In the name of God 1
Heat Transfer Jami Institute of Technology Heat Exchangers part 2 Mehdi Rasti 2
The Counter-Flow Heat Exchanger In contrast to the parallel-flow exchanger, this configuration provides for heat transfer between the hotter portions of the two fluids at one end, as well as between the colder portions at the other. For this reason, the change in the temperature difference, ΔT = Th Tc, with respect to x is nowhere as large as it is for the inlet region of the parallel-flow exchanger. Note that the outlet temperature of the cold fluid may now exceed the outlet temperature of the hot fluid. 3
The Counter-Flow Heat Exchanger 4 FIGURE: Temperature distributions for a counter-flow heat exchanger
The Counter-Flow Heat Exchanger for the counter-flow exchanger the endpoint temperature differences must now be defined as: 5
The Counter-Flow Heat Exchanger Note that, for the same inlet and outlet temperatures, the log mean temperature difference for counter-flow exceeds that for parallel flow,. Hence the surface area required to effect a prescribed heat transfer rate q is smaller for the counterflow than for the parallel-flow arrangement, assuming the same value of U. Also note that Tc,o can exceed Th,o for counter-flow but not for parallel flow. 6
Special Operating Conditions It is useful to note certain special conditions under which heat exchangers may be operated. Figure below shows temperature distributions for a heat exchanger in which the hot fluid has a heat capacity rate, which is much larger than that of the cold fluid, 7
Special Operating Conditions Figure: Special heat exchanger conditions. Ch >> Cc or a condensing vapor 8
Special Operating Conditions For this case the temperature of the hot fluid remains approximately constant throughout the heat exchanger, while the temperature of the cold fluid increases. The same condition is achieved if the hot fluid is a condensing vapor. Condensation occurs at constant temperature, and, for all practical purposes. 9
Special Operating Conditions Conversely, in an evaporator or a boiler (Figure below), it is the cold fluid that experiences a change in phase and remains at a nearly uniform temperature ( ). The same effect is achieved without phase change if Ch<<Cc. Note that, with condensation or evaporation, the heat rate is given by below Equations: 10
Special Operating Conditions Figure: Special heat exchanger conditions. Ch << Cc or a evaporating liquid 11
Special Operating Conditions The third special case (below Figure) involves a counter-flow heat exchanger for which the heat capacity rates are equal (Ch=Cc). The temperature difference ΔT must then be constant throughout the exchanger, in which case 12
Special Operating Conditions Figure: Special heat exchanger conditions. A counter-flow heat exchanger with equivalent fluid heat capacities (Ch = Cc). 13
Example 1 A counter-flow, concentric tube heat exchanger is used to cool the lubricating oil for a large industrial gas turbine engine. The flow rate of cooling water through the inner tube (Di = 25 mm) is 0.2 kg/s, while the flow rate of oil through the outer annulus (Do=45 mm) is 0.1 kg/s. The oil and water enter at temperatures of 100 and 30C, respectively. How long must the tube be made if the outlet temperature of the oil is to be 60C? 14
Example 1 Known: Fluid flow rates and inlet temperatures for a counterflow, concentric tube heat exchanger of prescribed inner and outer diameter. Find: Tube length to achieve a desired hot fluid outlet temperature. 15
Example 1 Assumptions: 1. Negligible heat loss to the surroundings. 2. Negligible kinetic and potential energy changes. 3. Constant properties. 4. Negligible tube wall thermal resistance and fouling factors. 5. Fully developed conditions for the water and oil (U independent of x) 16
Example 1 Schematic: 17
Properties: unused engine oil (Th = 80C = 353 K): cp = 2131 J/kg K, μ = 3.25 10^(-2) N.s/m2, k = 0.138 W/m K. Water: (Tc = 35C): cp = 4178 J/kg K, μ = 725 10^(-6) N.s/m2, Example 1 k = 0.625 W/m K, Pr 4.85 18
Example 1 Analysis: The required heat transfer rate may be obtained from the overall energy balance for the hot fluid: The water outlet temperature is: 19
Example 1 Accordingly, use of to evaluate the water properties was a good choice. The required heat exchanger length may now be obtained from below Equation: where and: 20
Example 1 The overall heat transfer coefficient is: For water flow through the tube: 21
Example 1 Accordingly, the flow is turbulent and the convection coefficient may be computed from below Equation: Hence 22
Example 1 For the flow of oil through the annulus, the hydraulic diameter is: and the Reynolds number is: 23
Example 1 The annular flow is therefore laminar. Assuming uniform temperature along the inner surface of the annulus and a perfectly insulated outer surface, the convection coefficient at the inner surface may be obtained from Table below: 24
