Heat Exchanger Effectiveness Calculator
Find the effectiveness of a heat exchanger from its size and conductance, and the heat it moves and the outlets it gives.
What it does
Sizing a heat exchanger is awkward when you know the size of the machine but not the temperatures at the outlets, which is the usual situation. The effectiveness method was invented for exactly that case, and it reverses the usual problem: start with what the machine is and work out what it will do.
The Purdue lecture notes behind this page set out the two ways of analysing an exchanger. The log mean temperature difference method needs the outlet temperatures, which is fine when they are known and awkward when they are not. The effectiveness method instead takes the conductance of the exchanger and its size, which are always known, and produces the temperatures as the answer.
The key idea is the maximum possible heat rate. The California State University notes put it as the smaller of the two capacity rates multiplied by the difference between the two inlet temperatures, and the effectiveness as the actual heat rate divided by that maximum. The number of transfer units is the conductance times the area over the smaller capacity rate, and it is an indicator of the physical size of the exchanger, while the capacity rate ratio is the smaller rate over the larger one.
This page runs that method for parallel flow and counter current. Give it the overall heat transfer coefficient, the area, the capacity rate of each stream and the two inlet temperatures, and it returns the capacity rate ratio, the number of transfer units, the effectiveness, the heat rate, the largest heat rate that was possible and the two outlet temperatures.
The limits are worth stating. The correlations are for a simple two stream exchanger, and a shell and tube with several passes or a cross flow arrangement needs its own set of equations. The capacity rate of each stream is a constant, which is true for a stream of liquid or gas with no phase change, and a stream that boils or condenses has an infinite capacity rate instead.
How to use
- Choose metric or imperial for the conductance, the area, the capacity rates and the temperatures.
- Pick the flow arrangement, parallel or counter current.
- Enter the overall heat transfer coefficient U, which is the conductance of the whole surface including the films and the wall.
- Enter the heat transfer area.
- Enter the capacity rate of each stream, which is the mass flow rate times the specific heat.
- Enter the two inlet temperatures, then read the effectiveness, the heat rate and the outlet temperatures.
Formula
- NTU
- Number of transfer units, the conductance times the area over the smaller capacity rate.
- Cr
- Capacity rate ratio, the smaller capacity rate over the larger one.
- Cmin
- Smaller of the two heat capacity rates, in watts per kelvin.
- Qmax
- Largest heat rate the exchanger could move, Cmin times the difference between the inlet temperatures.
When the two capacity rates are equal the counter current expression has a zero in the denominator, and the limit of the formula is the effectiveness NTU over one plus NTU. A stream that changes phase has an infinite capacity rate, which pushes the capacity rate ratio to zero and the effectiveness towards one minus the exponential of minus NTU.
Worked example
Example: a 10 square metre exchanger with U of 500 W per m2 K in counter current
Defaults: metric units, counter current, U of 500 W per m2 K, an area of 10 m2, capacity rates of 5000 and 2000 W per K, and inlet temperatures of 80 and 20 degrees C.
- The conductance is U times A, which is 500 times 10 = 5000 W per K.
- The smaller capacity rate is 2000 W per K, so the number of transfer units is 5000 over 2000 = 2.5000, and the capacity rate ratio is 2000 over 5000 = 0.40000.
- For counter current the effectiveness is (1 - e to the minus 2.5 times 0.6) over (1 - 0.4 times e to the minus 1.5), which is 0.8530, so 85.30 percent.
- The largest possible heat rate is 2000 W per K times the 60 K inlet difference, which is 120,000.0 W.
- The real heat rate is the effectiveness times that, which is 102,360.2 W.
- The hot stream gives up 102,360.2 over 5000 = 20.472 K, so it leaves at 59.528 degrees C, and the cold stream gains 102,360.2 over 2000 = 51.180 K, so it leaves at 71.180 degrees C.
- The tool shows a capacity ratio of 0.40000, 2.5000 transfer units, 85.30 %, 120,000.0 W, 102,360.2 W, 59.528 and 71.180 degrees C and a conductance of 5,000.0 W per K.
