Chapter 16 Gas Turbines – Thermal Engineering

Chapter 16

Gas Turbines

16.1 ❐ INTRODUCTION

The gas turbine is the most satisfactory way of producing very large power in a self-contained and compact unit. The gas turbine obtains its power by utilising the energy of a jet or burnt gases and air, the velocity of the jet being absorbed as it flows over several rings of moving blades which are fixed to a common shaft. The gas turbine requires an air compressor which is driven off its own shafting. This absorbs a considerable proportion of the power produced and thus lowers the overall efficiency, which is of the order of 20%–30%.

16.2 ❐ FIELDS OF APPLICATION OF GAS TURBINE

The following are the major fields of application of gas turbines:

  1. Aviation: A gas turbine is used in aviation because it is self-contained, light weight not requiring cooling water, and generally fits into the overall shape of the aircraft structure. It is also suitable for turbo-jet and turbo-propeller engines.
  2. Power generation: It is used for electric power generation because of its simplicity, lack of cooling water, quick installation, and starting. It is also suitable for supercharging, road transport, and railways.
  3. Oil and gas industry: It is used in the oil and gas industry due to cheaper supply of fuel and low installation cost.
  4. Marine propulsion: It is used in marine propulsion due to its light weight and it does not require cool water.
16.3 ❐ LIMITATIONS OF GAS TURBINES

The limitations of gas turbines are as follows:

  1. They are not self-starting.
  2. Low efficiency at part loads
  3. Higher rotor speeds
  4. Low overall plant efficiency
16.4 ❐ COMPARISON OF GAS TURBINES WITH IC ENGINES

16.4.1 Advantages

  1. Perfect balancing of the rotating parts is possible.
  2. Mechanical efficiency is very high (≈95%) compared with 85% of IC engine because of its large number of sliding parts.
  3. Torque on the turbine shaft is continuous and uniform; therefore, flywheel is not required.
  4. Work developed is more because expansion of gases up to atmospheric pressure is possible.
  5. The weight of the gas turbine per kW of power developed is less, and therefore, they are superior for use in aircraft.
  6. The compression, combustion, and expansion take place in different units. Therefore, these units can be designed, tested, and developed individually.
  7. The gas turbine can be driven at much higher speeds.
  8. The maximum pressure used in gas turbines is very low; therefore, its components can be made lighter.
  9. The lubrication and ignition systems are much simpler.
  10. Cheaper fuels can be used in gas turbine.
  11. The installation and maintenance costs are less.
  12. The exhaust is free from smoke and is less polluting.

16.4.2 Disadvantages

  1. The thermal efficiency of a simple gas turbine is low.
  2. There is very poor thermal efficiency at part loads.
  3. The temperature of gases supplied to the turbine is limited to 1100 K as there are no metals which can be used for blades to sustain this temperature at high centrifugal stresses.
  4. The manufacturing of blades is difficult and costly as Ni-Cr alloy is used.
  5. The control of fuel is difficult with wide operating speeds and output.
  6. A special cooling system is required for the blades.
  7. The starting is more difficult than IC engines.
  8. It produces five times exhaust gases than an IC engine for same output.
  9. Speed reduction device is always required due to higher operating speed.
  10. The life of blades and combustion chamber is low due to high temperature.
16.5 ❐ ADVANTAGES OF GAS TURBINES OVER STEAM TURBINES
  1. Boiler, along with accessories, is not required.
  2. The capital and running cost is less.
  3. Less space is required for the plant.
  4. Starting the plant is more easy and quick.
  5. The weight of the plant per kW generated is less.
  6. Control of plant is much easier.
  7. The plant can be installed anywhere as cooling water is not required.
  8. Gas turbine plant is used as a best type for peak load plant.
16.6 ❐ CLASSIFICATION OF GAS TURBINES

Gas turbines may be classified as follows:

  1. Constant pressure combustion (or continuous combustion) gas turbine
    1. Open cycle constant pressure gas turbine
    2. Closed cycle constant pressure gas turbine
  2. Constant volume combustion (or explosion) gas turbine

16.6.1 Constant Pressure Combustion Gas Turbine

In this type, the fuel is burned at constant pressure and the cycle used is the Joule cycle. The turbine belongs to the reaction type using oil fuel and is fitted with an axial flow rotary air compressor which is coupled to the rotor shaft. The heat is supplied to the working fluid at constant pressure.

The open and closed cycle gas turbines are shown in Figs 16.1(a) and 16.1(b), respectively. A simple open cycle constant pressure gas turbine consists of a compressor, combustion chamber, and turbine. The air is taken from the atmosphere by the compressor and is compressed to the combustion pressure of 1 to 4 atm. It is then forced into the combustion chamber. A part of this air is used as combustion air for the oil which enters the burner, and the remainder is forced through the annular space between the wall of the combustion chamber and the burner jacket. The air receives heat from the burner jacket and also mixes with the products of combustion chamber and the burner jacket. This raises the temperature and volume of the air. The use of a very large quantity of air in excess of the combustion air prevents the temperature of the mixture from reaching values which are too high for the metal of the rotor blades. It also prevents the burner from becoming too hot.

The high pressure and high temperature gases now enter the turbine and flow through the blade rings, where they continuously expand, and the pressure energy is converted into kinetic energy, which in turn is absorbed by the rotor. On leaving the turbine, the spent gases pass away to the exhaust. The speed of the turbine is generated by controlling the fuel oil supply and the safety valve which by-passes some of the mixture. If the pressure becomes too high, the net power developed by the turbine is utilised in driving the generator but a considerable amount of the total power produced by the rotor is absorbed in driving the compressor. The turbine is started by the electric starter motor.

In the closed cycle constant pressure gas turbine, the compressed fluid coming out from the compressor is heated in the heater by an external source at constant pressure and the high pressure and temperature fluid develop the work while passing through the turbine. The fluid coming out from the turbine is cooled to its original temperature in the cooler using an external cooling source before passing into the compressor. Thus, the working fluid is continuously circulated in the closed cycle. Such a turbine has now become obsolete.

Figure 16.1 Constant pressure combustion gas turbine: (a) Open cycle turbine, (b) Closed cycle turbine

16.6.2 Constant Volume Combustion Gas Turbine

The constant volume combustion gas turbine is shown in Fig. 16.2. Air is taken from the atmosphere and is compressed by an axial flow compressor driven by a separate steam turbine. The compressed air coming out of the compressor at about 3 bar is passed to the constant volume combustion chamber. The fuel is injected into the combustion chamber in the current of air by a separate fuel pump. The air-fuel mixture formed in the combustion chamber is then ignited by means of a spark plug. The combustion takes place at constant volume in the combustion chamber and the pressure inside the chamber increases to nearly 12 bar. The high pressure and temperature gases are fed to the gas turbine through the water-cooled nozzle. The gases passed through the turbine develop the useful work. The exhaust gases are fed to the steam boiler to generate steam for the steam turbine driving the compressor.

Figure 16.2 Constant volume combustion gas turbine

The major disadvantage of this turbine is the intermittent combustion which impairs its smooth functioning. The absolute necessity of a separate steam turbine to recover the heat from exhaust gases further discourages its use in practice.

16.7 ❐ COMPARISON OF OPEN AND CLOSED CYCLE GAS TURBINES

The comparison of open and closed cycles gas turbines is given in Table 16.1.

 

Table 16.1 Comparison of open and closed cycles gas turbines

16.8 ❐ POSITION OF GAS TURBINE IN THE POWER INDUSTRY

Constant pressure gas turbines are now used for generating electricity, driving locomotives and aeroplane propellers, producing the air stream for wind tunnels, and jet propulsion of air craft. In some types of aircraft, the exhaust gases from the main engines are used for driving a gas turbine which, in turn, drives the supercharger. Gas turbines driven by exhaust gases are also used for supplying a pressure charge to diesel engines.

Figure 16.3 Gas turbine utilising exhaust heat

Overall efficiencies of constant pressure gas turbine plants vary between 18%–25%. This is based on a turbine efficiency of 75%–78%. The efficiency of this type of turbine can be improved by using heat from exhaust gases to preheat the compressed air before it enters the combustion chamber. The hot mixture of air and burnt gases is then expanded through the turbine from which they are exhausted into the heat exchanger. The available heat is now given to the high pressure air from the compressor. Such a system is shown in Fig. 16.3.

16.9 ❐ THERMODYNAMICS OF CONSTANT PRESSURE GAS TURBINE: BRAYTON CYCLE

16.9.1 Efficiency

The Brayton cycle is shown on p-V and T-s diagrams in Fig. 16.4. During this cycle, air is drawn into the compressor at point 1 and then compressed isentropically along the process 1-2. Process 2-3 represents the burning of the oil at constant pressure p2. Process 3-4 is the isentropic expansion of the gases through the turbine. Process 4-1 represents the exhausting and cooling of the gases at constant pressure p1.

For 1 kg of air,

Compressor work, wc = w1−2 = h2 h1 = cp (T2 T1)

Heat supplied, q = q2−3 = h3 − `h2 = cp (T3T2)

Turbine work, wt = w34 = h3 h4 = cp (T3T4)

Net work done, wnet = wtwc = cp [(T3T4) − (T2T1)]

Figure 16.4 Brayton cycle for constant pressure gas turbine: (a) p-V diagrams, (b) T-s diagrams

Let be the pressure ratio and

For the adiabatic compression process 1-2, we have

and for the adiabatic expansion process 3-4, we have

But p1 = p4 and p2 = p3

Thermal efficiency,

Figure 16.5 Variation of thermal efficiency with pressure ratio

The variation of thermal efficiency with rp is shown in Fig. 16.5.

16.9.2 Specific Output

Specific output

For maximum specific output,

or

Figure 16.6 Variation of specific output with pressure ratio

Hence, for maximum work output, the temperature after compression must be equal to the exhaust gas temperature, which depends on the expansion ratio. The variation of specific output with rp is shown in Fig. 16.6.

