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

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

an obstruction

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