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, η_c = 0.88, η_t = 0.90

Refer to Fig. 16.40.

Isentropic process 1−2:

or T₂ = 300 × 1.6842 = 505.3 K

Compressor efficiency, 

or T_2′ = 533.3 K

Combustion process 2′−3:

ṁ_f × C.V. = ṁ_a × c_pa (T₃ − T_2′) + ṁ_f × c_pg (T₃ − T_2′)

Let c_pa = 1.005 kJ/kgK and c_pg = 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 T_4′ = 803.3 K

Compressor work, w_c = c_pa (T_2′− 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, w_net = w_t − w_c = 474.47 − 234.47 = 240 kJ/kg

Heat supplied, q_s = ṁ_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; c_pa = 1 kJ/kgK, c_pg = 1.05 kJ/kgK, γ_a = γ_g = 1.4, C.V. of fuel = 44,000 kJ/kg, η_c = 0.85, η_t = 0.90, η_comb = 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: r_p = 6, T₁ = 273 + 20 = 293 K, T₃ = 273 + 850 = 1223 K, c_pa = 1 kJ/kgK

c_pg = 1.05 kJ/kgK, γ_a = γ_g = 1.4, CV = 44,000 kJ/kg,

η_c = 0.85, η_t = 0.90, η_comb = 0.95, ṁ_a= 5 kg/s

Refer to Fig. 16.40

Isentropic process 1−2:

or 

or T_2′ = 523.43 K

Combustion process 2′−3

ṁ_f × C.V. × η_comb = (ṁ_a × c_pa + ṁ_f × c_pg) (T₃ − T_2′)

Isentropic process 3−4:

Compressor work, w_c = c_pa (T_2′ − T₁) = 1 × (523.43 − 293) = 230.43 kJ/kg

Turbine work, w_t = 

Net work per kg of air, w_net = w_t − w_c = 411.19 − 230.43 = 180.76 kJ/kg

Heat supplied, 

1. Thermal efficiency of plant, 
2. Net power developed,P = ṁ_a × w_net = 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, c_p = 1.005 kJ/kgK.

Solution

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

Given:  T₆ = 925 K, η_t = 0.88, η_c = 0.85, γ = 1.4, c_p = 1.005 kJ/kgK.

Refer to Fig. 16.41.

L.P. compressor:

or T_2′ = 364.7 K

∴ T_4′ = 364.7 K for the H.P. compressor.

H.P. turbine:

Figure 16.41 T-s diagram for gas turbine with intercooling

or T_7′ = 560.3 K

Heat exchanger effectiveness, 

or 

or T_5′ = 511.4 K

Heat supplied, q_s = c_p (T₆ − T_5′) = 1.005 (925 − 511.4) = 415.67 kJ/kg

Compressor work, w_c = 2c_p (T_2′ − T₁) = 2 × 1.005 (364.7 − 290) = 150.147 kJ/kg

Turbine work, w_t = c_p (T₆ − T_7′) = 1.005 (925 − 500.3) = 366.52 kJ/kg

Net work, w_net = w_t − w_c = 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, r_p = 6, w_t = 2.5 w_c

Refer to Fig. 16.4 (b)

Process 1−2:

T₂ = T₁(r_p)^{(γ − 1)/γ} = 300(6)^0.4/1.4 = 500.55 K

w_c = c_p (T₂ − T₁) = 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 w_t = 2.5 × 201.55 = 503.87 kJ/kg

Net work, w_net = w_t − w_c = 503.87 − 201.55 = 302.32 kJ/kg

Heat supplied, q_s = c_p (T₃ − T₂) = 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,

r_p = 4, η_c = 0.75, T₃ = 273 + 750 = 1023 K

Refer to Fig. 16.40,

Process 1−2:

T₂ = T₁ (r_p)^{(γ − 1)/γ} = 298(4)^0.4/1.4 = 442.8 K

or T_2′ = 491K

Process 3−4:

T_4′ = 1023 − 334.6 η_t

Compressor work, w_c = c_p (T_2′ − T₁) = 1.005(491 − 298) = 193.96 kJ/kg

Turbine work, w_t = c_p (T₃ − T_4′) = 1.005(1023 − 1023 + 334.6 η_t)

= 336.27 η_t

Net work, w_net = w_t − w_c = 336.27 η_t − 193.96

Heat supplied, q_s = c_p (T₃ − T_2′) = 1.005(1023 − 491) = 534.66 kJ/kg

Cycle efficiency, 

or 336.27 η_t − 193.96 = 0

or η_t = 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.

