Effectiveness-NTU Calculator — Heat Exchanger Effectiveness, NTU, q and Outlet Temperatures

What is the effectiveness-NTU method?

With this effectiveness-NTU calculator you can carry out design and performance calculations of heat exchangers. With the design calculation you can determine the heat transfer area (A) for a given set of conditions. With the performance calculation you can find the outlet temperatures (T_co and T_ho) of the fluids and the actual heat transfer rate (q).

Similar to the LMTD (log mean temperature difference), the effectiveness-NTU method is a method used to analyse heat exchangers. This one is preferred when the outlet temperatures of the fluids are unknown, since in those cases the LMTD method requires a cumbersome iterative solution. You can read more about that approach with our LMTD calculator.

The term effectiveness (ε) is a dimensionless indicator that relates the actual heat transfer rate (q) to the maximum possible heat transfer rate (q_max) that could occur for a particular heat exchanger and a particular set of fluids. The effectiveness is given by the ratio of these heats:

ε = q / q_max

The actual heat rate q is determined from the fluid properties as:

q = C_c · (T_co − T_ci) = C_h · (T_hi − T_ho)

where:

• T_ci and T_co — inlet and outlet temperatures of the cold fluid;
• C_c — heat capacity rate of the cold fluid (C = ṁ · c_p);
• T_hi and T_ho — inlet and outlet temperatures of the hot fluid; and
• C_h — heat capacity rate of the hot fluid.

To calculate the maximum possible heat transfer rate (q_max), we use the fluid with the minimum heat capacity rate (C_min) and the largest possible temperature difference in the exchanger — the difference between the two inlet temperatures:

q_max = C_min · (T_hi − T_ci)

where C_min is the smaller of C_h and C_c. The fluid with the minimum heat capacity rate is the one that would experience the maximum temperature change.

How to use the effectiveness-NTU calculator

1. Choose the unit system: metric (SI) or American (Imperial).
2. Pick the heat exchanger type (counter flow, parallel flow, shell-and-tube, or one of the cross-flow arrangements).
3. Enter the heat capacity rates of the hot and cold fluids, C_h and C_c. Each one equals the mass flow rate times the specific heat, C = ṁ · c_p.
4. Enter the two inlet temperatures, T_hi and T_ci.
5. Enter the NTU (number of transfer units). If you don't know the NTU yet, open the advanced option and enter the overall conductance UA instead — the calculator will compute NTU = UA / C_min.

The calculator returns the effectiveness ε, the actual heat transfer rate q (in W, kW and BTU/h), the maximum heat transfer rate q_max, and the two outlet temperatures T_ho and T_co.

How do I determine the effectiveness of a heat exchanger?

The effectiveness depends on two dimensionless groups and the flow arrangement:

• The number of transfer units, NTU = U·A / C_min, which describes the "size" of the exchanger.
• The capacity ratio, C_r = C_min / C_max, which compares the two fluid streams (0 ≤ C_r ≤ 1).

Once you know ε, the heat transfer rate follows directly from q = ε · q_max, and the outlet temperatures from:

T_ho = T_hi − q / C_h   |   T_co = T_ci + q / C_c

How do I calculate the NTU of a heat exchanger?

The number of transfer units is defined as:

NTU = U · A / C_min

where U is the overall heat transfer coefficient, A is the heat transfer area, and C_min is the smaller heat capacity rate. For a design problem you usually solve the effectiveness relations for NTU and then find the required area as A = NTU · C_min / U.

NTU formulas for different types of heat exchangers

For a counter-flow exchanger the NTU can be obtained explicitly from the effectiveness (with C_r < 1):

NTU = 1/(C_r − 1) · ln[(ε − 1) / (ε·C_r − 1)]

For a parallel-flow exchanger:

NTU = −ln[1 − ε·(1 + C_r)] / (1 + C_r)

When C_r = 0 (a boiler or condenser, where one fluid changes phase) all arrangements share NTU = −ln(1 − ε).

Effectiveness formulas for different types of heat exchangers

The calculator implements the standard ε-NTU relations:

• Counter flow (C_r < 1): ε = [1 − exp(−NTU(1 − C_r))] / [1 − C_r·exp(−NTU(1 − C_r))]; and ε = NTU/(1 + NTU) when C_r = 1.
• Parallel flow: ε = [1 − exp(−NTU(1 + C_r))] / (1 + C_r).
• Shell-and-tube (one shell pass): ε = 2 / {(1 + C_r) + √(1 + C_r²) · [(1 + e) / (1 − e)]}, with e = exp(−NTU·√(1 + C_r²)).
• Cross flow, both fluids unmixed (approximation): ε = 1 − exp{(1/C_r)·NTU^0.22·[exp(−C_r·NTU^0.78) − 1]}.
• Cross flow, C_max mixed, C_min unmixed: ε = (1/C_r)·[1 − exp(−C_r·(1 − exp(−NTU)))].
• Cross flow, C_min mixed, C_max unmixed: ε = 1 − exp{−(1/C_r)·[1 − exp(−C_r·NTU)]}.
• All exchangers with C_r = 0: ε = 1 − exp(−NTU).

Advantages over the LMTD method

The main advantage of the effectiveness-NTU method is that it does not require the outlet temperatures to be known in advance. The LMTD method needs all four terminal temperatures, so when only the inlet temperatures are given it forces an iterative trial-and-error solution. The ε-NTU method gives the answer directly, which makes it ideal for rating (performance) problems.

Example: Using the effectiveness-NTU calculator

Consider a counter-flow heat exchanger with C_h = 1000 W/°C, C_c = 1250 W/°C, hot inlet T_hi = 80 °C, cold inlet T_ci = 20 °C and NTU = 1.5.

• C_min = 1000 W/°C, C_max = 1250 W/°C, so C_r = 0.8.
• ε = [1 − exp(−1.5·0.2)] / [1 − 0.8·exp(−1.5·0.2)] ≈ 0.636.
• q_max = 1000 × (80 − 20) = 60 000 W; q = 0.636 × 60 000 ≈ 38.2 kW.
• T_ho = 80 − 38178/1000 ≈ 41.8 °C; T_co = 20 + 38178/1250 ≈ 50.5 °C.

Switch the calculator to the American (Imperial) system to work in °F, BTU/(h·°F) and BTU/h.

FAQs

What is the difference between effectiveness and efficiency?

Effectiveness compares the actual heat transfer with the maximum thermodynamically possible heat transfer for the given inlet temperatures, q_max = C_min·(T_hi − T_ci). It is not the same as a thermal efficiency; ε can range from 0 to 1 and reaches 1 only for an infinitely large exchanger.

Which fluid sets C_min?

C_min is simply the smaller of the two heat capacity rates C_h = ṁ_h·c_p,h and C_c = ṁ_c·c_p,c. The fluid with the minimum capacity rate undergoes the larger temperature change, and it is the one used to define q_max and NTU.

Why does counter flow give the highest effectiveness?

For the same NTU and capacity ratio, a counter-flow arrangement keeps a more uniform temperature difference along the exchanger, so it extracts more heat than parallel flow or cross flow. That is why counter flow is the benchmark configuration.

What happens when C_r = 0?

C_r = 0 corresponds to a fluid changing phase (an evaporator or condenser), where its temperature stays constant and its effective heat capacity rate is enormous. In that limit every flow arrangement collapses to the same relation, ε = 1 − exp(−NTU).

Does the result depend on the unit system?

The effectiveness ε and the capacity ratio C_r are dimensionless, so they are the same in any consistent unit system. Only the heat transfer rate and outlet temperatures carry units — the calculator reports q in W, kW and BTU/h so you can use whichever you need.