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.
