Answer:
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
[tex]n_{th} = 1 - \frac{Q_{out} }{Q_{in}} \\= 1 - \frac{25834}{45012.6} \\= 42.6 %[/tex]
