Question

the net work output and the thermal efficiency for the Carnot and the simple ideal Rankine cycles with steam as the working fluid are to be calculated and compared. Steam enters the turbine in both cases at 5 MPa as a saturated vapor and the condenser pressure is 50 kPa. in the Rankine cycle, the condenser exit state is saturated liquid and in the Carnot cycle, the boiler inlet state is saturated liquid. Draw the T-s diagrams for both cycles.

Answer 1

a) Rankine

Net work output = 719.1 KJ/kg

Thermal Eff = 0.294

a) Carnot

Net work output = 563.2 KJ/kg

Thermal Eff = 0.294

For T-s diagrams see attachments

Step-by-step explanation:

Part a Rankine Cycle

The obtained data from water property tables:

`P_(L,sat liquid) = 50 KPa \\v_(1) = 0.00103m^3/kg\\\\ h_(1) = 340.54KJ/kg\\\\P_(H) = 5000KPa\\h_(2) = h_(1) + v_(1) *(P_(H) - P_(L) )\\h_(2) =350.54 + (0.00103)*(5000 - 50)\\\\h_(2) = 345.64KJ/kg\\\\P_(H,satsteam) = 5000KPa\\s_(3) = 5.9737KJ/kgK\\\\h_(3) = 2794.2KJ/kg\\\\s_(3) = s_(4) = 5.9739KJ/kgK\\P_(L)= 50KPa\\\\h_(4)= 2070KJ/kg\\\\`

Heat transferred from boiler

`q_(b) = h_(3)-h_(2)\\q_(b)=2794.2-345.64\\\\q_(b) =2448.56KJ/kg\\\\`

Heat transferred from condenser

`q_(c) = h_(4)-h_(1)\\q_(b)=2070-340.54\\\\q_(b) =1729.46KJ/kg\\\\`

Thermal Efficiency

`u_(R) = 1- (q_(c))/(q_(b))\\\\u_(R) = 1 - (1729.46)/(2448.56)\\\\u_(R) =0.294`

Net work output

`w_(R) = q_(b)-q_(c)\\w_(R) = 2448.56-1729.46\\\\w_(R)=719.1KJ/kg`

Part b Carnot Cycle

The obtained data from water property tables:

`P_(H,sat-steam) = 5000KPa\\T_(3) = 263.94 C\\s_(3) = 5.9737KJ/kgK\\\\h_(3) = 2794.2KJ/kg\\\\T_(2,sat-liquid) = T_(3) = 263.94C\\s_(2) = 2.920KJ/kgK\\\\h_(2) = 1150KJ/kg\\\\P_(L) = 50KPa\\s_(1)=s_(2) = 2.920KJ/kgK\\\\h_(1) = 989KJ/kg\\\\s_(3) = s_(4) = 5.9737KJ/kgK\\P_(L) = 50KPa\\\\h_(4) = 2070KJ/kg`

Heat transferred from boiler

`q_(b) = h_(3)-h_(2)\\q_(b)=2794.2-1150\\\\q_(b) =1644.2KJ/kg\\\\`

Heat transferred from condenser

`q_(c) = h_(4)-h_(1)\\q_(b)=2070-989\\\\q_(b) =1081KJ/kg\\\\`

Thermal Efficiency

`u_(C) = 1- (q_(c))/(q_(b))\\\\u_(C) = 1 - (1081)/(1644.2)\\\\u_(C) =0.343`

Net work output

`w_(C) = q_(b)-q_(c)\\w_(C) = 1644.2-1081\\\\w_(C)=563.2KJ/kg`