Question #242679

Air enters a gas turbine system with a velocity of 105 m/s and has a specific volume of 0.8m3 /kg. The inlet area of the gas turbine system is 0.05m2. At exit the air has a velocity of 135m/s and has a specific volume of 1.5 m3 /kg. In its passage through the turbine system, the specific enthalpy of air is reduced by 145kJ/kg and the air also has a heat transfer loss of 27 kJ/kg. Determine: (a) The mass flow rate of the air through the turbine system in kg/s? (b) The exit area of the turbine system in mạ ? (c) The power developed by the turbine system in kW?


1
Expert's answer
2021-09-27T10:12:02-0400

Q=AV=0.05×105m3/sQ=AV=0.05\times105 m^3/s

a.

V=0.8m3/kgm=QV=6.5625kg/sV= 0.8 m^3/kg \\ m= \frac{Q}{V}=6.5625 kg/s

b.

g1a1v1=g2a2v2        atsteady      stafe11.8×0.05×105=11.5×a2×135a2=0.0729m2g_1 a_1 v_1=g_2 a_2 v_2 \; \;\;\;at steady \;\;\;stafe\\ \frac{1}{1.8}\times0.05\times105= \frac{1}{1.5} \times a_2 \times135 \\ a_2=0.0729m^2

c.

δqδw=dH+d(ke)27δw=145+12(13521052)×10327δw=145+3.6δw=1453.627δw=114.4ks/kgδq – δw =dH + d(ke)\\ -27 - δw = -145 + \frac{1}{2}(135^2 – 105_2)\times10^-3\\ -27 - δw = -145 +3.6\\ δw= 145 -3.6 -27\\ δw= 114.4 ks/kg\\

powerdeveloped=114.4×6.5625kwP=750.75kwpower developed = 114.4\times6.5625 kw\\ P=750.75 kw


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!
LATEST TUTORIALS
APPROVED BY CLIENTS