LECTURE NOTES
ON
APPLIED THERMODYNAMICS
BY
Dr. T.R.Seethram
CHAPTER 3
GAS POWER CYCLES
3.1. Introduction:- Two important applications of thermodynamics are power generation
and refrigeration. Both are usually accomplished by systems that operate on
thermodynamic cycles. Hence thermodynamic cycles are usually divided into two general categories, viz., “power cycles” and “refrigeration cycles”. Power or refrigeration cycles are further classified as “gas cycles” and “vapour cycles”. In the case of gas cycles, the working substance will be in gaseous phase throughout the cycle, where as in vapour cycles, the working substance will be in liquid phase in one part of the cyclic process and will be in vapour phase in some other part of the cycle.
Thermodynamic cycles are also classified as “closed cycles” and “open cycles”. In closed cycles, the working fluid is returned to its original state at the end of each cycle of operation and is recirculated. In an open cycle, the working substance is renewed at the end of each cycle instead of being re-circulated. In
automobile engines, the combustion gases are exhausted and replaced by fresh air-fuel mixture at the end of each cycle. Though the engine operates in a mechanical cycle, the working substance does not go through a complete thermodynamic cycle.
3.2. Basic Considerations in the Analysis of Power Cycles:- The cycles encountered in
actual devices are difficult to analyse because of the presence of friction, and the absence of sufficient time for establishment of equilibrium conditions during the cycle. In order to make an analytical study of a cycle feasible, we have to make some idealizations by neglecting internal irreversibilities and complexities. Such cycles resemble the actual cycles closely but are made up totally of internal reversible processes. These cycles are called ideal cycles.
3.3. Carnot Power Cycle:- The T-s and p-v diagrams for a Carnot power cycle are
shown in Fig.3.1. The cycle consists of two reversible adiabatic and two reversible isothermal processes, The working of the cycle is as follows:
Process 1-2:Reversible isothermal heating of the working substance from state1to state 2. Process 2-3 :- Isentropic expansion of the working substance from state 2 to state 3. During this process work is done by the working substance on the surroundings.
Process 3-4:- Reversible isothermal cooling of the working substance from state 3 to state 4.
Process 4-1:- Isentropic compression of the working substance so that it comes back to its initial state. During this process work is done on the working substance by the
Expression for Thermal Efficiency of the Cycle
Net work output from the cycle = Wn = ∫ dW .
1-2-3-4-1
2 3 4 1
By first law for a cyclic process we have ∫dW = ∫ dQ = ∫dQ + ∫dQ + ∫dQ + ∫dQ 1-2-3-4-1 1-2-3-4-1 1 2 3 4
2 4
Or Wn = ∫TdS + 0 + ∫TdS + 0 = Th [S2 – S1] + Tc [S4 – S3] 1 3
Since S4 = S1 and S3 = S2, we have
Wn = (S2 – S1)[Th – Tc] ………(3.1)
Assuming that the working substance behaves as a perfect gas and since process 1 – 2 is isothermal we have
S2 – S1 = m R ln(p1 / p2).
Substituting this expression in Eq. (3.1) we have
Wn = m R ln(p1 / p2) [Th – Tc] ……….(3.2) s T 2 3 1 4 v p 3 2 1 4
Fig. 3.1: T-s and p-v diagrams for a Carnot power cycle
Th
2
External heat supplied per cycle = Qs = Q1 – 2 = ∫TdS = Th[S2 – S1]
1
(S2 – S1)[Th – Tc]
Thermal Efficiency = ηCarnot = Wn / Qs = ---
Th[S2 – S1]
Or ηCarnot = [Th – Tc] / Tc = 1 – Th / Tc ...(3.3)
Carnot cycle can be executed in a closed system (a piston and cylinder device or in a steady flow device. It can be seen from Eq. (3.3) that the thermal efficiency depends only on two temperatures Th and Tc and is independent of working substance. The Carnot
cycle is the most efficient cycle that can be executed between a heat source at
temperature Th and a heat sink at temperature Tc. But reversible isothermal heat transfer
process is difficult to achieve in practice, because, it would require very large heat
exchangers and it would take a very long time (a power cycle in a typical engine has to be completed in a fraction of a second). Therefore it is not practical to build an engine that would operate on a cycle that closely approximates a Carnot cycle.
The real value of the Carnot cycle comes from the fact that it is used as a standard against which the actual or other ideal power cycles are compared. It can be seen from Eq. (3.3) that the thermal efficiency of the Carnot power cycle increases with
increase in Th and with decrease in Tc. Hence in actual or other ideal cycles attempts are
made in increasing the average temperature at which heat is supplied or by decreasing the average temperature at which heat is rejected. It should also be noted that the source and sink temperatures that can be used in practice have their limitations. The highest
temperature in the cycle is limited by the maximum temperature the components of the engine can withstand and the lowest temperature is limited by the temperature of the cooling medium used in the cycle such as the atmospheric air, ocean, lake or a river.
3.4. Illustrative examples on Carnot cycle
Example 3.1:- A Carnot cycle using air as the working substance works between
temperature limits of 900 K and 300 K. The pressure limits are 60 bar and 1 bar. Determine (i) pressure at salient points of the cycle, (ii) the heat supplied per unit mass of air, (iii) net work output per unit mass of air, (iv)mean effective pressure and (v) thermal efficiency of the cycle
Given:- Refer to T – s diagram shown in Fig. E3.1. Tmax = T1 = T2 = 900 K ;
Tmin = T3 = T4 = 300 K ; pmax = p1 = 60 bar ; pmin = p3 = 1 bar.
Hence Cv = Cp / γ = 1.005 / 1.4 = 0.718 kJ/kg-K ; R = 1.005 – 0.718 = 0.287.kJ/(kg-K)
To find:- (i) p2 ; p4 ; (ii) q 1-2 ; (iii) wn (iv) mep;(v) η Carnot.
Solution:-
(i) Process 2 – 3 is isentropic. For a perfect gas undergoing isentropic process we have p2 / p3 = (T2 / T3)γ / (γ – 1)
Or p2 = p3 (T2 / T3)γ / (γ – 1)
= 1 x [ 900 / 300 ] 1.4 / 0.4 = 46.77 bar. Similarly for process 4 – 1 we have
p4 = p1 (T4 / T1)γ / (γ – 1) = 60 x [ 300 / 900 ] 1.4 / 0.4
= 1.283 bar
(ii) Applying first law for process 1 – 2 we have q 1-2 − w1-2 = (u2 – u1) = C (T – T ) = 0 because T = T s T T1 T3 q 1-2 q 3-4
2
Hence q1-2 = w1-2 = ∫pdv = p1v1 ln(p1/p2), as process 1-2 is isothermal.
1
For a perfect gas p1v1 = RT1. Hence
q1-2 = w1-2 = RT1 ln(p1/p2) = 0.287 x 900 x ln (60 / 46.77)
= 64.34 kJ/kg.
(iii) For process 3 -4 we have q3-4 = w3-4 = RT3 ln (p3 / p4)
= 0.287 x 300 x ln (1 / 1.283) = − 21.46 kJ/kg.
( negative sign indicates that during this process heat is rejected by air to the surroundings).
Net work output = wn = ∑w = ∑q = q1-2 + q3-4 = 64.34 – 21.46
= 42.88 kJ/kg (iv) mean effective pressure is given by
Net work output wn
mep = --- = --- Stroke volume of the piston (v2 – v1)
Now (v2 – v1) = [(RT2 / p2) – (RT1/p1)] = RT1 [ 1 / p2 – 1/p1]
= 287 x 900 x [ (1/ 46.77) – (1 / 60)] x (1 / 10 5) = 0.0122 m3/kg.
42.88 x 1000
Hence mep = --- = 35.15 x 10 5 N / m2 = 35.15 bar 0.0122
Example 3.2:- The maximum pressure and temperature in a Carnot gas power cycle are
limited to 20 bar and 400 C. The volumetric ratio of isentropic compression is 6 and volumetric ratio of isothermal expansion is 1.5. Assuming that air is the working substance and the volume of air at the beginning of isothermal expansion is 0.1 m3, determine (i) the minimum temperature in the cycle, (ii) change in entropy during
isothermal expansion process, (iii) thermal efficiency of the cycle, (iv) power output from the cycle if there are 200 cycles per minute and (v) mean effective pressure.
Given:- p1 = 20 bar ; T1 = 400 + 273 = 673 K = T2 ; V4 / V1 = 6 ; V2 / V1 = 1.5 ;
V1 = 0.1 m3; For air Cp = 1.005 kJ/(kg-K); γ = 1.4 ; R = 0.287 kJ/(kg-K); Cv = 0.718
kJ/(kg-K).
.
To find :- (i) T3 ; (ii) S2 – S1 ; (iii) η Carnot ; (iv) Wn ; (v) MEP.
(i) Since process 1 – 2 is isothermal and air is assumed to behave as a perfect gas, it follows that
p1V1 = p2V2 .Hence p2 = (V1 / V2) p1 = (1/1.5) x 20 = 13.33 bar.
Process 4-1 is isentropic. Hence T4= (V1 / V4) (γ – 1) T1 = (1/6) 0.4 x 673 = 328.7 K.
Also T3 = T4 = 328.7 K.
