ME161P – POWER PLANT DESIGN WITH RENEWABLE ENERGY 1T 2023-2024
PREPARED BY: ENGR. MANUEL B. RUSTRIA
WEEK 3 – HEAT EXCHANGERS
8.4 A surface condenser receives 200 t/h of steam at 40 °C with 10 % moisture. The cooling water enters
at 33 °C and leaves at 38 °C. The pressure inside the condenser is found to be 0.078 bar. Neglecting any
sub-cooling, determine (a) the condenser cooling water flow in kg/s; (b) air ejector capacity in kg/s; and
(c) the surface area of the condenser. Take Uo = 2, 970 W/m2-K.
[Ans. (a) 5, 748 kg/s, (b) 4. 623 kg/s, (c) 10, 150 m2]
SOLUTION: SURFACE CONDENSER PERFORMANCE EVALUATION
LET mw = be the cooling water mass flow rate, kg/s
ma = be the ejector capacity, kg/s
Ao = be the surface area of condenser, m2
ASSUMPTIONS:
1.
2.
3.
4.
5.
6.
Steady State Steady Flow (SSSF) condition exists
Constant water specific heat, cpw = 4.187 kJ/kg-K
Gas constant for air, R = 0.287 kJ/kg-K
Constant steam temperature, tsteam = tsh = 40 °C
Constant condenser pressure, psh = 0.078 bar = 0.0078 MPa
1 ton = 1, 000 kg
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SCHEMATIC DIAGRAM OF THE SYSTEM (JUST INSERT THE GIVEN VALUES IN THE FIGURE)
A. For the cooling water mass flow rate, mw
Applying energy balance in the condenser,
Ein = Eout
Energy released by the steam = Energy conducted through the condenser tube wall = Energy
absorbed by the water
or, Qsteam = Qwall = Qwater
ms(x2hfg2) = Uo(Ao)(LMTD) = mw(cpw)(Δtw)
or, mw = ms(x2hfg2)/ (cpw)(Δtw)
From Steam Tables, hfg2 = hfg at 40 °C = 2, 406.7 kJ/kg
2
Substituting values and applying dimensional analysis,
mw = [(200 ton/h x (1, 000 kg/1 ton) x (1h / 3, 600 s) (0.9 x 2, 406.7 kJ/kg)]/ [(4.187 kJ/kg-K x
(38 - 33) K]
mw = 5, 748.5 kg/s
(NOTE: 200 ton/h = 55.56 kg/s)
B. For the ejector capacity, ma
From the Ideal Gas equation,
paVa = maRaTa
or, ma = paVa/RaTa
Where:
1. From Dalton’s Law,
pa = air partial pressure = pc – pv
From Steam Tables, pv = vapor partial pressure = psat at 40 °C = 0.007384 MPaa = 7.384 kPaa
Also, pc = condenser pressure = total pressure = 0.078 bar = 7.8 kPaa
Then, pa = 7.8 kPaa - 7.384 kPaa = 0.416 kPaa
2. Va = air volume flow rate = steam volume flow rate = v2(ms)
Where:
v2 = steam specific volume = vf at 40 °C + x2(vfg at 40°C)
= [1.0078x10-3 + 0.90(19, 522x10-3)] m3/kg
= 17, 570.81x10-3 m3/kg = 17.57 m3/kg
Then, Va = (17.57 m3/kg) (55.56 kg/s) = 976.2 m3/s
3. Ta = air absolute temperature = 40 °C + 273 = 313 K
Substituting values,
ma = [0.416 kPa x (1 kN/m2/ 1 kPa)(976.23 m3/s)]/[0.287 kJ/kg-K)(1 kN-m/ 1 kJ)(313 K)]
= 4.52 kg/s
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C. For the surface area of condenser, Ao
From condenser energy balance,
ms(x2hfg2) = UoAo(LMTD)
or, Ao = ms(x2hfg2)/ Uo(LMTD)
Where:
LMTD = (Δtin – Δtout) / ln(Δtin/Δtout)
Also, from the temperature-length profile of condenser,
Δtin = tsteam – twin = 40 – 33 = 7 °C and Δtout = tsteam – twout = 40 – 38 = 2 °C
Thus, LMTD = (7 – 2) °C / ln(7 °C / 2 °C) = 4 °C
Substituting values,
Ao = [(55.56 kg/s)(0.9 x 2, 406.7 kJ/kg)(1, 000 J/1 kJ)]/[(2, 970 W/m2-K x 4 °C)] = 10, 152.4 m2
Summary of final answers:
a. mw = 5, 748.5 kg/s
b. ma = 4.52 kg/s
c. Ao = 10, 152.4 m2
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