Although heat exchangers usually are selected for unique sets of conditions, sometimes it is of interest to know how they will respond to changes in inlet temperature. For example, it might be necessary to predict the change in coldside outlet temperature resulting from a change in inlet boilerwater temperature or steam pressure. Because when one terminal temperature is changed, the others become unknowns, it is impractical to predict performance by means of the standard heattransferrate equation:
q = (UA)(∆T_{lm})F (1)
where:
q = heattransfer rate
U = overall heattransfer coefficient
A = heattransfersurface area
∆T_{lm} = logmeantemperature difference (LMTD) of the two fluids
F is the LMTD correction factor for noncounterflow operation. It also is the ratio of surface area required with purecounterflow operation to surface area actually required. It varies from unity (e.g., in platetype heat exchangers and steamwater applications) to about 0.75, below which surface requirements tend to be impracticably large. Without a rate equation not involving LMTD or F, new temperature profiles for rated heat exchangers would be found through tedious trial of leaving temperatures until the known values of UA were satisfied—that is, until:
q ÷ [(∆T_{lm})F] = UA (1a)
Fortunately, there is a rate equation not involving LMTD or F. It is based on the concepts of heattransfer effectiveness (ε) and number of heattransfer units (N_{tu}) and gives q as a function of the difference between the two entering temperatures. It is a consequence of the εN_{tu} method developed by W.M. Kays and A.L. London^{1} for the design of lightweight heat exchangers with complex geometries for use in aircraft and ships. This article outlines the method and provides examples illustrating its usefulness to HVAC designers.
THE εN_{TU} METHOD
Heattransfer effectiveness is defined as:
ε = (q_{a} ÷ q_{t}) = (actual heat transfer ÷ theoretical maximum heat transfer) (2)
Actual heattransfer rate is given by heat balance:
q_{a} = C_{h}(T_{1 } T_{2}) = C_{c}(t_{2 } t_{1}) (3)
where:
C_{h}, C_{c} = flowstreamcapacity rates of hot and cold fluids, respectively, defined as mass rate of flow times specific heat
T_{1}, T_{2} = hotside entering and leaving temperatures, respectively
t_{1}, t_{2} = coldside entering and leaving temperatures, respectively
The fluidcapacity rates and general temperature profile of all of the heat exchangers discussed in this article are represented by the following notation:
T_{1} → T_{2} C_{h}
t_{2} ← t_{1} C_{c}
From the general temperature profile, it is evident that fluidtemperature difference in a heat exchanger can be no greater than (T_{1}  t_{1}). Moreover, no more heat can be transferred than can be absorbed at the coldside capacity rate, C_{c}, and released at the hotside capacity rate, C_{h}. And the maximum theoretical heat transfer cannot exceed the theoretical value attainable if the fluid with minimum capacity rate, C_{min}, could experience the maximum temperature difference, (T_{1}  t_{1}). So, regardless of which side has the minimum capacity rate, the theoretical maximum heattransfer rate, q_{t}, is given by:
q_{t} = C_{min}(T_{1}  t_{1}) (4)
From equations 2, 3, and 4:
ε = [C_{h}(T_{1} – T_{2})] ÷ [C_{min}(T_{1} – t_{1})] = [C_{c}(t_{2} – t_{1})] ÷ [C_{min}(T_{1} – t_{1})]
When C_{h} is C_{min}:
ε = (T_{1} – T_{2}) ÷ (T_{1} – t_{1}) (5)
When C_{c} is C_{min}:
ε = (t_{2} – t_{1}) ÷ (T_{1} – t_{1}) (6)
Thus, given ε, T_{1}, and t_{1} and the fluidcapacity rates, T_{2} and t_{2} always can be found.
Combining equations 2 and 4 results in a rate equation not involving LMTD:
q_{a} = εC_{min}(T_{1} – t_{1}) (7)
Number of heattransfer units, or “heatexchanger size,” is defined as:
N_{tu} = UA ÷ C_{min} (dimensionless) (8)
To show the relationship between N_{tu} and ε, Kays and London^{1} introduced capacityrate ratio (C_{min} ÷ C_{max}) and plotted εvs.N_{tu} curves for a large number of heatexchanger types. Figures 1 and 2 show the curves for two exchanger types commonly used in HVAC design.
