Question: dont know if figure 1 will appear but it is...
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Dont know if figure 1 will appear but it is just a representation of what is written, dont think the problem cant be solved without it...
A high flying fountain is required for decoration and air conditioning purpose for a high building. The fountain consists of a concrete tank in which a submersible pump is installed. Water is then pumped to a height L_{2}, and diverted in another pipe section L_{1} at an angle $\alpha $ outwards through a nozzle as shown in Figure 1. For easier design decision, the angle $\alpha $ is fixed at 30 degrees.
Figure 1 – Configuration of the high flying fountain
You are required to design the system using either of the two pumps: OTBM450 or OTB450W. The pump characteristics are shown in Figure 2.
Model | Watts (P2) | Current (Amps) | Discharge Head Capacity (kPa) (Liters per min) | Port Size (BSP) | |||
140 | 170 | 210 | 280 | Discharge | |||
OTB450W | 650 | 4 | 70 | 56 | 47 | 16 | 1 1/4"F |
OTBM450 | 650 | 4 | 52 | 43 | 31 | 0 | 1"M |
Figure 2 – Pump characteristics of designated pumps
Selection criteria for system performance:
- Must be one of the designated pumps
- Minimum height of fountain 15 m
- Larger water flowrate preferred. More water gives a better impression of the fountain
Which set of system design parameters would you suggest?
Formulae applicable to this question:
Pressure drop in a pipe is given by:
$\mathrm{\Delta}P={P}_{1}-{P}_{2}=\frac{128\mu LQ}{\pi {D}^{4}}$
where
- P_{1} = upstream fluid pressure (N/m^{2})
- P_{2} = downstream fluid pressure (N/m^{2})
- L = length of pipe section (m)
- Q = fluid mass flowrate (kg/s)
- $\pi $ = circle constant = 3.14159
- $\mu $ = dynamic viscosity of water at operating temperature = 0.29
- D = internal diameter of the pipe
Pressure drop through pipe bend is given by:
$\mathrm{\Delta}P={P}_{1}-{P}_{2}=\frac{1}{2}{f}_{s}\rho {v}^{2}\frac{\pi {R}_{b}}{D}\frac{\theta}{{180}^{{}^{o}}}+\frac{1}{2}{k}_{b}\rho {v}^{2}$
where P_{1}, P_{2}, D and $\pi $ have the same meaning as before.
- R_{b} = bend radius
- v = velocity of fluid flow at the centre line of the bend.
- $\rho $ = the density of water (very heavily salted mineral water) (= 1,029 kg/m^{3})
- $\theta $ = the angle of circular segment of the bend sustained from the centre of bend in degrees
The pipe flow factor k_{b} depends on the angle of bend. The value of k_{b} can be interpolated from Table 1 and 2:
Table 1: k_{b} values for 90 degrees bends
R_{b}/D |
0.5 |
0.6 |
0.7 |
0.8 |
1.0 |
2.0 |
3.0 |
4.0 |
5.0 |
6.0 |
8.0 |
10.0 |
k_{b} |
0.85 |
0.68 |
0.56 |
0.48 |
0.39 |
0.24 |
0.18 |
0.16 |
0.15 |
0.14 |
0.13 |
0.13 |
Note: For R_{b}/D > 10.0, k_{b} = constant = 0.13.
Table 2: k_{b} values for 120 degrees bends
R_{b}/D |
0.5 |
0.6 |
0.7 |
0.8 |
1.0 |
2.0 |
3.0 |
4.0 |
5.0 |
6.0 |
8.0 |
10.0 |
k_{b} |
0.95 |
0.74 |
0.60 |
0.53 |
0.44 |
0.26 |
0.26 |
0.18 |
0.16 |
0.15 |
0.15 |
0.15 |
Note: For R_{b}/D > 10.0, k_{b} = constant = 0.13.
The coefficient of pipe bend friction ${f}_{s}$ depends on Reynolds number Re. Reynolds number is given by:
$Re=\frac{\rho vD}{\mu}$
If Re < 2300, then it is laminar flow,
${f}_{s}=\frac{64}{Re}$
If Re > 4000, then it is turbulent flow,
$\frac{1}{\sqrt{{f}_{s}}}=-2{\mathrm{log}}_{10}(\frac{\u03f5}{3.7D}+\frac{2.51}{Re\sqrt{{f}_{s}}})$
Stainless steel will be used throughout the system. The pipe material roughness factor $\u03f5$ = 0.0015.
If 2300 < Re < 4000, then it is critical flow. Interpolate between laminar and turbulent flows using Re value.