Reference table
| Area | Transfer units | Effectiveness | Heat rate | Cold outlet |
|---|---|---|---|---|
| 2.5 m2 | 0.6250 | 43.13 % | 51,753.0 W | 45.876 degrees C |
| 5 m2 | 1.2500 | 65.06 % | 78,066.4 W | 59.033 degrees C |
| 10 m2 | 2.5000 | 85.30 % | 102,360.2 W | 71.180 degrees C |
| 20 m2 | 5.0000 | 96.95 % | 116,342.5 W | 78.171 degrees C |
Limits
- The correlations are for a simple two stream exchanger, and a shell and tube with several passes or a cross flow arrangement needs its own equations.
- The capacity rate of each stream is treated as constant, which is true for a single phase stream with no big temperature change.
- A stream that boils or condenses has an infinite capacity rate, and the formula for that limit is different from the one used here.
- The overall heat transfer coefficient is taken as given, and in a real exchanger it changes with the flow conditions and fouls up over time.
Common mistakes
- Confusing the capacity rate with the mass flow rate. The capacity rate is the mass flow times the specific heat, in watts per kelvin.
- Using the heat capacity of the exchanger itself instead of the two streams, which are what set the maximum possible heat rate.
- Expecting the same answer for parallel and counter flow. The same exchanger gives 85.30 percent in counter current and 69.27 percent in parallel flow.
- Forgetting that the maximum possible heat rate depends on the smaller capacity rate, so the bigger stream cannot cool below the inlet of the smaller one.
- Reading the effectiveness as the efficiency of the exchanger. It is a fraction of what was possible, not of the energy that went in.
Try it yourself
Quick checks. Answers sit tucked under each question.
How much does the flow arrangement change the answer?
A lot. The same exchanger with the same capacity rates gives 85.30 percent in counter current and 69.27 percent in parallel flow, so the heat rate falls from 102,360.2 W to 83,125.9 W and the cold outlet drops from 71.180 to 61.563 degrees C.
What happens if the area is doubled?
The number of transfer units doubles to 5.0000 and the effectiveness rises to 96.95 percent, so the heat rate goes up to 116,342.5 W and the cold stream leaves at 78.171 degrees C. The gain is much smaller than the extra area, because the exchanger is already close to the limit set by the capacity rates.
What sets the largest possible heat rate?
The smaller capacity rate and the difference between the two inlet temperatures, which here is 2000 W per K times 60 K, so 120,000.0 W. The cold stream can never be heated past the hot inlet temperature, and the hot stream can never be cooled past the cold inlet temperature, and that is what puts the ceiling on the heat rate.
Questions
What is the effectiveness of a heat exchanger?
The heat rate it actually moves divided by the largest heat rate it could possibly move. The California State University notes give the maximum as the smaller capacity rate times the difference between the inlet temperatures.
When should I use this method instead of the log mean temperature difference?
When you know the size and the conductance of the exchanger but not the outlet temperatures, which is the normal design situation. The Purdue notes say the two methods are equivalent, and the LMTD is easier when the outlet conditions are given.
What is a transfer unit?
The conductance times the area divided by the smaller capacity rate, and it is a measure of how much heat transfer surface the exchanger has compared with the amount of heat the flow can carry.
Why is counter current better than parallel flow?
Because in counter current the two streams enter at opposite ends, so the temperature difference along the whole length stays large, while in parallel flow both streams start at the same end with the largest difference and it collapses quickly.
What if the two streams have the same capacity rate?
The counter current expression has a zero in the denominator, and the limit is the effectiveness NTU over one plus NTU. That case is worth knowing because it is where the two flow arrangements are closest.
What about a stream that boils?
Its capacity rate is effectively infinite because its temperature does not change as it gives up heat, which pushes the capacity rate ratio towards zero and the effectiveness towards one minus the exponential of minus NTU.
Sources
- California State University mechanical engineering notes: heat exchangers, the effectiveness and the NTU method
- Purdue University ME 418 lecture notes: heat exchanger analysis and design, LMTD and effectiveness methods
- Washington University thermal laboratory: heat exchangers and the effectiveness method