Answer to Question #169054 in Mechanical Engineering for Gh

Question #169054

Water vapor bc enters an insulated turbine steadily at 6 Mpa 700 defress celcius at a rate of 100 of/min and exits at 15 kpa. the isentropic turbine efficiency is 96% . the kinetic energy are negligible the surroundings ate T0= 300 K and P0= 1atm . Determine the (a) the actual power developed by turbine, in kw (b) the rate at which energy is destroyed within the turbine, in kw(c) the second law of efficiency of the turbine.


1
Expert's answer
2021-03-08T03:08:49-0500

To determine the power developed by the turbine, apply the 1st Law to a control volume surrounding the turbine.

 


dECV/dt + "\\sum" (h + 1/2V2 + gz)out x m - "\\sum" (h + 1/2V2 + gz)in x m = Qinto + Won, other,

where

dECV/dt = 0 (steady flow)

Δ(1/2V2) and Δ(gz) are assumed to be negligibly small compared to the specific enthalpy

Qinto= 0 (adiabatic flow)

mout = min = m (from conservation of mass)

Won, other / m = h2 - h1

Won, other = mcp(T2 - T1)

Won, other = 100 x 1.864 x (973 - 300) = 125.45 kW

The rate

dSCV / dt = "\\sum"in sm - "\\sum"out sm + "\\int" Qinto / T + σ ,

where

dSCV / dt = 0 (steady flow),

sm!

"\\sum"in sm - "\\sum"out sm = m (s1 - s2) (the mass flow rate is constant from COM),

"\\int" Qinto / T= 0 (adiabatic operation)

σ = m (s2 - s1)

The specific entropies may be found using thermodynamic property tables

σ = 100 x (8.1502 - 6.9029) = 1.24 kW



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