2. Water (op 4192 kj is flowing at a rate of 6.5 kg/s enters a parallel flow hea

Question

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.

Explanation / Answer

Cold fluid is water. Hot fluid is oil.

Cmin is obtained for cold fluid, that is water.

Cmin = mwater*Cpwater

Cmin = 6.5*4.192 = 27.248 kW/K

Cmin / Cmax = 0.2

Cmax = 27.248 / 0.2

Cmax = 136.24 kW/K

a)

Cmax = moil*Cpoil

136.24 = moil * 1.89

moil = 72.085 kg/s

b)

Max heat transfer qmax = Cmin (THotInlet - TColdInlet)

qmax = 27.248*(380 - 300)

qmax = 2179.84 kW

Heat transfer rate q = Effectiveness * qmax

q = 0.8*2179.84

q = 1743.872 kW

q = ChotFluid*(ThotInlet - ThotOutlet) = CcoldFluid*(TcoldInlet - TcoldOutlet)

1743.872 = 136.24 * (380 - ThotOutlet) = CcoldFluid*(300 - TcoldOutlet)

ThotOutlet = 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 / Cmin and q = UA*LMTD

Utilizing both these relations,

LMTD = q / (NTU*Cmin)

LMTD = 1743.872 / (3 * 27.248)

LMTD = 21.33 deg C

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.