In this case pressure drop degree in power turbine and work in it:

,

(J/kg)

where power turbine efficiency ^*_t.p. = 0,89–0,91 in transitional channel ; _p = = 0,98–0,99 – pressure drop coefficient.

Gas temperature at the exit from the turbine:

Т^*₅ = Т^*₄ – L_t.p./[k_gR_g/(k_g – 1)] = 1055.941 – 306023.710 / [1.33 ∙ ∙288 / (1.33 – 1)] = 792.290 (K)

Define the value of reduced velocity in the range of _n = 0,5–0,75 and gas velocity at the exit from the turbine, static temperature and gas pressure are determined:

(m/s), (K),

(Pa).

1.2. Calculation of main parameters of gas-turbine power plant

Equivalent specific power is equal to:

N_es = η_m∙L_t.p.(1+g_fuel) = 0.9925 ∙ 306026.71∙(1+0.0213) = 310200.991 (W)

where _m – power turbine mechanical efficiency (chosen in the range _m = 0,99–0,995).

GTPP cycle efficiency:

_е = L_t.p. /q₁ = 306026.710 / / 899449.501 = 0.340

Useful work coefficient:  = 1 – L_t.p./(L_t + L_t.p.) = 1 – 306026.710 /

/ (451591.652 + 306026.710) = 0.596

Specific fuel consumption: С_е = 3600g_fuel/N_es = 3600 ∙ 0.0213 /

/ 310200.991 = 0.000247 (W^-1)

For given power air expenses through the compressor are equal: G_air = N_е/N_еs = 8600 ∙ 10³ / 310200.991 = 27.724.

Nominal expenses of gaseous fuel per hour (nm³/h) for measured gas with density: _fuel = 0,682 кг/м³

(m³/kg).

According to results of calculations plot the graphs of changing of values p^* and T^* along GTPP channels as shown on Figure 1.

2. Gasdynamic calculation of gas-turbine power plant

2.1 Determination of diametrical sizes on the entry of the compressor

For modern GTPP with power turbine с_1a = 150...180 m/s. Assume that с_1a = 180 m/s. Reduced velocity λ_1а and flux density function q(λ_1a) are calculated

,

At k = 1,4; .

Channel area at the entry of the compressor is defined

(m²),

where m = 0,0403.

Relative diameter is chosen , assume that and external diameter of working wheel at the entry of the compressor is defined

(m).

Then hub diameter and mean diameter are calculated:

(m),

h₁ = (D_1t – D_1H)/2 = (0.522 – 0.288) / 2 = 0.177 (m),

(m),

where h₁ is vane height of compressor first stage at the entry.

2.2. Determination of stages and air compression work distribution in two-spool compressors

In case when ^*_c > 12–13 compressor is made as two-spool, that consists of low pressure compressor (LPC) and high pressure compressor (HPC). In this case it’s necessary to define LPC and HPC works in such way L_LPC = (0,4–0,45)L_c, L_HPC = L_c – L_LPC.

L_LPC = 0.425 ∙ 421131.357 = 178980.827 (J/kg)

L_HPC = 421131.357-178980.827 = 242150.53 (J/kg)

Then pressure ratio ^*_1l  is calculated according to the formula:

^*_1l =[1 + ^*_LPCL_LPC(k – 1)/(kRТ_н)]^{k/(k – 1)} = [1 + 0.8666 ∙ ∙178980.827 ∙ (1.4 – 1)/(1.4 ∙ 287 ∙ 288)]^{1.4/(1.4 – 1)} = 1.536^3.5 = 4.493

LPC efficiency assumed greater on 1–2 % than taken in thermodynamic calculation ^*_c, hence ^*_LPC = 1.01 ∙ 0.858 = 0.8666. Air pressure р^*_1l and temperature Т^*_1l after LPC are calculated by the formulas:

р^*_1l = ^*_1l р^*₁ = 4.493 ∙ 98791.875 = 443832.526 (Pa),

Т^*_1l = Т^*₁[1 + (^*_1l^{(k – 1)/k} – 1)/ ^*_LPC] = 288 ∙ [1 + (4.493^{(1.4 – 1)/1.4} –

- 1) / 0.8666] = 466.410 (K)

Then air speeds after LPC с_a1l are determined:

с_a1l = 0,5(с_1a + с_2a) = (180 + 140) / 2 = 160 (m/s), where

с_1a, с_2a – reduced velocities at the entry, exit from the compressor.

