# Thermodynamics - heat exchanger

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:

Given:

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 $$Tco$$

# 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:

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

where:

• 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 $$q_{max}$$

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

where:

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

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