Answer to Question #140245 in Mechanics | Relativity for Martin Navarro

Question

If 10 kg/min of air are compressed isothermally from P1=96 kPa and V1 =7.65 m^ 3/min to P2 =620 kPa , find the work, the change of entropy and the heat for (a) non flow process and b) a steady flow process with v1= 15 m/s and v2 =60 m/s .

Answer 1

Solution

a) for non flow process

This is isothermal process so

Change in internal energy

$\Delta U=0$

$P_1V_1=nT_1\\T_1=\frac{P_1V_1}{n}$

Where n=0.3452Kg

After putting value of pressures and n then

$T_1=255.9K$

n = m / M

= ( 10 kg )/ ( 28.97 kg/kgmol )

= 0.3452 kgmol

Work done

W=nRT$\ln{\frac{P_1}{P_2}}$

W=0.3452$\times$ 8.314$\times$ 255.9$\ln{\frac{96KPa}{620KPa}} \\$

W=-1370KJ

Now heat

$Q=\Delta U+\Delta W\\Q=-1370KJ$

Now entropy change

$\Delta S=\frac{Q}{T}=\frac{-1370KJ}{255.9}$

$\Delta S=-5.354KJ/kelvin$

b) now for steady flow

Work done

W_s=0.3452$\times$ 8.314$\times$ 255.9$\ln{\frac{96KPa}{620KPa}} \\$

$W_s=-1370KJ/min$

$\Delta H=0$

$\Delta KE=\frac{10\times(60^2-15^2) }{2}=14$ KJ/min

$\Delta PE=0$

Total heat

$Q=\Delta H+\Delta KE+\Delta PE+$ W_s

Q=-1353KJ/min

Now entropy change

$\Delta S=\frac{Q}{T}=\frac{-1353KJ/min}{255.9Kelvin}$

$\Delta S=-5.267$ KJ/min-Kelvin