In rerating an exchanger at new fluid temperatures, too little information is available to accurately predict the new U value. For moderate changes in average fluid temperature, however, the variation in U is unlikely to exceed the normal range of error expected in detailed heattransfer calculations or even sophisticated computer programs. In most cases, the U value at the cataloged rating can be assumed valid for the new rating without serious error.^{2} With N_{tu} assumed constant, the actual U value does not have to be known because it is implicit in the N_{tu}. (See Equation 8.) This is especially applicable to hydronic systems because of the relatively small change in the heattransfer properties of water.
If there is reason to believe that a U value will be significantly different at new conditions, a design value can be used for a new N_{tu}, and a new ε can be found from curves such as those in figures 1 and 2. Figure 3 is useful for estimating variations in U over a range of average film temperatures. Changes in flow rates can, of course, affect U values significantly. Their effect can be calculated by methods found in standard texts. For this article, however, fluidcapacity rates are assumed to be constant.
EXAMPLES
In the following examples, English engineering units are used. Water is assumed to have a specific heat of 1.0 and a weight of 8.33 lb per gallon. In most of the examples, a “hydronic constant” of 500 (8.33 multiplied by 1.0 multiplied by 60 [minutes per hour]) is used to obtain fluidcapacity rate in British thermal units per hour per degree Fahrenheit (Btuh/°F). The heattransfer coefficients were chosen for illustrative purposes only and are not necessarily the values one would calculate using the detailed procedures given by Kern^{4} and others.^{5}
Shellandtube heat exchanger, T_{1} and T_{2} unknown. Consider a shellandtube heat exchanger with a catalog rating of:
Boiler water:
200 gpm 210°F → 180°F C_{min} = (200)(500) = 100,000 Btuh/°F
Process water:
300 gpm 160°F ← 140°F C_{max} = (300)(500) = 150,000 Btuh/°F
A systemdesign change requires the leaving temperature, t_{2}, to be 180°F, with a return temperature, t_{1}, of 150°F. Find the new T_{1} to satisfy this requirement. The new temperature profile to be completed is:
T_{1} → T_{2}
180°F ← 150°F
Assuming there will not be a significant change in U, find ε from the catalog rating, and use it for the new rating.
From Equation 5:
ε = (210 – 180) ÷ (210 – 140) = 0.43
From Equation 3:
q_{a} = (150,000)(180 – 150) = 4,500,000 Btuh
and:
(T_{1} – T_{2}) = (q_{a} ÷ C_{h}) = 4,500,000 ÷ 100,000 = 45°F
From Equation 5:
(T_{1} – t_{1}) = [(T_{1} – T_{2}) ÷ ε] = 45 ÷ 0.43 = 104.7°F
T_{1} = 150 + 104.7 = 254.7°F
T_{2} = 254.7 − 45 = 209.7°F
The new temperature profile is:
255°F → 210°F
180°F ← 150°F
Note that while U was assumed constant, its actual value did not have to be known.
Although the heattransfer rate went from 3,000,000 Btuh ([100,000][210  180]) to 4,500,000 Btuh, there was no increase in heattransfersurface area. That followed from the assumption of a constant N_{tu}. Thus, from Equation 1a:
UA = q ÷ [(ΔT_{lm})F] = constant
For a change in q, there must be a proportional change in LMTD so that UA remains constant.
Semiinstantaneous water heater, new steam pressure. Consider a semiinstantaneous water heater with the following catalog rating:
25psig steam in shell 267°F ↔ 267°F
100 gpm in tubes 120°F ← 50°F
The notation ↔ indicates isothermal condensation of saturated steam.
If this water heater has 40 sq ft of heattransfer surface, what steam pressure would be required to produce a coldside outlet temperature of 140°F at a maximum U value of 450 Btuh per square foot per degree Fahrenheit (Btuh/sq ft/°F)?
From Equation 3, noting that C_{h} is C_{max}:
C_{max} = [q_{a} ÷ (T_{1} – T_{2})]
But for an isothermal process, T_{1} minus T_{2} equals 0; thus, C_{max} is infinitely large, and C_{min} divided by C_{max} equals 0.