(ii) mass of air = m = (p1V1) / (RT1)
20 x 10 5 x 0.1
= --- = 1.035 kg 287 x 673
Change in entropy for process 1-2 for a perfect gas is given by S2 – S1 = m [ Cv ln (T2 / T1) + R ln (V2 / V1) ] = 1.035 x [ 0 + 0.287 x ln (1.5) ] = 0.1204 kJ/kg-K. (iii) η = (T – T ) / T = [673 – 328.7] / 673 = 0.5116 = 51.16 %. s T T1 T3 p V 1 2 3 4 1 2 4 3 Isothermal Process Isentropic process
(iv) Heat supplied per cycle = Qs = Q1-2 = W1-2 = mRT1 ln(V2 / V1)
= 1.035 x 0.287 x 673 x ln (1.5) = 81.06 kJ/cycle.
Net work output per cycle = Wn = η Carnot Qs = 0.5116 x 81.06
= 41.47 kJ/cycle.
Power output = P = Wc Nc, where Nc = Number of cycles executed per unit time.
Hence P = 41.47 x 200 /60 = 138 kJ/s (kW). Net work output per cycle in J/cycle Wn
(v) MEP = --- = --- Swept volume per cycle in m3 / cycle (V3 – V1)
Now for process 2-3 we have
V3 = V2 (T2 / T3) 1 / (γ – 1) = 1.5 x 0.1 x (673 / 328.7) 2.5
= 0.9 m3. 41.47 x 1000
Hence MEP = --- = 51837.5 N / m2 (0.9 – 0.1)
Example 3.3:- In an air-standard Carnot cycle, 110 kJ/kg of heat is transferred to the
working fluid at 1110 K. Heat is rejected at 273 k. The minimum pressure in the cycle is 1 bar. Find (i) thermal efficiency, (ii) mean effective pressure.
Solution: The T – s diagram for the cycle is shown in Fig. E3.3.
Given :- T1 = T2 = 1110 K ; T3 = T4 = 273 K; q 1-2 = 110 kJ/kg ; p3 = 1 bar.
To find:- (i) η Carnot ; (ii) MEP
(T1 – T3) (1110 – 273) (i)η Carnot = --- = --- = 0.754 = 75.4 %. T1 1110 (ii) MEP = wn / (v3 – v1)
wn = η Carnot q1-2 = 0.754 x 110 = 82.94 kJ/kg
Applying I law for process 1 – 2 we have q 1-2 = w1-2 = RT1 ln (v2 / v1)
Hence v2 / v1 = exp [q1-2 / RT1] = exp [ 110 / (0.287 x 1110)]
= 1.4124.
Also v3 = RT3 / p3 = 287 x 273 / (1 x 10 5)
= 0.7835 m3 / kg.
Process 2-3 is isentropic. Hence T2 v2 (γ – 1) = T3 v3 (γ – 1)
Or v2 = (T3 / T2) 1 / (γ – 1) v3 = (273 / 1110) 2.5 x 0.7835 = 0.0235 m3 / kg Therefore v1 = 0.0235 / 1.4124 = 0.0166 m3 / kg. 82.94 x 1000 Therefore MEP = --- (0.7835 – 0.0166) = 1.082 x 10 5 N/m2 = 1.082 bar.
3.5 Air Standard Cycles: In gas power cycles, the working fluid will be in gaseous
phase throughout the cycle. Petrol engines (gasoline engines), diesel engines and gas turbines are familiar examples of devices that operate on gas cycles. All these devices are called “Internal combustion engines” as the fuel is burnt within the boundaries of the system. Because of the combustion of the fuel, the composition of the working fluid changes from a mixture of air and fuel to products of combustion during the course of the cycle. However, considering that air is predominantly nitrogen which hardly undergoes any chemical reaction during combustion, the working fluid closely resembles air at all times.
The actual gas power cycles are complex. Hence actual gas cycles are
approximated by ideal cycles by making the following assumptions called “air standard
assumptions”.
Air standard assumptions:- (i) The working fluid is air which continuously circulates in
a closed loop.
(iii) All the processes that make up the cycle are internally reversible.
(iv) The combustion process is replaced by a heat addition process from an external source.
(v) The exhaust process is replaced by a heat rejection process that restores the working substance to its original state.
(vi) Changes in kinetic and potential energies of the working substance is very small and hence negligible.
A cycle which is analyzed making use of these assumptions is called an “air standard
cycle”. The air standard assumptions make the thermodynamic analysis very simple without significantly deviating from the actual cycle. This simplified model will help to study qualitatively the influence of major parameters on the performance of the cycle.
3.5. Air standard Otto cycle: Otto cycle is the ideal cycle for spark ignition engines.
The cycle is named after Nikolaus A Otto, a German who built a four – stroke engine in 1876 in Germany using the cycle proposed by Frenchman Beau de Rochas in 1862.The p – V and T – s diagrams for an Otto cycle are shown in Fig. 3.2.The cycle consists of the
following processes.
Process 1 – 2: Isentropic compression of air from state 1 to state 2. During this process work is done on air by the surroundings.
Process 2 -3 : Constant volume of heating of air from state 2 till the maximum permissible temperature is reached.
V
p
s T
Fig. 3.2: p-V and T-s diagrams for Otto cycle.
3 2 4 1 Isentropic Process 2 1 3 4 Constant volume process
Process 3 – 4: Isentropic expansion of air from state 3 to state 4. During this process work is done by air on the surroundings.
Process 4 – 1: Constant volume cooling of air till the air comes back to its original state.
3.5.1. Expressions for Net work output and thermal efficiency
Net work output per unit mass of air = wn = w1-2 + w2-3 + w3-4 + w4-1
By I law for the cycle we have w1-2 + w2-3 + w3-4 + w4-1 = q1-2 + q 2-3 + q 3-4 + q 4-1.
Also q 1-2 = q 3-4 = 0 as processes 1-2 and 3-4 are isentropic.
Hence wn = q 2-3 + q 4-1. ………..(3.4)
Since both the processes 2-3 and 3-4 are at constant volume, applying I law for these two processes we have
q2-3 = (u3 – u2) = Cv[T3 – T2] ………(3.5)
and q4-1 = (u1 – u4) = Cv[T1 – T4] ………(3.6)
It should be noted that q4-1 will be negative (T1 < T4) as heat is transferred from the
working substance to the surroundings.
Hence the net work output in terms of temperatures at the four salient points of the cycle is given by
wn = Cv[T3 – T2] + Cv[T1 – T4] ……….(3.7)
Thermal efficiency is given by
Net work output wn
η Otto = --- = --- Heat Supplied q2-3 Cv[T3 – T2] + Cv[T1 – T4] [T4 – T1] η Otto = --- --- = 1 − --- Cv[T3 – T2] [T3 – T2] ………(3.8) Eq. (3.8) gives the expression for thermal efficiency of the Otto cycle in terms of the temperatures at the salient points of the cycles. It is possible to express the net work output and thermal efficiency of the Otto cycle in terms two parameters namely (i) the
comprssion ratio, Rc and the maximum cycle temperature ratio, t. The compression ratio
is defined as the ratio of volume of air befor compression to the volume of air after compression; i.e., Rc = V1 / V2 and the maximum cycle temperature ratio is defined as the ratio of the maximum temperature in the cycle to the minimum temperature in the cycle; i.e., t = T3 / T1.
Now Process 1 – 2 is isentropic. Hence T1V1(γ – 1) = T2V2(γ – 1)
Or T2 / T1 = (V1/V2) (γ – 1) = Rc(γ – 1) ……….(3.9)
Similarly we have for ptocess 3-4,
T3 / T4 = (V4/V3) (γ – 1) = (V1/V2) (γ – 1) = Rc(γ – 1) ………... (3.10)
Therefore T2 T3 [T3 – T2]
---- = --- = --- = Rc(γ – 1)
T1 T4 [ T4 – T1]
Suibstituting this in Eq.(3.8) we have
1
ηOtto = 1 − --- ………(3.11)
Rc(γ – 1)
It can be seen from Eq. (3.11) that the thermal efficiency of the Otto cycle depends only on the compression ratio Rc. The efficiency increases with increase in Rc. The increase is
steep at low values of Rc , but becomes flatter as Rc exceeds 8 as shown in Fig. 3.3.
2 4 6 8 10 12 14 16 Rc 0.6 0.4 0.2
Fig.3.3: Effect of Rc on thermal efficiency of Otto cycle
For very high compression ratios (Rc >10), the temperature of the air fuel mixture in an
actual petrol engine will be so high as to cause pre-ignition of the fuel leading to
“knocking” of the engine and hence should not be used. Therefore the compression ratio cannot be increased arbitrarily to have higher efficiency. Between the same temperature limits the thermal efficiency of the Otto cycle is less than that for a Carnot cycle.In order to know how far the Otto cycle deviates from Carnot cycle, a parameter called “relative
efficiency” is defined as the ratio of the thermal efficiency of the Otto cycle to the thermal efficiency of a Carnot cycle working between the same temperature limits. i.e.,
Air standard efficiency of Otto cycle
Relative efficiency of Otto cycle = --- ……..(3.12)
Efficiency of Carnot cycle working between same temperature limits.
