Thermodynamics Cycles to Calculate the Thermodynamics Efficiency (Coursework Sample)

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Abstract
This 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 one
Efficiency of a Carnot cycle
T
41
3 2
ABS
* Efficiency
n = 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 = 0.667
n = 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 - TiTH
T
Efficiency
200400600Temperature 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 0.8
0.6
0.4
0.2
200 400 600 80 Temperature
Question 2
Efficiency of a Rankine cycle
600/ 1
5 30 MPa
4 MPa350/ 6
4
310 kPa7
n = 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 – h7
From the tables
h1 = h at 6000C and 30MPa
= 3445 kJ / kg
h2 = hg at 4 MPa
= 2801 kJ / kg
h6 = h at 3500C
= 200 kPa
By interpolation h6= 3072 + (3277-3072)(400-300) × (350 - 300)
h6= 3174.5 kJ/ kg
h3 = hfat 10 kPa
= 1920 kJ / kg
h7 = hg at 10 kPa
= 2584 kJ / kg
n = (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 3
Efficiency of a Rankine cycle
52MPa1
4
310 kPa2
n = Net work doneheat supplied to the boiler
W = work of turbine – work of the feed pump
n = h1-h2- (h4-h3)(h1-h4)
= (h7-h2)-(h4-h3)(h1-h3)-(h4-h3)
h1 = hg at 2 MPa = 2799 kJ / kg
h2 = ?
S1 = S2 = Sgat 2 MPa
Find X
Sgat 2 MPa = 6.340 kJ / kgK
S2 = 6.340 kJ / kgK
S2 = Sf + x Sfgat 0.1 bar
= 0.649 + 7.5x = 6.340
X = 0.7588
h2= hf+ 0.7588 (hfg) at 0.1 bars
= 192 + 0.7588 x 2392
= 2007 kJ/ kg
h3 = hfat 0.1 bars
= 192 kJ / kg
h4 = h5 = hfat 20 bars
h4 = hfat 20 bars
= 909 kJ / kg
For the feed pump
Work 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 / kg
W3-4 = 0.001 (20 – 0.1) x 105
= 1990 J = 1.99 kJ
Therefore;
Efficiency (n) = 2799-2007- (1.990)(2799-192)
= 792-1.9902.607-1.990