Air enters a compressor operating at steady state at a pressure of 1?, a temperature of 290 K, and a velocity of 6 m? through an inlet with an area of 0.1 m2. At the exit, the pressure is 7?, the temperature is 450 K and the velocity is 2 m?. Heat transfer from the compressor to its surroundings occurs at a rate of 180? m?. Employing the ideal gas model, calculate the power input to the compressor in? W.

Answer: - 119.56Kw.

Explanation:

Bringing out the parameters from the equation;

Inlet pressure P1 1bar = 10N/m

Outlet pressure P2 7bar = 70N/m

T1 is 290K, T2 is 450K

Velocity V1 is 6m/s, V2 is 2m/s

Heat transfer rate is 180kj/min. which can be written as 180kj/60secs.

Area A is 0.1m³.

From the steady state energy equation

∆H+g (z2-z1) + 1/2 (V²2-V²1) = Q/m-W/m

Solving for W with g (z2-z1) = 0, we have

∆H+0+1/2 (V²2-V²1) = Q/m-W/m

W=Q+m[∆H + (V²2-V²1/2) ] ... eq. 1

m is the mass flow rate given as

A. V/v,

Where W is the power, V is the velocity, and v is the specific volume

A is Area.

But the specific volume is derived from the ideal gas law;

v = (R/M) * T/P1, where R is universal gas constant given as 8314N. m

M is the molecular mass of air given as 28.97kg. k.

STEP 1: Find the specific volume from above formula:

Specific volume, v = [ (8314N. m/28.97kg. k) * 290K] / 10N/m² = 0.8324kg/m³

STEP 2: Find mass flow rate, m = A. V/v = 0.1m³ * 6m/s / 0.8324kg/m³

m = 0.7209kg/s

STEP 3: Find the specific enthalpy from air property table as

@ T1 of 290k and P1 of 0.1Mpa, specific enthalpy is 290.6kj,/kg

@T2 of 450k and P2 of 0.7Mpa, specific enthalpy is 452.3kj/kg.

Therefore, ∆H = H2 - H1 = 290.6kj/kg - 452.3kj/kg = - 161.7kj/kg.

STEP 4: Calculate change in K. E

1/2 (V²2-V²1) = 6²-2² / 2 = 16J/kg or 0.016Kj/kg ...

STEP 5: Calculate power in KW using eq. 1.

W = (180kj/60secs) + [ (0.7209kj/s (-161.7-0.016) kj/kg]

W = 3kj/secs - 116.557kj. s

W = - 119.56KW.