3.5.2. Condition for optimum work output from an Otto cycle
Net work output per unit mass of air is given by Eq. (3.7) :
wn = Cv[T3 – T2] + Cv[T1 – T4] ……….(3.7)
= Cv T1 [ T3 / T1 – T2 / T1 – T4 / T1 + 1 ]
T3 / T1 is the ratio of maximum temperature in the cycle to the minimum temperature in
the cycle and is called “maximum cycle temperature ratio” and is denoted by ‘t’.
t
Now T4 / T1 = (T4 / T3) x (T3 / T1) = --- ………(3.13)
Rc(γ – 1)
Hence wn = CvT1 [ t – Rc(γ – 1) – t / Rc(γ – 1) + 1] ………...(3.14)
For given values of ‘t’ and T1, wn depends only on Rc.
Hence for optimum output, dwn / dRc = 0.
i.e., dwn / dR = Cv T1 [−(γ – 1) Rc(γ – 2) − t (1 – γ)Rc− γ ] = 0
or Rc 2(γ – 1) = t
or Rc = t 1 / 2(γ – 1) = Rc*………..(3.15)
If this value of Rc is substituted in Eq. (3.14) we get the expression for maximum work
output as
3.13
Or (wn)Maximum = Cv T1 [ √ t – 1 ] 2 ……….(3.16)
Thermal efficiency corresponding to maximum work output is therefore given by
(ηOtto) * = 1 − 1 / Rc* (γ – 1) = 1 – [1 / √ t ] ………(3.17) 3.5.3. Illustrative examples on Otto cycle
Example 3.4:- An ideal Otto cycle has a compression ratio of 8. The conditions at the
beginning of compression stroke are 100 kPa and 17 C. If the heat added during the cycle is 800 kJ/kg find (i) temperatures and pressures at salient points of the cycle, (ii) net work output per unit mass of air, (iii)thermal efficiency of the cycle, (iv) mean effective pressure, (v) compression ratio corresponding to maximum work output, (vi) maximum work output and (vii) thermal efficiency corresponding to maximum work output.
Solution: The p – V diagram for the cycle is shown below.
(v) Rc* ; (vi) (wn) maximum
(i) Process 1-2 is isentropic. Hence T1V1(γ – 1) = T2V2(γ – 1)
Or T2 = T1 (V1/V2) (γ – 1) = 290 x 8 0.4 = 666 K. V p Vs = V1 – V2 Vc Vs = Stroke Volume = V1 – V2. Vc = ClearanceVolume =V2=V3 Given:- Rc = V1 / V2 = 8. T1 = 17 + 273 = 290 K. p1 = 100 kPa. Heat supplied = q2-3 = 800 kJ/kg. To find:- (i) T2, p2, T3,p3, T4, p4
(ii) wn; (iii) ηOtto; (iv) MEP;
4 1
3
3.14 Also p1V1 / T1 = p2V2 / T2 . Or p2 = p1 (V1/V2) (T2 / T1) = 100 x 8 x (666 / 290) = 1837.24 kPa = 18.3724 bar. Now q2-3 = Cv(T3 – T2) or T3 = T2 + q2-3 / Cv Hence T3 = 666 + 800 / 0.718 = 1780.2 K. Also p3V3 / T3 = p2V2 / T2 and V3 = V2. Hence p3 = p2 (T3 / T2) = 18.3724 x (1780.2 / 666) = 49.11 bar.
Process 3-4 is isentropic. Hence T4 = T3 (V3 / V4) (γ – 1) = T3 (V2 / V1) (γ – 1)
= 1780.2 x (1/8) 0.4 = 775 K.
Also p4V4 / T4 = p1V1 / T1 and V4 = V1.
Hence p4 = p1 (T4 / T1) = 100 x (775 / 290)
= 267.24 kPa.
(ii)Heat rejected per unit mass of air = q4-1 = Cv (T1 – T4) = 0.718 x (290 – 775)
= − 348.23 kJ/kg
(Negative sign for q4-1 indicates that heat is transferred from air to the surroundings
during this process).
Hence Net work output per unit mass of air = wn = 800 – 348.23 = 451.77 kJ/kg.
(iii)Thermal efficiency = η Otto = wn / q2-3 = 451.77 / 800 = 0.565 = 56.5 %.
Check for thermal efficiency: η Otto = 1 – 1 / Rc(γ – 1) = 1 – 1 / 8 0.4
= 0.565 = 56.5 %
v1 = RT1 / p1 = 287 x 290 / (100 x 1000) = 0.8323 m3/kg.
Net work output in J/kg wn 451.77 x 1000
MEP = --- = --- = --- Stroke volume in m3 / kg (v1 – v2) [0.8323 – 0.8323 / 8]
= 6.203 x 105 N/m2 = 6.203 bar.
(v) Maximum cycle temperature ratio = t = Tmaximu / Tminimum = T3 / T1 = 1780.2 / 290
= 6.138.
For maximum work output, Rc = Rc* = t 1/ 2 (γ – 1) = 6.138 1 / (2 x 0.4)
= 9.66
(vi) Net work output corresponding to Rc * is maximum and is given by
(wn)maximum = CvT1 [√ t − 1] 2
= 0.718 x 290 x [√ 6.138 – 1] 2 = 454.54 kJ/kg.
Example 3.5:- An air standard Otto cycle is to be designed according to the following
specifications.
Pressure at the start of the compression process = 101 kPa ; Temperature at the start of compression process = 300 K; Compression ratio = 8;
Maximum pressure in the cycle = 8.0 MPa;
Find (i) the net work output per unit mass of air, (ii) cycle efficiency, and (iii)MEP.
Solution: Refer to p-V diagram of the cycle shown in example 3.4.
Given:- p1 = 101 kPa ; T1 = 300 K; V1 / V2 = Rc = 8 ; p3 = 8 MPa ;
Assume Cv = 0.718 kJ/kg – K ; γ = 1.4.
To find:- (i) wn ; (ii) η Otto ; (iii) MEP
(i)Process 1-2 is isentropic. Hence T2 = T1 (V1 / V2) (γ – 1) = 300 x (8) 0.4
= 689.2 K. Also p2 = p1 (V1/ V2) γ = 101 x 8 1.4 = 1856.3 kPa.
For process 2-3 we have p2V2 / T2 = p3V3 / T3 and V2 = V3.
Hence T3 = T2 (p3 / p2) = 689.2 x (8 x 103 / 1856.3)
= 2970 K.
Process 3 -4 is isentropic. Hence T4 = T3 (V3 / V4) (γ – 1) = T3 (V2 / V1) (γ – 1)
= 2970 x (1 / 8) 0.4 = 1292.9 K.
Heat supplied per unit mass of air = q 2-3 = Cv(T3 – T2) = 0.718 x (2970 – 689.2)
= 1637.6 kJ/kg.
Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (300 – 1292.9)
= − 712.9 kJ/kg.
(Negative sign for q 4-1 indicates that heat is transferred from air to the surroundings)
Net work output per unit mass of air = wn = ∑q = 1637.6 – 712.9 = 924.7 kJ/kg.
1 1 (ii) Thermal efficiency = η Otto = 1 − --- = 1 − ---
Rc (γ – 1) 8 0.4
= 0.565 = 56.5 %.
Thermal efficiency can also be calculated from the formula wn 924.7
η Otto = --- = --- = 0.565 = 56.5 %
q2-3 1637.6
(ii)Specific volume at state 1 = v1 = RT1 / p1 = 287 x 300 / 101 x 10 3
= 0.8525 m3 / kg. Hence v2 = v1 / 8 = 0.8525 / 8 = 0.10656 m3/kg. wn 924.7 x 1000 MEP = --- = --- = 12.396 x 10 5 N/m2 = 12.396 bar. (v1 – v2) (0.8525 – 0.10656)
Example 3.6:- From the p – V diagram of an engine working on Otto cycle, it is found
that the pressure in the cylinder after 1/8th of the compression stroke is completed is 1.4 bar. After 5/8th of the compression stroke is completed, the pressure is found to be3.5 bar. The maximum cycle temperature is limited to 1000 C.If the compression process is
according to the law pV1.35= constant, find (i) the compression ratio, (ii) work output per unit mass of air, and (iii) thermal efficiency. Assume the minimum temperature in the cycle to be 27 C.
Solution: Refer to p – V diagram shown below.
To find:- (i) Rc ; (ii) wn ; (iii) η
(i) V1 – Va = (1/8) (V1 – V2) or Va = V1 – (1/8)(V1 – V2)
= (7/8)V1 + (1/8)V2
Hence Va / V2 = (7/8)Rc + 1/8 ………..(a)
Similarly Vb / V2 = (3/8)Rc + 5/8 ………..(b)
From Eqs. (a) and (b) we have
Va (7/8)Rc + 1/8 7Rc + 1 --- = --- = --- ……….(c) Vb (3/8)Rc + 5/8 3Rc + 5 1 2 3 4 V p Vb Va V1 pVn = Const a b Given:- V1 – Va = (1/8)Vs V1 – Vb = (5/8)Vs. pa = 1.4 bar; pb = 3.5 bar; T3 = 1000 + 273 =1273 K. T1 = 27 + 273 = 300 K Compression index = n = 1.35
For the compression process we have paVan = pbVbn
Or Va / Vb = (pb/pa) 1/n = (3.5 / 1.4) 1/1..35
= 1.971 Substituting this value in Eq. (c) we have 7Rc + 1 --- = 1.971 3Rc + 5 1.971 x 5 – 1 Or Rc = --- = 8.146. (7 – 1.971 x 3)
(ii) For process 1-2 we have T2 = T1 Rc(n – 1) = 300 x 8.146 0.35
= 625 K.
q2-3 = Cv(T3 – T2) = 0.718 x (1273 – 625)
= 465.26 kJ/kg.
