Calculate the pressure drop in, and the power required to operate, a condenser consisting of 400 tubes, 4.5 m long and 10 mm internal diameter. The coefficient of contraction at the entrance of the tubes is 0.6, and 0.04 m³/s of water is to be pumped through the condenser.

Question

Accepted Answer

Flow of water through each tube = (0.04/400) = 0.0001 m³/s.
Cross-sectional area of each tube = (π/4)(0.01)² = 0.0000785 m²

Water velocity =(0.0001/0.0000785) = 1.273 m/s.
Entry pressure drop is:

[latex]\Delta P_{f}=-\frac{ ho u^{2}}{2} (\frac{1}{C_{c}}-1 )^{2}[/latex] (from equation 3.78)

= (1000 × 1.273²/2)[(1/0.6) - 1² = 360 N/m²

Re = ρud/μ = (1000 × 1.273 × 0.01)/(1.0 ×[latex]10^{-3}[/latex]) = 1.273 × [latex]10^{4}[/latex]
If e is taken as 0.046 mm from Table 3.1, then e/d D 0.0046 and from Fig. 3.7:

R/ρu² = 0.0043.

Table 3.1. Values of absolute roughness e

	(ft)	(mm)
Drawn tubing	0.000005	0.0015
Commercial steel and wrought-iron	0.00015	0.046
Asphalted cast-iron	0.0004	0.12
Galvanised iron	0.0005	0.15
Cast-iron	0.00085	0.26
Wood stave	0.0006-0.003	0.18-0.9
Concrete	0.001-0.01	0.3-3.0
Riveted steel	0.003-0.03	0.9-9.0

The pressure drop due to friction is:

[latex]\Delta P_{f}[/latex] = 4(R/ρu²)(l/d)(ρu²) (equation 3.18)

= (4 × 0.0043(4.5/0.01)(1000 × 1.2732 )) = 12,540 N/m²
Total pressure drop across one tube = (12,540 + 360) = 12,900 N/m² or [latex]\underline{\underline{12.9 ext{ kN/m²}}}[/latex]

If tubes are connected in parallel, power required =(pressure drop)× (volumetric flowrate) = (12,900 ×0.04)= [latex]\underline{\underline{516 W}}[/latex]

Question 8.15: Calculate the pressure drop in, and the power required to op...

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