Heat Exchangers: Parallel Flow and Counterflow

Set the two inlet temperatures, the two flows and the size of the exchanger, U·A. The page solves the same exchanger twice, once with the streams running the same way and once running opposite ways, and draws both streams' temperatures along the wall.

Worked example — Heat Exchangers: Parallel Flow and Counterflow

Capacity rate, hot stream	418.0 W/K
Capacity rate, cold stream	418.0 W/K
Number of transfer units	2.000
Capacity ratio	1.0000
The most heat any exchanger could move	29.26 kW
Effectiveness, counterflow	66.67%
Effectiveness, parallel flow	49.08%
Heat moved, counterflow	19.51 kW

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Heat Exchangers: Parallel Flow and Counterflow

A heat exchanger passes heat from a hot stream to a cold one through a wall, without mixing them. Each stream's capacity rate C = ṁc_p is the heat it carries per degree. The most heat any exchanger could move is set by the smaller of the two, Q_max = C_min(T_h,in − T_c,in), and the effectiveness ε is the fraction of that the exchanger actually moves. Its size enters only as the number of transfer units, NTU = UA/C_min.

What each symbol means

C: capacity rate of a stream, ṁc_p, in W/K. Water at 0.1 kg/s carries 418 W/K
NTU: number of transfer units, UA/C_min: the exchanger's size measured against the smaller stream
ε: effectiveness, the heat moved as a fraction of C_min(T_h,in − T_c,in)
Cr: capacity ratio C_min/C_max, from 0 for a condensing vapour to 1 for equal flows

Assumes: Steady flow, no heat lost to the surroundings, and one overall coefficient U along the whole wall, which is when the effectiveness–NTU relations are exact. Both streams are liquid water with c_p = 4.18 kJ/(kg K), within 0.6% of the measured value between 10 and 90 °C and within 1% from 0 to 99 °C. c_p enters through the capacity rates, so an error in it moves NTU, and the kilowatts, by about the same fraction.
Breaks when: When U changes along the wall, as it does where a stream starts to boil or its flow turns laminar partway. When a stream changes phase partway along the wall, so that part of it is heated and part boils, because its capacity rate is then no longer ṁc_p. And for shell-and-tube and cross-flow units, which need their own effectiveness formulas and, unless one stream's temperature hardly changes, recover less than counterflow at the same NTU.

With equal flows at NTU 2, a parallel-flow exchanger four times larger recovers less than one point more 🖖

Press Equal flows: 0.1 kg/s of water each way, NTU 2. Run the two streams the same way and the exchanger recovers 49.08% of the heat it could; run them in opposite directions and it recovers 66.67%. Now press Four times larger. Counterflow climbs to 88.89%. Parallel flow reaches 50.00%, and no size will take it further. Both streams enter at the same end and head for the same temperature, 45 °C, which is what the two would reach if they were simply mixed. As they near it the difference that drives the heat through the wall runs out, and the far end of the exchanger does almost nothing. With equal flows that ceiling is exactly half. In counterflow the incoming hot water meets the cold water that is about to leave, and the incoming cold water meets hot water that has already given up most of the heat it will give up, so there is a difference to work with all the way along.

In counterflow the cold water can leave hotter than the hot water does 🖖

At Equal flows the counterflow outlets are 56.67 °C for the cold water and 33.33 °C for the hot. The stream that came in cold leaves 23.33 degrees warmer than the stream that came in hot. In the drawing the two counterflow lines are parallel, because with equal flows the difference across the wall is the same 23.33 degrees at every point along it. Parallel flow can never do this. Its two lines close on each other from either side of 45 °C with the cold one always below, so at any size the cold outlet stays under the hot outlet: 44.36 °C against 45.64 °C here. Press Twice the hot flow and counterflow still sends the cold water out 11.33 degrees above the hot.

When one stream barely changes temperature, it stops mattering which way it runs 🖖

Press Large hot stream: hot water at 5 kg/s against 0.1 kg/s of cold, fifty times the capacity, at NTU 2. The hot water cools by less than 1.5 degrees, and counterflow and parallel flow recover 86.16% and 85.29%, 0.87 percentage points apart. As the exchanger grows at this ratio the gap rises toward 1.96 points, the distance from 100% to parallel flow's ceiling of 98.04%, and never passes it. A condensing vapour, which gives up heat without changing temperature at all, is the limit: both arrangements then recover exactly 1 − e^−NTU, 86.47% at NTU 2. Short exchanger is the other way to make the direction irrelevant: at NTU 0.2 neither stream changes much, and the two recover 16.67% and 16.48%. The choice of direction pays when both streams change temperature a lot.

Problem solved in full

How much larger, to go from 80% to 90%? 4 steps
A counterflow exchanger with equal flows recovers 80% of the heat. How much larger must it be to recover 90%? And what would a parallel-flow exchanger of that larger size recover?

Open this problem in the tool
1. With equal flows, C_r = 1, the counterflow effectiveness is NTU/(1 + NTU). Solved for NTU:
  
  $ε = \frac{N T U}{1 + N T U} \Rightarrow N T U = \frac{ε}{1 - ε}$
2. 80% needs NTU 4, and 90% needs NTU 9.
  
  $N T U_{80%} = \frac{0 . 80}{0 . 20} = 4 N T U_{90%} = \frac{0 . 90}{0 . 10} = 9$
3. NTU is UA/C_min and the flows have not changed, so UA has to grow by the same factor. With the same wall, that is 2.25 times the area.
  
  $\frac{( U A ) _{90%}}{( U A ) _{80%}} = \frac{9}{4} = 2.25$
4. Parallel flow at NTU 9 with equal flows recovers (1 − e⁻¹⁸)/2:
  
  $ε_{p a r} = \frac{1 - e ^{- 2 \times 9}}{2} = 0.49999999$
Answer

2.25 times the area, and parallel flow at that size would recover 50.00%. The last ten points cost more wall than the first eighty did. Press Equal flows and set U·A to 1,672 W/K and then to 3,762: the counterflow row reads 80.00% and then 90.00%, and the parallel-flow row reads 49.98% and then 50.00%.

$U A = 4 \times 418 = 1, 672 W/K \to 9 \times 418 = 3, 762 W/K$

References (3)

Both effectiveness formulas the page solves, and the balanced-counterflow result that the temperature difference is the same all along the wall: J. H. Lienhard V and J. H. Lienhard IV, A Heat Transfer Textbook, 6th edition. Phlogiston Press, Cambridge MA, 2024. §3.3, eqns (3.20) and (3.21); Example 3.2.
The specific heat of liquid water, 4.18 kJ/(kg K), and how far it moves between 10 and 90 °C: "Isobaric Properties for Water", in P. J. Linstrom and W. G. Mallard (eds.), NIST Chemistry WebBook, NIST Standard Reference Database Number 69. National Institute of Standards and Technology, Gaithersburg MD. Computed from W. Wagner and A. Pruss, the IAPWS-95 formulation.
Where the effectiveness–NTU method was first set out in full: W. M. Kays and A. L. London, Compact Heat Exchangers. McGraw-Hill, New York, 1955.

By the 42Math team · Reviewed against published sources. Corrections welcome.