16.9.3 Maximum Work Output

Maximum work output,

 

(wnet)max = cp [(T3T4) − (T2T1)]
= cp [T3 + T1T4T2]

16.9.4 Work Ratio

16.9.5 Optimum Pressure Ratio for Maximum Specific Work Output

wnet = wtwc = cp (T3T4) − cp (T2T1)

wnet will be maximum, when

Let

From Eq. (16.3), we have

Combining with Eq. (16.7), we get

16.10 ❐ CYCLE OPERATION WITH MACHINE EFFICIENCY

16.10.1 Maximum Pressure Ratio for Maximum Specific Work

It is not possible to achieve isentropic compression and expansion in the actual Brayton cycle because of inevitable losses due to friction and turbulence in the compressor and turbine. Therefore, the temperature at the end of compression and expansion are higher than that in an ideal cycle for the same pressure ratio. The actual Brayton cycle is shown in Fig. 16.7.

Isentropic efficiency of compressor,

Isentropic efficiency of turbine,

Net specific work available,

wnet = wtwc = cp (T3T4′) −cp (T2′T1)

For specific work output to be maximum for given temperature limits,

Figure 16.7 Actual Brayton cycle with machine efficiencies

16.10.2 Optimum Pressure Ratio for Maximum Cycle Thermal Efficiency

Heat supplied, q = cp (T3 - T2) = cp (T3 - T2)

The thermal efficiency of the cycle is dependent only on rp for the fixed values of T1 and T3. The condition for the maximum value of thermal efficiency for given temperature limits is

Let

Taking -ve sign

16.11 ❐ OPEN CYCLE CONSTANT PRESSURE GAS TURBINE

The open cycle constant pressure gas turbine plant is shown diagrammatically in Fig. 16.8. The fuel is burned with air coming out from the compressor into the combustion chamber. The products of combustion from the combustion chamber are passed through the turbine and developed the required power. Part of the power developed is used to run the compressor. The gases leaving the turbine are exhausted to atmosphere.

Let a = mass rate of flow of air

f = mass rate of flow of fuel

Compressor work,  Wc = a cpa (T2T1)

Turbine work,    Wt = (a + f) cpg (T3T4)

where cpa, cpg = specific heat of air and gases, respectively

Network Wnet = WtWc = (a + f) cpg (T3T4′) − acpa (T2′T1)

Heat supplied, Q = f × C.V.

where CV = calorific value of fuel

Figure 16.8 Open cycle constant pressure gas turbine: (a) Schematic diagram, (b) T-s diagrams

Thermal efficiency,

16.12 ❐ METHODS FOR IMPROVEMENT OF THERMAL EFFICIENCY OF OPEN CYCLE CONSTANT PRESSURE GAS TURBINE

The thermal efficiency of open cycle constant pressure gas turbine can be improved by the following methods:

  1. Regeneration
  2. Intercooling
  3. Reheating
  4. Combination of above

16.12.1 Regeneration

In this method, the heat of the exhaust gases is used to heat the air coming out from the compressor, thus reducing the mass of fuel supplied in the combustion chamber. A schematic diagram of such a plant is shown in Fig. 16.9 (a).

  1. Without Machine Efficiencies

    The T-s diagram without machine efficiencies is shown in Fig. 16.9 (b).

    For 1 kg of air

    Compressor work,   wc = w1 2 = cp (T2T1)

    Heat supplied,    qs = q5 3 = cp (T3T5)

    Turbine work,    wt =w3 4 =cp (T3T4)

    Net work,     wnet = wtwc = cp (T3T4) – cp (T2T1)

    Figure 16.9 Open cycle gas turbine with regeneration: (a) Schematic diagram, (b) T-s diagram without machine efficiencies, (c) T-s diagram with machine efficiencies

    Thermal efficiency,

    In an ideal regenerator, T4 = T5

    Figure 16.10 Variation of thermal efficiency with pressure ratio for open cycle gas turbine with regeneration

    The variation of thermal efficiency with pressure ratio is shown in Fig. 16.10. It may be seen that thermal efficiency increases with an increase in and decreases with increase in rp.

  2. Considering Machine Efficiencies

    The T-s diagram is shown in Fig. 16.9 (c).

    Figure 16.11 Actual regeneration process

  3. Effectiveness of Regenerator

    In actual practice, the rise in temperature from T2′ to T5 = T4′ is not possible. The actual temperature of air will be T5′ < T5. Hence, actual thermal efficiency [see Fig. (16.11)] of the cycle becomes,

    The effectiveness of the regeneration is given by

    where a, g = mass rate of flow of air and exhaust gases, respectively

    cpa, cpg = specific heats of air and exhaust gases, respectively.

16.12.2 Intercooling

The work consumed by the compressor can be reduced by compressing the air in two stages and incorporating the intercooler in between, as shown in Fig. 16.12 (a). The corresponding T-s diagram is shown in Fig. 16.12 (b).

Figure 16.12 Gas turbine plant with intercooling: (a) Schematic diagram, (b) T-s diagram

For perfect cooling, T3 = T1 and if nc1 = nc2 = nc then

wnet = wtwc = cp (T5T6) – [cp (T2T1) + cp (T4T3)]

For perfect intercooling,T3 = T1 and if hc1 = ηc2 = ηc, then

16.12.3 Reheating

A considerable increase in power output can be achieved by expanding the gases in two stages with a reheat combustion chamber between the two as shown in Fig. 16.13 (a). The corresponding T-s diagram is shown in Fig. 16.13 (b).

wt = cpg (T3T4′) + cpg (T5T6′)
= cpg ηt1 (T3T4) + cpg ηt2 (T5T6)

where

Let

Figure 16.13 Gas turbine plant with reheating: (a) Schematic diagram, (b) T-s diagram

If ηt1 = ηt2 = ηt and T5 = T3, then

16.12.4 Reheat and Regenerative Cycle

This cycle is shown in Fig. 16.14.

wc = cp (T2T1)

qs = q7 – 3 + q4 – 5 = cp (T3T7) + cp (T5T4)

wt = w3 – 4 + w5 – 6 = cp (T3T4) + cp (T5T6)

Figure 16.14 Gas turbine plant with reheat and regeneration: (a) Schematic diagram, (b) T-s diagram

Net work wnet = wtwc

= cp [(T3T4) + (T5T6) – (T2T1)]

Let so that rp = r1r2

So that c = c1c2

Now,

Now, T7 = T4 and T5 = T3

16.12.5 Cycle with Intercooling and Regeneration

The schematic arrangement and T-s diagram of the cycle are shown in Fig. 16.15.

wc = w1 – 2 + w3 – 4 = cp [(T2T1) + (T4T3)]

qs = q7 – 5 = cp (T5T7)

wt = w5 – 6 = cp (T5T6)

wnet=wtwc = cp [(T5T6) – (T2T1) – (T4T3)]

Let so that c = c1c2 and T3 = T1

Figure 16.15 Gas turbine plant with inter cooler and regeneration: (a) Schematic diagram, (b) T-s diagram

Now, qs = cp (T5T7)

16.12.6 Cycle with Intercooling and Reheating

The schematic arrangement of this cycle is shown in Fig. 16.16 (a) and the corresponding T-s diagram in Fig. 16.16 (b).

wc = cp [(T2T1) + (T4T3)]

qs = cp [(T5T4) + (T7T6)]

wt = cp [(T5T6) + (T7T8)]

wnet = wtwc = cp [(T5T6) + (T7T8) – (T2T1) – (T4T3)]

Now, T3 = T1 and T7 = T5 for perfect intercooling and reheating, respectively. Then,

Thus, c = c1c2

For maximum work output,

or

Figure 16.16 Gas turbine cycle with intercooler and reheating: (a) Schematic diagram, (b) T-s diagram

16.12.7 Cycle with Intercooling, Regeneration and Reheating

The schematic arrangement of the cycle and corresponding T-s diagram are shown Fig. 16.17.

wc = cp [(T2T1) + (T4T3)]

qs = cp [(T5T9) + (T7T6)]

wt = cp [(T5T6) + (T7T8)]

wnet = wt wc = cp [(T5T6 + T7T8) − (T2T1 + T4T3)]

T7 = T5

Let T1=T3 and T5=T7 so that c1=c2

Figure 16.17 Gas turbine cycle with intercooling, regeneration and reheating: (a) Schematic diagram, (b) T-s diagram

Similar to the case of 16.12.6, it can be shown that

Example 16.1

Calculate the indicated mean effective pressure and efficiency of a Joule cycle if the temperature at the end of combustion is 2000 K and the temperature and pressure before compression are 350 K and 1 bar, respectively. The pressure ratio is 1.3 and assume cp = 1.005 kJ/kgK.

Solution

Refer to Fig. 16.18.

p1 = 1 bar, T1 = 350 K, T3 = 2000 K

rp = 1.3, cp =1.005 kJ/kgK

wc = cp (T2T1) = 1.005 (377.24 − 350) = 27.376 kJ/kg

wt = cp (T3T4) = 1.005 (2000 − 1855.56) = 145.162 kJ/kg

Net work done, wnet = wt wc = 145.162 − 27.376 = 117.786 kJ/kg

qs = cp (T3T2) = 1.005 (2000 − 377.24) = 1630.874 kJ/kg

Thermal efficiency

Figure 16.18 p-v diagram

Example 16.2

A gas turbine operates on a pressure ratio of 6. The inlet air temperature to the compressor is 300 K and the temperature of air entering to the turbine is 580°C. If the volume rate of air entering the compressor is 250 m3/s, calculate the net power output of the cycle. Also, calculate the efficiency.

