A steam turbine operates on rankine cycle receives

Question

estion: A steam turbine operating on the Rankine cycle, receives steam from the boiler at 3.5 MPa and 350...

A steam turbine operating on the Rankine cycle, receives steam from the boiler at 3.5 MPa and 350C and exhaust to the condenser at 10 kPa. The condensate is then returned to the boiler by the feed pump. Determine: i) the dryness fraction of the steam entering the condenser and ii) the Rankine efficiency.

Answer 1

BOILER--------------------(3)high pressure vapor⇒-----------------------------TURBINE

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⇑(2)high pressure water-------------------------------------------(4)low pressure steam⇓

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PUMP-----------------------⇐(1)low pressure water------------------------CONDENSER

Noting the clockwise sequence above (with states and pressures of

the water between the components), first establish the pump work

as [P₂ − P₁] ÷ ρ_water or [3500 − 10] kPa ÷ 1000 kg/m³ which goes to

3.49 kJ/kg. A unit mass of working fluid is assumed here since only

efficiency is sought.

Next guide on h₂ = h₁ + w_in. A Properties Table Of Saturated Water By Pressure

at 10 kPa (or 0.01 MPa) gives h₁ as h_f = 191.8 kJ/kg. Then evaluate h₂ as

(191.8 + 3.49) or 195.29 kJ/kg.

For heat input, go to a Properties Table Of Superheated Steam under 3.5 MPa and

across 350°C to gain (h₃, s₃) as (3104.0 kJ/kg, 6.6579 kJ/kg•K). Then heat input is

q_B = h₃ − h₂ or (3104.0 − 195.29) or 2908.71 kJ/kg.

A T-s diagram shows that entropy at State 4 is s₄ equal to s₃ or 6.6579 kJ/kg•K. Then

build s₄ = s_f + x₄s_fg; this enumerates to 6.6579 kJ/kg•K = (0.6491 + 7.5019x₄) kJ/kg•K

[with values for s_f & s_fg taken from the Properties Table Of Saturated Water By Pressure

across from 0.01 MPa]. This will isolate the dryness fraction (or quality) as x₄ = 0.8009704208.

Now return to Properties Table Of Saturated Water By Pressure to gain (at P = 0.01 MPa)

h₄ = h_f + x₄h_fg or 191.8 kJ/kg + (0.8009704208)(2392.8 kJ/kg) equal to 2108.362023 kJ/kg.

Then work output from the turbine is w_T = h₃ − h₄ or (3104.0 − 2108.362023) or 995.637977 kJ/kg.

As a result, the efficiency is (including pump work) η = (w_T − w_P) ÷ q_B

or (995.637977 − 3.49) kJ/kg ÷ 2908.71 kJ/kg which reduces to

0.3410955293 (or roughly 34.1%).