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Energy Engineering - Fundamentals of Chemical Processes

Exercise 2: Solution

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Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 1 EXERCISE 2 Analysis of a combustion boiler supplied by natural gas In a power plant, a boiler is fueled by natural gas (NG) , with the following composition (molar basis): CH 4 80% C2H6 10% C3H8 5% N2 5% The natural gas enters the boiler at 25°C and 1 atm and is burned with air, at 50°C , 1 atm and relative humidity of 25%. The air is fed with 30% excess. The exhaust gases exit the boiler at 150°C , and do not contain any hydrocarbon species (i.e., the combu stion is complete). In the steam production section, the incoming liquid water is pumped at 28 bar and 90°C (the effect of pressure on the enthalpy content of water can be neglected) and is converted to saturated steam at 28 bar and 238°C. Assuming that the heat losses amount to 3% of the LHV of the natural gas, calculate the amount of steam produced per Nm 3 of NG and the efficiency of the boiler. DATA: Natural Gas : T IN = 25°C, P = 1 atm ∆�������,�� 4 0 = −802 ������� ��� ∆�������,�2�6 0 = −1429 ������� ��� ∆�������,�3�8 0 = −2043 ������� ��� Air : TIN = 50°C, P = 1 atm, R.H. = 25% a 50°C, P SAT H2O (50°C) = 12.349 kPa Water/Steam : h INH2O (90°C) = 376.92 kJ/kg H2O hOUT H2O (238°C) = 2804 kJ/kg H2O Exhaust Gases : TOUT = 150°C, P = 1atm Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 2 Reference state: ideal gas, 1 bar, 298 K ������̃�,������= �������+ �������∙������+ �������∙������2+ �������∙������3 [J/mol/K] T in [K] Species a b c d O2 28.11 -3.680 x 10 -6 -17.46 x 10 -6 -10.65 x 10 -9 N2 31.15 -13.57 x 10 -3 26.80 x 10 -6 -11.68 x 10 -9 H2O 32.24 1.924 x 10 -3 10.55 x 10 -6 -3.596 x 10 -9 CO 2 19.80 73.44 x 10 -3 -56.02 x 10 -6 17.15 x 10 -9 SOLUTION As we are not given extensive in formation on the inlet flowrates, we have to a ssume a base of calcul us. It is convenient in this case to assume 100 mol/s of Natural Gas (NG) as basis for material balances. In this way we can directly evaluate the i nlet molar flowrates for each component present in the NG stream . Specie s �̇������ ��������� [mol/s] CH 4 80 C2H6 10 C3H8 5 N2 5 The stoichiometric amount of oxygen �̇�2�� can be evaluated from the combustion stoichiometry for eac h component. We are tol d that the combust or is working with 30% air excess , and from here we can evaluate the flow rates o f O2 and N 2 effectiv ely entering the process. ������������ 4+ 2�2→ ������� 2+ 2������2� (1) ������2������6+ 7 2�2→ 2������� 2+ 3������2� (2) ������3������8+ 5�2→ 3������� 2+ 4������2� (3) �̇�2�� = 2�̇�� 4 ������� + 7 2�̇�2�6+ 5�̇�3�8 ������� = 220 ��� ������ �̇�2������� = �̇�2�� ∙(1+ �% 100 )= 286 ��� ������ �̇�2������� = 79 21 ∙�̇�2������� = 1075 .90 ��� ������ The air fed to the system is h umid , with a relative humidity (RH) of 25%. Given the air flowrate , we can evaluat e the amount of water : Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 3 �������2�= �.������.(50° ������)∙��2��������� (50° ������) � = 0.25 ∙12 .349 101 .325 = 0.0305 �̇�2�������� = �̇�������� ������������� ������� ∙ �������2� 1− �������2�= 42 .80 ��� ������ Now the inlet gases are fully known and it is possible to determine the outlet composition . W e are working at complete fuel c onsumption and he nce there are no hydrocarbons at the outlet : this m eans that the extent of reaction for each combustion reaction (1 -3) is e qual to the molar stream of the c orrespo nding fuel. The combust ion produ cts can be evaluated using the reaction stoichiomet ry: �̇�2��� = �̇�2������� − 2�̇�� 4 ������� − 7 2�̇�2�6 ������� − 5�̇�3�8 ������� �̇�2��� = �̇�2������� = �̇�2 �� ������� + �̇�2 ������������� ������� �̇�� 2��� = �̇�� 4 ������� + 2�̇�2�6 ������� + 3�̇�3�8 ������� �̇�2���� = �̇�2�������� + 2�̇�� 4 ������� + 3�̇�2�6 ������� + 4�̇�3�8 ������� Now the ou tlet stream composition is fully known. Consi dering a control volume comprising both the combustion streams as well as the water stream s, we can w rite the entha lpic balance as : ∑ �̇������������� ∙∫ ������̃�,������������� �������� ��������� ������������������� ������=1 + �̇��������� ∙������������������ �� − ∑ �̇��������� ∙∫ ������̃�,������������� ���� ��������� �������� ������=1 + �̇���� = �̇�2�(ℎ̂�2��������� − ℎ̂�2��������� ) By integrat ing the pol ynomial form of the C p provided in the te xt we get : ∫ ������̃�,������������� � ��������� = �������∙(������− ��������� )+ ������� 2 ∙(������2− ���������2 )+ ������� 3∙(������3− ���������3 )+ ������� 4 ∙(������4− ���������4 ) In this case , since the combustion streams and the water streams are not integr ated from the material point of view (i.e. they are complete ly segregate d) and he nce it is possi ble to use different reference states. In fac t: • Com bustion gases: T and P where the ΔH°C is known , (T Ref = 25°C, P Ref = 1 atm), inlet composition and state ; • Water streams: pure water at the triple point (T Ref = 0.01°C, P Ref = 611.73 Pa), liquid state . This is the reference for which the massive enthalp ies are provid ed in the t ext. The natural gas lower heating value (LHV) is calculated based on the molar fraction of each hydrocarbon specie s pre sent in the NG stream . Note that the LHV of a fuel is equ al to the combustion entha lpy ΔH°C with opposite sig n. ������������������ �� = ∑ ������������∙������������������ ������ �� ������ = −(�������� 4∙∆�������,�� 4 0 + �������2�6∙∆�������,�2�6 0 + �������3�8∙∆�������,�3�8 0 ) The therm al dispersion �̇���� amo unt to 3% o f the LHV of the NG. Sin ce we are con sidering the dispe rsion a s heat produc ed leaving the syst em and not “enter ing ” in the water streams , we will consi de r it with a negative sign according to our sign conve ntion. Prof. Gianpiero Groppi – Exercises – Fundamentals of Chemical Processes – A.A. 202 3-202 4 4 �̇���� = −3% ∙������������������ �� Note that sin ce the ΔH°C is provided at the same temperature of t he inlet NG stream , the enthalpic contribution of this stream to the energy balance is null. The enthalpic contribution at the combustion gas inlet will take into account the n on ly the species entering with the AIR flow . Th en, the enthalpic bala nce can be solved with respect to the mass of water fed into the steam pro duction cycle , resu ltin g in : �̇�2�= 33 .46 ������� ������ �̇�2� �̇�� 4 ������� = 33 .46 100 ∙1000 22 .414 ������� �� 3= 14 .93 ������� �� 3 The efficien cy of the combustion s ystem is evaluated as: ������= �̇�2�∙(ℎ̂�2��������� − ℎ̂�2��������� ) �̇�� 4 ������� ∙������������������ �� = 91 .6%