An existing heat exchanger has a cross section as shown in Figure 13.3 with nine 1-in-OD tubes inside a 5-in-ID pipe. For a 5-ft length of heat exchanger, what flow rate of water at 60 F can be achieved in the shell side of this unit for a pressure drop of 3 psi?
An energy-equation analysis using equation (6-10) will follow the same steps as in Example 2 in this chapter, yielding, as the governing equation,
dtδQ−dtδWs=∬c.s.(e+ρP)ρ(v⋅n)dA+∂t∂∭c.v.eρdV+dtδWμ (6-10)
0=ρP2−P1+ghL
The equivalent diameter for the shell is evaluated as follows:
Flow area=4π(25−9)=4πin⋅2
Wetted perimeter=π(5+9)=14πin.
thus
Deq=414π4π=1.142in.
Substituting the proper numerical values into the energy equation for this problem reduces it to
0=−1.94slugs/ft33lbf/in2(144in.2/ft2)+2ffvavg2ft2/s2(1.142/12)ft5ftgg
or
0=−223+105ffvavg2
As ff cannot be determined without a value of Re, which is a function of υavg, a simple trial-and-error procedure such as the following might be employed:
1. Assume a value for ff.
2. Calculate υavg from the above expression.
3. Evaluate Re from this value of υavg.
4. Check the assumed value of ff using equation (13-15a).
ff1=−3.6log10[Re6.9+(3.7De)10/9] (13-15a)
5. If the assumed and calculated values for ff do not agree, repeat this procedure until they do.
Employing this method, we find the velocity to be 23.6 fps, giving a flow rate for this problem of 2.06ft3/min(0.058m3/s).
Notice that in each of the last two examples in which a trial-and-error approach was used, the assumption of ff was made initially. This was not, of course, the only way to approach these problems; however, in both cases a value for ff could be assumed within a much closer range than either D or υavg.