The temperature profile to be completed is:
T_{1 } ↔ T_{2} C_{max} = ∞
140°F ← 50°F C_{min} = (100)(500) = 50,000 Btuh/°F
Because the design U value for the new conditions will not necessarily equal the value at the catalog rating, N_{tu} must be calculated.
From Equation 8:
N_{tu} = [(UA) ÷ C_{min}] = [(450)(40) ÷ 50,000] = 0.36
From Figure 1, using the curve for C_{min} divided by C_{max} equals 0, ε equals 0.32.
From Equation 6:
T_{1} – t_{1} = [(t_{2} – t_{1}) ÷ ε] = [(140 – 50) ÷ 0.32] = 281.3°F
Thus:
T_{1} = 50 + 281.3 = 331.3°F
From standard steam tables, the steam pressure at 331°F is 89 psig. The new temperature profile is:
Steam, 89 psig 331°F ↔ 331°F
Water, 100 gpm 140°F ← 50°F
Rating an exchanger for use with a propyleneglycol solution. A shellandtube heat exchanger with 80 sq ft of heattransfer surface is available for service hot water for a small solar heating system. The design calls for a 50percentbyvolume propyleneglycol solution to enter the shell from the solar panels at a rate of 100 gpm and an average temperature of 150°F. Service water in the tubes will enter from a hotwater storage tank at an average temperature of 110°F. The specific heat of the 50percent glycol solution (taken at 120°F) is 0.87. The density is 64.1 lb per cubic foot, or 8.6 lb per gallon. The overall heattransfer coefficient is an estimated 250 Btuh/sq ft/F. What will the heattransfer rate and outlet temperatures be?
The temperature profile to be completed is:
100 gpm, 50percent PG:
150°F → T_{2} C_{max} = (100)(0.87)(8.6)(60) = 44,892 Btuh/°F
50 gpm, water:
t_{2} ← 110°F C_{min} = (50)(500) = 25,000 Btuh/°F
Capacity ratio = C_{min} ÷ C_{max} = 25,000 ÷ 44,892 = 0.56
From Equation 8:
N_{tu} = UA ÷ C_{min} = (250)(80) ÷ 25,000 = 0.80
From Figure 1:
ε = 0.50
From Equation 7:
q_{a} = 0.50(25,000)(150 – 110) = 500,000 Btuh
From Equation 3:
t_{2} = t_{1} + q ÷ C_{min} = 110 + 500,000 ÷ 25,000 = 130°F
T_{2} = T_{1} – (q_{a} ÷ C_{max}) = 150 – (500,000 ÷ 44,892) = 139°F
Thus, the design temperature profile will be:
150°F → 139°F
130°F ← 110°F
Plateandframe heat exchanger, new t_{1}. A plateandframe heat exchanger with 4,600 sq ft of plate surface was ordered for an indirect freecooling system designed to provide water to a conditioned space at a rate of 2,400 gpm and a temperature of 60°F when furnished with coolingtower water at a rate of 1,800 gpm and a temperature of 56°F. An increase in airconditioning load requires a deliveredwater temperature of 58°F and a return temperature of 67°F, based on an overall U factor of 650 Btuh/sq ft/°F. If the flow rates are to be maintained, what does the temperature of the coolingtower water (t_{1}) need to be?
The new temperature profile to be completed is:
67°F → 58°F C_{max} = (2,400)(500) = 1,200,000 Btuh/°F
T_{2} ← t_{1} C_{min} = (1,800)(500) = 900,000 Btuh/°F
The cooling load is:
(1,200,000)(67 – 58) = 10,800,000 Btuh
Because there is no operating experience to verify the original design temperatures, use Equation 8 to establish effectiveness:
N_{tu} = [(UA) ÷ C_{min}] = [(650 × 4,600) ÷ 900,000] = 3.3
C_{min} ÷ C_{max} = 900,000 ÷ 1,200,000 = 0.75
From Figure 2:
ε = 0.84
From Equation 6:
t_{1} = T_{1} – [(t_{2} – t_{1}) ÷ ε]
From Equation 3:
(t_{2} – t_{1}) = (q_{a} ÷ C_{min}) = (10,800,000 ÷ 900,000) = 12°F
Thus:
t_{1} = 67 – (12 ÷ 0.84) = 52.7°F maximum inlet temperature
t_{2} = 52.7 + 12.0 = 64.7°F
The new temperature profile is:
67°F → 58°F
64.7°F ← 52.7°F
Note that a “temperature cross” (t_{2} greater than T_{2}) occurs. This is possible only with counterflow exchangers and is one of the reasons why platetype exchangers are ideal for closeapproach applications, such as free cooling.
