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An insulated open feedwater heater (mixing box) is used in a steam power plant to raise the temperature of the water going into the boiler which increases the thermal efficiency. The water exits the feedwater heater as a saturated liquid at P6 with a mass flow rate of m6. The two inlets to the feedwater heater are both at a pressure of P6. The first inlet has a temperature of T3 and second at a temperature of T5. Determine the following given the properties listed below.
--Given Values--

P6 (Bar) = 5
m6 (kg/s) = 72
T3 (C) = 281
T5 (C) = 65

1. Determine the specific enthalpy (kJ/kg) at the feedwater heater exit.
Your Answer =
2. Determine h3, the specific enthalpy (kJ/kg) at the first feedwater heater inlet.
Your Answer =
3. Determine h5, the specific enthalpy (kJ/kg) at the second feedwater heater inlet.
Your Answer =
4. Determine m3, the mass flow rate (kg/s) at the first feedwater heater inlet.
Your Answer =
5. Determine m5, the mass flow rate (kg/s) at the second feedwater heater inlet.
Your Answer =

Answer :

Answer:

1. 640.09 KJ/kg

2. 3025.232 KJ/kg

3. 272.09 KJ /kg

4. m3 = 9.624 kg /s  

5. m5 = 62.37 kg/s

Explanation:

part a

P6 = 5 bar @saturated liquid

Use table A-5

h_exit@Psat 5bar = h_f,5bar = 640.09 KJ/kg

part b

T3 = 281 C  ... super-heated steam

P6 = 500 KPa    ... Tsat,@500 KPa = 151.83 C

Use table A-6

P = 0.5 MPa ......

T = 250 C , h = 2961 KJ/kg

T = 300 C , h = 3064.6 KJ/kg

Interpolate:

m = (3064.6  - 2961) / (300 - 250) = 2.072

c = 3064.6 - 2.072*300 = 2443

h3 = 2.072*281 + 2443 = 3025.232 KJ/kg

part c

T5 = 65 C   .... compressed liquid

P6 = 500 KPa    ... Tsat,@500 KPa = 151.83 C

Use Table A-4

h5@65 C, f = 272.09 KJ /kg

part d and e

Continuity Equation

m3 + m5 = m_exit = 72 kg /s   .... Eq 1

Energy balance:

m3*h3 + m5*h5 = m_exit * h_exit   .... Eq 2

3025.232*m3 + 272.09*m5 = 72*640.09 = 46086.48   .... Eq 2

Solve above two equation simultaneously for

m3 = 9.624 kg /s

m5 = 62.376

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