Solution

Given: T₁ = T₃ = 300 K, T₅ = 1200 K, , η_c1 = η_c2 = 0.85, η_t = 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, w_c = w_c1 + w_c2 = c_p [(T₂ − T₁) + (T_4′ − T₁)]

Turbine: 

Turbine work, 

Net work done, w_net = w_t − w_c

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.

Solution

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

1. η_c = 0.86, η_t = 0.88, η_alt = 0.98, (η_mech)_c = (η_mech)_t = 0.97  Figure 16.43 Open cycle gas turbine: (a) Schematic, T-s diagramCompressor work,For maximum expansion work,p₅ = p_i = 3.873 barT₅ =  = 1600  = 1086.7 Kη_t1 = or T_5′ = T₄ − η_t1 (T₄ − T₅)= 1600 − 0.88 (1600 − 1086.7) = 1158.5 KT₇ = T₆ p₇ = p₁ = 1 bar, p₆ = p_i = 2.873 barT₇ = 1500  = 1018.8 Kη_t2 = or T_7′ = 1500 − 0.88 (1500 − 1018.8) = 1076.5 KTurbine work,w_t = w_t1 + w_t2= 1.005 [(1600 − 1158.5) + (1500 − 1076.5)] × 0.97= 843.245 kgNet work, w_net = w_t − w_c= 843.245 − 405.2 = 438.045 kJ/kgHeat exchanger effectiveness = or 0.75 = or T_3′ = 977.15 KHeat supplied = 1.005 [(1600 − 977.15) + (1500 − 1158.5)] = 969.17 kJ/kgThermal efficiency,η_th =  = 0.452 or 45.2%
2. Work ratio =  = 0.5195.
3. Specific power output1 × w_net = 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 c_p= 1.005 kJ/kgK for air.

Solution

Given: r_p1 = 2.5, r_p2 = 2.1, ṁ_a = 10 kg/s, p₁ = 1.013 bar, T₁ = 20 + 273 = 293 K, ∆T₂ = 60°C, η_i1 = η_i2 = 0.9, γ = 1.4, c_p = 1.005 kJ/kgK.

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

1.  = 1.299or T₂ = 380.68 Kη_i1 = or 0.9 = or T_2′ = 390.42 K or 117.42°CTemperature at the end of first stage is 117.42°C.Temperature at the inlet of second stage= 117.42 – 60 = 57.42°CT₃ = 57.42 + 273 = 330.42 K Figure 16.44 T-s diagramor T₄ = 408.44 Kη_i2 = 0.9 = Temperature at the end of second stage,T_4′ = 417.11 K or 144.1°C
2. P_c1 = ṁ_ac_p (T_2′ – T₁) = 10 × 1.005 (390.42 – 293) = 979.07 kJ/sP_c2 = ṁ_ac_p (T_4′ – T₃) = 10 × 1.005 (417.11 – 330.42) = 877.23 kJ/sTotal compressor power,P_c = P_c1 + P_c2 = 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 c_pa= 1.005 kJ/kgK and c_pg = 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.

Solution

Given: T₁ = 300 K, p₁ = 1 bar, r_p = 20, T₃ = 1650 K, h_c = h_t = 0.90, ∆p₂ = 3%, h_alt = 0.97, c_pa = 1.005 kJ/kgK, c_pg = 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

η_c = 

or 0.9 = 

or T_2′ = 747.8 K

p₂ = 20 × 1 = 20 bar

p₃ = 0.97 × 20 = 19.4 bar

Combustion process 2′–3:

ṁ_f × CV = ṁ_f × c_pa (T₃ – T_2′) + ṁ_f × c_pg (T₃ – T_2′)

= (ṁ_a × c_pa + ṁ_f × c_pg) (T₃ – T_2′)

Figure 16.45 T-s diagram

or ṁ_f × 42 × 10³ = (ṁ_a × 1.005 + ṁ_f × 1.128) (1650 – 747.8)

= 906.7 ṁ_a + 1017.7 ṁ_f

or 40982.3 ṁ_f = 906.7 ṁ_a

Turbine process 3–4′:

γ_g = , R = c_pg – c_vg, c_vg = c_pg – R = 1.128 – 0.287 = 0.841 kJ/kgK

γ_g =  = 1.34

T₄ =  = 777.5 K

η_t = 

or 0.9 = 

or T_4′ = 864.75 K

Specific power output:

w_net = w_t – w_c = c_pg (T₃ – T_4′) – c_pa (T_4′ – T₁)

= 1.128 (1650 – 864.75) – 1.005 (747.8 – 300)

= 435.72 kJ/kg of air

Overall efficiency, η_overall =  =0.4281 or 42.81%

Specific fuel consumption = 

=  = 0.1828 kg/kWh