Design of a new exchanger. To illustrate the εN_{tu} method's value in design, consider the calculation of surface area for a Utube exchanger with 10psig steam condensing in the shell and water at a rate of 450 gpm heated from 165°F to 210°F in the tubes. Assume a U value of 500 Btuh/sq ft/°F.
From standard steam tables, at 10 psig, the saturatedsteam temperature is 240°F. Thus, the temperature profile to be satisfied is:
Steam at 10 psig:
240°F ↔ 240°F C_{h} = C_{max} = ∞ (See second example.)
Water, 450 gpm:
210°F ← 165°F C_{c} = C_{min} = (450)(500) = 225,000 Btuh/°F
For steam to water:
C_{min} ÷ C_{max} = 0
From Equation 6:
ε = (t_{2} – t_{1}) ÷ (T_{1} – t_{1}) = (210 – 165) ÷ (240 – 165) = 0.60
From Figure 1:
N_{tu} = 0.91
From Equation 8:
(UA) = (N_{tu})(C_{min}) = (0.91)(225,000) = 204,750 Btuh/°F
Thus:
A = 204,750 ÷ 500 = 410 sq ft
The duty can be found from Equation 3 or Equation 7. From Equation 7:
q_{a} = 0.60(225,000)(240 – 165) = 10,125,000 Btuh
COMPARISON WITH THE LMTD METHOD
It is interesting to compare the calculation above with the LMTD method. The LMTD of the terminaltemperature differences, (T_{1}  t_{2}) and (T_{2}  t_{1}), is found from:
ΔT_{lm} = (ΔT_{l} – ΔT_{s}) ÷ ln(ΔT_{l} ÷ ΔT_{s})
where:
ΔT_{l} = the larger terminaltemperature difference
ΔT_{S} = the smaller terminaltemperature difference
The terminaltemperature differences are:
(T_{2} – t_{1}) = (240 – 165) = 75°F = ΔT_{l}
(T_{1} – t_{2}) = (240 – 210) = 30°F = ΔT_{s}
Thus:
ΔT_{lm} = [(75 – 30) ÷ ln(75 ÷ 30)] = 49.1°F
As noted earlier, LMTD correction factor for noncounterflow operation, F, is 1.00 in steamtowater applications because one fluid stream is isothermal. Thus, from Equation 1:
A = [10,125,000 ÷ (500)(49.1)] = 412 sq ft
If this had been a watertowater application, several additional steps would have been required to evaluate F. These are not necessary with the εN_{tu} method, which is based on thermodynamic principles requiring no correction factor.
CONCLUSION
The εN_{tu} method not only eliminates calculation of LMTD and its correction factor, it can predict a heat exchanger's performance from one set of conditions to another, even when final temperatures and LMTD are not known. It is convenient for answering the whatif questions that often accompany heatexchanger applications. In predicting how an exchanger in a hydronic system will respond to a change in inlet temperature, the heattransfer product implicit in the N_{tu} of the cataloged rating usually can be assumed to remain constant over the extrapolation. Although there is a loss of accuracy, it usually is not significant, thanks to the relative stability of the thermal properties of water.
ACKNOWLEDGEMENT
While assuming full responsibility for any errors that may be found in this article, the author wishes to acknowledge the patient reading and constructive criticism volunteered by Kenneth M. Elovitz, PE, Esq.
REFERENCES

Kays, W.M., & London, A.L. (1955). Compact heat exchangers. New York: McGrawHill.

Guyer, E. (Ed.). (1989). Handbook of applied thermal design. New York: McGrawHill.

Potter, P.J. (1949). Steam power plants. New York: The Ronald Press Co.

Kern, D.Q. (1950). Process heat transfer. New York: McGrawHill.

McAdams, W.H. (1933). Heat transmission. New York: McGrawHill.
The president of James Breese & Co., Jim Breese represents manufacturers of heattransfer equipment and provides design services to facilities engineers. He has a bachelor's degree in mechanical engineering from the University of New Mexico.
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