Question

5. 20% A 2.4-L gasoline fuelled internal combustion engine produces a maximum power of 220 kW when the fuel is burned stoichiometrically with air. The thermal efficiency of the engine at this maximum power is 43%. Methanol (CH3OH) is to be used as an alternative biofuel for this engine. The stoichiometric air-fuel ratio for gasoline is 14.6. Assume a combustion efficiency of 98%. The lower heating values of gasoline and methanol are 44 MJ/kg and 20 MJ/kg, respectively. Consider the molecular weight of the following substances are; C:12, 0:16, H:1, N:14 kg/kmol Determine: The maximum power from this engine if methanol is used under stoichiometric combustion with the same thermal efficiency. The rates of gasoline and methanol consumption for the maximum power operation. The heat produced from the complete combustion of methanol at 25°C and 1 atm. (1) Useful Relations A/F". W ine LHoline Archanol Wethanol LHV ethanol AF poline 0-MAMA LHV xn and tales of formation

Answer 1

Given data:

Pout = 220kW

StochiometricA/F, = 14.6

b.p nenlo = 43% = - mr * LHV * nc ---- ------ (1)

Combustion efficiency = $\eta_c= 98\%$

LHV gasoline = LHV, = 44MJ/kg

From equation-1:

$m_f=\frac{b.p}{\eta_{th}*LHV*\eta_c}= \frac{220*10^3}{0.43*44*10^6*0.98}= 0.01186 kg/sec$

We have,

$\frac{\dot W_{g}}{\dot W_m}= \frac{LHV_{g}}{LHV_m}*\frac{A/F_m}{A/F_g}-----------(2)$

Let us calculate $A/F_m$

Complete combustion of methanol:

CH3OH + O2 = CO2 + H2O

Balancing the equation at reactant and product side

$2CH_3OH+3O_2=2CO_2+4H_2O$

molecular weights are: C:12, H:1, O:16, N:14

From the reactant side

of methanol/mol

of oxygen/mol

So the ogygen- methanol ratio =$\frac{96}{44}= 2.1818$

Air contains 23.2% of oxygen by mass.

SO mass of air required is:

$\frac{2.1818}{0.232}= 9.404 kg$ of air per kg of methanol

Hence, the

Substituting in equation-2, We get

$\frac{\dot W_{g}}{\dot W_m}= \frac{44*10^6}{20*10^6}*\frac{9.404}{14.6} = 1.4171$

${\dot W_m} = \frac{\dot W_g}{1.4171}=\frac{0.01186}{1.4171}= 8.34*10^{-3}kg/sec$

Maximum power by methanol with same thermal efficiency

$P_{out}|_m=\eta_{th}*LHV_m*\eta_e * \dot W_m\newline P_{out}|_m= 0.43*20*10^6*0.98*8.3692*10^{-3} = 70.5356KW$

b) Rates for maximum power output are:

$\dot W_g= 0.01186kg/sec$

$\dot W_m= 8.3692*10^{-3}kg/sec$

c)The heat produced in the combustion (Q):

The balance equation for complete combustion is :

$2CH_3OH+3O_2=2CO_2+4H_2O$

$Q=\sum_{prod}N_ih_i - \sum_{react}N_ih_i$

From the given Table:

Reactants:

$N_{methanol}= 2 \newline h_{methanol}= -238.86MJ/kmol= kJ/mol$

$N_{oxygen}= 3 \newline h_{oxygen}= 0 kJ/mol$

$\sum_{react}N_ih_i= N_{methanol}*h_{methanol}+ N_{ogygen}*h_{oxygen} \newline \sum_{react}N_ih_i= 2*-238.86 + 3* 0 =-477.72 kJ/mol$

$N_{CO_2}= 2 \newline h_{CO_2}= -393.52 kJ/mol$

$N_{H_2O}= 4 \newline h_{H_2O}= -285.84kJ/mol$

$\sum_{prod}N_ih_i= N_{CO_2}*h_{CO_2}+ N_{H_2O}*h_{H_2O} \newline \sum_{prod}N_ih_i= 2*-393.52 + 4* -285.84 =-1930.4 kJ/mol$

We have,

$Q=\sum_{prod}N_ih_i - \sum_{react}N_ih_i$