Solution

The T-s diagram shown in Fig. 16.19

 

rp = 6, T1 = 300 K, T3 = 273 + 580 = 853 K

 

1 = 250 m3 /s

Compressor work wc = cp (T2T1) = 1.005 (500.55 − 300) = 201.55 kJ/kg

Turbine work wt = cp (T3T4) = 1.005 (853 − 511.23) = 343.48 kJ/kg

Net work output wnet = wt wc = 343.48 − 201.55 = 141.93 kJ/kg

Heat input qs = cp (T3T2) = 1.005 (853 − 500.55) = 354.21 kJ/kg

Thermal efficiency

Figure 16.19 T-s diagram

Example 16.3

A gas turbine has a perfect heat exchanger. Air enters the compressor at a temperature and pressure of 300 K and 1 bar and discharges at 475 K and 5 bar. After passing through the heat exchanger, the air temperature increases to 655 K. The temperatures of air entering and leaving the turbine are 870°C and 450°C, respectively. Assuming no pressure drop through the heat exchanger, compute (a) the output per kg of air, (b) the efficiency of cycle, and (c) work required to drive the compressor.

Solution

The p – V and T-s diagrams are shown in Fig. 16.20.

T1 = 300 K, p1 = 1 bar,

T2 = 475 K, p2 = 5 bar,

T5 = 655 K, T3 = 870 + 273 = 1143 K,

T4 = 450 + 273 = 723 K

or γ = 1.4, cp = 1.005 kJ/kg

Compressor work wc = cp (T2 - T1) = 1.005 (475 - 300) = 175.9 kJ/kg

Turbine work wt = cp (T3 - T4) = 1.005 (1143 - 723) = 422.1 kJ/kg

Heat input qs = cp (T3 - T5) = 1.005 (1143 - 655) = 490.44 kJ/kg

  1. Net work output, wnet = wtwc = 422.1 − 175.9 = 246.2 kJ/kg
  2. Thermal efficiency,
  3. Compressor work, wc = 175.9 kJ/kg

    Figure 16.20 Diagrams for a gas turbine: (a) p-V diagram, (b) T-s diagram

Example 16.4

A closed cycle regenerative gas turbine operating with air has the following data: p1 = 1.4 bar,T1 = 310 K, rp = 5, Tmax = 1050 K, effectiveness of generator = 100%, net output = 3 MW. Calculate (a) the mass flow rate of air per minute and (b) the thermal efficiency.

Solution

The cycle in T-s diagram is shown in Fig. 16.21.

Refer to Fig. 16.21

Figure 16.21 T-s diagram for closed cycle regenerative gas turbine

For 100% regenerative effectiveness,

 

T4 = T5 = 662.65 K
  1.     wnet = ṁcp (T3T4 ) − ṁcp (T2T1)

     

    3000 = ×1.005 [(1050 − 662.65) − (491.2 − 310)]

     

    = 14.48 kg/s or 868.8 kg/min
  2. Heat input     Q = ṁcp (T3T5)

     

    = 14.48 × 1.005 (1050 − 662.65) = 5636.87 kJ/s

     

    Thermal efficiency

Example 16.5

The following data refers to a turbine with regenerator:

T1 = 290 K, T2 = 460 K, T3 = 900°C, T4 = 467°C

Calculate (a) the pressure ratio, (b) the specific work output, (c) the efficiency of cycle, and (d) the compressor work. Assume ηmech = 100%.

Figure 16.22 T-s diagram for a gas turbine with regenerator

Solution

The T-s diagram of the cycle is shown in Fig. 16.22.

  1. or rp = 5

  2. Work output, wnet =cp [(T3T4) − (T2T1)]
    = 1.005[(1173 − 740) − (460 − 290)]
    = 264.3 kJ/kg
  3. Heat input, qs = cth (T3T5) = cp (T3T4)
    = 1.005 (1173 − 740) = 435.165 kJ/kg

    Thermal efficiency

  4. Compressor work, wc = cp (T2T1) = 1.005 (460 − 290) = 170.85 kJ/kg

Example 16.6

The ratio of net work to turbine work of an ideal gas turbine plant is 0.563. The temperature of air at the inlet to the compressor is 300 K. Calculate the temperature drop across the turbine if thermal efficiency of the unit is 35%. Assume a mass flow rate of 10 kg/s, cp = 1 kJ/kgK and γ = 1.4.

Solution

Thermal efficiency

or

or rp = 4.51

Now

Compressor work wc = cp (T2T1) = 1 × (461.54 − 300) = 161.54 kJ/kg

or wt = 369.66 kJ/kg = cp · ∆T

ΔT = 369.66°C

Example 16.7

In a Brayton cycle, the pressure ratio in the compressor is rp. The minimum and maximum temperatures are T1 and T3. The air is expanded in two stages, each turbine having the same pressure ratio, with reheat to T3 between the stages. Show that the specific work output will be maximum when

Solution

The T-s diagram is shown in Fig. 16.23.

Refer to Fig. 16.23.

Turbine work wt = cp (T3T4) + cp (T5T6)

Figure 16.23 T-s diagram for Brayton cycle with two-stage turbine

= cp (T3T4) + cp (T3T6)

Compressor work

= cpT1 (c − 1)

Specific work output

For maximum specific work output

Example 16.8

In a gas turbine, air is taken in at 1 bar pressure and 30°C. It is compressed to 6 bar with an isentropic efficiency of 87%. Heat is added by the combustion of fuel in combustion chamber to raise the temperature to 700°C. The efficiency of the turbine is 85%. The calorific value of fuel used is 43.1 MJ/kg. For an air flow of 80 kg/min, calculate (a) the air/fuel ratio of turbine gases, (b) the final temperature of exhaust gases, (c) the net power developed, and (d) the overall thermal efficiency of plant.

Assume cpa = 1.005 kJ/kgK, γa = 1.4, cpg = 1.147 kJ/kgK, γg = 1.33.

Figure 16.24 T-s diagram for gas turbine with machine efficiencies

Solution

The T-s diagram is shown in Fig. 16.24.

p1 = 1 bar, T1 = 273 + 30 = 303 K

p2 = 6 bar, ηc = 0.87, T3 = 273 + 700 = 973 K

ηt = 0.85, CV = 43.1 MJ/kg

  1. f × c.v = acpg (T3T2)         [∵a + fa ]
  2. wnet = acpg (T3T4′) − acpa (T2′T1)

Example 16.9

In a gas turbine plant, air enters the compressor at 1 bar and 7°C. It is compressed to 4 bar with an isentropic efficiency of 82%. The maximum temperature at the inlet to the turbine is 800°C. The isentropic efficiency of turbine is 85%. The calorific value of fuel used is 43100 kJ/kg. The heat losses are 15% of the calorific value. Calculate (a) the compressor work, (b) the heat supplied, (c) the turbine work, (d) the net work, (e) the thermal efficiency, (f) the air-fuel ratio, (g) the specific fuel consumption, and (h) the ratio of compressor to turbine work. Assume cpa = 1.005 kJ/kgK, γa = 1.4, cpg = 1.147 kJ/kgK, γg = 1.33.

Solution

p1 = 1 bar, T1 = 280 K, ηc = 0.82, p2 = 4 bar, T3 = 1073 K, ηt = 0.85, ηcomb = 0.85.

The T-s diagram is shown in Fig. 16.25.

Refer to Fig 16.25

  1. The compressor work

     

    wc = cpa (T2′T1) = 1.005 (446 − 280) = 166.82 kJ/kg

     

    Figure 16.25 T-s diagram for gas turbine with machine efficiencies

  2. The heat supplied
  3.  

    T4′ = T3 − ηt (T3T4) = 1073 − 0.85 (1073 − 760.7) = 807.5 K

     

    The turbine work

     

    wt = cpg (T3T4′) = 1.147 (1073 −807.5) = 304.53 kJ/kg

     

  4. The net work output

     

    wnet= wtwc = 304.153 −166.83 = 137.7 kJ/kg

     

  5. The thermal efficiency
  6. f × C.V. (1 - 0.15) = (a + f ) Q × ηcomb
  7. The specific fuel consumption

Example 16.10

In a closed circuit gas turbine plant, the working fluid at 38°C is compressed with an adiabatic efficiency of 82%. It is then heated at constant pressure to 650°C. The fluid expands down to initial pressure in a turbine with an adiabatic efficiency of 80%. The pressure ratio is such that work done per kg is maximum. cpa = 1.005 kJ/kgK, cpg = 1.147 kJ/kgK. Calculate the pressure ratio for maximum work output and the corresponding cycle efficiency.

Solution

Refer to Fig 16.25.

For maximum specific work,

Example 16.11

A constant pressure closed cycle turbine plant works between a temperature range of 800°C and 30°C. The isentropic efficiencies of compressor and turbine are 80% and 90%, respectively. Find the pressure ratio of the cycle for maximum thermal efficiency and maximum specific output of the cycle.

Solution

T1 = 273 + 30 = 303 K, T3 = 273+ 800 = 1073 K, ηc = 0.8, ηt = 0.9

Pressure ratio for maximum specific output

Pressure ratio for maximum thermal efficiency

Example 16.12

A gas turbine plant operates between 5°C and 839°C. Find the following:

  1. Pressure ratio at which cycle efficiency equals to Carnot cycle efficiency
  2. Pressure ratio at which maximum work is obtained
  3. Efficiency corresponding to maximum work output

Solution

The efficiency of Carnot cycle

For simple gas turbine cycle,

  1. or

  2. For maximum work output

    or

Example 16.13

A simple open cycle gas turbine has a compressor turbine and a free power turbine. It develops electrical power output of 250 MW. The cycle takes in air at 1 bar and 288 K. The total compressor pressure ratio is 14. The turbine inlet temperature is 1500 K. The isentropic efficiency of compressor and turbine are 0.86 and 0.89, respectively. The mechanical efficiency of each shaft is 0.98. Combustion efficiency is 0.98 while combustor pressure loss is 0.03 bar. The alternator efficiency is 0.98. Take calorific value of fuel equal to 42,000 kJ/kg, cpa = 1.005 kJ/kgK and cpg = 1.15 kJ/kgK. Calculate (a) the air-fuel ratio, (b) the specific work output, (c) the specific fuel consumption, (d) the mass flow rate of air, and (e) the thermal efficiency of the cycle.

