Question

8. Air at 25C, 100 kPa and air at 50C, 200 kPa at 1 to 1 volume ratio are mixed inside an adiabatic compressor to 55C, 500 kPa and pushed through an isothermal nozzle with inlet area a quarter the size of the outlet area and reaching 10 m/s at 300 kPa. How much work is required if we want to get 5 kg of this mixture

Answer 1

Workdone, w = 68.935 kJ

Step-by-step explanation:

m1h1 + m2h2 + Q = m3h3 - Wc

Final mass of mixture,

m3 = m1 + m2 = 5kg

p1V1 = m1R1T1

p2V2 = m2R2T2

Since they are at 1:1,

V1 = V2 and R1 = R2

Comparing equations,

p1/p2 = (m1/m2) x (T1/T2)

m1/m2 = (100/200) x ((50 + 273)/(25+273))

m1/m2 = 0.542

m1 = 0.542m2

m3 = m1 + m2

5 = m2 + 0.542m2

m2 = 3.243kg

m1 = 1.757 kg

Workdone is given as,

Wc = m3h3 - m1h1 - m2h2

h = Cp x T, since air is an ideal gas

Cp = 1.005kJ/kGK

Wc = (5 x 1.005 x (55+273)) - (1.757 x 1.005 x 298) -(3.243 x 1.005 x 323)

Wc = 68.935 kJ