52 - Heat Exchangers - Effectiveness-NTU Method

Example 1: Effectiveness NTU Method 

Water (\(C_P=4180\) J/kg.^oC) is to be heated by solar-heated hot air (\(C_P=1010\) J/kg.^oC) in a double-pipe counter flow heat exchanger. Air enters the heat exchanger at 90^oC at a rate of 0.3 kg/s while water enters at 22^oC at a rate of 0.1 kg/s. The overall heat transfer coefficient based on the inner side of the tube is given to be 80 W/m\(^2\).^oC. The length of the tube is 12 m and the inner diameter of the tube is 1.2 cm. Determine the outlet temperature of the water and the air. (AU-May-2017)

Solution: 

Data:

Cold fluid: water, \(C_{P\text{c}}=4180\) J/kg.,^oC \(\dot{m}_{\text{c}}=0.1\) kg/s

Hot fluid: air, \(C_{P\text{h}}=1010\) J/kg.,^oC \(\dot{m}_{\text{h}}=0.3\) kg/s

\(A=\pi D L=3.142\times0.012\times12=0.4524\) m\(^2\)

\(U=80\) W/m\(^2\).^oC.

Rate of heat transfer is given by <p><span class="math display">\[\begin{align*} \text{For the cold fluid (water):} \\ Q &= \dot{m}_{\text{c}}C_{P\text{c}}(T_{\text{c,out}}-T_{\text{c,in}}) = 0.1\times4180\times(T_{\text{c,out}}-22) \tag*{(1)} \\ \text{For the hot fluid (air):} Q &= \dot{m}_{\text{h}}C_{P\text{h}}(T_{\text{h,in}}-T_{\text{h,out}}) = 0.3\times1010\times(90-T_{\text{h,out}}) \tag*{(2)} \\ \text{For the <a class="autolink" title="Heat Exchanger" href="https://msubbu.academy/mod/page/view.php?id=4527">heat exchanger</a>:} \\ Q &= UA\Delta T_{\text{lm}} = 80\times0.4524\times\left[\frac{(90-T_{\text{c,out}})-(T_{\text{h,out}}-22)}{\ln\left(\frac{90-T_{\text{c,out}}}{T_{\text{h,out}}-22} \right)} \right] \tag*{(3)} \end{align*}\]</span></p>

The above 3 equations contain the unknowns \(Q,T_{\text{h,out}}\) and \(T_{\text{c,out}}\). Solving them involves a trial and error calculation.

The steps are:

1. Assume values for \(T_{\text{h,out}}\) or \(T_{\text{c,out}}\). Obtain the other (\(T_{\text{c,out}}\) or \(T_{\text{h,out}}\)) from heat balance using Eqns.(1) or (2).

2. Obtain \(\Delta T_{\text{lm}}\), and calculate \(Q\) from Eqn.(3).

3. Using the value of \(Q\), calculate \(T_{\text{c,out}}\) and \(T_{\text{h,out}}\) from Eqns.(1) and (2).

4. Compare the outlet temperatures determined in the step-3, with the values assumed in step-1.

5. If the calculated values of \(T\) are different from the assumed values, then repeat the calculations, until a specified convergence is achieved.

Clearly, such a computation is very tedious. The calculation may be simplified by using \(\varepsilon\)-NTU method.

Calculation steps as per \(\varepsilon\)-NTU method:

Step-1: (Calculation of \(C\) and \(N\)) \[\begin{aligned} C_{\text{air}} &= \dot{m}_{\text{h}}C_{P\text{h}} = 0.3\times1010=303 \text{ W/$^{\circ}$C} = C_{\text{min}} \\ C_{\text{water}} &= \dot{m}_{\text{c}}C_{P\text{c}} = 0.1\times4180=418 \text{ W/$^{\circ}$C} = C_{\text{max}} \end{aligned}\] \[C = \frac{C_{\text{min}}}{C_{\text{max}}} = \frac{303}{418} = 0.725\] and, \[N = \text{NTU} = \frac{UA}{C_{\text{min}}} = \frac{80\times0.4524}{303} = 0.12\]

Step-2: (Calculation of \(\varepsilon\))

For the countercurrent heat exchanger, \[\begin{aligned} \varepsilon &= \frac{1-\exp\left[-N(1-C) \right]}{1-C\exp\left[-N(1-C)\right]} \\ &= \frac{1-\exp[-0.12\times(1-0.725)]}{1-0.725\times\exp[-0.12\times(1-0.725)]} = 0.109 \end{aligned}\] Step-3: (Calculation of \(Q\)) \[\begin{aligned} Q &= \varepsilon C_{\text{min}}(T_{\text{h,in}}-T_{\text{c,in}}) \\ &= 0.109\times0.303\times(90-22) = 2246 \text{ W} \end{aligned}\]

Step-4: (Calculation of outlet temperatures)

From Eqns.(1) and (2), we get \[\begin{aligned} T_{\text{c,out}} &= T_{\text{c,in}} + \frac{Q}{\dot{m}_{\text{c}}C_{P\text{c}}} = 22 + \frac{2246}{0.1\times4180} = {27.4}^{\circ}C \\ T_{\text{h,out}} &= T_{\text{h,in}} - \frac{Q}{\dot{m}_{\text{h}}C_{P\text{h}}} = 90 - \frac{2246}{0.3\times1010} = {82.6}^{\circ}C \end{aligned}\]