[IES 2004]

Solution

p1 = 1 bar, T1 = 288 K, p2 = 14 bar, p3 = 14 − 0.03 = 13.97 bar, T3 = 1500 K, P = 250 kW, ηc = 0.86, ηt = 0.89, ηmech = 0.98, ηcomb = 0.98, ηa = 0.98, CV = 42,000 kJ/kg, cpa = 1.005 kJ/kgK, cpg = 1.15 kJ/kgK

Figure 16.26 T-s diagram for gas turbine with pressure loss in combustion chamber

The T-s is shown in Fig. 16.26.

T2 = 288 (14)0.4/1.4 = 612.15 K

Compressor efficiency,

or     T2′T1 = 376.92 K

or     T2′ = 664.92 K

Turbine efficiency,

Actual net work done, wnet = Turbine work, wt − Compressor work, wc

Effective heat supplied

qe = cpg ( T3T2′)
= 1.15 (1500 − 664.92) = 960.34 kJ/kg

Actual heat supplied,

  1. Air/fuel ratio,
  2. Specific work output = 618.72 kJ/kg of air.
  3. Specific fuel consumption =
  4. Mass flow rate of air, ma × Specific work output
    = Electric power developed × Alternator efficiency
  5. Thermal efficiency of the cycle

Example 16.14

An open cycle gas turbine employs a regenerative arrangement. The air enters the compressor at 1 bar and 288 K and is compressed to 10 bar with a compression efficiency of 85%. The air is heated in the regenerator and the combustion chamber till its temperature is raised to 1700 K and during the process pressure falls by 0.2 bar. The air is then expanded in the turbine and passes to regenerate which has 75% effectiveness and causes a pressure drop of 0.2 bar. The isentropic efficiency of the turbine is 86%. By sketching the gas turbine system and showing the process on T-s diagram, calculate the thermal efficiency and the power output if mass flow rate of air is 100 kg/s. Take mechanical and alternator efficiency as 98% each. Take cpg = 1.15 kJ/kgK and cpa = 1.005 kJ/kgK.

[IES 2005]

Figure 16.27 Open cycle gas turbine: (a) Gas turbine system, (b) T-s diagram

Solution

The gas turbine system and T-s diagram are shown in Fig. 16.27.

Work consumed by the compressor,

wc = cpa (T2′T1)
= 1.005 (603.3 − 288) = 316.876 kJ/kg

Work done by the turbine

wt = cpg (T4T5′) = 1.15(1700 − 1040.34)
= 758.61 k/kg

Net work done = wt – wc = 758.61 − 316.876 = 441.73 kJ/kg

Regenerator effectiveness,

T3 = 0.75(1040.34 − 603.3) + 603.3 = 931.08 K

Heat supplied = cpa (T4T3) = 1.005(1700 − 931.08) = 772.76 kJ/kg

Thermal efficiency =

Power developed = 100 × 441.73 × 0.98 × 0.98 = 42423.75 kW

Example 16.15

Air enters the compressor of a gas turbine at 100 kPa, 300 K with a volumetric flow rate of 5 m3/s. The air is compressed in two stages to 1200 kPa with intercooling to 300 K between stages at a pressure of 350 kPa. The turbine inlet temperature is 1400 K and the expansion occurs in two stages with reheat to 1340 K between the stages at a pressure of 350 kPa. The compressor and turbine stage efficiencies are 87% and 85%, respectively. Draw the schematic diagram of the cycle and indicate the process on T-s diagram. Determine (a) the thermal efficiency of the cycles, (b) the back work ratio, (c) the net power developed in kW. Assume effectiveness of the regenerator as 80% and cp = 1.0045 kJ/kgK for air and gas.

[IES 2006]

Solution

The schematic diagram of the cycle and T-s diagram are shown in Fig. 16.28.

p1 = pa = 100 kPa, T1 = 300 K, 1 = 5 m3/s, p2 = p3 = p7 = p8 = 350 kPa,
p4 = p6 = 1200 kPa, T1 = T3 = 300 K, T6 = 1400 K, T8 = 1340 K,
ηc1 = ηc2 = 0.87, ηt1 = ηt2, ɛ = 0.85, cpa = cpg 1.0045 kJ/kgK

Compressor 1st stage:

Work done, wc1 = cpa (T2′T1) = 1.0045 (448.4 − 300) = 149.07 kJ/kg

Figure 16.28 Gas turbine with intercooling, reheating, and regeneration: (a) Gas turbine system, (b) T-s diagram

Compressor 2nd stage:

Work done, wc2 = cpa (T4′T3) = 1.0045 (445.50 − 300) = 146.16 kJ/kg

Total compressor work, wc = wc1 + wc2 = 295.23 kJ/kg

Turbine Ist stage:

or T7′ = 1400 − 0.85 (1400 − 984.55) = 1046.87 K

Work done, wt1 = cpg (T6T7′) = 1.0045 (1400 −1046.87) = 354.72 kJ/kg

Turbine 2nd stage:

or T9′ = 1340 − 0.85 (1340 − 936.82) = 997.3 K

Work done, wt2 = cpg (T8T9′) = 1.0045 (1340 − 997.3) = 344.24 kJ/kg

Total turbine work wt = wt1 + wt2 = 698.96 kJ/kg

Net work wnet = wt – wc = 698.96 − 295.23 = 403.73 kJ/kg

Regenerator effectiveness,

or

or T5′ = 886.94 K

Heat supplied, qs = cpg [(T6T5′) + (T8T7′)]

= 1.0045 [(1400 − 886.94) + (1340 − 1046.87)]
= 809.82 kJ/kg
  1. Thermal efficiency,
  2. Work ratio =
  3. p11 = ṁRT1

    or

    Power developed = × wnet = 580.72 × 403.73 = 234454 kW

Example 16.16

Air enters the compressor of a gas turbine plant operating on Brayton cycle at 1 bar, 27°C. The pressure ratio in the cycle is 6. If Wt = 2.5 Wc, where Wt and Wc are the turbine and compressor work, respectively, calculate the maximum temperature and the thermal efficiency of the cycle. Take γ = 1.4.

Solution

p1 = 1 bar, T1 = 27 + 273 = 300 K, p2 = 6 bar, γ = 1.4

Refer to Fig. 16.4, we get

Thermal efficiency =

Example 16.17

In an air-standard regenerative gas turbine cycle, the pressure ratio is 5. Air enters the compressor at 1 bar, 300 K and leaves at 490 K. The maximum temperature in the cycle is 1000 K. If the effectiveness of the regenerator and the isentropic efficiency of the turbine are each 80%, determine the cycle efficiency. Assume γ =1.4 for air.

Solution

The cycle on the T-s diagram is shown in Fig. 16.29 assuming compressor efficiency to be 100%.

Figure 16.29 T-s diagram for gas turbine with regeneration

Turbine efficiency, =

or

or T4′ = 705.11 K

Regenerative effectiveness =

or

or T5′ = 659.116 K

wc = cp (T2T1) = 1.005 × (475.14 − 300)
= 176.01 kJ/kg
wt = cp (T3) = 1.005 (1000 − 705.11)
= 296.36 kJ/kg

Net work, wnet = wtwc = 296.36 − 176.01

= 120.35 kJ/kg

Heat supplied, qs = cp (T3T5′)

= 1.005 × (1000 − 659.116) = 342.58 kJ/kg

Thermal efficiency =

16.13 ❐ EFFECTS OF OPERATING VARIABLES

The operating variables for the performance of open cycle gas turbine are as follows:

  1. Pressure ratio
  2. Compressor inlet temperature
  3. Turbine inlet temperature
  4. Efficiency of the compressor
  5. Efficiency of the turbine

These variables affect the thermal efficiency and the work ratio of the gas turbine.

16.13.1 Effect of Pressure Ratio

For the simple Brayton cycle,

Thermal efficiency,

Work ratio,

The thermal efficiency for various values of γ and rp are given in Table 16.2.

The variation of thermal efficiency with pressure ratio for different values of ratio of specific heats is shown in Fig. 16.30. It may be observed that the thermal efficiency increases as the pressure ratio and the ratio of specific heats increases.

The work ratios for different values of rp and T3/T1 ratio are given in Table 16.3.

 

Table 16.2 Effect of pressure ratio on thermal efficiency

Figure 16.30 Variation of thermal efficiency with pressure ratio of a gas turbine

 

Table 16.3 Effect of pressure ratio and temperature ratio on work ratio (γ = 1.4)

Figure 16.31 Variation of work ratio with pressure ratio of a gas turbine

The variation of work ratio with pressure ratio for different values of temperature ratio is shown in Fig. 16.31. It may be observed that the work ratio decreases as the pressure ratio increases. However, the work ratio increases with increase in temperature ratio.

16.13.2 Effect of Efficiencies of Compressor and Turbine on Thermal Efficiency

From Eq. (16.11), we have

where

Figure 16.32 Variation of thermal efficiency with pressure ratio

Figure 16.33 Variation of thermal efficiency with temperature ratio

Let then

For

The effect of machine efficiencies on thermal efficiency has been shown in Fig. 16.32. The thermal efficiency initially increases, becomes maximum, and then starts decreasing. Figure 16.33 shows the effect of temperature ratio on thermal efficiency. It may be observed that the thermal efficiency increases as temperature ratio increases.

16.14 ❐ MULTI-SHAFT SYSTEMS

It is possible to use two turbines in parallel by using a single compressor and a single combustion chamber. The arrangement for such a system is shown in Fig. 16.34. For 1 kg of air through the compressor, a certain fraction x is required to pass through turbine I in order to drive the compressor. Then (1 − x) kg passes through turbine II to produce the power output. This is advantageous in that the speed of the compressor would not be affected by the load. A further advantage in this respect can be had by the use of two combustion chambers as shown in Fig. 16.35.

In such an arrangement, the inlet temperatures of the two turbines, as well as their speeds, are separately controllable.

Figure 16.34 Two turbine in parallel

Figure 16.35 Two turbines of parallel with two combustion chambers

Figure 16.36 Two turbines in series

16.15 ❐ MULTI-SHAFT SYSTEM TURBINES IN SERIES

The arrangement shown in Fig. 16.36 has two shafts with two turbines in series. It gives advantage of independent shaft speeds.

