2 Water op 4192 kj is flowing at a rate of 65 kgs enters a p

2. Water (op 4192 kj is flowing at a rate of 6.5 kg/s enters a parallel flow heat exchanger kg K to be heated by an oil (c 1.89 kg-K The effectiveness of the heat exchanger for this case is 0.80 and Cmin/Cmax ratio is 0.20. Cmin is obtained for the cold fluid. a) What is the mass flow rate of oil (hot fluid)? b) If the inlet temperature for the hot fluid is 380Kand the inlet temperature for the cold fluid is 300 K, what would be the outlet temperature of the hot fluid? c) Calculate the heat transfer rate. flow.

Solution

Cold fluid is water. Hot fluid is oil.

C_min is obtained for cold fluid, that is water.

C_min = m_water*Cp_water

C_min = 6.5*4.192 = 27.248 kW/K

C_min / C_max = 0.2

C_max = 27.248 / 0.2

C_max = 136.24 kW/K

a)

C_max = m_oil*Cp_oil

136.24 = m_oil * 1.89

m_oil = 72.085 kg/s

b)

Max heat transfer q_max = C_min (T_HotInlet - T_ColdInlet)

q_max = 27.248*(380 - 300)

q_max = 2179.84 kW

Heat transfer rate q = Effectiveness * q_max

q = 0.8*2179.84

q = 1743.872 kW

q = C_hotFluid*(T_hotInlet - T_hotOutlet) = C_coldFluid*(T_coldInlet - T_coldOutlet)

1743.872 = 136.24 * (380 - T_hotOutlet) = C_coldFluid*(300 - T_coldOutlet)

T_hotOutlet = 367.2 deg C

c)

As found above, q = 1743.872 kW

d)

For parallel flow heat exchanger (figure 2), we see that for effectivenss = 0.8 and Cmin / Cmax = 0.2, we get corresponding NTU = 3

NTU = UA / C_min and q = UA*LMTD

Utilizing both these relations,

LMTD = q / (NTU*C_min)

LMTD = 1743.872 / (3 * 27.248)

LMTD = 21.33 deg C