60% sulphuric acid is to be pumped at the rate of 4000 cm³/s through a lead pipe 25 mm diameter and raised to a height of 25 m. The pipe is 30 m long and includes two right-angled bends. Calculate the theoretical power required. The density of the acid is 1531 kg/m³ and its kinematic viscosity is

Question

60% sulphuric acid is to be pumped at the rate of 4000 cm³/s through a lead pipe 25 mm diameter and raised to a height of 25 m. The pipe is 30 m long and includes two right-angled bends. Calculate the theoretical power required. The density of the acid is 1531 kg/m³ and its kinematic viscosity is 4.25 × [latex]10^{-5}[/latex] m²/s. The density of water may be taken as 1000 kg/m³.

Accepted Answer

Cross-sectional area of pipe = (π/4)(0.025)² = 0.00049 m²
Velocity, u = (4000 × [latex]10^{-6}[/latex]/0.00049) = 8.15 m/s.
Re = ρud/μ = ud/(μ/ρ) = (8.15 × 0.025)/(4.25 × [latex]10^{-5}[/latex]) = 4794
If e is taken as 0.05 mm from Table 3.1, e/d = 0.002 and from Fig. 3.7, R/ρu² = 0.0047.

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

Head loss due to friction is given by:

[latex]h_{f}[/latex] = 4(R/ρu² )(l/d)(u²/g) (equation 3.20)

= (4 × 0.0047)(30/0.025)(8.15²/9.81) = 152.8 m and Δz = 25.0 m

From Table 3.2, 0.8 velocity heads (u²/2g) are lost through each 90° bend so that the loss through two bends is 1.6 velocity heads or (1.6 × 8.15²)/(2 × 9.81) = 5.4 m.

Table 3.2. Friction losses in pipe fittings

	Number of pipe diameters	Number of velocity heads (u² /2g)
45° elbows (a)*	15	0.3
90° elbows (standard radius) (b)	30-40	0.6-0.8
90° square elbows (c)	60	1.2
Entry from leg of T-piece (d)	60	1.2
Entry into leg of T-piece (d)	90	1.8
Unions and couplings (e)	Very small	Very small
Globe valves fully open	60-300	1.2-6
Gate valves: fully open	7	0.15
[latex]\frac{3}{4} [/latex] open	40	1
[latex]\frac{1}{2} [/latex] open	200	4
[latex]\frac{1}{4} [/latex]open	800	16

* See Figure 3.17.

Total head loss = (152.8 + 25 + 5.4) = 183.2 m.
Mass flowrate = (4000 × [latex]10^{-6}[/latex] × 1.531× 1000) = 6.12 kg/s.
From equation 8.61 the theoretical power requirement = (6.12 × 183.2 × 9.81) = 11,000 W or 11.0 kW.

Question 8.17: 60% sulphuric acid is to be pumped at the rate of 4000 cm³/s...

Related Answered Questions

In a single-stage compressor: Suction pressure = 101.3 kN/m². Suction temperature = 283 K. Final pressure = 380 kN/m². If each new charge is heated 18 deg K by contact with the clearance gases, calculate the maximum temperature attained in the cylinder. ...

Verified Answer:

Verified Answer:

Verified Answer:

Calculate the power required to pump oil of density 850 kg/m³ and viscosity 3 mN s/m² at 4000 cm³/s through a 50 mm pipeline 100 m long, the outlet of which is 15 m higher than the inlet. The efficiency of the pump is 50%. What effect does the nature of the surface of the pipe have on the ...

Verified Answer:

600 litres/s of water at 320 K is pumped in a 40 mm i.d. pipe through a length of 150 m in a horizontal direction and up through a vertical height of 10 m. In the pipe there is a control valve which may be taken as equivalent to 200 pipe diameters and other pipefittings equivalent to 60 pipe ...

Verified Answer:

A pump developing a pressure of 800 kN/m² is used to pump water through a 150 mm pipe 300 m long to a reservoir 60 m higher. With the valves fully open, the flowrate obtained is 0.05 m³/s. As a result of corrosion and scaling the effective absolute roughness of the pipe surface increases by a ...

Verified Answer:

Verified Answer:

Verified Answer:

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. ...

Verified Answer:

Verified Answer: