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Water vapor at 5 bar, 320 8 C enters a turbine operatingat steady state with a volumetric flow rate of 0.65 m 3 /s andexpands adiabatically to an exit state of 1 bar, 160 8 C. Kineticand potential energy effects are negligible. Determine forthe turbine (a) the power developed, in kW, (b) the rate ofentropy production, in kW/K, and (c) the isentropic turbineefficienc

Respuesta :

Answer:

371.3/435.12=0.8524

Explanation:

P1 = 5 bar

T1 = 320 deg. C

AV = 0.63 m^3/s

Q = 0 (adiabatic)

P2 = 1 bar

T2 = 160 deg. C

EK = EP = 0

(a) Power developed.

h1 = h(P1,T1) = 3106.0715

h2 = h(P2,T2) = 2796.1667

ρ1 = 1.8463 kg/m^3

s1 = s(P1,T1) = 7.5322 kJ/kg.K

s2 = s(P2,T2) = 7.6601 kJ/kg.K

h2s = (P2,s1) = 2742.5234

Mass flow rate

m_dot = AVρ = 1.163169 kg/s

Energy balance for steam turbine at steady state,

0 = Q - W + m_dot*(h1 - h2)

W = m_dot*(h1 - h2) = 360.47 kW

(b)

Entropy balance for control volume

o = Q/T + m_dot(s1-s2) + σ

σ = m_dot(s2-s1) = 1.163169*(7.6601 - 7.5322)

σ = 0.1488 kW/K

(c)

ηs = h1 - h2 / h1 - h2s

ηs = 3106.0715 - 2796.1667 / 3106.0715 - 2742.5234

ηs = 0.8524

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