Thermodynamics and Fluids II - Essay Example

Question According to Rao (2004), combustion between fuel and air result in carbon dioxide, water and nitrogen) .In solving this problem, two assumptions is made. The assumptions are:Each mole of oxygen in the combustion is accompanied by 3.76 moles of nitrogen which is inert and does not undergo chemical reaction.Fuel mixture is modeled as an ideal gas.The chemical equation for the combustion readsCH18+x (O2+3.76N2) CO2+H2O+ y3.76N2Balancing the chemical equation, it gives2CH18+11 (O2+3.76N2) 2CO2+ 18H2O+41.

36N2From the balanced chemical equation above, it can be deduced that 11kmol of air requires 2kmol octane for the combustion.Mole fraction of fuel = Mfuel/MTotalMole fraction of fuel = 2kmol/13kmol =0.1538Mole fraction for air (O2) = 11/13 =0.8462Mass fraction for octane =2(30)/[2(30)+11(32)] =0.1456Mass fraction for air (oxygen) =11(32)/[11(32)+2(30)] =0.8544.Air –fuel ratio is = (11*4.76)/2 =26.18kmol (air)/ kmol (fuel).

Question 2 According to Chih, Rankin cycle is thermodynamic cycle which converts heat from steam into work. It does not involve any internal irreversibility and consist of the following four processes.1-2 isentropic compression in a pump2-3 Constant heat addition in the boiler.3-4 isentropic expansion in a turbine.4-1 P=Constant heat rejection in the condenser.It is assumed that the components of the engine are joined by conduits for easy transportation of working fluid fro exit of one component to the entrance of the next component.

It is also assumed that all mass flow and energy are steady so that the steady state configuration equation is applicable. At state 1. Water as saturated liquid is compressed isentropicaly to the operation pressure of the boiler. The temperature increases slightly during isentropic compression. At state 2, water enters as compressed liquid and leave as superheated at state 3.The boiler together with section the section where the steam is superheated is called steam generator. Superheated vapor at state 3 enters the turbine where it expands isentropically and produces work by rotating the shaft connected to the steam generator.

The temperature and pressure drop at this process (state 4).Steam is condensed at constant pressure in the condenser which is basically large heat exchanger, by rejecting heat to a cooling medium such as lake. Steam as saturated liquid leaves the condenser and enters the pump. This completes the cycle. Rankin cycle can be analyzed as steady flow process. It is assumed that potential energy and kinetic energy changes are small and can be neglected.q-w=h2-h1In this case the assumption is that the boiler and condenser do not involve any work and the turbine and pump is considered isentropic.

W (pump in)=h2-h1 =v(p2-p1)H=hfap1 and w1=v1In the boiler (w=0) = qin= h3-h2In the turbine(q=0)=wturbine=h3-h4 qout=h4-h1η=wnet/qin1-qout/qin. At p1 =12.5Mpa and T1=350degrees,s1=5.7130Kj/kg.k and h1=2826.6kj/kg.At state 2 , P2=1.0Mpa and T2=350degrees s2=7.3029 and h2=3515.2The exit enthalpy of steam for the isentropic process h2s is determined from the requirement that the entropy of the steam remain constant (s2s=1)X2=(7.3029-5.7130)/7.3029 = 0.2177 340.49+ 0.2177(2826.6)*(kj/kg) =955.8408.kj/kg η = ( h1-h24)/(h1-hs) = (3515.2-2826.2)/(3515.2-955.8408) =689/2559.3592 =0.26921*100 =26.

92%References Chih Wu (2006). Thermodynamics and heat powered cycles : a cognitive engineering approach. New York: Nova Science Publishers. 511. Y. V. C. Rao (2004). An Introduction to Thermodynamics. N/A: Sangam Books Ltd. 345.