A problem from Tianjin University’s version of Fundamentals of Chemical Engineering (11–15), page 258 – solution required

The flow velocity is calculated as follows: (Pa-P2)/pg = 0.5 + 3.5u12/g

Sorry, there was a mistake; it should be p1 instead of p2 in the above formula. The corrected formula is: (Pa-P1)/pg=0.5+3.5u12/g

Oh, so you establish the Bernoulli equation in this way: at the water surface and at the vacuum gauge location at the inlet, Z1 + P0/pg = Z2 + (P0 – P1)pg + u2/2g + ∑hf. Is that means that the installation height of the centrifugal pump, 2.5 meters, is taken as Z2? Once the flow velocity is determined, the flow rate can be calculated; but how should the flow velocity at point P2 after leaving the pump outlet be determined? The diameter of the pipe at the pump outlet is not given in the question. So how do we find He?

If the loss due to dynamic head is not considered, how is then the head loss in the pump outlet pipeline calculated?

My calculation process is as follows: (Pa—P1)/pg = 2.5 + 3.5u12/g; from this, the inlet flow velocity u1 is determined to be 2.07 m/s. The flow rate Q is then calculated using the pipe diameter, resulting in Q = 0.0104 m3/s. He = (p2 – P1)/10 – 0.5 = (220 + 60 – 100)/10 – 0.5 = 25.5 m. The effective power is given by pgQHe/0.68 = 9.8*1000*0.0104*25.5 = 3899 W; therefore, the correct answer is C

I’m sorry, there was a mistake; it should be He=(p2-p1)/10+0.5=(220+60-100)/10+0.5=26.5 m. The final result is 4037 W

The standard answer is A. If that person could provide some guidance as well, I would be extremely grateful

I’m ashamed – I got the pressure values wrong; here’s the correction. My calculation process is as follows: (Pa–P1)/pg = 2.5 + 3.5u1²/g. With Pa = 0 (gauge pressure) and P1 = –60, the inlet flow velocity u1 is calculated to be 3.16 m/s. The flow rate Q is then determined using the pipe diameter, giving Q = 0.0159 m³/s. He = (p2–P1)/10 + 0.5 = (220 + 60)/10 + 0.5 = 28.5 m. The effective power is calculated as pgQHe/0.68 = 10*1000*0.0159*28.5/0.68 = 6658 W

: (Pa—P1)/pg = 2.5 + 3.5u12; Pa = 0 (gauge pressure in atmospheres), P1 = -60. Using these values, u1 is calculated to be 1.0 meter/second, and ∑hf = 3.5u12 – it’s not 3.5u12/g. Could there be an issue with the data given in the problem?

The data is correct; the unit is KPa.

∑hf=3.5u12, not 3.5u12/g; was g omitted in the question?

The head can be calculated using the vacuum level at the pump inlet p1, the outlet pressure p2, and the vertical distance h0 between the two pressure measurement points (it can be assumed here that the diameters of the inlet and outlet pipes are the same). H = h0 + (p1 + p2) / (ρg) = 0.5 + (60,000 + 220,000) / (1,000 × 9.81) = 28.54 m. The flow velocity can be determined using the pump’s installation height z, the flow resistance in the suction pipe expressed as ∑hf, and the pressure at the pump inlet, p1 (which is an absolute pressure of 101.3 – 60 = 41.3 KPa). The formula for calculating the installation height is: z = (p0 – p1)/ρg – (u1)²/2g – ∑hf/g = (p0 – p1)/ρg – 0.5(u1)²/g – 3.5(u1)²/g. Therefore, 4(u1)²/g = (p0 – p1)/ρg – z. The formula for calculating the square of the velocity can be written as: (u1)² = [(p0 – p1)/ρg – z]g/4 = (9.81/4) = 10.095. Hence, u1 = 3.177 m/s. The flow rate is given by: Q = (π/4) × (d)² × u1 = 0.785 × (0.08)² × 3.177 = 0.01596 m³/s. The power required by the pump is: Pa = QHρg/η = 0.01596 × 28.54 × 1000 × 9.81/0.68 = 6564 W = 6.6 KW

Isn’t the installation height specified as z=2.5m in the question? Why is z=2 used in the calculation of u1 squared? There is also H=0.5+. . . =29.04m

=m/s/s, = m*m/s/s, = m – this is consistent with the unit of flow resistance ∑hf, which is also m; it seems there’s an issue with the question

This question was asked by someone from 09; maybe that person has passed away by now, but I also couldn’t solve it

I agree with what was said above; I calculated it several times and got 6.27 as the result!

This post was last edited by zly541881 on 2015-7-1 10:26. Choose A. According to the formula: ΔP/ρg = Δu²/2g + Δz + hf/g (key step); where Δu = u1 – u0, and u0 = 0. Thus, (60×10³) / (10³×9.81) = u1²/(2×9.81) + 2.5 + 3.5. Solving this gives u1 = 2.98 m/s. The flow rate Q is given by Q = u1×A = 2.98×πd²/4 = 0.015 m³/s； The head HT = ΔP/ρg + h0 = (220 + 60)×1000/1000×9.81 + 3 = 31.5 m; the effective power Ne = HQρg = 31.5×0.015×1000×9.81 = 4.635 Kw ； Shaft power N=Ne/68%=6.8kw.

In the calculation upstairs, “Head HT = ΔP/ρg + h0 = (220 + 60) × 1000/1000 × 9.81 + 3 = 31.5 m”; the +3 should be +0.5. Thus, H = 29, Ne = 4.26 kW, and N = Ne/0.68 = 6.26 kW. Option B should be chosen

[Analysis] A (should be 6.27). Given the installation height, Hg = (P0 – P1) / pg – u1²/2g – 3.5u1²/g. Thus, 2.5 = 60*1000/1000/9.81 – 0.408u1², which gives u1 = 2.98 m/s. Then, Q = 0.785*0.08²*2.98 = 0.015 m³/s. Using the energy equation for pumps, p1 (gauge) + He = p2 (gauge) + 0.5; hence, -60000/1000/9.81 + He = 220000/1000/9.81 + 0.5, which yields He = 29.0 m. The power required is given by N = QHp/102h = 0.015*29*1000/102/0.68 = 6.27 KW. [Comment] Note that the resistance value given in the problem is hf, not the static head Hf; there is a difference of the gravitational acceleration between them! An easy question! Work hard right away!