A 2-liter displacement four-cylinder engine, operating at 2000 rev/min and 30 pe

Question

A 2-liter displacement four-cylinder engine, operating at 2000 rev/min and 30 percent of its maximum power at that speed, has the following exhaust composition (in present by volume or mole percent): CO_2, 11%; H_2 O, 11.5%; CO, 0.5%, H_2, 0%; O_2, 2%; unburned hydrocarbons (expressed as CH_2, i.e., with a molecular weight of 14), 0.5%; N_2, 74.5% The fuel is (CH_2) with a beating value of 44 MJ/kg. The atomic weights of the elements are C = 12, H = 1. O = 16, N = 14. The heating values of CO and HC are 10 and 44 MJ/kg, respectively. Is the engine a diesel or spark-ignition engine? Is there enough oxygen in the exhaust to burn the fuel completely? Briefly explain. Calculate the fraction of the input fuel energy (m_f, Q_uv) which exits the engine unburned as (1) CO and (2) unburned HC. An inventor claims a combustion efficiency of 100 percent can be achieved, what percentage improvement in engine specific fuel consumption would result?

Explanation / Answer

(a) The comparatively higher speed(2000 rpm) and high CO and unburnt HC at exhaust indicate that the engine is SI type. Since 2% O2 is available at exhaust there is enough O2 to burn 0.5% CO and 0.5% unburnt HC. However, due to availability of limited time for combustion the combustion is not completed. The combustion efficiency is normally improved by using excess air which is manifested by unreacted O2 in the exhaust.

(b) We write the combustion equation as follows:

k(CH2)n+m(O2+3.76N2) = 0.11 CO2 + 0.115 H2O+0.005CO+0.02O2+0.005CH2+0.745N2

Balancing the two sides for C

nk=0.11+0.005+0.005

nk=0.12

Hence equivalent of 0.12 kg-moles CH2 is used per 1 kg-mol of exhaust gas

Hence 0.12x14 kg = 1.68 kgs of fuel used per kg-mol exhaust gs

1.68x44 = 73.92 MJ energy in input by fuel per kg-mol exhaust gas

(1). Fraction lost as CO =(0.005x28x10)/73.92

=0.019

(2) Fraction lost as CH2 = (0.005x14x44)/73.92

=0.042

c) With no CO and CH2 at exhaust the fraction reduction in loss = 0.019+0.042 = 0.061

Keeping all other losses same the output will improve by 0.061 (6.1%)

Hence percentage improvement in engine specific fuel consumptio = 6.1 %

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