For process 3-4 we have T4 = T3 (V3 / V4) (γ – 1) = 1273 x (1 / 8.146) 0.4
= 550.1 K.
q4-1 = Cv(T1 – T4) = 0.718 x (300 – 550.1)
= − 179.6 kJ/kg.
(Negative sign for q 4-1 indicates that during the process 4-1 heat is transferred from air to
the surroundings)
Since process 1-2 is not isentropic q1-2 is not equal to zero and therefore we have to find
q1-2.
Applying first law for process 1-2, we have, per unit mass of air
q1-2 – w1-2 = (u2 – u1) ………(d)
Now w1-2 = (p1v1 – p2v2) / (n – 1) = R(T1 – T2) /(n – 1)
R(T1 – T2) q1-2 = --- + Cv(T2 – T1) = (T2 – T1) [Cv – R /(n– 1)] (n – 1) Hence q1-2= (625 – 300) x [ 0.718 – 0.287 / 0.35] = − 33.15 kJ/kg wn = ∑w = ∑q = q1-2 + q2-3 + q3-4 + q4-1 = − 33.15 + 465.26 + 0 − 179.6 = 252.5 kJ/kg. Wn 252.5 Thermal efficiency = η = --- = --- = 0.5427 = 54.27 %. q2-3 465.26
Example 3.7:- Derive an expression for thermal efficiency in terms of compression ratio
and maximum cycle temperature ratio for a cycle which is similar to Otto cycle except that the compression process is isothermal.Compare the efficiency and work output of this cycle with that of an Otto cycle having the same compression ratio of 6 and same maximum cycle temperature ratio of 5..
Solution: The p – V diagram for the cycle is shown in Fig. E3.7.
Isothemal process Isentropic process 1 4 2 3 V p
Solution:
Net work output per unit mass = wn= w1-2 + w2-3 + w3-4 + w4-1.
But w2-3 = w4-1 = 0 as both the processes 2-3 and 4-1 are constant volume processes.
Since process 1-2 is isothermal ( for a perfect gas this is equivalent to process according to the law pv= constant)
w1-2 = p1v1 ln (v2/v1) = − RT1 ln Rc
where Rc = v1/v2.
(p3 v3 – p4v4) R(T3 – T4)
Since process 3-4 is isentropic, w3-4 = --- = ---
(γ – 1) (γ – 1)
RT1 (T3 / T1 – T4 / T1)
= ---
(γ – 1) Now T3 / T1 = t = maximum cycle temperature ratio,
And T4 / T1 = (T4/T3) x (T3/T1) = (V3/V4)(γ – 1) t = t / Rc(γ – 1) RT1 [ t – t / Rc(γ – 1) ] Hence w3-4 = --- (γ – 1) RT1 [ t – t / Rc(γ – 1) ] Therefore wn = − RT1 ln Rc + --- (γ – 1) ={ RT1 / (γ – 1)}{ [ t – t / Rc(γ – 1) ] − (γ – 1) ln Rc}………..(a) Heat supplied = q2-3 = Cv(T3 – T2) = Cv(T3 – T1) = CvT1(T3/T1 – 1) = CvT1(t – 1) Thermal efficiency = η = wn / q2-3 { RT1 / (γ – 1)}{ [ t – t / Rc(γ – 1) ] − (γ – 1) ln Rc} = --- CvT1(t – 1)
Since Cv = R / (γ – 1), the above expression simplifies to
{t [ 1 – 1 / Rc(γ – 1) ] − (γ – 1) ln Rc}
η = --- ( t – 1)
(ii) Given:- Rc = 6 ; t = 5 ; To find (a) wn / (wn)Otto ; and (b) η / ηOtto
(a)Work output for an Otto cycle in terms of T1, Rc and t is given by Eq.(3.14) as
(wn)Otto = CvT1 [t – Rc(γ – 1) – t / Rc(γ – 1) +1]
= 0.718 T1 x [ 5 – 6 0.4 – 5 / 6 0.4 + 1]
= 1.0844 T1
For the given cycle, from Eq. (a)
wn ={ RT1 / (γ – 1)}{ [ t – t / Rc(γ – 1) ] − (γ – 1) ln Rc} 0.287 x T1 = --- x { [ 5 – 5 / 60.4] – 0.4 x ln 6 } 0.4 = 1.321 T1 Hence (wn)Otto 1.0844 T1 --- = --- = 0.821. wn 1.321 T1
(b) Thermal efficiency of Otto cycle = η otto = 1 – 1 / 6 0.4
= 0.5116 = 51.16 % For the given cycle thermal efficiency is
{t [ 1 – 1 / Rc(γ – 1) ] − (γ – 1) ln Rc} η = --- ( t – 1) { 5 x [ 1 – 1/ 60.4] – 0.4 x ln 6} = --- = 0.4603 = 46.03 %. (5 – 1)
3.6. Diesel Cycle:- The diesel cycle is the ideal cycle for compression ignition engines
(CI engines). CI engine was first proposed by Rudolph Diesel in 1890. The diesel engine works on the principle of compression ignition. In such an engine, only air is compressed and at the end of the compression process, the fuel is sprayed into the engine cylinder containing high pressure air, so that the fuel ignites spontaneously and combustion occurs. Since only air is compressed during the compression stroke, the possibility of auto ignition is completely eliminated in diesel engines. Hence diesel engines can be designed to operate at much higher compression ratios (between 12 and 24). Also another benefit of not having to deal with auto ignition is that fuels used in this engine can be less refined (thus less expensive).
The p – V and T – s diagrams for an air-standard diesel cycle are shown in Fig. 3.4. The diesel cycle is similar to Otto cycle except that the heating process takes place at constant pressure in a diesel cycle. The various processes involved in an ideal diesel cycle are as follows.
Process 1-2: Isentropic compression of air from state 1 to state 2. During this process work is done on air by the surroundings.
Process 2-3: Constant pressure heating of air till the maximum permissible temperature is reached.
Process 3-4: Isentropic expansion of air from state 3 to state 4. during this process work is done by air on the surroundings.
Process 4-1: Constant volume cooling of air so that air comes back to its original state to complete the cycle.
V p 2 3 4 1 Isentropic Process s T 2 1 Constant Pressure Process Constant volume process 3 4
3.6.1.Expressions for net work output and thermal efficiency
Heat supplied per unit mass of air = q2-3 = Cp (T3 – T2).
Heat rejected per unit mass of air = q4-1 = Cv (T1 – T4)
Net work ouput per unit mass of air = wn = ∑w= ∑q
= q1-2 + q2-3 + q3-4 + q4-1.
But q1-2 = q3-4 = 0 as these two processes are isentropic.
Hence w n = q2-3 + q4-1
= Cp (T3 – T2).+ Cv (T1 – T4) ………(3.18)
Therefore thermal efficiency of the diesel cycle is given by
wn Cp (T3 – T2).+ Cv (T1 – T4) η Diesel = --- = --- q2-3 Cp (T3 – T2). (Cv/Cp) (T4 – T1) = 1 − --- ………(3.19) (T3 – T2).
Temperatures T2, T3 and T4 can be expressed in terms of T1, the compression ratio Rc and
the cut off ratio ρ (ρ = V3 / V2) as follows.
Process 1-2 is isentropic. Hence T2 = T1(V1/V2)(γ – 1)
= Rc(γ – 1) T1.
For process 2-3, p2V2 / T2 = p3V3 / T3 and p2 = p3.
Hence T3 = (V3 / V2) T2
= ρ Rc(γ – 1) T1.
Proocess 3-4 is isentropic. Hence T4 = T3 (V3 / V4) (γ – 1).
But V4 / V3 = Expansion ratio = (V4 / V2)(V2 / V3) = (V1 / V2)(V2 / V3)
Therefore T4 = ( ρ / Rc) (γ – 1) T3
= ( ρ / Rc) (γ – 1) ρ Rc(γ – 1) T1
= ρ γ T1.
Substituting the expressions for T2,T3, and T4 in Eq. (3.19)
( ρ γ − 1 ) η Diesel = 1 − --- γ [ρ Rc(γ – 1) − Rc(γ – 1) ] ( ρ γ − 1 ) or η Diesel = 1 − --- ………(3.20) γ Rc(γ – 1) [ρ −1]
Now substituting the expressions for T2,T3, and T4 in terms of T1 in Eq. (3.18) we get
wn = CvT1[γ (ρ Rc(γ – 1) – Rc(γ – 1)) − ρ γ + 1 ]
Or wn = CvT1[ γ Rc(γ – 1) (ρ – 1) − ρ γ + 1] …………..(3.21)
1 Rc
ηDiesel
ρ
It can be seen from Eq. (3.20) that the thermal efficiency of the diesel cycle depends on the compression ratio Rc and the cut-off ratio ρ. The effect of these parameters on thermal
efficiency is shown in Fig. 3.5. It can be seen from this figure that for a given value of the cut-off ratio the thermal efficiency increases with Rc and for a given value of Rc, the
efficiency deceases with increase in the cut-off ratio.The increase in cut-off ratio results in decrease in the expansion ratio which in turn decreases the work done during
expansion process. At the same time increase in cut-off ratio also results in increase in heat supplied. The overall effect is that the efficiency decreases.
