ICAM2Heat Exchanger Problems 1Autumn 2002
Problem 1
A hot fluid at 100ºC enters a double-pipe heat exchanger and is cooled to 75ºC. A cooler fluid at 5ºC enters the exchanger and is warmed to 50ºC. Determine the LMTD for both counter-flow and parallel flow configurations.
Counter-flow:ΔT1 = 100 - 50 = 50ºCΔT2 = 75 - 5 = 70ºC
59.4ºC
Parallel-flow:ΔT1= 100 - 5 = 95ºCΔT2 = 75 - 50 = 25ºC
52.4ºC
Problem 2
A hot fluid at 120ºC enters a double-pipe heat exchanger and is cooled to 65ºC. A cooler fluid enters the exchanger at 38ºC and is to be warmed to 65ºC. Determine the LMTD for both counter-flow and parallel flow configurations.
Counter-flow:ΔT1 = 120 - 65 = 55ºCΔT2 = 65 - 38 = 27ºC
39.4ºC
Parallel-flow:ΔT1= 120 - 38 = 82ºCΔT2 = 65 - 38 = 27ºC
0ºC
A parallel-flow heat exchanger would have to have an infinite area in order that both fluids leave at the same temperature. This, of course, is not physically possible.
Problem 3
Water at a mass flow rate of 68 kg/min is heated from 35ºC to 75ºC by an oil having a heat capacity of 1.9 kJ/kg K. The fluids are used in a counter-flow, double-pipe heat exchanger and the oil enters the exchanger at 110ºC and leaves at 75ºC. If the overall heat transfer coefficient is 320 W/m2 K, calculate the heat exchanger area.
Data: Heat capacity of water = 4.18 J/kg K.
But
15.82 m2
Problem 4
Steam passes through a turbine into a condenser. Liquid water from the condensed steam is used to heat ethylene glycol. The water is available at 90ºC, with a mass flow rate of 2300 kg/h. The ethylene glycol has a temperature of 30ºC and a mass flow rate of 5500 kg/h. It is proposed to use a double-pipe heat exchanger, made of 2x1¼ inch standard type-M copper tubing with soldered fittings, which is 6 m long. Determine the outlet temperature of the ethylene glycol under counter-flow conditions.
Assumptions and simplifications:
- Steady-state conditions exist.
- Fluid properties are constant and evaluated at the average of the inlet temperatures (90+30)/2 = 60ºC.
- Neglect heat loss to the surroundings.
- Neglect resistance to heat transfer due to conduction through the wall.
Fluid properties:
From tables of properties:
For waterρ = 0.985 x 103 kg/m3k = 0.651 W/m K
at 60ºC: C = 4.184 x 103 kJ/kg Kα = 1.554 x 10-7 m2/s
ν = 0.478 x 10-6 m2/sPr = 3.02
For ethylene-ρ = 1.087 x 103 kg/m3k = 0.260 W/m K
glycol at 60ºC: C = 2.562 x 103 kJ/kg Kα = 0.932 x 10-7 m2/s
ν = 4.75 x 10-6 m2/sPr = 51
Tube sizes:
Annulus (a)
Pipe (p)
5.1 cm3.5 cm 3.3 cm
Arrangement of Heat Exchanger:
We choose to route the ethylene glycol through the annular passage, which has the greater area, as this stream has the higher mass flow rate.
Annulus (a)
Pipe (p)
3.1DaDpo
For heat transfer: 0.039 m
For fluid friction:0.016 m
Velocities and Reynolds numbers:V = m/ρAandRe = VD/ν
0.76 m/s5.2 × 104 [water]
1.30 m/s1.07 × 104 [e.g.]
Nusselt Numbers:Nu = 0.023 (Re)0.8(Pr)nwhere n = 0.3 (cooling)
n = 0.4 (heating)
Water:Nu = 0.023(5.2×104)0.8(3.02) = 190
E.G.:Nu = 0.023(1.07×104)0.8(51) = 185
Local coefficients of heat transfer:Nu = hD/ksoh = Nu(k/D)
Water:hpi = (190)(0.651)/(0.033) = 3748 W/m2 K
E.G.:hpo = (185)(0.260)/(0.035) = 1233 W/m2 K
Overall coefficient of heat transfer:
so U0 = 914 W/m2 K
LMTD method - Parameter R1.46
Area:A = ΠDL = Π(0.035)(6) = 0.66 m2
Inlet temperatures:Thi = 90ºCTci = 30ºC
Outlet temperatures:
78.7ºC
But also, 37.7ºC
Use heat balance to check calculation:
Water:qh = (2300)(4184)(90-78.7)/3600 = 30206 W
E.G.qc = (5500)(2562)(37.7-30)/3600 = 30139 W
To increase the outlet temperature of ethylene glycol:
- Use bigger tube diameters to increase the surface area - but this will decrease velocity
- Increase the mass flow rates - this will increase velocity
- Use several exchangers in series.
Calculation of pressure drops and pumping costs is necessary to complete the design.
NTU method:Using the above data
0.2250.683
For a counter-flow heat exchanger:
0.189
but sincemhCh < mcCc then 0.189
so Tho = 90-(0.189)(60) = 78.7ºC
but also
so Tco = 30+(0.683)(90-78.7) = 37.7ºC