Asteam power plant operates on an ideal reheat rankine cycle between the pressure limits of 15 mpa and 10 kpa. the mass flow rate of steam through the cycle is 12 kg/s. steam enters both stages of the turbine at 500c.
if the moisture content of the steam at the exit of the low-pressure turbine is not to exceed 10%, determine
(a) the pressure, in kpa ,at which reheating takes place,
(b) the total rate of heat input, in kw, in the boiler,
(c) the thermal efficiency of the cycle. also, show the cycle on a t-s diagram with respect to the saturation lines.

a) 2161 KPa

b) 45,012.6 kW

c) 42.6 % and see attachment for T-s diagram

Explanation:

Assumptions

1. SS − SF

2. heat loss to the surroundings is negligible

3. KE = P E → 0

4. properties are constant

Part a)

The values obtained from water property tables at respective state points are tabulated in the attachment.

Pump

wp,in = h2 − h1 = ν1(P2 − P1)

= (0.00101 m3/kg)*(15000 − 10) kPa

= 15.14 kJ/kg

h2 = h1 + wp,in

= 191.81 kJ/kg + 15.14 kJ/kg = 206.95 kJ/kg

State 6:

h6 = (1 − x)hf + xhg

= 0.1(191.81 kJ/kg)+0.9(2583.9 kJ/kg)

= 2344.7 kJ/kg

s6 = (1 − x)sf + xsg

= 0.1(0.6492 kJ/kg · K)+0.9(8.1488 kJ/kg · K)

= 7.3988 kJ/kg · K

State 5

To find the pressure at 5 we need to interpolate in the super heated tables. First the entropy at a pressure  above the desired value and then at a pressure below the desired value.

@500 ◦C, @2 MPa −→ s = 7.4337 kJ/kg · K

@500 ◦C, @2.5 MPa −→ s = 7.3254 kJ/kg · K

These values of s bound our value of s5 = 7.3988 kJ/kg · K.

Interpolate to find the pressure, P5

(7.4337 − 7.3988) /  (7.4337 − 7.3254)

= 0.32225

P3 = 2000 kP a + 0.32225 × (2500 − 2000) kPa

= 2161 kPa

h5 = 0.32225 × (3468.3 kJ/kg)+0.67775 × (3462.8 kJ/kg) = 3464.8 kJ/kg

State 4

To find h4 we need to interpolate within the super heated tables

@2 MPa, s = 6.348 kJ/kg · K −→ h = 2802.7 kJ/kg

@2.5 MPa, s = 6.348 kJ/kg · K −→ h = 2848.9 kJ/kg

To find the actual value of h4 we need to interpolate between these two pressures to find the value at  P4 = 2161 kPa

h4 = 0.678 * (2802.7 kJ/kg)+0.322 *(2848.9 kJ/kg) = 2817.6 kJ/kg

Part b

The heat input to the boiler is given as :

Q_in = m_flow*[(h3 − h2)+(h5 − h4)]

= (12 kg/s)*[(3310.8 − 206.95) + (3464.8 − 2817.6)] kJ/kg = 45,012.6 kW

Part c

Q_out = m_flow*[(h6-h1)]

= (12 kg/s)*(2344.7 - 191.8) KJ/kg

= 25,835 KW

can u provide the graph and if its not possible

then i can tell u the segment where the

average time interval starts degrading is the segment where the car is slowing down