Further measurements on the MD-6K-N pump used by Tadahiko Mizuno

First issue: 10/01/2015
Publication: 10/01/2015
Translation: 12/01/2015

After reading recently a comment (posted in the mail-archive vortex-l by Dave Roberson) regarding our recent calorimetry measurement of the magnetic drive pump Iwaki MD6, followed by a post that tried to explain the large difference between the power transferred to water during our tests (4.3W) and that estimated by Jed Rothwell (0.25W), we decided to respond with a new post rather than just putting a simple answer in the comments, in the hope that shedding light on these measures can be helpful to anyone willing to address these issues.

Roberson in his comment asserted that the different obtained values were due to the pipe diameter difference (5mm in our test, 10mm in the Mizuno’s pipe).

In particular, he writes:

…Now our favorite skeptic claims that he is using .5 cm pipe instead of the 1 cm pipe used by Mizuno and does not realize that he is making a major error. But, the area of that pipe is reduced by a factor of 4 since it is exactly 1/2 the inner diameter of the original. With a factor of 4 reduction in area comes an increase in the velocity of the water flowing through it by that factor 4 in order to achieve the same mass flow rate. Every thing else being equal you find that the energy imparted upon the water that is sped up from rest to a velocity that is 4 times that from the first case yields the square of that factor. In which case it is 4^2 which is 16 times.

According to this comment, since the pipe used by GSVIT has a diameter half that used by Mizuno, its section is 4 times smaller, and the speed of water at the output 4 times higher; since the kinetic energy depends on the square of the velocity, the kinetic energy associated with water at the output (that is transformed into heat by friction) is 16 times higher, the exact relationship between the two measurements (4.3W against 0.25W).

The commentator also ended by thanking the Mathematics that in his view had solved the problem and congratulated Jed.

…I love it when the math holds up so well.
Congratulations Jed, you got it right.

Unfortunately we have to disenchant him. His reasoning is typical of a person who has scientific knowledge but, almost certainly due to a lack of experience in the field, has analyzed the problem only at it surface. We hope that these notes and the further test that we have added can convince him (and those who make use of his approach) that the reality may seem simple, but almost always the simplest systems, if not observed with the necessary depth, can deceive and lead to erroneous conclusions.

First of all let’s be clear that he would have to be aware of his error by the following simple calculation based on the very few data provided by the pump manufacturer and already published in our previous post:

  • maximum pump flow = 8 litres / min
  • maximum pump head = 1 meter H2O

Obviously the head decreases with increasing flow rate, and the maximum flow is measured at 0 head, as per original data sheet.

We considered, however, applying more conservative constrains, to have simultaneously the maximum flow rate and the maximum head.

The power transferred to the water under these conditions is [by standard theory]:

P = q * ρ * g * h


  • q è la flow capacity = [(8 x 10^-3) /60] m3/s
  • ρ è la density of water @20°C = 998 kg/m3
  • g è la standard acceleration gravity = 9.81 m/s2
  • h è la differential head = 1 m

so that P = [(8 x 10^-3) /60] x 998 x 9.81 x 1 = 1.3W

from this calculation it is clear, however, that this is not the right way to explain a 4W difference.

If we calculate the kinetic energy associated with the water jet exiting the discharge the pump [as done by Roberson], assuming as he does that the flow rate remains the same when passing from 5mm to 10mm [that is wrong], we can write:

  1. water speed @10mm = (8 x 10^-3/60)/[(5 x 10^-3)^2 x π] = 1.7 m/s
  2. water speed @ 5mm = (8 x 10^-3/60)/[(2.5 x 10^-3)^2 x π] = 6.8 m/s
  3. power in the first case = ½ x (8 x 10^-3/60) x 1000 x (1.7)^2 = 0.19W
  4. power in the second case = ½ x (8 x 10^-3/60) x 1000 x (6.8)^2 = 3.07W

Note that this power (ref. 4) is still significantly lower than the measured 4.3W, but it is certainly much higher than the actual value since we considered that the pump could simultaneously provide the maximum flow (8 liters / h) and the maximum head (1 m). The output speed calculated in the case of the 5mm pipe is in fact too high (6.8 m/s !) and corresponds (with a 100% nozzle efficiency) to 2.3 meters head, whereas the pump actually has a maximum head of 1 meter.

For verification, as shown in Figure I, we connected again 5mm pipe to the pump to measure the actual output speed of the water.