16.16 ❐ GAS TURBINE FUELS

The various fuels used for gas turbines are as follows:

  1. Gaseous fuels: Natural gas, blast furnace, and producer gas
  2. Liquid fuels: Liquid fuels of petroleum origin such as distillate oils or residual oils
  3. Solid fuels: Pulverised coal and crushed coal
16.17 ❐ BLADE MATERIALS

Materials used in the manufacture of the blades consist of mainly three different elements, namely iron, nickel, and cobalt with chromium which forms one of the major alloying elements, since it gives high resistance to oxidation. Other alloying elements that have been used include most of the metals of the periodic table.

Materials working under high stress and high temperature have a particular rate of creep. It means elongation under stress is not a fixed quantity as it is in case of normal temperature. Thus, elongation continues to increase with time and the blades gradually take up the original gap provided at their tips. Therefore, contact with casing results in failure. The repeated heating and cooling of the material affect the physical properties of the material.

16.17.1 Selection

The turbine blades are the most conditioned members of a gas turbine. They must withstand the following:

  1. Effects of high operating temperature
  2. Centrifugal tensile stresses due to rotational speeds of the order of 8000 to as high as 30,000 r.p.m.
  3. Bending stress due to equivalent impulse load of the gases acting at a certain distance from fixing of the cantilever blade.
  4. Hot erosive and corrosive effects due to high temperature combustion products e.g., CO2 and CO with O2.

Turbine blade material must, therefore, be selected from considerations of the working conditions.

16.17.2 Requirements of Blade Material

The material should possess the following characteristics:

  1. Maximum strength at high temperature
  2. High creep strength
  3. High resistance of corrosion
  4. Maximum erosion resistance
  5. Structural stability when exposed to varying temperatures
  6. Castability or forgability
  7. Weldability, if welding is used in the manufacture
  8. Machinability
  9. Absence of embrittlement in the service
16.18 ❐ COOLING OF BLADES

The thermodynamic analysis of the gas cycle shows that the efficiency of the cycle depends on the range of maximum temperature of the gas. If maximum temperature is increased, efficiency of the cycle is also increased. Any increase in the maximum temperature of gas results in an increase in the blade temperature, thereby inducing more thermal stresses in the blade material. Thus, this limits the capacity of gas temperature when blades are cooled properly.

16.18.1 Advantages of Cooling

  1. Specific thrust of jet engine increases by 32% for gas temperature increase from 800°C to 1100°C at the expense of 7% specific fuel consumption.
  2. If the maximum temperature is increased to 1600°C in place of present practice of 900°C, the specific fuel consumption will be increased by 50% and there is an increase in specific power output by 200%.

16.18.2 Different Methods of Blade Cooling

  1. Internal air cooling: It is done by supplying cold air through hollow blade. It employs an internal baffle or deflector to direct the flow over the hotter portions on the internal surface (Fig. 16.37).
  2. Film cooling: It involves supplying of thin film of cold compressed air through a narrow slit on the blade’s surface to form a cold boundary layer over the blades (Fig. 16.38).
  3. Water cooling: Circulation of water is maintained through a section of the blade from the root towards the tip.
  4. Rim cooling: The rim is cooled by circulating the cooling fluid and thus the blade temperature is reduced by conduction.

Figure 16.39 shows the distribution of temperature over the blade height.

Figure 16.37 Internal cooling of blade

Figure 16.38 Air film cooling of blade

Figure 16.39 Distribution of temperature over blade height

Example 16.18

A gas turbine unit receives air at 100 kPa and 300 K and compresses it adiabatically to 620 kPa. The fuel has heating value of 44180 kJ/kg and fuel/air ratio is 0.017 kg fuel per kg air. The isentropic efficiencies of the compressor and the turbine are 88% and 90%, respectively. Calculate the compressor work, the turbine work, and the thermal efficiency.

Solution

The T-s diagram is shown in Fig. 16.40.

Given: p1 = 100 kPa, T1 = 300 K, p2 = 620 kPa, C.V. = 44180 kJ/kg,

F:A = 0.017 kg fuel/kg air, ηc = 0.88, ηt = 0.90

Refer to Fig. 16.40.

Isentropic process 1−2:

or T2 = 300 × 1.6842 = 505.3 K

Compressor efficiency,

or T2′ = 533.3 K

Combustion process 2′−3:

f × C.V. = a × cpa (T3T2′) + f × cpg (T3T2′)

Let cpa = 1.005 kJ/kgK and cpg = 1.15 kJ/kgK

0.017 × 44180 = (1 × 1.005 + 0.017 × 1.15) (T3 − 533.3)
= 1.0246 (T3 − 533.3)

Figure 16.40 T-s diagram for gas turbine

or T3 = 1266.4 K

Isentropic process 3−4:

Turbine efficiency,

or

or T4′ = 803.3 K

Compressor work, wc = cpa (T2′T1) = 1.005 (533.3 − 300) = 234.47 kJ/kg

Turbine work,

= (1.005 + 0.017 × 1.15) (1266.4 − 803.3) = 477.47 kJ/kg

Net work, wnet = wtwc = 474.47 − 234.47 = 240 kJ/kg

Heat supplied, qs = f × C.V. = 0.017 × 44180 = 751.06 kJ/kg

Thermal efficiency,

Example 16.19

The pressure ratio of an open cycle constant pressure gas turbine plant is 6. The temperature range of the plant is 20°C and 850°C. Using the following data; cpa = 1 kJ/kgK, cpg = 1.05 kJ/kgK, γa = γg = 1.4, C.V. of fuel = 44,000 kJ/kg, ηc = 0.85, ηt = 0.90, ηcomb = 0.95, find (a) the thermal efficiency of the plant, (b) the net power developed if circulation of air is 5 kg/s, (c) the air-fuel ratio, and (d) the specific fuel consumption.

Solution

Given: rp = 6, T1 = 273 + 20 = 293 K, T3 = 273 + 850 = 1223 K, cpa = 1 kJ/kgK

cpg = 1.05 kJ/kgK, γa = γg = 1.4, CV = 44,000 kJ/kg,

ηc = 0.85, ηt = 0.90, ηcomb = 0.95, a= 5 kg/s

Refer to Fig. 16.40

Isentropic process 1−2:

or

or T2′ = 523.43 K

Combustion process 2′−3

f × C.V. × ηcomb = (a × cpa + f × cpg) (T3T2′)

Isentropic process 3−4:

Compressor work, wc = cpa (T2′T1) = 1 × (523.43 − 293) = 230.43 kJ/kg

Turbine work, wt =

Net work per kg of air, wnet = wtwc = 411.19 − 230.43 = 180.76 kJ/kg

Heat supplied,

  1. Thermal efficiency of plant,
  2. Net power developed,
    P = a × wnet = 5 × 180.76 = 903.8 kW
  3. Specific fuel consumption,

Example 16.20

Determine the efficiency of a gas turbine plant fitted with a heat exchanger of 75% effectiveness. The pressure ratio is 4:1 and the compression is carried out in two stages of equal pressure ratio with intercooling back to initial temperature of 290 K. The maximum temperature is 925 K. The isentropic efficiency of the turbine is 88% and the isentropic efficiency of each compressor is 85%. For air γ = 1.4, cp = 1.005 kJ/kgK.

Solution

The T-s diagram is shown in Fig. 16.41.

Given: T6 = 925 K, ηt = 0.88, ηc = 0.85, γ = 1.4, cp = 1.005 kJ/kgK.

Refer to Fig. 16.41.

L.P. compressor:

or T2′ = 364.7 K

T4′ = 364.7 K for the H.P. compressor.

H.P. turbine:

Figure 16.41 T-s diagram for gas turbine with intercooling

or T7′ = 560.3 K

Heat exchanger effectiveness,

or

or T5′ = 511.4 K

Heat supplied, qs = cp (T6T5′) = 1.005 (925 − 511.4) = 415.67 kJ/kg

Compressor work, wc = 2cp (T2′T1) = 2 × 1.005 (364.7 − 290) = 150.147 kJ/kg

Turbine work, wt = cp (T6T7′) = 1.005 (925 − 500.3) = 366.52 kJ/kg

Net work, wnet = wtwc = 366.52 − 150.147 = 216.376 kJ/kg

Efficiency of the gas turbine,

Example 16.21

Air enters the compressor of a gas turbine plant operating on Brayton cycle at 1 bar and 27°C. The pressure ratio in the cycle is 6. Assuming the turbine work 2.5 times the compressor work, calculate the maximum temperature in the cycle and cycle efficiency. Take γ = 1.4

Solution

Given: p1 = 1 bar, T1 = 273 + 27 = 300 K, rp = 6, wt = 2.5 wc

Refer to Fig. 16.4 (b)

Process 1−2:

T2 = T1(rp)(γ − 1)/γ = 300(6)0.4/1.4 = 500.55 K
wc = cp (T2T1) = 1.005(500.55 − 300) = 201.55 kJ/kg

Process 3−4:

Now 0.40266 T3 = 2.5 × 201.55

or T3 = 1251.36 K

Maximum temperature is 1251.36 K.

Turbine work wt = 2.5 × 201.55 = 503.87 kJ/kg

Net work, wnet = wtwc = 503.87 − 201.55 = 302.32 kJ/kg

Heat supplied, qs = cp (T3T2) = 1.005(1251.36 − 500.55)

= 754.564 kJ/kg

Cycle efficiency,

Example 16.22

A gas turbine cycle takes in air at 25°C and atmospheric pressure. The compression pressure ratio is 4. The isentropic efficiency of the compressor is 75%. The inlet temperature to turbine is limited to 750°C. What turbine efficiency would give overall cycle efficiency zero percent?