3.6.2.Illustrative examples on diesel cycle
Example 3.8:- An air standard diesel cycle has a compression ratio of 14. The air
condition at the beginning of compression is 1 bar and 27 C.The maximum temperature in the cycle is 2500 C. Determine (i) temperature and pressure at salient points of the cycle, (ii) net work output per unit mass of air, (iii) thermal efficiency, (iv) specific air consumption in kg/kWh, and (v) MEP.
Solution:
(i) Process 1-2 is isentropic. Hence T2 = T1 (V1/V2) (γ – 1) = 300 x 14 0.4
= 862.13 K. Also p2 = p1(V1/V2) γ = 1 x 14 1.4 = 40.23 bar 1 2 3 4 q2-3 q4-1 V p Given:- Rc = V1/V2 = 14; p1= 1 bar; T1 = 27 + 273 = 300 K ; T3 = 2500 + 273 = 2773 K ; Assume:γ = 1.4;Cp = 1.005 kJ/kg-K Cv = 0.718 kJ/kg-K To find:- (i) T2,p2, p3, T4, p4 ;
(ii) wn ; (iii) η Diesel ; (iv) SACin
For process 2-3 we have p2V2 / T2 = p3V3 / T3 and p2 = p3.
Hence cut – off ratio = ρ = V3 / V2 = T3 / T2 = 2773 / 862.13
= 3.216.
Expansion ratio = V4 / V3 = Rc / ρ = 14 / 3.216 = 4.353.
Process 3-4 is isentropic. Hence T4 = T3 (V3 / V4) (γ – 1)
= 2773 x (1 / 4.353) 0.4 = 1539.7 K.
Also p3V3 γ = p4V4 γ or p4 = p3 (V3 / V4) γ
Hence p4 = 40.23 x (1/ 4.353) 1.4
= 5.131 bar.
(ii) Heat supplied per unit mass of air = q2-3 = Cp(T3 – T2)
= 1.005 x (2773 – 862.13) = 1920.4 kJ/kg
Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (300 – 1539.7)
= − 890.11 kJ/kg
(Negative sign for q4-1 indicates that heat is transferred from air to the surroundings
during this process).
Net work output per unit mass of air = wn = 1920.4 – 890.11
= 1030 kJ/kg wn 1030
(iii) Thermal efficiency = η Diesel = --- = ---
q2-3 1920.11
= 0.5364 = 53.64 %. .
.
Air consumption in kg/h m x 3600 Then Specific air consumption in kg/kWh = --- = --.---
Net work output in kW m x wn
3600 3600
= --- = --- = 3.495 kg/kWh. wn 1030
(v) Specific volume at the beginning of compression = v1 = RT1 / p1
287 x 300 Hence v1 = --- = 0.861 m 3/kg 1 x 10 5 Therefore v2 = v1 / Rc = 0.861 / 14 = 0.0615 m3 / kg. Wn 1030 x 1000
Mean effective pressure = MEP = --- = --- (v1 – v2) ( 0.861 – 0.0615)
= 12.883 x 10 5 N / m2 = 12.883 bar
Example 3.9:- A diesel cycle has a compression ratio of 16. The temperature before
compression is 300 K and after expansion it is 900 K. Determine (i) Net work input per unit mass of air, (ii) the air standard efficiency and (iii) MEP if the minimum pressure in the cycle is 1 bar.
Given:- Rc = V1 / V2 = 16 ; T1 = 300 K ; T4 = 900 K ;
To find :- (i) wn ; (ii) η Diesel ; (iii) MEP.
Solution: Refer to p-v diagram of example 3.8.
(i)Process 1-2 is isentropic. Hence T2 = T1(V1 / V2) (γ – 1) = 300 x 16 0.4
Or T2 = 909.4 K
Process 3-4 is isentropic. Hence T3 = T4 (V4/V3) (γ – 1) = T4 [V1/V3] (γ – 1)
Or T3 = T4 [(V1/V2) (V2/V3)] (γ – 1) = T4 Rc(γ – 1) (T2 / T3) (γ – 1)
Or T3γ = T4 [Rc T2](γ – 1)
Heat supplied per unit mass of air = q2-3 = Cp(T3 – T2) = 1.005 x (1993.27 – 909.4)
= 1089.3 kJ/kg.
Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (900 – 300)
= 430.8 kJ/kg Net work output per unit mass of air = wn = 1089.3 – 430.8
= 658.5 kJ/kg. wn 658.5
(ii) Thermal efficiency = ηDiesel = --- = --- = 0.6045
q2-3 1089.3
= 60.45 %
(iii)Specific volume of air at the beginning of compression = v1 = RT1 / p1 287 x 300 Hence v1 = --- = 0.861 m3 / kg. 1 x 105 Therefore v2 = v1 / 16 = 0.861 / 16 = 0.0538 m3/kg. wn 658.5 x 1000 MEP = --- = --- (v1 – v2) (0.861 – 0.0538) = 8.158 x 10 5 N/m2 = 8.158 bar.
Example 3.10:- In an air standard diesel cycle, the pressure at the end of expansion is
240 kPa and temperature is 550 C. At the end of compression process, the pressure is 4.2 MPa and temperature is 700 C. Determine (i) the compression ratio, (ii) the cut-off ratio, (iii) heat supplied per unit mass of air, and (iv) cycle efficiency.
Solution: The p – V diagram for the cycle is shown in Fig. E3.10.
Given:- p4 = 240 kPa ; T4 = 550 + 273 = 823 K ; p2 = 4.2 x10 3 kPa ;
T2 = 700 + 273 = 973 K.
(i) Process 3-4 is isentropic. Hence T3 = T4 (p3 /p4) (γ – 1) / γ = T4 (p2 /p4) (γ – 1) / γ
Or T3 = 823 x (4.2 x103 / 240) 0.286
= 1866 K Cut-off ratio = ρ = V3 / V2 = T3 / T2 = 1866 / 973
= 1.92.
(ii) Expansion ratio = Re = V4 / V3 = (p3 / p4) 1 / γ = (4.2 x 103 / 240) 1/ 1.4
= 7.725
Compression ratio = Rc = V1/V2 = Re ρ = 7.725 x 1.92
= 14.832
(iii) Heat supplied per unit mass of air = q2-3 = Cp(T3 – T2)
= 1.005 x (1866 – 973) = 897.5 kJ/kg.
(iv) For process 1-2 we have, T1 = T2 (V2 / V1) (γ – 1) = 973 x (1/14.832) 0.4
= 331 K
Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (331 – 823)
= − 353.3 kJ/kg
(Negative sign for q4-1 indicates that heat is transferred from air to the surroundings
during this process).
Net work output = wn = 897.5 – 353.3 = 544.2 kJ/kg.
544.2
Hence thermal efficiency = η Diesel = wn / q2-3 = ---
897.5 = 0.6063 = 60.63 %.
Example 3.11:- An oil engine works on diesel cycle with a compression ratio of 20. Heat
addition takes place up to 10 % of the stroke. Initial pressure and temperature of air are 1 bar and 27 C.The bore and stroke of the engine are 16 cm and 20 cm respectively. The
compression process is according to the law pV 1.32= constant and the expansion process is according to the law pV 1.30= constant. Find (i) the pressure and temperature at salient points of the cycle, (ii) the net work output per unit mass of air, (iii) MEP, (iv) thermal efficiency of the engine, (v) relative efficiency with respect to air standard efficiency.
Solution:
(i) Stroke volume = V1 – V2 = (π / 4) d 2 L = (π / 4) x 0.16 2 x 0.20 = 4.02 x 10 − 3 m3.
Hence 20 V2 – V2 = 4.02 x 10 − 3 Or V2 = 2.116 x 10− 4 m3.
And V3 = V2 + 0.1 (V1 – V2).
Hence cut-off ratio = ρ = V3 / V2 = 1 + 0.1 (V1/V2 – 1) = 1 + 0.1 x (20 – 1)
= 2.9. Expansion ratio = Re = V4 / V3 = Rc / ρ = 20 / 1.9 = 6.9. Now T2 = T1 Rc (nc – 1) = 300 x 20 0.32 = 782.44 K T3 = T2 ρ = 782.44 x 2.9 = 2269.1 K T4 = T3 (1 / Re)(ne – 1) = 2269.1 / 6.9 0.3 = 1271.6 K. p2 = p1 Rcnc = 1 x 20 1.32 = 52.163 bar = p3. 1 2 3 4 q4-1 V p pV ne = Constant pV nc = Constant Given:- Rc = V1/V2 = 20 V3 – V2 = 0.1(V1 – V2) p1 = 1 bar; T1 = 27 + 273 = 300 K; nc = 1.32 ; ne = 1.30 bore = d = 0.16 m ; stroke = L = 0.20 m
To find:- (i) p2, p3,p4,T2,T3,T4 ; (ii) wn ; (iii) MEP ; (iv) η engine ;
(v) η engine / η Diesel
p4 = p3 (1/Re) ne = 52.163 / 6.9 1.30 = 4.235 bar.
