pump:** \[ \boxed{5.21\,\text{kJ/kg}} \] --- **Summary Table** | Quantity | Value | |------------------|-----------------------| | \( h_1 \) | 191.81 kJ/kg | | \( h_2 \) | 197.39 kJ/kg | | \( s_1 \) | 0.6492 kJ/kgK | | \( s_2 \) | 0.6667 kJ/kgK | | \( E_{\text{destroyed}} \) | 5.21 kJ/kg |