Solution

Given: T1 = 273 + 25 = 298 K, p1 = 1.013 bar,

rp = 4, ηc = 0.75, T3 = 273 + 750 = 1023 K

Refer to Fig. 16.40,

Process 1−2:

T2 = T1 (rp)(γ − 1)/γ = 298(4)0.4/1.4 = 442.8 K

or T2′ = 491K

Process 3−4:

T4′ = 1023 − 334.6 ηt

Compressor work, wc = cp (T2′T1) = 1.005(491 − 298) = 193.96 kJ/kg

Turbine work, wt = cp (T3T4′) = 1.005(1023 − 1023 + 334.6 ηt)

= 336.27 ηt

Net work, wnet = wtwc = 336.27 ηt − 193.96

Heat supplied, qs = cp (T3T2′) = 1.005(1023 − 491) = 534.66 kJ/kg

Cycle efficiency,

or 336.27 ηt − 193.96 = 0

or ηt = 0.5768 or 57.68%

Example 16.23

In a Brayton cycle gas turbine power plant, the minimum and maximum temperatures of the cycle are 300 K and 1200 K. The compression is carried out in two stages of equal pressure ratio with intercooling of the working fluid to the minimum temperature of the cycle after the first stage of compression. The entire expansion is carried out in one stage only. The isentropic efficiency of both compressors is 0.85 and that of the turbine is 0.9. Determine the overall pressure ratio that would give the maximum net work per kg working fluid. Derive the expression you use. γ = 1.4.

[IES, 1995]

Solution

Given: T1 = T3 = 300 K, T5 = 1200 K, , ηc1 = ηc2 = 0.85, ηt = 0.90, γ = 1.4

The T−s diagram is shown in Fig. 16.42.

Figure 16.42 T-s diagram for gas turbine

Compressor 1:

Compressor 2:

Compressor work, wc = wc1 + wc2 = cp [(T2T1) + (T4′T1)]

Turbine:

Turbine work,

Net work done, wnet = wtwc

For maximum net work,

Example 16.24

In an open-cycle gas turbine plant, air enters at 15°C and 1 bar, and is compressed in a compressor to a pressure ratio of 15. The air from the exit of compressor is first heated in a heat exchanger which is 75% efficient by turbine exhaust gas and then in a combustor to a temperature of 1600 K. The same gas expands in a two-stage turbine such that the expansion work is maximum. The exhaust gas from HP turbine is reheated to 1500 K and then expands to LP turbine. The isentropic efficiencies of compressor and turbine may be taken as 86% and 88%, respectively. The mechanical efficiencies of compressor and turbine are 97% each. The alternator efficiency is 98%. The output of turbo-alternator is 250 MW. Sketch the system and show the process on T-s diagram, Determine (a) the cycle thermal efficiency, (b) the work ratio, (c) the specific power output, and (d) the mass flow rate of air.

[IES, 2007]

Solution

The schematic is shown in Fig. 16.43(a) and T-s diagram in Fig. 16.43(b).

  1. ηc = 0.86, ηt = 0.88, ηalt = 0.98, (ηmech)c = (ηmech)t = 0.97

    Figure 16.43 Open cycle gas turbine: (a) Schematic, T-s diagram

    Compressor work,

    For maximum expansion work,

    p5 = pi = 3.873 bar

    T5 = = 1600 = 1086.7 K

    ηt1 =

    or T5′ = T4ηt1 (T4T5)

    = 1600 − 0.88 (1600 − 1086.7) = 1158.5 K

    T7 = T6

    p7 = p1 = 1 bar, p6 = pi = 2.873 bar

    T7 = 1500 = 1018.8 K

    ηt2 =

    or T7′ = 1500 − 0.88 (1500 − 1018.8) = 1076.5 K

    Turbine work,

    wt = wt1 + wt2
    = 1.005 [(1600 − 1158.5) + (1500 − 1076.5)] × 0.97
    = 843.245 kg

    Net work, wnet = wtwc

    = 843.245 − 405.2 = 438.045 kJ/kg

    Heat exchanger effectiveness =

    or 0.75 =

    or T3′ = 977.15 K

    Heat supplied = 1.005 [(1600 − 977.15) + (1500 − 1158.5)] = 969.17 kJ/kg

    Thermal efficiency,

    ηth = = 0.452 or 45.2%

  2. Work ratio = = 0.5195.
  3. Specific power output

    1 × wnet = 1 × 438.045 = 438.045 kW

  4. Mass flow rate of air =

    = = 559.3 kg/s

Example 16.25

A gas turbine utilises a two-stage centrifugal compressor. The pressure ratios for the first and second stages are 2.5 and 2.1, respectively. The flow of air is 10 kg/s, this air being drawn at 1.013 bar and 20°C. If the temperature drop in the intercooler is 60°C and the isentropic efficiency is 90% for each stage, calculate:(a) the actual temperature at the end of each stage and (b) the total compressor power. Assume γ = 1.4 and cp= 1.005 kJ/kgK for air.

[IES, 2012]

Solution

Given: rp1 = 2.5, rp2 = 2.1, a = 10 kg/s, p1 = 1.013 bar, T1 = 20 + 273 = 293 K, ∆T2 = 60°C, ηi1 = ηi2 = 0.9, γ = 1.4, cp = 1.005 kJ/kgK.

The T–s diagram is shown in Fig. 16.44.

  1. = 1.299

    or T2 = 380.68 K

    ηi1 =

    or 0.9 =

    or T2′ = 390.42 K or 117.42°C

    Temperature at the end of first stage is 117.42°C.

    Temperature at the inlet of second stage

    = 117.42 – 60 = 57.42°C
    T3 = 57.42 + 273 = 330.42 K

    Figure 16.44 T-s diagram

    or T4 = 408.44 K

    ηi2 =

    0.9 =

    Temperature at the end of second stage,

    T4′ = 417.11 K or 144.1°C
  2. Pc1 = acp (T2′T1) = 10 × 1.005 (390.42 – 293) = 979.07 kJ/s

    Pc2 = acp (T4′T3) = 10 × 1.005 (417.11 – 330.42) = 877.23 kJ/s

    Total compressor power,

    Pc = Pc1 + Pc2 = 979.07 + 871.23 = 1850.3 kW

Example 16.26

An open cycle gas turbine takes in air at 300 K and 1 bar and develops a pressure ratio of 20. The turbine inlet temperature is 1650 K. The polytropic efficiency of the compressor and the turbine each is 90%. The pressure loss in the combustor is 3% and the alternator efficiency is 97%. Take cpa= 1.005 kJ/kgK and cpg = 1.128 kJ/kgK for air and gas, respectively. The calorific value of fuel is 42 MJ/kg. Determine the overall efficiency, the specific power output, the fuel to air ratio and the specific fuel consumption.

[IAS, 2010]

Solution

Given: T1 = 300 K, p1 = 1 bar, rp = 20, T3 = 1650 K, hc = ht = 0.90, ∆p2 = 3%, halt = 0.97, cpa = 1.005 kJ/kgK, cpg = 1.128 kJ/kgK, C.V. = 42 MJ/kg

The T–s diagram is shown in Fig. 16.45.

Compressor process 1–2′:

or T2 = 300 × 2.3535 = 706 K

ηc =

or 0.9 =

or T2′ = 747.8 K

p2 = 20 × 1 = 20 bar
p3 = 0.97 × 20 = 19.4 bar

Combustion process 2′–3:

f × CV = f × cpa (T3T2′) + f × cpg (T3T2′)
= (a × cpa + f × cpg) (T3T2′)

Figure 16.45 T-s diagram

or f × 42 × 103 = (a × 1.005 + f × 1.128) (1650 – 747.8)

= 906.7 a + 1017.7 f

or 40982.3 f = 906.7 a

Turbine process 3–4′:

γg = , R = cpgcvg, cvg = cpgR = 1.128 – 0.287 = 0.841 kJ/kgK

γg = = 1.34

T4 = = 777.5 K

ηt =

or 0.9 =

or T4′ = 864.75 K

Specific power output:

wnet = wtwc = cpg (T3T4′) – cpa (T4′T1)
= 1.128 (1650 – 864.75) – 1.005 (747.8 – 300)
= 435.72 kJ/kg of air

Overall efficiency, ηoverall = =0.4281 or 42.81%

Specific fuel consumption =

= = 0.1828 kg/kWh

Summary for Quick Revision

  1. Gas turbines may be classified as: (i) constant pressure combustion (a) open cycle type (b) closed cycle type and (ii) constant volume combustion
  2. Brayton cycle:
    1. Compressor work, wc = cp (T2T1)
    2. Heat supplied, qs = cp (T3T2)
    3. Turbine work, wt = cp (T3T4)
    4. Net work, wnet = wtwc
    5. Thermal efficiency, ηth =
    6. Specific output =
    7. For maximum specific output, T2 = T4
    8. Maximum work output, (wnet)max =
    9. Work ratio, Rw
    10. Optimum pressure ratio for maximum specific work output,
    11. rp =

      and

  3. Cycle operation with machine efficiencies:
    1. Optimum pressure ratio for maximum specific work output

      Isentropic efficiency of compressor,

      Isentropic efficiency of turbine,

      For specific work output to be maximum for given temperature limits,

      Pressure ratio,

    2. Optimum pressure ratio for maximum cycle efficiency,
  4. Open cycle constant pressure gas turbine:
    Wc = a cpa (T2′T1)
    Wt = (a + f) cpg (T3T4′)
    Wnet = wtwc
    Q = f × C.V.

    ηth =

  5. The thermal efficiency of open cycle gas turbine can be improved by: Regeneration, intercooling, reheating, and combination of above.
    1. Regeneration.
      1. Without machine efficiencies: T5 = T4

        ηth = 1 – , c = , α =

      2. With machine efficiencies.

        ηth = 1 –

      3. Effectiveness of regenerator, =
    2. Intercooling:

      For minimum compressor work.

      wc = cpa (T2′T1) + cpa(T4′T3)

      Intermediate pressure for minimum compressor work,

      pi =

      For T3 = T1 and ηc1 = ηc2, pi =

      (wc)min =

    3. Reheating: For ηt1 = ηt2 = ηt and T5 = T3, pi = for maximum wt.

      (wc)max = 2cpg ht T3

Multiple-choice Questions

  1. Consider the following statements:
    1. Intercooling is effective only at lower pressure ratios and high turbine inlet temperatures.
    2. There is very little gain in thermal efficiency when intercooling is used without the benefit of regeneration.
    3. With higher value of γ and cp of the working fluid, the net power output the Brayton cycle will increase.

