Let's break down the **problem** and see how far we can go. --- ### **Given Data** - **Pipeline diameter:** \( D = 0.6 \) m - **Initial T:** \( T_1 = 25^\circ C = 298.15 \) K - **Initial P:** \( P_1 = 3 \) MPa - **Mass flow rate:** \( \dot{m} = 125 \) kg/s - **Final P:** \( P_2 = 2 \) MPa (before next pump) - **After compression:** \( P_3 = 3 \) MPa (to match initial) - **After cooling:** \( T_4 = 25^\circ C \) (restored to initial T at 3 MPa) - **Gas:** Methane, \( M = 16 \) g/mol \( = 0.016 \) kg/mol - \( C_p = 36.8 \) J/mol·K --- ## **a) Find the temperature and velocity of the gas just before it enters the pumping station (i.e., at 2 MPa).** ### **Step 1: Adiabatic Expansion (from 3 MPa, 25°C to 2 MPa, ?)** **Assume:** - Flow is adiabatic (no heat exchange) - Ideal gas - \( C_p, C_v \) are constant #### **1. Find \( C_p \) and \( C_v \) per kg** \[ C_p = 36.8\, \text{J/mol·K} \] \[ C_p (\text{per kg}) = \frac{36.8}{0.016} = 2300\, \text{J/kg·K} \] \[ C_v = C_p - R \] \[ R = \frac{8.314\, \text{J/mol·K}}{0.016\, \text{kg/mol}} = 519.6\, \text{J/kg·K} \] \[ C_v = 2300 - 519.6 = 1780.4\, \text{J/kg·K} \] #### **2. Find \( \gamma \):** \[ \gamma = \frac{C_p}{C_v} = \frac{2300}{1780.4} \approx 1.292 \] #### **3. Adiabatic Relation:** \[ \frac{T_2}{T_1} = \left(\frac{P_2}{P_1}\right)^{\frac{\gamma-1}{\gamma}} \] \[ P_1 = 3\, \text{MPa},\quad P_2 = 2\, \text{MPa} \] \[ T_1 = 298.15\, \text{K} \] \[ \frac{T_2}{298.15} = \left(\frac{2}{3}\right)^{\frac{1.292-1}{1.292}} \] \[ \frac{\gamma-1}{\gamma} = \frac{0.292}{1.292} \approx 0.226 \] \[ \frac{T_2}{298.15} = (0.6667)^{0.226} \approx e^{0.226 \cdot \ln(0.6667)} \approx e^{0.226 \times (-0.405)} \approx e^{-0.0916} \approx 0.9125 \] \[ T_2 = 0.9125 \times 298.15 \approx 272\, \text{K} \] **→** **The temperature at the end of the 40 mile pipeline is \( \approx 272\, \text{K} = -1^\circ C \)** --- ### **Step 4: Velocity at \( P_2 = 2 \) MPa, \( T_2 \approx 272\, \text{K} \)** **Find density:** \[ \rho = \frac{P M}{R_u T} \] where \( R_u = 8.314 \) J/mol·K, \( M = 0.016 \) kg/mol \[ P_2 = 2 \times 10^6\, \text{Pa} \] \[ T_2 = 272\, \text{K} \] \[ \rho = \frac{2 \times 10^6 \times 0.016}{8.314 \times 272} \] \[ \rho = \frac{32,000}{2,262.208} \approx 14.15\, \text{kg/m}^3 \] **Area:** \[ A = \frac{\pi}{4} D^2 = \frac{\pi}{4} \times (0.6)^2 = 0.2827\, \text{m}^2 \] **Velocity:** \[ \dot{m} = \rho v A \implies v = \frac{\dot{m}}{\rho A} \] \[ v = \frac{125}{14.15 \times 0.2827} \approx \frac{125}{4.004} \approx 31.2\, \text{m/s} \] --- ## **Summary of a):** - **Temperature before station (\( T_2 \))**: **~272 K** (**-1°C**) - **Velocity before station (\( v_2 \))**: **~31.2 m/s** --- ## **b) Work of Compressor, Outlet T, and Heat Load on Cooler** ### **Compressor (2 MPa, 272 K) → (3 MPa, ?K) [adiabatic]** #### **1. Outlet T after adiabatic compression:** \[ \frac{T_3}{T_2} = \left(\frac{P_3}{P_2}\right)^{\frac{\gamma-1}{\gamma}} \] \[ \frac{T_3}{272} = \left(\frac{3}{2}\right)^{0.226} \] \[ \ln(1.5) = 0.405 \] \[ 0.226 \times 0.405 = 0.0916 \] \[ e^{0.0916} = 1.0959 \] \[ T_3 = 1.0959 \times 272 \approx 298.1\, \text{K} \] **So, after adiabatic compression, the temperature is restored to ~298 K (25°C).