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2 years ago

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Air enters a compressor operating at steady state at 176.4 lbf/in.^2, 260°F with a volumetric flow rate of 424 ft^3/min and exit

s at 15.4 lbf/in.^2, 80°F. Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in hp

1 answer:

Lera25 [3.4K]2 years ago

6 0

Answer:

$W_s =$ 283.181 hp

Explanation:

Given that:

Air enters a compressor operating at steady state at a pressure $P_1$ = 176.4 lbf/in.^2 and Temperature $T_1$ at 260°F

Volumetric flow rate V = 424 ft^3/min

Air exits at a pressure $P_2$ = 15.4 lbf/in.^2 and Temperature $T_2$ at 80°F.

Heat transfer occurs at a rate of 6800 Btu/h from the compressor to its surroundings; since heat is released to the surrounding; then:

$Q_{cv}$ = -6800 Btu/h = - 1.9924 kW

Using the steady state energy in the process;

$h_2 - h_1 + g(z_2-z_1)+ \dfrac{1}{2}(v^2_2-v_1^2) = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}$

where;

$g(z_2-z_1) =0$ and $\dfrac{1}{2}(v^2_2-v_1^2) = 0$

Then; we have :

$h_2 - h_1 = \dfrac{Q_{cv}}{m}- \dfrac{W_s}{m}$

$h_2 - h_1 = \dfrac{Q_{cv} - W_s}{m}$

${m}(h_2 - h_1) ={Q_{cv} - W_s}$

$W_s ={Q_{cv} + {m}(h_2 - h_1)$ ----- (1)

Using the relation of Ideal gas equation;

P₁V₁ = mRT₁

Pressure $P_1$ = 176.4 lbf/in.^2 = ( 176.4 × 6894.76 ) N/m² = 1216235.664 N/m²

Volumetric flow rate V = 424 ft^3/min = (424 × 0.0004719) m³ /sec

= 0.2000856 m³ /sec

Temperature = 260°F = (260°F − 32) × 5/9 + 273.15 = 399.817 K

Gas constant R=287 J/kg K

Then;

1216235.664 N/m² × 0.2000856 m³ /sec = m × 287 J/kg K × 399.817 K

$m = \dfrac { 1216235.664 N/m^2 \times 0.2000856 m^3 /sec } {287 J/kg K \times 399.817 K }$

m = 2.121 kg/sec

The change in enthalpy:

$m(h_1-h_2) = m * C_p * \Delta T= m* C_p * ( T_1 -T_2)$

$= 2.121* 1.005* ( 399.817 -299.817)$

= 213.1605 kW

From (1)

$W_s ={Q_{cv} + {m}(h_2 - h_1)$

$W_s =$ - 1.9924 kW + 213.1605 kW

$W_s =$ 211.1681 kW

$W_s =$ 283.181 hp

The power input is $W_s =$ 283.181 hp

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An ideal Otto cycle has a compression ratio of 9.2 and uses air as the working fluid. At the beginning of the compression proces

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(a) The amount of heat transferred to the air, $q_{out}$ is 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is 308.07 kJ/kg

(c) The thermal efficiency is 58.8%

(d) The Mean Effective Pressure, MEP, is 393.209 kPa

Explanation:

(a) The assumptions made are;

$c_p$ = 1.005 kJ/(kg·K), $c_v$ = 0.718 kJ/(kg·K), R = 0.287 kJ/(kg·K),

Process 1 to 2 is isentropic compression, therefore;

$T_{2}= T_{1}\left (\dfrac{v_{1}}{v_{2}} \right )^{k-1} = 300.15\times 9.2^{0.4} = 729.21 \, K$

From;

$\dfrac{p_{1}\times v_{1}}{T_{1}} = \dfrac{p_{2}\times v_{2}}{T_{2} }$

We have;

$p_{2} = \dfrac{p_{1}\times v_{1}\times T_{2}}{T_{1} \times v_{2}} = \dfrac{98\times 9.2\times 729.21}{300.15 } = 2190.43 \, kPa$

Process 2 to 3 is reversible constant volume heating, therefore;

$\dfrac{p_3}{T_3} =\dfrac{p_2}{T_2}$

p₃ = 2 × p₂ = 2 × 2190.43 = 4380.86 kPa

$T_3 = \dfrac{p_3 \times T_2}{p_2} =\dfrac{4380.86 \times 729.21}{2190.43} = 1458.42 \, K$

Process 3 to 4 is isentropic expansion, therefore;

$T_{3}= T_{4}\left (\dfrac{v_{4}}{v_{3}} \right )^{k-1}$

$1458.42= T_{4} \times \left (9.2 \right )^{0.4}$

$T_4 = \dfrac{1458.42}{(9.2)^{0.4}} = 600.3 \, K$

$q_{out} = m \times c_v \times (T_4 - T_1) = 0.718 \times (600.3 - 300.15) = 215.5077 \, kJ/kg$

The amount of heat transferred to the air, $q_{out}$ = 215.5077 kJ/kg

(b) The net work output, $W_{net}$ , is found as follows;

$W_{net} = q_{in} - q_{out}$

$q_{in} = m \times c_v \times (T_3 - T_2) = 0.718 \times (1458.42 - 729.21) = 523.574 \, kJ/kg$

$\therefore W_{net} = 523.574 - 215.5077 = 308.07 \, kJ/kg$

(c) The thermal efficiency is given by the relation;

$\eta_{th} = \dfrac{W_{net}}{q_{in}} \times 100= \dfrac{308.07}{523.574} \times 100= 58.8\%$

(d) From the general gas equation, we have;

$V_{1} = \dfrac{m\times R\times T_{1}}{p_{1}} = \dfrac{1\times 0.287\times 300.15}{98} =0.897\, m^{3}/kg$

The Mean Effective Pressure, MEP, is given as follows;

$MEP =\dfrac{W_{net}}{V_1 - V_2} = \dfrac{W_{net}}{V_1 \times (1- 1/r)}= \dfrac{308.07}{0.897\times (1- 1/9.2)} = 393.209 \, kPa$

The Mean Effective Pressure, MEP = 393.209 kPa.

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