    Of these statements,

    1. I, II, and III are correct
    2. I and II are correct
    3. I and III are correct
    4. II and III are correct
  2. In a gas turbine cycle with two stages of reheating, working between maximum pressure p1 and minimum pressure p4 the optimum reheat reassure would be
    1. (p1p4)1/2 and (p1p2)2/3
    2. and
    3. and
    4. and
  3. Intercooling in gas turbines
    1. decreases net output but increases thermal efficiency
    2. increases net output but decreases thermal efficiency
    3. decreases both net output and thermal efficiency
    4. increases both net output and thermal efficiency
  4. Figure 16.46 shows four plots A, B, C, and D of thermal efficiency against pressure ratio.

    Figure 16.46

    The curve which represents that of a gas turbine plant using Brayton cycle (without regeneration) is the one labelled

    1. A
    2. B
    3. C
    4. D
  5. The optimum intermediate pressure pi for a gas turbine plant operating between pressure limits p1 and p2 with perfect intercooling between the two stages of compression (with identical isentropic efficiency) is given by
    1. p1 = p2p1
  6. In a gas turbine cycle, the turbine output is 600 kJ/kg, the compressor work is 400 kJ/kg and the heat supplied is 1000 kJ/kg. The thermal efficiency of this cycle is:
    1. 80%
    2. 60%
    3. 40%
    4. 20%
  7. In a single-stage open-cycle gas turbine, the mass flow through the turbine is higher than the mass flows through compressor, because
    1. the specific volume of air increases by use of intercooler
    2. the temperature of air increases in the reheater
    3. the combustion of fuel takes place in the combustion chamber
    4. the specific heats at constant pressure for incoming air and exhaust gases are different
  8. A gas turbine develops 120 kJ of work while the compressor absorbed 60 kJ of work and the heat supplied is 200 kJ. If a regenerator which would recover 40% of the heat in the exhaust were used, then the increase in the overall thermal efficiency would be:
    1. 10.2%
    2. 8.6%
    3. 6.9%
    4. 5.7%
  9. A gas turbine works on which one of the following cycles?
    1. Brayton
    2. Rankine
    3. Stirling
    4. Otto
  10. Reheating in a gas turbine
    1. increases the compressor work
    2. decreases the compressor work
    3. increase the turbine work
    4. decreases the turbine work
  11. Consider the following statements relating to a closed gas turbine cycle:
    1. The cycle can employ monatomic gas like helium instead of air to increase the cycle efficiency if other conditions are the same.
    2. The efficiency of heat exchanger increases with the use of helium.
    3. The turbine blades suffer higher corrosion damages.
    4. Higher output can be obtained for the same size.

    Which of these statements are correct?

    1. I, II, and III
    2. I, II, and IV
    3. II, III, and IV
    4. I, III, and IV
  12. The efficiency of a simple gas turbine can be improved by using a regenerator, because the
    1. work of compression is reduced
    2. heat required to be supplied is reduced
    3. work output of the turbine is increased
    4. heat rejected is increased
  13. In a regenerative feed heating cycle, the economic number of the stages of regeneration
    1. increases as the initial pressure and temperature increase
    2. decreases as the initial pressure and temperature increase
    3. is independent of the initial pressure and temperature
    4. depends only on the condenser pressure
  14. Match List I (Blades) with List II (Features) and select the correct answer using the codes given below the Lists:
    List I
    (Blades)
    List-II
    (Features)
    A. Ceramic blades 1. High creep strength
    B. Steam turbine blades 2. Forged and machined
    C. Alloy steel blades 3. Precision cast
    D. Compressor blades 4. Thick at mid chord
    4. Thin trailing edge

    Codes:

              A B C D

    1.   2  1  5  4
    2.   3  4  5  1
    3.   2  4  3  5
    4.   3  2  1  5
  15. Consider the following statements about modification in a gas turbine power plant working on a simple Brayton cycle:
    1. Incorporation of the regeneration process increases specific work output as well as thermal efficiency.
    2. Incorporation of regeneration process increases thermal efficiency but specific work output remains unchanged.
    3. Incorporation of intercooling process in a multi-stage compression increases specific work output but the heat input also increases.
    4. Incorporation of intercooling process in a multi-stage compression system increases specific work output, the heat addition remains unchanged.

    Which of the above statements are correct?

    1. 1 and 3
    2. 1 and 4
    3. 2 and 3
    4. 2 and 4
  16. A power plant, which used a gas turbine followed by steam turbine for power generation, is called
    1. Topping cycle
    2. Bottoming cycle
    3. Brayton cycle
    4. Combined cycle
  17. Consider the following statements:
    1. The speed of rotation of the moving elements of gas turbines is much higher than those of steam turbines.
    2. Gas turbine plants are heavier and larger in size than steam turbine plants.
    3. Gas turbines require cooling water for its operations.
    4. Almost any kind of fuel can be used with gas turbines.

    Which of the statements given above are correct?

    1. I and II
    2. I and III
    3. I and IV
    4. III and IV
  18. The thermal efficiency of a gas turbine cycle with regeneration in terms of T3 (maximum temperature), T1 (minimum temperature), rp (pressure ratio) and is given by
  19. Consider the following features for a gas turbine plant:
    1. Intercooling
    2. Regeneration
    3. Reheat

    Which of the above features in a simple gas turbine cycle increase the work ratio?

    1. I, II, and III
    2. Only I and II
    3. Only II and III
    4. Only I and IV
  20. Consider the following statements:

    For a large aviation gas turbine, an axial flow compressor is usually preferred over centrifugal compressor because

    1. the maximum efficiency is higher
    2. the frontal area is lower
    3. the pressure rise per stage is more
    4. the cost is lower

    Which of the statements given above are correct?

    1. I and IV
    2. Only I and II
    3. I, II, and III
    4. II, III, and IV
  21. In a reaction turbine, the heat drop in fixed blade is 8 kJ/kg and the total heat drop per stage is 20 kJ/kg. The degree of reaction is
    1. 40%
    2. 66.7%
    3. 60%
    4. 25%
  22. Consider the following statements:

    Which of the following increase the work ratio in a simple gas turbine plant?

    1. Heat exchanger
    2. Inter cooling
    3. Reheating

    Select the correct answer using the code given below:

    1. 1 and 2
    2. 2 and 3
    3. 1 and 3
    4. 1, 2, and 3
  23. Optimum pressure ratio for maximum specific output for ideal gas turbine plant operating at initial temperature of 300 K and maximum temperature of 1000 K is closer to
    1. 4
    2. 8
    3. 12
    4. 16
  24. The cycle shown in Fig. 16.47 represents a gas turbine cycle with intercooling and reheating.

    Figure 16.47

    Match List-X (Units) with List-Y (Processes) and select the correct answer using the codes given below the Lists:

    List-X List-Y
    A. Intercooler I. 1–2
    B. Combustor II. 2–3
    C. Reheater III. 3–4
    D. High pressure compressor IV. 4–5
    V. 6–7

    Codes:

            A   B    C    D

    1.  I    II    IV  III
    2.  II   IV  V   III
    3.  III  IV  V   II
    4.  II    V  IV   I
  25. Which one of the following is the correct statement? In a two-stage gas turbine plant with intercooling and reheating,
    1. both work ratio and thermal efficiency improve
    2. work ratio improves but thermal efficiency decreases
    3. thermal efficiency improves but work ratio decreases
    4. both work ratio and thermal efficiency decrease

Explanatory Notes

  1. 6. (a) Thermal efficiency = = 0.2 or 20%
  2. 8. (d) Thermal efficiency = 0.3 or 30%

    Heat in exhaust gas = 200 – 120 = 80 kJ

    Heat recovered in regenerator = 80 × 0.4 = 32 kJ

    Heat supplied = 200 – 32 = 168 kJ

    New thermal efficiency 0.357 or 35.7%

    Increase in efficiency = 35.7 – 30 = 5.7%

  1. 21. (c) Degree of reaction
  2. 23. (b) Optimum pressure ratio for maximum specific output of a gas turbine,

Review Questions

  1. List the fields of application of gas turbines.
  2. What are the limitations of a gas turbine?
  3. List four advantages of gas turbine over an IC engine.
  4. List four disadvantages of gas turbine over an IC engine.
  5. Enumerate five advantages of gas turbine over steam turbine.
  6. How do you classify gas turbines?
  7. Make a comparison of open and closed cycle gas turbines.
  8. Explain the position of the gas turbine in the power industry.
  9. Write the formula for the thermal efficiency of the Brayton cycle.
  10. Define specific output and work ratio.
  11. Write the expressions for maximum work output and work ratio of the Brayton cycle.
  12. Draw the schematic arrangement of open cycle constant pressure combustion gas turbine.
  13. What do you understand by regeneration?
  14. Define effectiveness of a regenerator.
  15. What is the intermediate pressure for intercooling to have minimum compressor work?

Exercises

16.1 A Brayton cycle operates with air between 1 bar, 300 K and 5 bar, 1000 K. The air is compressed in two stages with perfect intercooling. Similarly, in the turbine, expansion occurs in two stages with perfect reheating. Calculate the optimum intermediate pressure, net work output, and the fraction of turbine output that has to be put back to the compressor.

[Ans. 2.236 bar, 257.3 kJ/kg, 0.378]

16.2 A gas turbine unit operates at a mass flow of 30 kg/s. Air enters the compressor at a pressure of 1 bar, 288 K and is discharged from the compressor at a pressure of 10.5 bar. Combustion occurs at constant pressure and results in a temperature rise of 420 K. If the flow leaves of the turbine at a pressure of 1.2 bar, determine the net power output from the unit and also the thermal efficiency.

