2. The natural gas volume flow rate in the gas pipeline at NC is equal to 150cdot 10^3Nm^3/h Gas temperature is 3^circ C gauge pressure is 5 atg. Determine the gas mass flow rate (kg/h) and the actual gas volume flow rate under operating conditions (m^3/h) of air exnende fu

To determine the gas mass flow rate and the actual gas volume flow rate under operating conditions, we need to use the ideal gas law and the given parameters.Given:- Natural gas volume flow rate: - Gas temperature: - Gauge pressure: 5 atgFirst, let's convert the gauge pressure to absolute pressure:Absolute pressure = Gauge pressure + Atmospheric pressureAssuming standard atmospheric pressure is 14.7 psi (101.325 kPa):Absolute pressure = 5 atg + 14.7 psi = 19.7 psi = 136.5 kPaNext, we can use the ideal gas law to find the mass flow rate:PV = nRTWhere:P = Absolute pressure (136.5 kPa)V = Volume flow rate (150 x 10^3 Nm^3/h)n = Number of molesR = Universal gas constant (8.314 J/(mol·K))T = Temperature (3°C = 276.15 K)Rearranging the equation to solve for n:n = (PV) / (RT)n = (136.5 kPa x 150 x 10^3 Nm^3/h) / (8.314 J/(mol·K) x 276.15 K)n = 9.03 x 10^6 mol/hNow, we can calculate the mass flow rate using the molar mass of natural gas (approximately 16.04 g/mol):Mass flow rate = n x Molar massMass flow rate = 9.03 x 10^6 mol/h x 16.04 g/molMass flow rate = 1.45 x 10^8 g/hMass flow rate = 1.45 x 10^5 kg/hTherefore, the gas mass flow rate is 1.45 x 10^5 kg/h.To find the actual gas volume flow rate under operating conditions, we need to account for the gas temperature and pressure changes. We can use the ideal gas law again:PV = nRTRearranging the equation to solve for V:V = (nRT) / PV = (9.03 x 10^6 mol/h x 8.314 J/(mol·K) x 276.15 K) / 136.5 kPaV = 1.50 x 10^5 m^3/hTherefore, the actual gas volume flow rate under operating conditions is 1.50 x 10^5 m^3/h.