(ii) Since both the compression process 1-2 and the expansion process 3-4 are not
isentropic, the heat transfer during these processes will not be equal to zero.It is therefore necessary to calculate the heat transfer during these processes to determine the total external heat supplied during the cycle.
Now (p1v1 – p2v2) R(T1 – T2) 0.287 x (300 – 782.44) w1-2 = --- = --- = --- (nc – 1) (nc – 1) (1.32 – 1) = − 431.2 kJ/kg. w2-3 = p2 (v3 – v2) = p3v3 – p2v2 = R(T3 – T2) = 0.287 x (2269.1 – 782.44) = 425.2 kJ/kg. (p3v3 – p4v4) R (T3 – T4) 0.287 x (2269.1 – 1271.6) w3-4 = --- = --- = --- (ne – 1) (ne – 1) (1.30 – 1) = 951.4 kJ/kg.
w4-1 = 0 as process 4-1 is at constant volume.
Hence net work out put = wn = − 431.2 + 425.2 + 951.4 = 945.3 kJ/kg.
(iii) specific volume at state 1 = v1 = RT1 / p1 = 287 x 300 / 1 x105
= 0.861 m3/kg. Therefore v2 = v1 / Rc = 0.861 / 20 = 0.04305 m3/kg. wn 945.3 x 1000 MEP = --- = --- (v1 – v2) [0.861 – 0.04305] = 11.597 x 10 5 N/m2 = 11.597 bar. (iv) q2-3 = Cp(T3 – T2) = 1.005 x (2269.1 – 782.440 = 1494 kJ/kg. q4-1 = Cv(T1 – T4) = 0.718 x ( 300 – 1271.6) = − 698.3 kJ/kg.
q1-2 = w1-2 + (u2 – u1) = w1-2 + Cv(T2 – T1)
= − 431.2 + 0.718 x (782.44 – 300) = − 84.32 kJ/kg Applying I law for the cycle we have
q1-2 + q2-3 + q3-4 + q4-1 = wn
Or q3-4 = wn − (q1-2 + q2-3 + q4-1)
= 945.3 − (− 84.32 + 1494 − 698.3) = 233.92 kJ/kg
Total heat supplied during the cycle = qs = q2-3 + q3-4 = 1494 + 233.92 = 1727.9 kJ/kg
945.3 Thermal efficiency of the engine = η Engine = wn / qs = ---
1727.9 = 0.547 = 54.7 %. (v) Air standard efficiency of the diesel cycle is given by (ργ – 1) (2.9 1.4 – 1) ηDiesel = 1 – --- = 1 − ---
γ Rc(γ – 1) (ρ – 1) 1.4 x 20 0.4x (2.9 – 1)
= 0.6098 = 60.98 %
Relative efficiency = η Engine / ηDiesel = 0.547 / 0.6098 = 0.897 = 89.7 %.
Example 3.12:- In an air standard diesel cycle, air is compressed isentropically from 26
C and 105 kPa to 3.7 MPa. The entropy change during heat rejection process is − 0.6939 kJ/kg-K. Determine (i) heat supplied per unit mass of air, (ii)thermal efficiency, (iii) maximum temperature in the cycle, and (iv) temperature at the start of the heat rejection process.
Given:- T1 = 26 + 273 = 299 K ; p1 = 105 kPa ; p2 = 3.7 x 10 3 kPa ; s1 – s4 = − 0.6939
kJ/(kg-K).
To find :- (i) q2-3 ; (ii) η Diesel ; (iii) T3 ; (iv) T4
The p-v and T-s diagram for the air standard diesel cycle are shown in Fig. E3.12. It follows from the T-s diagram that s2 – s3 = s1 – s4
Process 2-3 is at constant pressure. Therefore
s3 – s2 = Cp ln (T3 / T2). Or T3 / T2 = exp {(s3 – s2) / Cp}
Or T3 / T2 = exp {0.6939 / 1.005} = 1.995
Process 1-2 is isentropic. Hence T2 = T1 (p2 / p1) (γ – 1) / γ
= 299 x {3.7 x 10 3 / 105 }0.286 = 828 K Hence T3= 1.995 x 828 = 1651.86 K
(i) Heat supplied per unit mass of air = q2-3 = Cp(T3 – T2) = 1.005 x (1651.86 – 828)
= 828 kJ/kg. (ii)For process 4-1 which is at constant volume we have s1 – s4 = Cv ln (T1 / T4)
Hence T4 = T1 exp {(s4 – s1) / Cv}
= 299 x exp {0.6939 / 0.718} = 784.9 K.
Heat rejected per unit mass of air = q4-1 = Cv(T1 – T4) = 0.718 x (299 – 784.9)
= − 349.4 kJ/kg.
Net work output per unit mass of air = wn = 828 – 349.4 = 478.6 kJ/kg
Thermal efficiency = ηDiesel = wn / q2-3 = 478.6 / 828
= 0.578 = 57.8 %.
(iii) Maximum temperature in the cycle = T3 = 1651.86 K.
3.7. Dual - Combustion Cycle or Semi-Diesel Cycle In practice, the p-V diagrams
taken from oil engines indicate that the combustion do not take place at constant pressure as in a diesel cycle, but is found to take place partly at constant volume and partly at constant pressure. Hence for most oil engines the ideal cycle is taken as one in which heating takes place partly at constant volume and partly at constant pressure. Such a cycle is called as “dual combustion or semi-diesel cycle”. The p-V and T-s diagrams for a dual combustion cycle are shown in Fig.3.6.
Process 1-2:- Isentropic compression of air from state 1 to state 2. During this process work is done on air by the surroundings.
Process 2-3:- Constant volume heating of air from state 2 to state 3.
Process 3-4:- Constant pressure heating of air till maximum permissible temperature is reached.
Process 4-5:- Isentropic expansion of air from state 4 to state 5. During this process work is done by air on the surroundings.
Process 5-1:- Constant volume cooling of air till the air comes back to its original state to complete the cycle.
3.7.1.Expressions for net work output and thermal efficiency
Heat supplied per unit mass of air = qs = q2-3 + q3-4 = Cv(T3 – T2) + Cp (T4 – T3)
Heat rejected per unit mass of air = q5-1 = Cv (T1 – T5)
V p 1 2 3 4 5 s = const.
Fig. 3.6 : p-V and T-s diagrams for dual combustion cycle.
2 1 3 Constant volume process Constant pressure process 4 5
Net work out put per unit mass of air = wn = ∑w = ∑q = qs + q5-1
Or wn = Cv(T3 – T2) + Cp (T4 – T3) + Cv (T1 – T5)
Cv(T3 – T2) + Cp (T4 – T3) + Cv (T1 – T5)
Hence thermal efficiency = η Dual = wn / qs = ---
Cv(T3 – T2) + Cp (T4 – T3)
Cv (T5 – T1)
= 1 − --- ………(3.22) Cv(T3 – T2) + Cp (T4 – T3)
The temperatures T2, T3, T4, and T5 can be expressed in terms of T1 as follows.
Process 1-2 is isentropic. Hence T2 = (V1 / V2) (γ – 1) T1 = Rc(γ – 1) T1.
For process 2-3, p2V2 / T2 = p3V3 / T3, and V2 = V3.
Hence T3 = (p3 / p2) T2 = Rp Rc(γ – 1) T1.
For process 3-4, p3V3 / T3 = p4V4 / T4, and p3 = p4.
Hence T4 = (V4 / V3) T3 = ρ Rp Rc(γ – 1) T1.
For process 4-5 we have T5 = (V4 / V5) (γ – 1) T4 = {(V4 / V3)(V3 / V1)} (γ – 1) T4
= {ρ / Rc}(γ – 1) ρ Rp Rc(γ – 1) T1
= ρ γ Rp T1
Substituting these expressions in Eq. (3.22) we have T1 [ρ γ Rp − 1] η Dual = 1 − T1[{Rp Rc(γ – 1) − Rc(γ – 1) } + γ{ ρ Rp Rc(γ – 1) − Rp Rc(γ – 1)}] [ρ γ Rp − 1] η Dual = 1 − --- ……….(3.23) Rc (γ – 1) [ (Rp – 1) + γ Rp (ρ – 1) ]
It can be seen from Eq. (3.23) that the air standard efficiency of the dual combustion cycle depends on the compression ratio Rc, the pressure ratio Rp and the cutoff ratio ρ.
The effect of these parameters on efficiency is shown in Fig.3.7. It can bessen from this figure that for given values of Rp and ρ , the efficiency increase with Rc, rapidly for small
efficiency decreases with increase in the cut – off ratio and for given values of Rcand ρ,
the efficiency increases with increase in Rp.
It can also be seen from Eq.(3.23) that this expression reduces to that of the diesel cycle if Rp = 1 and to that of Otto cycle if ρ = 1. It is also clear from this expression that the
efficiency of the dual combustion cycle is intermediate between those of an Otto cycle and a Diesel cycle which have the same compression ratio and same cut-off ratio.
