Question 13.2: A heat exchanger is required, which will be able to handle 0......

Once again, applying equation (6-10), we see that a term-by-term evaluation gives

{\frac{\delta Q}{d t}}-{\frac{\delta W_{\mathrm{s}}}{d t}}=\iint_{\mathrm{c.s.}}\left(e+{\frac{P}{\rho}}\right)\rho(\mathbf{v}\cdot\mathbf{n})\,d A+{\frac{\partial}{\partial t}}\iiint_{\mathrm{c.v.}}e\rho\,d V+{\frac{\delta W_{\mu}}{d t}}    (6-10)

{\frac{\delta Q}{d t}}=0\qquad{\frac{\delta W_{s}}{d t}}=0\qquad{\frac{\delta W_{\mu}}{d t}}=0

\iint_{ c.s.}\!\rho \left(e+\frac{P}{\rho}\right)\!\left({\bf v}\cdot{\bf n}\right)d A=\rho A~v_{\mathrm{avg}}\left(\frac{v_{2}^{2}}{2}+g y_{2}+\frac{P_{2}}{\rho}+u_{2}-\frac{v_{1}^{2}}{2}-g y_{1}-\frac{P_{1}}{\rho}-u_{1}\right)

{\frac{\partial}{\partial t}}\iiint_{\mathrm{c.v.}}\rho e\,d V=0

and the applicable equation for the present problem is

0={\frac{P_{2}-P_{1}}{\rho}}+g h_{L}

The quantity desired, the diameter, is included in the head-loss term but cannot be solved for directly, as the friction factor also depends on D. Inserting numerical values into the above equation and solving, we obtain

0=-\frac{103,000\,\mathrm{Pa}}{1000\,\mathrm{kg/m^{3}}}+2f_{f}\Big(\frac{0.0567}{\pi D^{2}/4}\Big)^{2}\frac{\mathrm{m^{2}}}{\mathrm{s}^{2}}\cdot\frac{122}{D}\frac{m}{\mathrm{m}}\frac{g}{g}

or

0=-103+1.27{\frac{f_{f}}{D^{5}}}

The solution to this problem must now be obtained by trial and error. A possible procedure is the following:

1. Assume a value for f_{f}.

2. Using this f_{f}, solve the above equation for D.

3. Calculate Re with this D.

4. Using e/D and the calculated Re, check the assumed value of f_{f}.

5. Repeat this procedure until the assumed and calculated friction-factor values agree.

Carrying out these steps for the present problem, the required pipe diameter is 0.132 m (5.2 in).