Thermodynamics Cycle to Calculate Thermodynamics Efficiency - 1100 Words

Thermodynamics Cycles to Calculate the Thermodynamics Efficiency (Coursework Sample) Content: Name:Tutor:Course:Date:AbstractThis paper focuses on the thermodynamics cycles to calculate the thermodynamics efficiency. The paper will focus on the three questions where the first question comprises of the carnot cycle while the other two questions comprise of the rankine cycles. The calculations are made based on the cycle types and the information given to establish the efficiency and the work done of the cycles.Question oneEfficiency of a Carnot cycleT413 2ABS * Efficiencyn = 1 - Q1Q2 where; Q1=heat supplied,Q2=heat rejectedbut Q2=T2 (SB-SA)andQ1=T1 (SB-SA)therefore; n = 1 - T2 (SB-SA)T1 (SB-SA)n = 1 - T2T1 = 1 - 300900 = 0.667n = 66.7 % * The fixed lower temperature T1 and lowering the higher temperature T4.The higher the temperature, the higher the quality of the energy and the more the energy and the work done.n (TH) = 1 - TiTHTEfficiency200400600Temperature 800S * Fixed TH at 900 K and increasing the lower temperature TL.The lower the temperature, the lower the quality of the energy required to do work. When the temperature TL is lower, the efficiency is high since TL and the efficiency are inversely proportional as shown in the graph below.Efficiency 0.80.60.40.2200 400 600 80 TemperatureQuestion 2Efficiency of a Rankine cycle600/ 15 30 MPa4 MPa350/ 64310 kPa7n = h1-h2+(h6-h7 )h1-h3+(h6-h2)Q1 = heat of expansion feed pump work (can be neglected)W = W 1-2 + W 6-7= h1 h2 + h6 h7From the tablesh1 = h at 6000C and 30MPa= 3445 kJ / kgh2 = hg at 4 MPa= 2801 kJ / kgh6 = h at 3500C= 200 kPaBy interpolation h6= 3072 + (3277-3072)(400-300) (350 - 300)h6= 3174.5 kJ/ kgh3 = hfat 10 kPa= 1920 kJ / kgh7 = hg at 10 kPa= 2584 kJ / kgn = (h1-h2)+(h6-h7)h1-h3+ (h6-h2)n = (3445-2801)+(3174.5-2584)(3445-192)+(3174.5-2801)= 1234.53626.5= 0.3404= 34.04 %Question 3Efficiency of a Rankine cycle52MPa14310 kPa2n = Net work doneheat supplied to the boilerW = work of turbine work of the feed pumpn = h1-h2- (h4-h3)(h 1-h4)= (h7-h2)-(h4-h3)(h1-h3)-(h4-h3)h1 = hg at 2 MPa = 2799 kJ / kgh2 = ?S1 = S2 = Sgat 2 MPaFind XSgat 2 MPa = 6.340 kJ / kgKS2 = 6.340 kJ / kgKS2 = Sf + x Sfgat 0.1 bar= 0.649 + 7.5x = 6.340X = 0.7588h2= hf+ 0.7588 (hfg) at 0.1 bars= 192 + 0.7588 x 2392= 2007 kJ/ kgh3 = hfat 0.1 bars= 192 kJ / kgh4 = h5 = hfat 20 barsh4 = hfat 20 bars= 909 kJ / kgFor the feed pumpWork done from 3 4= Vf34DP= Vf(P4 P3)Where Vf is the specific volume at 0.1 bars and on page 10 of the steam tables, by GFC Rogers and Y.R 0.1000 x 10-2= 0.00100 M3 / kgW3-4 = 0.001 (20 0.1) x 105= 1990 J = 1.99 kJTherefore;Efficiency (n) = 2799-2007- (1.990)(2799-192)= 792-1.9902.607-1.990