3.7.2. Illustrative examples on dual combustion cycle
Example 3.13:- The compression and expansion ratios of an oil engine working on a
dual cycle are 9 and 5 respectively. The initial pressure and temperature are 1 bar and 30 C. The heat added at constant pressure is twice that added at constant volume. Determine the thermal efficiency and the mean effective pressure.
Given : Rc = V1 / V2 = 9 ; Re = V5 / V4 = 5 ; p1 = 1 bar ; T1 = 30 + 273 = 303 K ;
q 3-4 = 2 q 2-3 ;
To find :- (i) η Dual ; (ii) MEP
Solution: Refer to p – V diagram shown in Fig. E3.13.
(i) Process 1-2 is isentropic. Hence T2 = T1 Rc(γ – 1) = 303 x 9 0.4 = 729.7 K.
ρ Rp
Rc
η Dual
Compression ratio (V1 / V2) 9
Cut off ratio = V4 / V3 = --- = --- = ---
Expansion ratio (V5 / V4) 5
= 1.8.
For process 3-4, p3V3 / T3 = p4V4/T4 and p3 = p4.
Hence T4 = (V4 / V3) T3 = 1.8 T3 ……….(a) Also q3-4 = 2 q 2-3 i.e., Cp (T4 – T3) = 2Cv(T3 – T2) or (γ / 2) [1.8 T3 – T3] = T3 – T2 T2 729.7 Or T3 = --- = --- [1 – 0.8(γ / 2)] [1 – 0.8 x 1.4 / 2] = 1658.7 K. Hence T4 = 1.8 x 1658.7 = 2985.4 K
Process 5-1 is isentropic. Hence T5 = T4 (V4 / V5) (γ – 1) = 2985.4 x (1 / 5) 0.4
= 1568.2 K.
Heat supplied per unit mass of air = qs = q2-3 + q3-4= 3q 2-3 = 3 Cv (T3 – T2)
= 3 x 0.718 x (1658.7 – 729.7)
= 2003.85 kJ/kg.
Heat rejected per unit mass of air = q5-1 = Cv(T1 – T5) = 0.718 x (303 – 1568.2)
= − 909.7 kJ/kg.
(Negative sign for q5-1 indicates that during the process heat is transferred from air to the
surroundings).
wn 1094.15
Thermal efficiency = η Dual = --- = --- = 0.546 = 54.6 %.
qs 2003.85
287 x 303
Specific volume of air before compression = v1 = RT1 / p1 = ---
1 x 10 5 = 0.8666 m3 / kg.
Specific volume after compression = v2 = v1 / Rc = 0.8666 / 9
= 0.0963 m3 / kg.
wn 1094.15 x 1000
Mean effective pressure = MEP = --- = --- (v1 – v2) (0.8666 – 0.0963)
= 14.2 x 10 5 N/m2 = 14.2 bar.
Example 3.14:- The maximum and the compression pressures in a dual cycle are 64 bar
and 32 bar respectively. The compression curve is polytropic with index n = 1.35. The pressure in the cycle after 1/3rd of the compression stroke is completed is 1.65 bar. If 60 percent of the energy addition occurs at constant volume while 40 percent occurs at constant pressure, find (i) the compression ratio, (ii) the suction pressure, (iii)work output if the expansion index is 1.34, and (iv) thermal efficiency.
2 3 4 5 1 V p Given:- p3 = p4 = 64 bar p2 = 32 bar; nc = 1.35 ; ne= 1.34 ; pa = 1.65 bar ; q2-3 = 0.6 qs; q3-4 = 0.4 qs ; V1 – Va = (1/3) (V1 – V2)
To find:- (i) Rc ; (ii) p1 ; (iii) wn ;
(iv) η Dual
pVnc = const
pVne = const
a
Solution: (i) V1 – Va = (1/3) (V1 – V2) or Va = V1 – (1/3) (V1 – V2)
Hence Va / V2 = (2/3)(V1/V2) + 1/3 ……..(a)
Now paVanc = p2V2nc or Va / V2 = (p2/pa) 1 / nc
Or Va / V2 = (32 / 1.65) 1 / 1.35 = 8.99
Substituting this in Eq.(a) and solving for (V1 / V2) we have
(8.99 x 1/3) Compression ratio = V1 / V2 = --- = 13
(2/3)
(ii)For process 1-2 we have p1 = (V2 / V1) nc p2 = (1 /13) 1.35 x 32
= 1.003 bar.
(iii) It is not possible to calculate the temperatures at staes 2, 3, 4 and 5 unless temperature at state 1 is known. Since T1 is not given it is assumed as 300 K.
Hence for process 1-2, T2 = T1 (V1 / V2) (nc – 1) = 300 x 13 0.35
= 736.21 K.
For process 2-3 we p2V2 / T2 = p3V3 / T3 and V2 = V3.
Hence T3 = T2(p3 / p2) = 736.21 x (64 / 32) = 1472.42 K. q2-3 0.6 qs Now --- = --- = 1.5 q3-4 0.4 qs Therefore q2-3 = 1.5 q3-4 Or Cv (T3 – T2) = 1.5 Cp (T4 – T3) Or T4 = T3 + (1/ 1.5)(1 / γ) [T3 – T2] = 1472.42 + (1/1.5) x (1/ 1.4) x [1472.42 – 736.21] = 1823 K.
Compression ratio 13
Expansion ratio = V5 / V4 = --- = --- = 10.5
Cut-off ratio 1.24 For the expansion process 4-5 we have, T5 = T4 (V4/V5) (ne– 1)
= 1823 x (1/10.5) 0.34 = 819.6 K.
Since the index for compression process and expansion process are not equal to γ, these processes are not isentropic. Therefore there will be heat transfers during these processes which have to be determined to know the total heat supplied during the cycle.
(p1v1 – p2v2) R (T1 – T2) 0.287 x (300 – 736.21)
Now w1-2 = --- = --- = ---
(nc – 1) (nc – 1) 0.35
= − 356.5 kJ/kg.
w2-3 = 0 as the process is at constant volume.
w3-4 = p3(v4 – v5) = p4v4 − p5v5 = R(T4 – T5) = 0.287 x (1823 – 1472.42) = 100.3 kJ/kg. (p4v4 – p5v5) R (T4 – T5) 0.287 x (1823 – 819.6) w4-5 = --- = --- = --- (ne – 1) (ne – 1) 0.34 = 844 kJ/kg.
w5-1 = 0 as the process is at constant volume.
Net work output per unit mass of air = wn = w1-2 + w2-3 + w3-4 + w4-5 + w5-1
= − 356.5 + 0 + 100.3 + 844 + 0 = 587.8 kJ/kg.
(ii) Now for process 4-5 by I law q4-5 = w4-5 + Cv(T5 – T4)
Total heat supplied per unit mass of air = qs = q2-3 + q3-4 + q4-5 = Cv(T3 – T2) + Cp(T4 – T3) + q4-5 = 0.718 x(1472.42 – 736.21) + 1.005 x(1823 – 1472.42) + 122.55 = 1004.25 kJ/kg 587.80
Thermal efficiency = η Dual = wn / qs = --- = 0.5853 = 58.53 %
1004.25
Example 3.15:- . A diesel engine works between the temperatures of 1250 C and 25 C.
The energy addition during combustion is 550 kJ /kg. A dual combustion cycle operates between the same temperature limits, and has the same total energy addition as for diesel cycle except that this energy is equally divided between the constant volume and constant pressure processes. Compare the efficiencies of the two cycles
(i) Analysis of diesel cycle:- Given: Tmax = T3 = 1250 + 273 = 1523 K ;
Tmin = T1 = 25 + 273 = 298 K ; qs = q 2-3 = 550 kJ/kg. 4 1 2 3 V p 5 1 2 3 4 V p
To find: η Diesel and compare it with η Dual.
Solution: q2-3= Cp (T3 – T2). Or T2 = T3 – q2-3 / Cp = 1523 – 550 / 1.005
= 975.74 K .
Cut – off ratio = ρ = V3 / V2 = T3 / T2 = 1523 / 975.74 = 1.56.
Compression ratio = Rc = V1 / V2 = (T2 / T1) 1 / (γ – 1) = (975.74 / 298) 1 / 0.4
= 19.4
1 [ρ γ – 1] Thermal efficiency = η Diesel = 1 − --- x ---
γ Rc(γ – 1) [ρ – 1]
1 [ 1.56 1.4 – 1] = 1 − --- x --- 1.4 x 19.4 0.4 [ 1.56 – 1] = 0.6635 = 66.35 %
(ii) Analysis of Dual combustion cycle: Given:- q2-3 = q3-4 = (1/2) x550 = 275 kJ/kg ;
Tmax = T4 = 1523 K ; Tmin = T1 = 298 K ;
To find: η Dual and compare it with η Diesel
q 3-4 = Cp(T4 – T3) or T3 = T4 – q3-4 / Cp = 1523 – 275 / 1.005 = 1249.4 K Similarly T2 = T3 – q2-3 / Cv = 1249.4 – 275 / 0.718 = 866.5 K. Compression ratio = V1 / V2 = (T2 / T1) (γ – 1) = (866.5 / 298) 0.4 = 14.4 Cut-off Ratio = V4 / V3 = T4 / T3 = 1523 / 1249 = 1.22
Compression ratio 14.4 Expansion ratio = V5 / V4 = --- = ---
Cut –off ratio 1.22
= 11.8
For expansion process 4-5 we have T5 = T4 (V4 / V5) (γ – 1) = 1523 x (1/11.8) 0.4
= 567.4 K.