[Ans. 5338.74 kW, 42.55%]

16.3 In a regenerative gas turbine cycle, air enters the compressor at a temperature and pressure of 30°C and 1.5 bar and discharges at 220°C and 5.2 bar. After passing through the regenerator, the air temperature is 395°C. The temperature of air entering and leaving the gas turbine are 900°C and 510°C, respectively. Assuming no pressure drop through the regenerator, determine (a) the output per kg of air, (b) the efficiency of cycle, and (c) the work required to drive the compressor. Take cp = 1.005 kJ/kgK.

[Ans. 201 kJ/kg, 39.6%, 190.95 kJ/kg]

16.4 A Brayton cycle works between 1 atm, 27°C and 5 atm, 977°C. There are two stages of compression with perfect intercooling and two stages of expansion. The work output of first expansion stage being used to drive the two compressors, where the interstage pressure is optimised for the compressor. The air from the first stage turbine is again heated to 977°C and expanded. Calculate the power output of free power turbine and cycle efficiency without and with a perfect heat exchanger and compare them. Calculate the percentage improvement in the efficiency because of the addition if heat exchangers.

[Ans. 350.8 kJ/kg, 33.96%, 69.2%, 50.9%]

16.5 The following data apply to a gas turbine set employing a separate power turbine, regenerator, and intercooler between two-stage compression:

Isentropic efficiency of compression in each stage = 80%

Isentropic efficiency of compressor turbine = 88%

Isentropic efficiency of power turbine =88%

Turbine to compressor transmission efficiency = 98%

Pressure ratio in each stage of compression = 3:1

Temperature after intercooler = 297 K

Air mass flow rate = 15 kg/s

Regenerator effectiveness =80%

Regenerator gas-side pressure loss = 0.1 bar

Maximum turbine temperature = 1000 K

Ambient temperature = 327 K

Ambient pressure = 1 bar

Calorific value of fuel = 43.1 MJ/kg

Calculate (a) the net power output, (b) specific fuel consumption, and (c) overall thermal efficiency. Assume that the pressure losses in the air-side of the regenerator and combustion chamber are accounted for in the compressor efficiency. Take cpa = 1.005 kJ/kgK, cpa = 1.147 kJ/kgK, γα = 1.4 and γg = 1.33.

[Ans. 1811.67 kW, 0.32 kg/kWh, 26.1%]

16.6 In a gas turbine plant, air is compressed from 1 bar, 15°C through a pressure ratio of 4:1. It is then heated to 650°C in a combustion chamber and expanded back to atmospheric pressure of 1 bar in a turbine. Calculate the cycle efficiency and the work ratio if a perfect heat exchanger is used. The isentropic efficiencies of turbine and compressor are 85% and 80%, respectively. Take cpa = 1.00 kJ/kgK.

[Ans. 31.6%, 0.316]

16.7 In a gas turbine plant, the pressure ratio through which air at 15°C is compressed is 6. The same air is then heated to a maximum permissible temperature of 750°C, first in a heat exchanger having effectiveness of 75%, and then in the combustion chamber. The same air at 750°C is expanded in two stages such that the expansion work is maximum. The air is reheated to 750°C after the first stage. Determine (a) the cycle thermal efficiency, (b) the work ratio, and (c) the net shaft work per kg of air. The machine efficiencies may be assumed to be 80% and 85% for the compressor and turbine, respectively. Take cpa = 1.00 kJ/kgK.

[Ans. 33.6%, 0.398, 160 kJ/kg]

16.8 In a gas turbine plant, the air at 283 K and 1 bar is compressed to 4 bar with compression efficiency of 80%. The air is heated in the regenerator and in the combustion chamber till its temperature is raised to 973 K, and during the process, the pressure falls by 0.1 bar. The air is then expanded in the turbine and passes to the regenerator which has 75% effectiveness and causes a pressure drop of 0.14 bar. If the isentropic efficiency of the turbine is 85%, determine the thermal efficiency of the plant.

[Ans. 23.15%]

16.9 A gas turbine works on Brayton cycle between 27°C and 827°C. Determine the maximum net work per kg and the cycle efficiency. Take cpa = 1.005 kJ/kgK.

[Ans. 252.34 kJ/kg, 47.77%]

16.10 Atmospheric air enters the compressor of an open cycle constant pressure gas turbine at a pressure of 1 bar, 293 K. The pressure of air after compression is 4 bar. The isentropic efficiencies of compressor and turbine are 80% and 85%, respectively. The air-fuel ratio used is 90:1. If the mass flow rate of air is 3 kg/s, then calculate (a) the power developed and (b) the thermal efficiency of the cycle. Assume cpa = cpg = 1.0 kJ/kgK and γa = γg = 1.4. Calorific value of fuel = 41800 kJ/kg.

[Ans. 252.84 kW, 18.14%]

16.11 In a constant pressure open cycle gas turbine, air enters at 1 bar, 20°C and leaves the compressor at 5 bar. Temperature of gases entering the turbine is 680°C, pressure loss in the combustion chamber is 0.1 bar, ηc = 0.85, ηt = 0.80, ηcombustion = 0.85, ga = 1.4 and cpa = cpg = 1.024 kJ/kg.K. Calculate (a) the quantity of air circulated if power developed is 1065 kW, (b) the heat supplied per kg of air circulated, and (c) the thermal efficiency of the cycle. Neglect the mass of fuel as compared to that of air.

[Ans. 13.43 kg, 552.9 kJ/kg, 14.34%]

16.12 An open cycle constant pressure gas turbine draws air at 1.01 bar, 15°C and pressure ratio of 7:1. The compressor is driven by the H.P. turbine and L.P. turbine drives a separate power shaft. The isentropic efficiencies of compressor, H.P. and L.P. turbines are 0.82, 0.85, and 0.85, respectively. If the maximum cycle temperature is 610°C, calculate (a) the pressure and temperature of gases entering the power turbine, (b) the net power developed (c) the work ratio, and (d) the thermal efficiency of the plant. Assume cpa = 1.005 kJ/kgK, γα = 1.4; cpg = 1.15 kJ/kgK, γg = 1.333. Neglect mass of fuel.

[Ans. 1.636 bar, 654.4 K; 72.1 kW; 0.25; 18.8%]

16.13 A gas turbine employs a heat exchanger with a thermal ratio of 0:72. The turbine operates between 1.01 bar and 4.04 bar and ambient temperature is 20°C. The isentropic efficiencies of the compressor and turbine are 80% and 85%, respectively. The pressure drop on each side of the heat exchanger is 0.05 bar and in the combustion chamber 0.14 bar. Assume ηcombustion = 1.0 and. C.V. of fuel = 41800 kJ/kg. Calculate the increase in thermal efficiency due to heat exchanger.

16.14 A gas turbine has a pressure ratio of 6:1 and a maximum cycle temperature of 600°C. The isentropic efficiencies of the compressor and turbine are 82% and 85%, respectively. Calculate the power output when the air enters the compressor at 15°C at the rate of 15 kg/s. Take cpa = 1.005 kJ/kgK, γα = 1.4; cpg = 1.11 kJ/kgK, γg = 1.333.

[Ans. 920 KW]

16.15 A gas turbine plant operates between the temperatures of 1200 K and 320 K and corresponding pressures of 6 bar and 1.2 bar. The ratio of turbine and compressor efficiencies is 0.95. Determine the two efficiencies if the plant operates at optimum pressure ratio for maximum specific output. Assume γ = 1.4.

[Ans. ηt = 79.72%, ηc = 83.90%]

16.16 The maximum and minimum temperatures occurring in a closed cycle gas turbine plant are 927°C and 37°C. If the pressure at the outlet and the inlet of the compressor are 5 bar and 1 bar, respectively. Determine the compressor work, the turbine work, the heat supplied to the cycle, and the net work done in the cycle. Also determine the thermal efficiency of the cycle and the optimum pressure ratio for maximum power output for the given temperatures. Assume γ for air = 1.4 and cp = 1.005 kJ/kgK.

16.17 An open cycle constant pressure gas turbine plant operates with a pressure ratio of 6.5. The temperatures at the inlet to the compressor and turbine are 17°C and 820°C, respectively. Assume: specific heat at constant pressure of air and gas to be 1.005 kJ/kgK and 1.079 kJ/kgK, respectively; ratio of specific heats for air and gas to be 1.4; calorific value of fuel 44350 kJ/kg; efficiency of compressor and turbine to be 0.86 and 0.92, respectively. Calculate: (i) the power of the plant for air circulation of 6 kg/s, (ii) the thermal efficiency of plant, (iii) the air-fuel ratio, and (iv) the specific fuel consumption. Take the mass of the fuel into account.

16.18 A gas turbine power plant works between pressures of 1 bar and 5 bar and temperatures of 285 K and 1100 K. The intercooler cools the air at 2.3 bar to 285 K before the air is sent to the second stage compressor. The compressor air from the second-stage compressor passes through a regenerator having effectiveness of 0.72 and then through the combustion chamber. The heated air is then expanded in a high pressure turbine to 2.3 bar and is then reheated to 1100 K. The air is finally expanded in the low-pressure turbine to 1 bar. Assuming the compressor and turbine efficiencies to be 85%, determine: (i) the ratio of compression work to the turbine work, (ii) the power developed for an air flow of 3 kg/s, (iii) the thermal efficiency of the cycle, (iv) the heat rejected per second to the cooling water in the intercooler, and (v) the heat rejected per second to the atmosphere. Assume that all the components are mounted on the same shaft. Sketch the flow diagram of the turbine and represent the process on T-s plane. Also, assume cp = 1.005 kJ/kg.K and g = 1.4.

16.19 A gas turbine set takes in air at 27°C and 1 atm. The pressure ratio is 4 and the maximum temperature is 560°C. The compressor and turbine efficiencies are 83% and 85%, respectively. If the regenerator effectiveness is 0.75, determine the overall efficiency.

ANSWERS TO MULTIPLE-CHOICE QUESTIONS
  1. d
  2. a
  3. b
  4. b
  5. c
  6. d
  7. c
  8. d
  9. a
  10. c
  11. b
  12. b
  13. a
  14. a
  15. d
  16. d
  17. c
  18. d
  19. d
  20. b
  21. c
  22. b
  23. b
  24. b
  25. b