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Helium is used as the working fluid in a Brayton cycle with regeneration. The pressure ratio of the cycle is 8, the compressor inlet temperature is 300 K, and the turbine inlet temperature is 1800 K. The effectiveness of the regenerator is 75 percent. Determine the thermal efficiency and the required mass flow rate of helium for a net power output of 60 MW, assuming both the compressor and the turbine have an isentropic efficiency of (a) 100 percent and (b) 80 percent. The properties of Helium are cp = 5.1926 kJ/kg.K and k = 1.667.

Respuesta :

shallomisaiah19

Answer:

Explanation:

Find the temperature at exit of compressor

[tex]T_2=300 \times 8^{\frac{1.667-1}{1.667} }\\=689.3k[/tex]

Find the work done by the compressor

[tex]\frac{W}{m} =c_p(T_2-T_1)\\\\=5.19(689.3-300)\\=2020.4kJ/kg[/tex]

Find the actual workdone by the compressor

[tex]\frac{W}{m} =n_c(\frac{W}{m} )\\\\=1 \times 2020.4kJ/kg[/tex]

Find the temperature at exit of the turbine

[tex]T_4=\frac{1800}{8^{\frac{1.667-1}{1.667} }} \\\\=787.3k[/tex]

Find the actual workdone by the turbine

[tex]1 \times 5.19 (1800-783.3)\\=5276.6kJ/kg[/tex]

Find the temperature of the regeneration

[tex]\epsilon = \frac{T_5-T_2}{T_4-T_2} \\\\0.75=\frac{T_5-689.3}{783.3-689.3} \\\\T_5=759.8k[/tex]

Find the heat supplied

[tex]Q_i_n=c_p(T_3-T_5)\\\\=5.19(1800-759.8)\\\\=5388.2kJ/kg[/tex]

Find the thermal efficiency

[tex]n_t_h=\frac{W_t-W_c}{Q_i_n} \\\\=\frac{5276.6-2020.4}{5388.2} \\\\n_t_h=60.4[/tex]

60.4%

Find the mass flow rate

[tex]m=\frac{W_net}{P} \\\\\frac{60 \times 10^3}{5276.6-2020.4} \\\\=18.42[/tex]

Find the actual workdone by the compressor

[tex]\frac{W_c}{m} =\frac{(\frac{W}{m} )}{n_c} \\\\=\frac{2020.4}{0.8} \\\\=2525.5kg[/tex]

Find the actual workdone by the turbine

[tex]\frac{W_t}{m} =n_t(\frac{W}{m} )\\\\=0.8 \times5.19(1800-783.3)\\\\=4221.2kJ/kg[/tex]

Find the temperature of the compressor exit

[tex]\frac{W_t}{m} =c_p(T_2_a-T_1)\\2525.5=5.18(T_2_a-300)\\T_2_a=787.5k[/tex]

Find the temperature at the turbine exit

[tex]4221.2=5.18(1800-T_4_a)\\\\T_4_a=985k[/tex]

Find the temperature of regeneration

[tex]\epsilon =\frac{T_5-T_2}{T_4-T_2}\\\\0.75=\frac{T_5-787.5}{985-787.5}\\\\T_5=935.5k[/tex]

Chrisnando

Answer:

a) 60.4%;  18.42 kg/s

b) 37.8% ;    35.4 kg/s

Explanation:

a) at an isentropic efficiency of 100%.

Let's first find the exit temperature of the compressor T2, using the formula:

[tex](r_p) ^k^-^1^/^k = \frac{T_2}{T_1}[/tex]

Solving for T2, we have:

[tex] T_2 = 300 * (8)^1^.^6^6^7^-^1^/^1^.^6^6^7 = 689.3 K [/tex]

Let's now find the work dine by the compressor.

[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]

[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]

The actual work done by the compressor =

[tex] W_c = 1 * 2020.4 = 2020.4 KJ/kg [/tex]

Let's find the temperature at the exit of the turbine, T4

[tex](r_p) ^k^-^1^/^k = \frac{T_3}{T_4}[/tex]

Solving for T4, we have:

[tex]T_4 = \frac{1800}{(8)^1^.^6^6^7^-^1^/^1^.^6^6^7} = 783.3 K[/tex]

Let's find the work done by the turbine.

