Air entries a compressor in a steady flow at 140KPa, $17^oc$ and 70m/s and leaves at 350 KPa, $127^oc$ & 110m/s. The environment is at 100 KPa, $7^oc$. - (i) The actual amount of work required - (ii) The minimum work required - (iii) The irreversibility of the process -

Mumbai university > MECH > SEM 3 > THERMO

Marks: 10M

Year: Dec 2014

Given:

P1=140 kPa; v1 = 70m/s; T1 = 290K;

P2 = 350kPa; v2=110m/s; T2=400K

P0= 100kPa; T0= 280K;

Solution:

i) Finding work

Q-W = ∆Potential energy + ∆Kinetic energy + ∆Pressure energy

$W = mg(z2-z1) + 1/2m (v2^2 – v1^2) + (H2-H1)$

Assuming m=1kg and Q=0 (adiabatic process)

$-W= 9.8 × (0-0) + ½ × 1 × (110^2 – 70^2) + mCp (T2-T1)$

= 3600 + 1 × 1000 × ( 400-290)

W = -113600 J/kg

ii) Finding minimum work

$∆Availability= ∆H + ∆Heat transfer + ∆V^2/2$

$A2-A1 = H2-H1 + T0(S2-S1) + (V2^2-V1^2)/2$

$= Cp(T2-T1) + T0( (mR × ln(p1/p2) + mCp × ln(T2/T1) ) + (V2^2-V1^2)$

$= 1000 (400-290) + 280 (1 × 280 × ln(140/350) + 1 × 1000 × ln(400/290) ) + (110^2-70^2)$

= 97250 J/kg

So, minimum work = Wrev = A1-A2 = -97.25 kJ/kg

iii) Finding irreversibility

I = Wrev-W = -97.25 – (-113.6) = 16.35 kJ/kg

$(V2^2-V1^2)/2$