# NTU method in Heat exchanger: Formula, Example, Effectiveness [with Pdf]

Contents

## What is NTU method?

NTU (Number of transfer units) gives the heat transfer capacity of the heat exchanger. In the NTU method, the effectiveness of the heat exchanger is expressed in the form of three non-dimensional terms.

The NTU method is useful when minimum data is provided for calculating heat transfer.

## NTU method formula:

The method has different effectiveness formulae for parallel flow and counterflow heat exchangers.

1) For parallel-flow heat exchanger:

The effectiveness of The parallel-flow heat exchanger is given by,

\varepsilon _{PARALLEL}=\frac{1-exp[-NTU(1+R)]}{1+R}

Where,

R (specific heat ratio) = \frac{C_{min}}{C_{max}}

Where Cmin is the minimum between Cc and Ch while the Cmax is the maximum between the same.

Where,
C_{c}=\dot{m}C_{pc}
C_{h}=\dot{m}C_{ph}

And, NTU = \frac{UA}{C_{min}}

Where U = Overall heat transfer coefficient
A = Area of the heat exchanger.

2) For counter-flow heat exchanger:

The effectiveness of the counterflow heat exchanger is given by,

\varepsilon _{counter}=\frac{1-exp[-NTU(1+R)]}{1-R[-NTU(1-R)]}

Where, R = \frac{C_{min}}{C_{max}}

and, NTU = \frac{C_{UA}}{C_{min}}

## NTU special cases:

Case 1] For condenser and evaporators:

For evaporating and condensing liquid the value of specific heat is considered as infinity.

Therefore, Cmax= ∞

∴ R = \frac{C_{min}}{C_{max}}=\frac{C_{min}}{\infty} =0

For R=0, the effectiveness is given by,

\varepsilon = 1-exp(-NTU)

Case 2] For regenators:

For regenerators the specific heat of both fluid is same

\therefore R=\frac{C_{min}}{C_{max}}=1

∴ a] For parallel flow heat exchanger:-

\varepsilon _{PARALLEL}=\frac{1-exp(-2NTU)}{2}

b] For counter flow heat exchanger:-

\varepsilon _{counter}=\frac{NTU}{1+NTU}

## Effectiveness:

The effectiveness of the heat exchanger is in the ratio of actual heat transfer to the maximum heat transfer.

\therefore \varepsilon =\frac{Q_{actual}}{Q_{max}}

## Numerical on NTU method:

Hot process fluid with cp = 2.6 Kj/Kg.°C enters in a parallel-flow heat exchanger at 100 °C at a mass flow rate of 25000 Kg/hr while the cooling water with cp = 4.2 Kj/Kg.°C enters into the heat exchanger at 10°C at a mass flow rate of 45000 Kg/hr. The heat exchanger has a heat transfer area of 11 m² with an overall heat transfer coefficient equal to 1000 W/m².°C.
Find,

1] Hot fluid outlet temperature
2]Cold fluid outlet temperature

Solution:-

Given:-

Cph = 2.6 KJ/Kg °C
Thi = 100 °C
\dot{m}_{h} = 25000 Kg/hr
Cpc = 4.2 KJ/Kg °C
Tci = 10 °C
\dot{m}_{c} = 45000 Kg/hr
A = 11 m²
U = 1000 W/m² °C

Step-1) Find Cmin, Cmax:-

Ch = \dot{m}_{h} Cph = 25000 x 2.5 = 62500 KJ/hr °C

Cc = \dot{m}_{c} Cpc = 45000 x 4.2 = 189000 KJ/hr °C

∴ Cmax = Cc = 189000 KJ/hr °C and
Cmin = Ch = 62500 KJ/hr °C

Step-2) Find R & NTU:-

R = \frac{C_{min}}{C_{max}} =\frac{62500}{189000}= 0.3306

NTU = \frac{UA}{C_{min}}=\frac{1000\times 11}{62500}=0.176

NTU = 0.176

Step-3) Find effectiveness (ε):-

For parallelflow heat exchanger,

\epsilon_{PARALLEL}=\frac{1-exp[-NTU(1+R)]}{1+R}

\epsilon_{PARALLEL}=\frac{1-exp[-0.176(1+0.3306)]}{1+0.3306}

\epsilon_{PARALLEL}= 0.1569

Step- 5) Exit temperature of hot fluid:-

By formula of effectiveness,

\epsilon=\frac{C_{h}(T_{hi}-T_{ho})}{C_{min}(T_{hi}-T_{ci})}

0.1569=\frac{62500(100-T_{ho})}{62500(100-10)}

T_{ho}= 85.879 °C —Answer 1]

This is the exit temperature of the hot fluid.

Step-6) Exit temperature of cold fluid:-

By using the formula of effectiveness,

\epsilon=\frac{C_{c}(T_{co}-T_{ci})}{C_{min}(T_{hi}-T_{ci})}

0.1569=\frac{189000(T_{co}-10)}{62500(100-10)}

T_{co}= 14.66 °C —Answer 2]

This is the exit temperature of the cold fluid.

Pratik is a Graduated Mechanical engineer. He enjoys sharing the engineering knowledge learned by him with people.