5. Steam enters a turbine with an enthalpy of hA = 2.80 MJ/kg at 40 m/s. So-call

Question

5. Steam enters a turbine with an enthalpy of hA = 2.80 MJ/kg at 40 m/s. So-called wet steam (steam-water mixture) leaves the turbine at an enthalpy of hB = 1.73 MJ/kg. at 15 ins. The mass flow through the turbine is 0.8 kg s. a. [10] Compute the ideal power output. Show your work. b. [5] Testing shows that the heat loss to the surroundings is 500 W. Compute the real power output. Show your work. c. [5] Compute the efficiency of the turbine. Show your work. All conditions are considered uniform. State any other assumptions clearly.

Explanation / Answer

given

hA=2.80MJ/kg,

vA=40m/s

hB=1.73 MJ/kg

vB=15m/s

m=0.8kg/sec

energy at section A= m*hA + Kinetiv energy

= 0.8*2.80 * 10 ^6 + 0.8*40^2

=2241280 J/s

Energy at section B =m*hB + kinetic enegry =0.8*1.73*10^6 +0.8*15*15 =1384180 J/s

power ouput =energy at A -energy at B

= 857100 J/sec

losses =500 W =500 J/s

net power ouput = 857100 -500 = 856600 J/Sec

efficiency = net ouput/exect power output =856600 /857100 = 99.94 %

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