Heat rejected per unit mass of air = q5-1 = Cv (T1 – T5) = 0.718 x (298 – 567.4)
= − 193.7 kJ/kg
( 550 – 193.7)
Thermal efficiency = η Dual = --- = 0.6478 = 64.78 %.
550
Comparing the two efficiencies we have η Diesel > η Dual
Example 3.16:- In a dual cycle, two thirds of the total energy added occurs at constant
volume.. If the compression ratio is 15, and the maximum pressure in the cycle is 53 bar, compute(i)the temperatures at the salient points of the cycle, and (ii) thermal efficiency. Assume standard conditions of air at the start of the compression process. Assume the minimum temperature and pressure in the cycle to be 27 C and 1 bar.
Solution: (i) Process 1-2 is isentropic. Hence T2 = T1 (V1 / V2) (γ – 1)
Or T2 = 300 x 15 0.4 = 880.3 K.
Also p2 = p1(V1/V2) γ = 1 x 15 1.4= 44.3 bar.
For process 2-3 we have T3 = T2 (p3 / p2) = 880.3 x (53 / 44.3)
= 1053.2 K.
q2-3 = Cv(T3 – T2) = 0.718 x (1053.2 – 880.3) = 123.85 kJ/kg.
Hence qs = (3/2) x 123.85 = 185.775 kJ/kg.
Therefore q3-4 = (1 / 3) x 185.775 = 61.925 kJ/kg
Now q3-4 = Cp (T4 – T3). Or T4 = T3 + q3-4 / Cp = 1053.2 + 61.925 / 1.005 = 1114.82 K
Cut – off ratio = V4 / V3 = T4 / T3 = 1114.82 / 1053.2 = 1.06
5 1 2 V p 3 4 Given:- q2-3 = (2/3) qs ; q3-4 = (1/3) qs ; V1 / V2 = 15 ; pmax = p3 = p4 = 53 bar ; Tmin = T1 = 27 + 273 = 300 K ; pmin = p1 = 1 bar. To find:- (i) T2, T3, T4, T5 (ii) η Dual
Expansion ratio = V5 / V4 = 15 / 1.06 = 14.15.
Hence T5 = T4 (V4 / V5) (γ – 1) = 1114.82 x (1 / 14.15) 0.4
= 386.3 K.
Heat rejected = q5-1 = Cv(T1 – T5) = 0.718 x (300 – 386.3)
= 61.96 kJ/kg
Net work out put = wn = 185.775 – 61.96 = 123.815 kJ/kg.
Thermal efficiency = η Dual = 123.815 / 185.775 = 0.6665
= 66.65 %.
3.8. Comparison between Otto, Diesel and Dual combustion cycles:- The important
variables which are used as the basis for comparison of the cycles are compression ratio, peak pressure, heat supplied, heat rejected and the net work output. In order to compare the performance of the Otto, Diesel and Dual combustion cycles some of these variables have to be fixed.
3.8.1. Comparison with same compression ratio and heat supply: The comparison of
these cycles for the same compression ratio and same heat supply are shown in Fig. 3.8 on both p – V and T – s diagrams.In these diagrams, cycle 1-2-3-4-1 represents Otto
V p 4’ 4’’ 4 1 3’ 3’’ 3 2’’ 2 2 1 3 3’’ 3’ 2’’ 4’ 4” 4 T
Fig.3.8: Comparison with same compression ratio and heat supply 5 6 6” 6’
Cycle, cycle 1-2-3’-4’-1 represents diesel cycle and cycle 1-2”-3”-4”-1 represents the dual combustion cycle for the same compression ratio and heat supply.
From the T-s diagram, it can be seen that area 5236 = area 522”3”6” = area 523’6’ as this area represents the heat supply which is same for all the cycles.All the cycles start from the same initial point 1 and the air is compressed from state 1 to state 2 as the compression ratio is same. It is seen from the T-s diagram, that for the same heat supply, the heat rejection in Otto cycle (area 5146) is minimum and heat rejection in Diesel cycle (area 514’6’) is maximum. Consequently Otto cycle has the highest work output and efficiency. Diesel cycle has the least efficiency and dual cycle has the efficiency between the two.
3.8.2:- Same compression ratio and heat rejection:- Fig. 3.9 shows the comparison
between the Otto cycle and Diesel cycle on p-V and T-s diagrams.Cycle 1-2-3-4-1 represents the Otto cycle and cycle 1-2-3’-4-1 represents the Diesel cycle.
Since both the cycles start from the same state point 1 and the heat rejection is same for both the cycles state 4 is same for both the cycles. Also since the compression ratio is same for both the cycles the state after compression process (state 2) is same for both the cycles.It can be seen from the T-s diagram that the area representing the Otto cycle (area 1234) is more than that representing the Diesel cycle (area 123’4). Hence the work output for the Otto cycle is more than that for the Diesel cycle.Thermal efficiency of any power cycle can be expressed as
Heat supply – Heat rejection Heat rejection Thermal efficiency = --- = 1 − --- Heat supply Heat supply
V p 4 1 2 3’ 3 s T 2 1 3 3’ 4
Fig. 3.9: Same compression ratio and same heat rejection
Since the heat rejection is same for both the cycle and heat supply for Otto cycle (area 2365) is more than that for the Diesel cycle (area 23’65), thermal efficiency for the Otto cycle is higher than that for the diesel cycle.
3.8.3:Same peak pressure, peak temperature and heat rejection: Fig. 3.10 show the
comparison, on p-V and T-s diagrams, between Otto and Diesel cycles with same peak pressure, peak temperature and heat rejection.Cycle 1-2-3-4-1 represents the Otto cycle,
Cycle 1-2’-3-4-1 represents the Diesel cycle. It can be seen from the T-s diagram that area representing the Diesel cycle (area 12’34) is more than that representing the Otto cycle (area 1234) and hence Diesel cycle has higher work output than Otto cycle.It can also be seen that the heat supply for Diesel cycle (area 52’36) is more than that for Otto cycle (area 5236). Since the heat rejection is same for both the cycles, it follows that the thermal efficiency for the Diesel cycle is more than that for the Otto cycle for the same peak pressure, peak temperature and same heat rejection.
3.9. Stirling Cycle The Carnot power cycle has a low mean effective pressure because of
its low work output. Hence one of the modified forms of the cycle to produce higher mean effective pressure with the efficiency equal to Carnot efficiency is the Stirling cycle. The Stirling cycle consists of two isothermal and two constant volume processes. The heat rejection and addition take place at constant temperatures. The p-V and T-s diagrams for the Stirling cycle are shown in Fig. 3.11. It is clear from the T-s diagram that the amount of heat addition and rejection during constant volume processes is same.Hence the thermal efficiency of the cycle is given as:
V p 4 1 2’ 3 2 2’ 2 1 3 4 5 6 s T
Fig. 3.10: Same peak pressure, same peak temperature and same heat rejection
∑w ∑q ηStirling = --- = --- qs q3-4 q1-2 + q2-3 + q3-4 + q-41 = --- q3-4 Now q2-3 = −q4-1. q1-2 + q3-4 RT3 ln(V4/V3) + RT1 ln(V2/V1) ηStirling = --- = --- q3-4 RT3 ln(V4/V3)
Since V4 = V1 and V3 = V2, the above expression for efficiency reduces to
T3 – T1
ηStirling = ---
T3
This expression is same as that for a Carnot cycle working between the same temperature limits.
The Stirling cycle was used for hot air engines and became obsolete as Otto and Diesel cycle came into use. The design of stirling engines involves a major difficulty in the design of a heat exchanger to achieve the heat transfer processes at constant volume and to operate at high temperature continuously. However, with the development of new materials and intensive research on this engine, Stirling engine has staged a come back. Since the heat exchanger cannot have 100 % efficiency in practice, the thermal efficiency of a practical Stirling engine will be less than that for a Carnot engine working between the same temperature limits.
3.10. Gas Turbine Cycles
3.10.1. Assumptions for analysis of Ideal Gas Turbine Cycles :- Following assumptions
are made to analyse ideal gas turbine cycles:
(i) The working substance is air and air behaves as a perfect gas (ii) Expansion and compression processes are isentropic.
(iii) There are no pressure losses in the piping connecting the various components as well as in the heat exchangers.
(iv) Changes in kinetic and potential energies of the fluid are negligible (v) Flow through various components is one dimensional, steady and uniform
3.10.2. Brayton Cycle (Simple Gas Turbine Cycle):- Brayton cycle is the basic cycle
for the simple gas turbine power plant.The p-v and T-s diagram for this cycle is shown in Fig.3.12.It can be seen from these diagrams that the cycle consists of two isentropic processes and two constant pressure processes. The schematic for the cycle is shown in Fig. 3.13.
Process 1 – 2: Isentropic compression of air in the compressor. During this process work is done on air by the surroundings.
Process 2 – 3: Constant pressure heating of air in the heater till the maximum permissible temperature is reached.
v p
T
s
Fig.3.12: p-v and T – s diagrams for a Brayton cycle
1 2 3 4 2 3 1 4
