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 . employing the ideal gas model, calculate the power input to the compressor in .

-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.

answer: 5

explanation:

6-1