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3 years ago

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A frictionless piston-cylinder contains carbon dioxide gas (CO2) initially at 500oC and 2 MPa. The system is o3 cooled in an iso

baric process until the final temperature becomes 350 C and the final volume is 1 m . For this process, determine: (a) The reduced pressure and the reduced temperature of the initial state. (b) The initial volume of the piston-cylinder (in m3). (c) The mass of CO2 in the piston-cylinder (in kg). (d) The total boundary work for the process (in kJ). (e) The amount of heat transfer during the cooling process (in kJ).

1 answer:

TiliK225 [7]3 years ago

8 0

Answer:

(a) The reduced pressure is 0.2711 MPa

The reduced temperature is 2.54 K

(b) The initial volume of the piston is approximately 0.806 m³

(c) The mass of CO₂ is approximately 16.9884 kg

(d) The work done, W is approximately 388.023 kJ

(e) The heat transfer is approximately -2,650.1904 kJ

Explanation:

The initial temperature of the piston-cylinder, T₁ = 500°C = 773.15 K

The initial pressure of the gas, P₁ = 2 MPa

The final temperature of the gas, T₂ = 350°C

The final volume of the gas = 1 m³

(a) For an isobaric process, we have;

The reduced pressure,

$P_r = \dfrac{P}{P_c}$

The critical pressure of carbon dioxide, $P_c$ = 7.3773 MPa

$P_r = \dfrac{2 \, MPa}{7.3773 \, MPa} \approx 0.2711 \, MPa$

The reduced pressure, $P_r$ = 0.2711 MPa

The critical temperature, $T_c$ = 304.13 K

The reduced temperature, $T_r$ , is given by the following formula;

$T_r = \dfrac{T}{T_c}$

Therefore, $T_r$ = (773.15 K)/(304.13 K) = 2.54216947 K

The reduced temperature, $T_r$ ≈ 2.54 K

(b) The initial volume of the piston, V₁ = (V₂/T₂) × T₁

∴ V₁ = (1 m³/773.15) × 623.15 = 0.80598848865 m³ ≈ 0.806 m³

The initial volume of the piston, V₁ ≈ 0.806 m³

(c) The number of moles of CO₂ in the cylinder, 'n', is given according to the following formula;

n = P·V/(T·R)

The universal gas constant, n = (2 × 10⁶Pa × 1 m³)/(623.15 K × 8.3145 J/(mol·K)) ≈ 386.0124 moles

The mass of CO₂ ≈ 386.0124 moles × 44.01 g/mol = 16.9884 kg

(d) The work done, W = P·( $V_f - V_i$ )

W = 2 × 10⁶ × (1 - 0.80598848865) = 388023.0227

The work done, W ≈ 388.023 kJ

(e) The heat transfer dQ = m· $c_p$ ×(T₂ - T₁)

$c_p$ for CO₂ ≈ 1.04 kJ/(kg·K)

∴ dQ = 16.9884 × 1.04 × (350 - 500) = -2,650.1904 kJ

Therefore, the heat transfer = dQ = -2,650.1904 kJ

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