After the calculation of critical c_cr1l and reduced _1la velocities after LPC, flux density function q(_1la)

(m/s) ,

_1la = с_1la / /с_cr1l = 160 / 395.216 = 0.405 q(_1la) = 1,577_1la(1 –

- 0,166²_1la)^2,5 = 1,577 ∙ 0.405 (1 – 0,166 ∙ ∙0.405²)^2,5 = 0.596, channel area after LPC F_1l is defined:

(m²).

Tip D_1lt and hub D_1lH diameters after LPC are determined in such way:

D_1t = D_1lt = 0.522 m;

(m)

h_1l = (D_1lt – D_1lH) / 2 = (0.522 – 0.448) / 2 = 0.037 (m),

D_1lm = D_1lH + h_1l = 0.448 + 0.037 = 0.485 (m).

After that air velocities after HPC are defined с_2a = 140 m/s.

After the calculation of critical c_cr2 and reduced 2_a velocities after HPC, flux density function q(_2a)

(m/s) ;

_2a = с_2a / /с_cr2 = 140 / 486.672 = 0.288;

q(2_a) = 1,577_2a(1 – 0,166²_2a)^2,5 = ,577 ∙ 0.288 (1 – 0,166 ∙ ∙0.288²)^2,5 = = 0.439, channel area after HPC F₂ is defined:

(m²).

Tip D_2t and hub D_2H diameters after HPC are determined in such way:

D_1lH = D_2H = 0.448 m;

(m)

h₂ = (D_2t – D_2H) / 2 = (0.482 – 0.448) / 2 = 0.017 (m),

D_1lm = D_2H + h₂ = 0.448 + 0.017 = 0.465 (m).

Let’s determine number of stages of LPC z_LPC and HPC z_HPC. Moreover angular velocity on HPC tip u_HPCt must be greater than LPC tip u_LPCt on 15–20 %, it means that u_HPCt = 1.15÷1.2u_LPCt = =1.175 ∙ 350 = 411.25 m/s, where u_LPCt = 250÷350 m/s (assume as 350 m/s) . Angular velocity near the hub of LPC working wheel is defined.

u_1H = u_LPCtD_1H/D_1t = 350 ∙ 0.288 / 0.522 = 193.1 (m/s) ,

Vane cascade density on diameter D_1H is assumed as (b/t)_1H = 1,5…2,0. Air flow spin on diameter D_1H is determined according to the formula:

w_u1H1=1,55с_1а/[1+1,5/(t/b)_H1] = 1.55 ∙ 180/[1 + 1.5 / 1.75] = =150.23 (m/s)

Work that is transferred by vanes to the air is calculated according to the Euler’s equation: L_st1=u_1Hw_u1H1=193.1 ∙ 150.23= =29009.413 (J/kg)

Effective work of last stage of LPC is determined as:

L_LPCz = u_1lHw_u1lH = 300.383 ∙ 141.714 = 42568.562 (J), where

u_1lH = u_LPC∙D_1lH/D_1t = 350 ∙ 0.448 / 0.522 = 300.383 m/s, and w_u1lH = 1,55с_1la/[1+1,5/(b/t)_1lH]=1.55∙160/[1+1.5/2]=141.714 m/s

Average value of compressor stage work:

(J/kg).

Number of LPC stages : .

Now we must round off the obtained value to the integer number, hence in my case there are 5 stages of LPC.

Let’s determine number of HPC stages.

At first let’s determine work near hub of HPC working wheel. Vane cascade density (b/t) at the entry of HPC is assumed as 1,6–1,9.

u_1lH = u_HPCtD_1lH/D_1lt = 411.25 ∙ 0.448 / 0.522 = 352.95 (m/s),

w_u1lH1l=1,55с_а1l/[1+1,5/(b/t)_1lH]=1.55∙160/[1+1.5/1.75]=133.54m/s

Effective work of first stage of HPC is determined as:

L_st1 = u_1lHw_u1lH1l = 352.95 ∙ 133.54 = 47132.943 (J).

Effective work of last stage of HPC is determined as:

L_stz = u_2Hw_u2 = 352.95 ∙ 116.85 = 41240.85 (J), where

u_2H = u_1lH =352.95 m/s, and

w_u2 = 1,55с_2a/[1+1,5/(b/t)_2H]=1.55∙140/[1+1.5/1.75]=116.85 m/s

Average value of compressor stage work:

(J/kg).

Number of HPC stages : . Let’s assume that according to my initial condition and previous calculation I have HPC which contain 5 stages.