Air at 10 degree C and 80 kPa enters the diffuser of a jet engine steadily with

Question

Air at 10 degree C and 80 kPa enters the diffuser of a jet engine steadily with a velocity of 200 m/s. The inlet area of the diffuser is 0.4 m^2.The air leaves the diffuser with a velocity that is very small compared with the inlet velocity. Determine the mass flow rate of the air and the temperature of the air leaving the diffuser. Air at 100 kPa and 280 K is compressed steadily to 600 kPa and 400 K. The mass flow rate of the air is 0.02 kg/s, and a heat loss of 16 kJ/kg occurs during the process. Assuming the changes in kinetic and potential energies are negligible, determine the necessary power input to the compressor. The power output of an adiabatic steam turbine is 5 MW, and the inlet and the exit conditions of the steam are as indicated in Fig. Compare the magnitudes of Delta h, Delta ke, and Delta pe. Determine the work done per unit mass of the steam flowing through the turbine. Calculate the mass flow rate of the steam.

Explanation / Answer

Solution:

(a) To determine the mass flow rate, we need to find the specific volume of the air first. This is determined from the ideal-gas relation at the inlet conditions:

Specific volume at inlet: v1 = RT1/P1 = (0.287kPa.m3/kg.K)(283K)/80kPa = 1.01 m3/kg

Mass flow rate: m =1/v1 x A1V1 = 1/(1.01 m3/kg)x (0.4m2)(200m/s) = 79.2 kg/s

Since the flow is steady, the mass flow rate through the entire nozzle will remain constant at this value.

(b) A nozzle normally involves

(i) no shaft or electrical work (w=0), (ii) negligible heat transfer (q 0), and (iii) a small (if any) elevation change between the inlet and exit (PE o). Then the conservation of energy relation on a unit mass basis for this single-stream steady-flow system reduces to:

Q – W = H + KE + PE

Or, 0 – 0 = H + KE + PE

Or, 0 = H2 – H1 + (V22 – V12)/2

Or, 0 = Cp(T2 – T1)+ (V22 – V12)/2

Or, T2 = T1 - (V22 – V12)/2Cp

Using specific heat at the anticipated average temperature of 300K (Cp = 1.005 kJ/kgK) we have:

T2 = 283K – [0 – (200)2 m2/s2]/(2x1005 J/kg.K)

Or, T2 = 283K + (40000 m2/s2)/(2x1005 m2/s2.K)

Or, T2 = 283K + 19.9K = 302.9 K

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