** #### **2. Work done by compressor (adiabatic):** For adiabatic compression: \[ w = C_p (T_3 - T_2) \] \[ w = 2300 \times (298.1 - 272) = 2300 \times 26.1 = 60,030\, \text{J/kg} \] \[ \dot{W} = \dot{m} \cdot w = 125 \times 60,030 = 7,503,750\, \text{W} = 7.5\, \text{MW} \] #### **3. After compressor, isobaric cooling to 25°C:** The gas leaves the compressor at 3 MPa, 298 K. It is cooled isobarically to 298 K, which means **no further temperature drop**. #### **4. Heat to be removed (isobaric cooling):** If the outlet temperature is exactly 298 K (same as initial), the heat removed is zero (since \( T_3 = T_4 \)), but in practice, if the compressor outlet was hotter, cooling would be needed. But in this case, the adiabatic compression exactly restores the temperature to 25°C, so **heat load is zero**. --- ## **Summary for b):** - **Work rate by compressor:** \( \dot{W} \approx 7.5 \) MW - **Temperature after compressor:** \( T_3 \approx 298 \) K (25°C) - **Heat load on gas cooler:** ~0 (since \( T_3 = T_4 \)) --- ## **c) Why can you NOT solve this problem exactly as stated?** **Key issues:** - The assumption of strictly adiabatic flow in a real 40 mile pipeline is not realistic. There will be heat exchange with the environment, frictional pressure losses, and possibly condensation/freeze-out at low temperatures (here, T drops below 0°C). - The pressure drop in the pipeline is likely dominated by friction, not just adiabatic expansion. The real temperature drop would be less extreme when frictional heating is included. - The velocity calculated (~31 m/s) is very high for a gas pipeline—real pipelines typically operate at lower velocities to reduce friction and erosion. - The density at 2 MPa, 272 K, and the mass flow imply a volumetric flow rate that may not be feasible in practice. - The assumption that the compression process is adiabatic is also an idealization; real compressors are not 100% efficient and heat up. --- ## **Final Answers** ### **a)** - **Temperature before station:** \( \boxed{272~\text{K}~(-1^\circ C)} \) - **Velocity before station:** \( \boxed{31.2~\text{m/s}} \) ### **b)** - **Work rate of compressor:** \( \boxed{7.5~\text{MW}} \) - **Temperature leaving compressor:** \( \boxed{298~\text{K}~(25^\circ C)} \) - **Heat load on cooler:** \( \boxed{0~\text{(since outlet T after compression = 25°C)}} \) ### **c) Why can't you solve this problem exactly as stated?** Because the assumptions (strictly adiabatic flow, ideal gas, no friction, no heat loss/gain, no consideration for practical pipeline operation) are not realistic for such a long pipeline. Real pipelines have significant heat exchange, frictional losses, and operate at more moderate velocities and temperature drops. Therefore, the simplified model gives physically unrealistic results (e.g., gas temperature dropping below freezing), and more detailed analysis including friction and heat transfer is required for a real solution. --- **If you need more details or a worked example with more realistic assumptions (including friction, heat transfer), let me know!**