A thin walled double pipe counter flow heat exchanger is to be used to cool oil (Cp = 2198 J/kg°K) from 150°C to 40°C at a rate of 2.27kg/s by water (Cp = 4187 J/kg°K) that enters at 21°C at a rate of 1.36 kg/s. The diameter of the tube is 0.127m and its length is 60m. Determine the overall heat transfer coefficient of this heat exchanger using (i) the LMTD method and (ii) the ε-NTU method

My attempt:


Mass flow rate of oil: $\dot{m}_{o} = 02.27 kg/s$

Mass flow rate of water: $\dot{w}_{o} 1.36 kg/s$

Specific heat of oil: $ C_{p,o} = 2798 J/Kg/.k $

Specific heat of water: $C_{p,w} =4187 J/kg.k $

$$C_o = m_oC_{p,o} = 2.27 \cdot 2198 = 4989.46J/s K$$ $$C_w=m_wC_{p,w} = 1.36 \cdot 4187 = 5694.32 J/s K$$ $$C_o = C{min} = 4989.46 J/Ks$$ $$C_i = C_{max} = 5694.32 J/Ks$$

From heat balance: Heat released by oil = heat gain by water

$$Co (Ti - To) = Cw (Tco - Tci)$$

$$Ti = 150 Degrees$$ $$To = 40 Degrees$$ $$Tci = 21 Degrees$$

$$ 4989.46 (150-40) = 5694.32 (Tco - 21)$$ $$Tco - 21 = ?$$

I am stuck on calculating $$Tco$$ How do i calculate $$Tco$$

Can anyone help please??


1 Answer 1


calculation of exit temperature

The heat transfer rate between $\dot{Q}$ oil and water should be equal to the change in heat capacity in the medium.

$$\dot{Q}= m_oC_{p,o}\delta T_{o} = - m_wC_{p,w}\delta T_{w}$$

you can solve this algebraically and obtain:

$$ 2.27 \cdot 2198 \cdot(150−40)= 1.36\cdot 4187 (T_{wo}−21)$$

$$ T_{wo} =\frac{2.27 \cdot 2198 \cdot(150−40)}{ 1.36\cdot 4187} +21 [^oC]$$

if my numerical calculations are correct this should be:

$$T_{wo} = 117[^oC]$$

LMTD method

For a counter flow heat exchanger the heat transfer rate $\dot{Q}$ is equal to:

$$\dot{Q} = kA\cdot\Delta T_{lm}$$


  • The $\Delta T_{lm}$ is the temperature difference for counterflow, which is given from the following equation.

$$\Delta T_{lm} = \frac{\Delta T_1-\Delta T_2}{\ln (\Delta T_1/\Delta T_2)}$$

For this particular example

  • $\Delta T_1 = T_{o,i}-T_{w,o} = 150-T_{w,o} = 33$ : temperature difference at one exit (at the center of the drawing)
  • $\Delta T_2 = T_{o,o}-T_{w,i} = 40 -21=19[C] $ : temperature difference at other Exit (bottom of the drawing).

Because $\dot{Q}= m_oC_{p,o}\delta T_{o}$, it is possible to calculate k for LMDT method as:

$$ k_{ry}= \frac{\dot{Q}\ln (\Delta T_1/\Delta T_2)}{A(\Delta T_1-\Delta T_2)} $$

$$ k_{ry}= \frac{\dot{Q}\ln (33/19)}{A(33-19)} =21642 \frac{1}{A} [\frac{W}{m^2*K}] = \approx 904[\frac{W}{m^2 K}]$$

A is the exchange area and its $A = 60\cdot 2\cdot \pi\cdot r_{tube}$.

NTU method

Maximum possible heat rate $q_{max}$

$$q_{max} = C_{min} (T_{o,i}- T_{w_i}$$


  • $ C_{min} = m_oC_{p,o} = 4989.46$
  • $T_{o,i}= 150 [^oC]$
  • $ T_{w_i}= 21 [^oC]$

The effectiveness is defined as: $$\epsilon = \frac{\text{Actual heat transfer}}{\text{max heat transferred}}$$

$$\epsilon = \frac{ m_oC_{p,o}\delta T_{o}}{4989.46(150-21)}$$

However $\epsilon$ for a parallel flow can be given from equation:

$$\epsilon = \frac{1- e^{- NTU \cdot (1-c)}}{1- c\cdot e^{-NTU \cdot (1-c)}}$$

where $c = \frac{C_{min}}{C_{max}} = 4989.46/5694.32=0.8762$

Substituting should get you $$\epsilon \approx 0.8527$$

solving for $NTU$ should get you:

$$NTU = \frac{\ln(\frac{\epsilon - 1}{c*\epsilon - 1})}{c - 1} = 4.3654$$

From there, because :

$$ NTU=\frac{UA}{C_{min}}$$ you can solve for U: $$U = =\frac{NTU \cdot C_{min}}{A}\approx 909[\frac{W}{m^2 K}]$$

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.