[tex]\frac{W_t}{m} = c_p(T_3 - T_4)[/tex]

[tex]\frac{W_t}{m} = 5.19(1800 - 783.3) = 5276.6 KJ/kg[/tex]

The actual work done by the turbine:

= 1 * 5276.6 = 5276.6 KJ/kg

Let's find the regeneration temperature, using the formula:

[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]

Substituting figures, we have:

[tex] 0.75 = \frac{T_r - 689.3}{783.3 - 689.3} [/tex]

[tex] T_r = [0.75(783.3 - 689.3)] + 689.3 = 759.8 [/tex]

Let's calculate the heat supplied.

[tex]Q = c_p(T_3 - T_r)[/tex]

[tex] Q = 5.19(1800 - 759.8) [/tex]

Q = 5388.2 kJ/kg

For thermal efficiency, we have:

[tex] n = \frac{W_t - W_c}{Q} [/tex]

Substituting figures, we have:

[tex] n = \frac{5276.6 - 2020.4}{5388.2} = 0.604 [/tex]

0.604 * 100 = 60.4%

For mass flow rate:

Let's use the formula:

[tex] m = \frac{W_n_e_t}{P} [/tex]

Wnet = 60MW = 60*1000

[tex] m = \frac{60*10^3}{5276.6 - 2020.4} = 18.42 [/tex]

b) at an isentropic efficiency of 80%.

Let's now find the work done by the compressor.

[tex] \frac{W_c}{m} = c_p(T_2 - T_1) [/tex]

[tex] \frac{W_c}{m} = 5.19(689.3 - 300) = 2020.4 KJ/kg[/tex]

The actual work done by the compressor =

[tex] W_c = \frac{2020.4}{0.8}= 2525.5 KJ/kg [/tex]

Let's find the work done by the turbine.

[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]

[tex] \frac{W_t}{m} = 5.19(1800 - 787.5) = 5276.6 KJ/kg[/tex]

The actual work done by the turbine:

= 0.8 * 5276.6 = 4221.2 KJ/kg

Let's find the exit temperature of the compressor T2, using the formula:

[tex]\frac{W_c}{m} = c_p(T_2 - T_1) [/tex]

[tex] 2525.5 = 5.19(T_2 - 300) [/tex]

Solving for T2, we have:

[tex] T_2 = \frac{2525.5 + 300}{5.19} = 787.5 [/tex]

Let's find the temperature at the exit of the turbine, T4

[tex] \frac{W_t}{m} = c_p(T_3 - T_4) [/tex]

[tex] 4221.2 = 5.19(1800 - T_4) [/tex]

Solving for T4 we have:

[tex] T_4 = 958 K[/tex]

Let's find the regeneration temperature, using the formula:

[tex] e = \frac{T_r - T_2}{T_4 - T_2}[/tex]

Substituting figures, we have:

[tex] 0.75 = \frac{T_r - 787.5}{985 - 787.5} [/tex]

[tex] T_r = [0.75(958 - 787.5)] + 787.5 = 935.5 K [/tex]

Let's calculate the heat supplied.

[tex]Q = c_p(T_3 - T_r)[/tex]

[tex] Q = 5.19(1800 - 935.5) [/tex]

Q =  4486.2 kJ/kg

For thermal efficiency, we have:

[tex] n = \frac{W_t - W_c}{Q} [/tex]

Substituting figures, we have:

[tex] n = \frac{4221.2 - 2525.2}{4486.2} = 0.378 [/tex]

0.378 * 100 = 37.8%

For mass flow rate:

Let's use the formula:

[tex] m = \frac{W_n_e_t}{P} [/tex]

Wnet = 60MW = 60*1000

[tex] m = \frac{60*10^3}{4221.2 - 2525.2} = 35.4 kg/s [/tex]

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