Answer to Question #101883 in Molecular Physics | Thermodynamics for moktar mitchell alame

Question

Air flows steadily at the rate of 0.4 kg/s through an air compressor, entering at
6 m/s with a pressure of 100 kPa and a specific volume of 0.85 m3
/kg. The air
exits at 4.5 m/s with a pressure of 690 kPa and a specific volume of
0.16 m3
/kg. The specific internal energy of the air leaving is 88 kJ/kg greater
than that of the air entering. Cooling water in the jacket surrounding the
cylinder absorbs heat from the air at the rate of 59 kJ/s. Calculate:
a. The power required to drive the compressor
b. The cross-sectional area of the inlet pipe
c. The cross-sectional area of the outlet pipe

Answer 1

$\dot m =0.4 \; kg/s$

$C_1 =6\; m/s, \quad p_1=100\times10^3 \;Pa, \quad v_1=0.85 \;m^3/kg$

$C_2 =4.5\; m/s, \quad p_2=690\times10^3 \;Pa, \quad v_2=0.16 \;m^3/kg$

$\Delta U=88\times 10^3 \; J/kg$

$\dot Q=-59\times 10^3 \; J/s$

a) $\dot W -? \;\;$

b) $A_1 -?$

c) $A_2 -?$

Solution:

$C$ - velocity, $p$ - pressure, $v$ - specific volume, $Z$ - height, $U$ - internal energy, $A$ - cross-sectional area.

a) The steady flow energy equation:

$\dot Q-\dot W= \dot m\Delta H+\frac{1}{2} \dot m(C^2_2-C_1^2)+\dot mg(Z_2-Z_1)$ (*)

$\Delta H=(U_2+ p_2v_2)-(U_1+p_1v_1)=\Delta U +p_2v_2-p_1v_1$

$\Delta H= 88\times 10^3+690\times10^3\times 0.16-100\times10^3\times0.85= 113.4\times 10^3 \;J/kg$

$\frac{1}2 \dot m(C_2^2-C_1^2)= \frac{1}2 \times0.4 \times (4.5^2-6^2)=-3.15\; J/kg$

$Z_1=Z_2$

(*): $-59\times 10^3 -\dot W=0.4\times 113.4\times 10^3-3.15$

$\dot W=-104.4 \;kW$, negative sign indicates that the work is done on the system.

b,c) The continuity of mass equation:

$\dot m=\frac{C_1A_1}{V_1}=\frac{C_2A_2}{V_2}$

$A_1=\frac{\dot m V_1}{C_1}=0.0566 \; m^2$

$A_2=\frac{\dot m V_2}{C_2}=0.0142 \;m^2$

Answers:

a) $\dot W=-104.4 \; kW$

b) $A_1=0.057 \;m^2$

c) $A_2=0.014 \; m^2$