Example 1 With, linear interpolation provides: And : 25
Example 1 The overall convection coefficient is then: and from the rate equation it follows that 26
Example 1 Comments: 1. The hot side convection coefficient controls the rate of heat transfer between the two fluids, and the low value of ho is responsible for the large value of L. Incorporation of heat transfer enhancement methods, could be used to decrease the size of the heat exchanger. 2. Because hi>>ho, the tube wall temperature will follow closely that of the coolant water. Accordingly, the assumption of uniform wall temperature, which is inherent in the use of Table 8.2 to obtain ho, is reasonable. 27
Example 2 The counter-flow, concentric tube heat exchanger of Example 1 is replaced with a compact, plate-type heat exchanger that consists of a stack of thin metal sheets, separated by N gaps of width a. The oil and water flows are subdivided into N/2 individual flow streams, with the oil and water moving in opposite directions within alternating gaps. It is desirable for the stack to be of a cubical geometry, with a characteristic exterior dimension L. 28
Example 2 Determine the exterior dimensions of the heat exchanger as a function of the number of gaps if the flow rates, inlet temperatures, and desired oil outlet temperature are the same as in Example 1. Compare the pressure drops of the water and oil streams within the plate-type heat exchanger to the pressure drops of the flow streams in Example 1, if 60 gaps are specified. 29
Example 2 Known: Configuration of a plate-type heat exchanger. Fluid flow rates, inlet temperatures, and desired oil outlet temperature. Find: 1. Exterior dimensions of the heat exchanger. 2. Pressure drops within the plate-type heat exchanger with N 60 gaps, and the concentric tube heat exchanger of Example 1. 30
Example 2 Schematic: 31
Example 2 Assumptions: 1. Negligible heat loss to the surroundings. 2. Negligible kinetic and potential energy changes. 3. Constant properties. 4. Negligible plate thermal resistance and fouling factors. 5. Fully developed conditions for the water and oil. 6. Identical gap-to-gap heat transfer coefficients. 7. Heat exchanger exterior dimension is large compared to the gap width. 32
Example 2 Properties: unused engine oil : ρ = 852.1 kg/m3 Water: 33
Example 2 Analysis: 1. The gap width may be related to the overall dimension of the heat exchanger by the expression a=l/n, and the total heat transfer area is Assuming a<<l and the existence of laminar flow, the Nusselt number for each interior gap is provided in Table 8.1 and is: 34
Example 2 The hydraulic diameter is Dh expressions yields for the water: 2a. Combining the preceding Likewise, for the oil: and the overall convection coefficient is 35
Example 2 From Example 1, the required log mean temperature difference and heat transfer rate are and q = 8524 W, respectively. which may be rearranged to yield: 36
Example 2 The size of the compact heat exchanger decreases as the number of gaps is increased, as shown in the figure below. 37
Example 2 2. For N = 60 gaps, the stack dimension is L = 0.131 m from the results of part 1, and the gap width is: a=l/n=0.131m/60=0.00218 m. The hydraulic diameter is Dh=0.00436 m, and the mean velocity in each waterfilled gap is: 38
Example 2 providing a Reynolds number of: For the oil-filled gaps: yielding a Reynolds number of: 39
Example 2 Therefore, the flow is laminar for both fluids, as assumed in part 1. below equations may be used to calculate the pressure drop for the water: Similarly, for the oil: 40
Example 2 For Example 1, the friction factor associated with the water flow for a smooth surface condition may be calculated by using below equation: The mean velocity is: 41
Example 2 and the pressure drop is: 42
Example 2 For the oil flowing in the annular region, the mean velocity is: And the pressure drop is 43
Example 2 Comments: 1. Increasing the number of gaps increases the UA product by simultaneously providing more surface area and increasing the heat transfer coefficients associated with the flow of the fluids through smaller passages. 2. The area-to-volume ratio of the N = 60 heat exchanger is L2(N - 1)/L3 = (N - 1)/L = (60-1)/0.131 m = 451 m2/m3. 44
Example 2 3. The pressure drops associated with use of the compact heat exchanger are significantly less than for a conventional concentric tube configuration. Pressure drops are reduced by 99.9% and 99.5% for the water and oil flows, respectively. 4. Fouling of the heat transfer surfaces may result in a decrease in the gap width, as well as an associated reduction in heat transfer rate and increase in pressure drop. 45
Example 2 5. Because hc > hh, the temperatures of the thin metal sheets will follow closely that of the water, and, as in Example 1, the assumption of uniform temperature conditions to obtain hc and hh is reasonable. 46
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