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Leya [2.2K]
3 years ago
15

A piston–cylinder device containing carbon dioxide gas undergoes an isobaric process from 15 psia and 80°F to 170°F. Determine t

he work and the heat transfer associated with this process, in Btu/lbm. The properties of CO2 at room temperature are R = 0.04513 Btu/lbm·R and cv = 0.158 Btu/lbm·R.
Engineering
1 answer:
drek231 [11]3 years ago
7 0

Answer:

See explanation

Explanation:

Given:

Initial pressure,

p

1

=

15

psia

Initial temperature,

T

1

=

80

∘

F

Final temperature,

T

2

=

200

∘

F

Find the gas constant and specific heat for carbon dioxide from the Properties Table of Ideal Gases.

R

=

0.04513

Btu/lbm.R

C

v

=

0.158

Btu/lbm.R

Find the work done during the isobaric process.

w

1

−

2

=

p

(

v

2

−

v

1

)

=

R

(

T

2

−

T

1

)

=

0.04513

(

200

−

80

)

w

1

−

2

=

5.4156

Btu/lbm

Find the change in internal energy during process.

Δ

u

1

−

2

=

C

v

(

T

2

−

T

1

)

=

0.158

(

200

−

80

)

=

18.96

Btu/lbm

Find the heat transfer during the process using the first law of thermodynamics.

q

1

−

2

=

w

1

−

2

+

Δ

u

1

−

2

=

5.4156

+

18.96

q

1

−

2

=

24.38

Btu/lbm

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Explanation:

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State 2:

P_2 = 800 KPa , constant volume process work done:

h_2 = h_1 + v_1 * ( P_2 - P_1)

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State 3:

P_3 = 800 KPa , saturated liquid

h_3 = 721 KJ/kg

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State 4:

P_4 = 10,000 KPa , constant volume process work done:

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P_5 = 10,000 KPa , T_5 = 550 C

h_5 = 3500 KJ/kg

s_5 = 6.760 KJ/kgK

State 6:

P_6 = 800 KPa , s_5 = s_6 = 6.760 KJ/kgK

h_6 = 2810 KJ/kg

State 7:

P_7 = 800 KPa , T_7 = 500 C

h_7 = 3480 KJ/kg

s_7 = 7.870 KJ/kgK

State 8:

P_8 = 10 KPa , s_8 = s_7 = 7.870 KJ/kgK

h_8 = 2490 KJ/kg

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- Use energy balance of steam bleed and cold feed-water:

                                        E_6 + E_2 = E_3

               flow(m_6)*h_6 + flow(m_2)*h_3 = flow(m_3)*h_3

                                    y*h_6 + (1-y)*h_3 = h_3

                                  y*2810 + (1-y)*192.8 = 721

Compute y:                          y = 0.2018

- Heat produced by the boiler q_b:

                             q_b = h_5 - h_4 +(1-y)*(h_7 - h_8)

                    q_b = 3500 -731.21 + ( 1 - 0.2018)*(3480 - 2810)

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                                       q_c = (1-y)*(h_8 - h_1)

                                 q_c= ( 1 + 0.2018)*(2810 - 192)

Compute q_c:               q_c = 1834.26 KJ/ kg

- Net power output w_net:

                                     w_net = q_b - q_c

                                w_net = 3303.58 - 1834.26

                                    w_net = 1469.32 KJ/kg

- Given out put P = 80,000 KW

                                     flow(m) = P / w_net

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                                     u = 1 - (q_c / q_b)

                                     u = 1 - (1834.26/3303.58)

                                     u = 44.48 %

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