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Air enters a compressor steadily at the ambient conditions of 100 kPa and 22°C and leaves at 800 kPa. Heat is lost from the compressor in the amount of 120 kJ/kg, and the air experiences an entropy decrease of 0.40 kJ/kg·K. Using constant specific heats, determine

(a) the exit temperature of the air,
(b) the work input to the compressor, and
(c) the entropy generation during this process.

Respuesta :

Chrisnando

Answer:

a) 358.8K

b) 181.1 kJ/kg.K

c) 0.0068 kJ/kg.K

Explanation:

Given:

P1 = 100kPa

P2= 800kPa

T1 = 22°C = 22+273 = 295K

q_out = 120 kJ/kg

∆S_air = 0.40 kJ/kg.k

T2 =??

a) Using the formula for change in entropy of air, we have:

∆S_air = [tex] c_p In \frac{T_2}{T_1} - Rln \frac{P_2}{P_1}[/tex]

Let's take gas constant, Cp= 1.005 kJ/kg.K and R = 0.287 kJ/kg.K

Solving, we have:

[/tex] -0.40= (1.005)ln\frac{T_2}{295} ln\frac{800}{100}[/tex]

[tex] -0.40= 1.005(ln T_2 - 5.68697)- 0.5968[/tex]

Solving for T2 we have:

[tex] T_2 = 5.8828[/tex]

Taking the exponential on the equation (both sides), we have:

[/tex] T_2 = e^5^.^8^8^2^8 = 358.8K[/tex]

b) Work input to compressor:

[tex] w_in = c_p(T_2 - T_1)+q_out[/tex]

[tex] w_in = 1.005(358.8 - 295)+120[/tex]

= 184.1 kJ/kg

c) Entropy genered during this process, we use the expression;

Egen = ∆Eair + ∆Es

Where; Egen = generated entropy

∆Eair = Entropy change of air in compressor

∆Es = Entropy change in surrounding.

We need to first find ∆Es, since it is unknown.

Therefore ∆Es = [tex] \frac{q_out}{T_1}[/tex]

[tex] \frac{120kJ/kg.k}{295K}[/tex]

∆Es = 0.4068kJ/kg.k

Hence, entropy generated, Egen will be calculated as:

= -0.40 kJ/kg.K + 0.40608kJ/kg.K

= 0.0068kJ/kg.k

Consider the expression for the change in entropy of the air.  

[tex]\to \Delta S_{air}=c_p \ \In \frac{T_2}{T_1} - R \In \frac{P_2}{P_1} \\\\[/tex]

Here, change in entropy of air in a compressor is[tex]\Delta S_{air}[/tex], specific heat at constant pressure is [tex]c_p[/tex], the inlet temperature is [tex]T_1[/tex], outlet temperature is [tex]T_2[/tex], the gas constant is R, inlet pressure is [tex]P_1[/tex], and outlet pressure is[tex]P_2[/tex].  

From the ideal gas specific heats of various common gases table, select the specific heat at constant pressure[tex]c_p[/tex] and gas constant (R) at air and temperature:

[tex]\to 22\ C \ \ or \ \ 295\ K\ \ as 1.005 \ \frac{kJ}{kg \cdot K} \ \ and \ \ 0.287\ \frac{kJ}{kg \cdot K}[/tex]

Substituting

Take exponential on both sides of the equation.  

[tex]\to T_2 = exp(5.8828) = 358.8\ K \\\\[/tex]

Hence, the exit temperature of the air is [tex]358.8\ K[/tex]. Apply the energy balance to calculate the work input.  

[tex]\to W_{in}=c_p(T_2-T_1 )+q_{out}[/tex]

Here, work input is [tex]w_{in}[/tex] initial temperature is [tex]T_1[/tex], exit temperature is [tex]T_2[/tex], and heat transfer outlet is [tex]q_{out}[/tex] 

Substituting

 

Hence, the work input to the compressor is

Express the entropy generated in the process.  

[tex]\to S_{gen} =\Delta S_{air}+\Delta S_{swr}[/tex]

Here, entropy generated is [tex]S_{gen}[/tex], change in entropy of air in a compressor is [tex]\Delta S_{air}[/tex], and change in entropy in the surrounding is [tex]\Delta S_{swr}[/tex].  

Finding the changes into the surrounded entropy.  

[tex]\to \Delta S_{swr} = \frac{q_{out}}{T_{swr}}[/tex]

Here, the heat transfer outlet is [tex]q_{out}[/tex] and the surrounded temperature is [tex]T_{swr}[/tex].  

Substituting

[tex]120\ \frac{kJ}kg } \ for\ q_{out} \ and\ 22\ C \ for\ T_{swr}[/tex]  

[tex]\Delta S_{swr} = \frac{ 120\frac{kJ}{kg}}{(22+273)\ K} =0.4068 \frac{kJ}{kg\cdot K}[/tex]

Finding the entropy generated process.  

[tex]S_{gen} = \Delta S_{air} +\Delta S_{swr}\\[/tex]

Substituting

[tex]-0.40 \frac{kJ}kg\cdot K} \ for \ \Delta S_{air},\ and\ 0.4068 \frac{kJ}{kg\cdot K}\ for\ \Delta S_{swr}\\\\[/tex]

[tex]S_{gen}=-0.40 \frac{kJ}{kg\cdot K} +0.4068 \frac{kJ}{kg\cdot K} = 0.0068 \frac{kJ}{kg\cdot K}[/tex]

Therefore, the entropy generated in the process is [tex]0.0068\ \frac{kJ}{kgK}[/tex].

Learn more:

brainly.com/question/16392696

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