Dans l'industrie pétrolière et gazière, comprendre la teneur énergétique des combustibles est crucial pour une production et une utilisation efficaces. Une métrique clé utilisée pour quantifier cette énergie est le Pouvoir Calorifique Brut Humide (PCBH). Ce terme fait référence à l'énergie totale transférée sous forme de chaleur lors de la combustion idéale d'un gaz saturé en eau à température et pression standard (STP), avec la stipulation cruciale que toute l'eau formée lors du processus de combustion apparaît à l'état liquide.
Pourquoi le PCBH est-il important ?
Comprendre le calcul du PCBH :
Le calcul du PCBH prend en compte les facteurs suivants :
Différences clés avec les autres pouvoirs calorifiques :
Conclusion :
Le PCBH est un paramètre crucial pour caractériser la teneur énergétique des combustibles gazeux saturés en eau dans l'industrie pétrolière et gazière. Comprendre le calcul et ses différences avec les autres pouvoirs calorifiques permet des évaluations énergétiques précises, une utilisation efficace du combustible et une tarification équitable dans les transactions de gaz. En fournissant une représentation réaliste de l'énergie totale disponible à partir du gaz saturé en eau, le PCBH permet une prise de décision éclairée dans divers aspects des opérations pétrolières et gazières.
Instructions: Choose the best answer for each question.
1. What does WGHV stand for? a) Wet Gross Heating Value b) Water Gross Heating Value c) Wet Gas Heating Value d) Water Gas Heating Value
a) Wet Gross Heating Value
2. What is the crucial factor differentiating WGHV from other heating values? a) The type of gas being analyzed. b) The temperature and pressure at which the combustion occurs. c) The state of water formed during combustion (liquid vs. vapor). d) The presence of impurities in the gas.
c) The state of water formed during combustion (liquid vs. vapor).
3. Why is WGHV important for gas trading contracts? a) It allows for accurate pricing based on the actual energy content of the fuel. b) It standardizes the measurement of gas volume. c) It facilitates the transportation of natural gas. d) It determines the composition of the gas stream.
a) It allows for accurate pricing based on the actual energy content of the fuel.
4. What is the difference between WGHV and GHV? a) WGHV considers the heat of condensation of water vapor, while GHV does not. b) GHV considers the heat of condensation of water vapor, while WGHV does not. c) WGHV only considers the heat released by the combustion of the gas, while GHV includes the heat of condensation of water vapor. d) GHV only considers the heat released by the combustion of the gas, while WGHV includes the heat of condensation of water vapor.
b) GHV considers the heat of condensation of water vapor, while WGHV does not.
5. Which of the following statements is TRUE regarding the WGHV calculation? a) It assumes the water formed during combustion remains as vapor. b) It is performed at a standard temperature and pressure (STP). c) It is used to analyze completely dry gas streams. d) It ignores the impact of water vapor on the energy content of the gas.
b) It is performed at a standard temperature and pressure (STP).
Scenario:
A natural gas stream contains 80% methane (CH4), 10% ethane (C2H6), and 10% water vapor (H2O). You need to determine the WGHV of this gas stream.
Instructions:
Use the following combustion reactions and standard enthalpy of formation data to calculate the heat released from each component's combustion:
Calculate the overall heat release per mole of the gas mixture.
Convert the heat release per mole to WGHV in kJ/m3, assuming the gas mixture behaves ideally at STP (0°C and 1 atm).
Exercise Correction:
**1. Heat Release from Combustion of Each Component:** - Methane: ΔH°(CH4) = -890 kJ/mol - Ethane: ΔH°(C2H6) = -1560 kJ/mol **2. Overall Heat Release per Mole of Mixture:** - Heat release from methane: 0.8 mol CH4 * (-890 kJ/mol) = -712 kJ - Heat release from ethane: 0.1 mol C2H6 * (-1560 kJ/mol) = -156 kJ - Total heat release: -712 kJ + (-156 kJ) = -868 kJ **3. WGHV in kJ/m3:** - Molecular weight of mixture: 0.8 * 16 g/mol + 0.1 * 30 g/mol + 0.1 * 18 g/mol = 20.2 g/mol - Density of ideal gas at STP: (20.2 g/mol) / (22.4 L/mol) = 0.902 g/L = 0.902 kg/m3 - WGHV: (-868 kJ/mol) / (0.902 kg/m3) = -962 kJ/kg = -962 kJ/m3 (since density is kg/m3) **Therefore, the WGHV of the natural gas stream is approximately -962 kJ/m3.**
Comments