EXAMPLE $16-8$ Using the Effectiveness$-$NTU Method

Repeat Example $16-4,$ which was solved with the LMTD method, using the

effectiveness$-$NTU method.

Solution The schematic of the heat exchanger is redrawn in Fig. $16-29,$

and the same assumptions are utilized.

Analysis In the effectiveness$-$NTU method, we first determine the heat

capacity rates of the hot and cold fluids and identify the smaller one:

$C_h=m_h^{}{c}_{ph}=(2k\frac{g}{s})(4\mathrm{.}31\frac{kJ}{k}g*C)=8\mathrm{.}62k\frac{W}{}C$

$C_c=m_c^{}{c}_{pc}=(1\mathrm{.}2k\frac{g}{s})(4\mathrm{.}18\frac{kJ}{k}g*C)=5\mathrm{.}02k\frac{W}{}C$

Therefore,

$C_{\text{m}\text{i}\text{n}}=C_c=5\mathrm{.}02k\frac{W}{}C$

and

$c=\frac{C_{\text{m}\text{i}\text{n}}}{C_{\text{m}\text{a}\text{x}}}=\frac{5\mathrm{.}02}{8\mathrm{.}62}=0\mathrm{.}582$

Then the maximum heat transfer rate is determined from Eq$.16-32$ to be

$Q_{\text{m}\text{a}\text{x}}^{}=C_{\text{m}\text{i}\text{n}}(T_{hin}-T_{cin})$

$=(5\mathrm{.}02\frac{kW}{}C)(160-20)C$

$=702\mathrm{.}8kW$

That is$,$ the maximum possible heat transfer rate in this heat exchanger is

$702\mathrm{.}8$ kW $.$ The actual rate of heat transfer is

$Q^{}=[(m^{})c_p(T_{\text{o}\text{u}\text{t}}-T_{in}){]}_{\text{w}\text{a}\text{t}\text{e}\text{r}}=(1\mathrm{.}2k\frac{g}{s})(4\mathrm{.}18\frac{kJ}{k}g*C)(80-20)C=301\mathrm{.}0kW$

Thus, the effectiveness of the heat exchanger is

$=\frac{(Q^{})}{Q_{\text{m}\text{a}\text{x}}^{}}=\frac{301\mathrm{.}0(kW)}{702\mathrm{.}8(kW)}=0\mathrm{.}428$

Knowing the effectiveness, the NTU of this counter$-$flow heat exchanger can

be determined from Fig. $16-26b$ or the appropriate relation from Table

$16-5.$ We choose the latter approach for greater accuracy:

NTU$=\frac{1}{c-1}ln(\frac{-1}{c-1})=\frac{1}{0\mathrm{.}582-1}ln(\frac{0\mathrm{.}428-1}{0\mathrm{.}4280\mathrm{.}582-1})=0\mathrm{.}651$

Then the heat transfer surface area becomes

NTU$=\frac{U{A}_s}{C_{\text{m}\text{i}\text{n}}}longrightarrow{A}_s=\frac{NTU{C}_{\text{m}\text{i}\text{n}}}{U}=\frac{(0\mathrm{.}651)(5020\frac{W}{}C)}{640\frac{W}{m^2}*C}=5\mathrm{.}11m^2$

To provide this much heat transfer surface area, the length of the tube

must be

$A_s=DL,$longrightarrow,$L=\frac{A_s}{D}=\frac{5\mathrm{.}11m^2}{(0\mathrm{.}015(m))}=108m$

Discussion Note that we obtained practically the same result with the

effectiveness$-$NTU method in a systematic and straightforward manner. $($ I want the solution by chart not equations $)$