Answer to Question #169054 in Mechanical Engineering for Gh

Question

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.

Answer 1

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

 

dE_CV/dt + $\sum$ (h + 1/2V² + gz)_out x m - $\sum$ (h + 1/2V² + gz)_in x m = Q_into + W_{on, other},

where

dE_CV/dt = 0 (steady flow)

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

Q_into= 0 (adiabatic flow)

m_out = m_in = m (from conservation of mass)

W_{on, other} / m = h₂ - h₁

W_{on, other} = mc_p(T₂ - T₁)

W_{on, other} = 100 x 1.864 x (973 - 300) = 125.45 kW

The rate

dS_CV / dt = $\sum$_in sm - $\sum$_out sm + $\int$ Q_into / T + σ ,

where

dS_CV / dt = 0 (steady flow),

sm!

$\sum$_in sm - $\sum$_out sm = m (s₁ - s₂) (the mass flow rate is constant from COM),

$\int$ Q_into / 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