Figure I - Pump with a 5mm pipe output

Figure I – Pump with a 5mm pipe output

Figure II shows that the jet has a length of 28cm and the height difference between the nozzle and the surface of the water, where the jet is falling, is 10cm.

Figure II - Jet Pump 28cm and 10cm height difference

Figure II – Jet Pump 28cm and 10cm height difference

From Physics, using classical formulas for uniformly accelerated motion, we can write:

(a) for the vertical motion: s = ½ x g x (t^2 )

0.1 = ½ x 9.81 x t^2 which immediately provides: t = 0.143 seconds

(b) for the horizontal motion: s = v x t

0.28 = v x 0.143 from which it is immediately obtained: v = 1.96 m/s

The flow rate is: Q = π x ((2.5 x 10^-3)^2) x 1.96 = 3.85 x 10^-5 m3/s, and then only 2.31 l/min

The power associated with the kinetic energy is then: P = ½ x (2.31 x 10^-3/60) x 1000 x (1.96^2) = 0.074W

The resulting power is very small and is lower than that calculated for the 10 mm pipe because in that case as well a 8 l / min flow rate was considered, while the 40cm long 5mm pipe used in this measurement (the same we used in our original test) set the pump working point very close to the working point in the Mizuno system (which adopted a 10mm pipe, but 16 meters long), as it was our intention.

The power associated with the real flow, in the Mizuno’s system, was about:

P = ½ x (2.3 x 10^-3/60) x 1000 x (0.49)^2 = 0.0046W.

It is thus demonstrated that the difference in power measured by us (4.3W) and that indicated by Rothwell (0.25W) is not caused by the pipe diameter.

These simple calculations would be sufficient to set the matter, but here we want to take the opportunity to clarify what we wrote privately to Rothwell about the matter: actually the pipe diameter has no influence on power transferred from the pump to the water, being in reality this power equal to the flow rate times the load loss (head) in the working point of the pump. As explained in the description of the previous test, the pipe diameter has been calculated in order to have a load loss (approximately) equal to that of the Mizuno’s system. Had we adopted a pipe of the same diameter as that used by Mizuno, but shorter, the working point of the pump would have been moved to higher flow and lower head, with the result of increasing (albeit slightly) the power transferred to water. On the other hand, it was impractical to use a 10mm pipe 16 meters long, because the heat loss to the environment would have been excessive thus affecting the measurement.

Figure 1 shows typical curves of a centrifugal pump.

Figure 1 - Typical curves of a centrifugal pump

Figure 1 – Typical curves of a centrifugal pump

As one can see the power consumption (heat transferred to the fluid) grows with decreasing head and is maximum at zero head. Therefore, the decision to adopt that very diameter was taken to avoid measuring a power greater than the real one. A proof of this assertion was obtained when running again the calorimetry test on the pump, increasing the diameter to 10mm, as shown in the following.

Test using a 10mm diameter pipe

Since it is not a test of special importance (it has only demonstration purposes as a response to some comments ) no insulation was applied. This fact, combined with the pipe larger surface area, explains why the delta temperature versus time characteristics is inclined more than in the previous test when the water temperature exceeds the room temperature substantially. However, in this case as well, the measured curve can be approximated by a straight line if the water temperature is kept close to the ambient one (a few °C): So the evaluation was performed in the range 18.1 °C – 23.0 °C, with 20.0 °C in the environment.

Figure 2 shows the new set-up, virtually identical to the previous one, save the paper box that supports the pump.

Figure 2 - Set-up

Figure 2 – Set-up

Figure 3 shows the temperature versus time during the test.

Figure 3 - Plot temperature / time during the test

Figure 3 – Plot temperature / time during the test

Figure 4 shows the system at the beginning of the period considered in the dissipated power calculation.

Figure 4 - T1 temperature

Figure 4 – T1 temperature

In Figure 5 one can see the system at the end of the period taken into account in the calculation of the dissipated power.

Figure 5 - T2 temperature

Figure 5 – T2 temperature


The power P transferred from the pump is:
P = 4183 x 0.800 x (T2T1) /t


  • 4183 the water specific heat @ 20°C [J/kg°C]
  • 0.800 the water mass to be heated [kg]
  • T2 is the final temperature [°C]
  • T1 is the initial temperature [C°]
  • t is the time between the two measurements [s]

Considering the values ​​taken from the two photographs (Figure 4 and Figure 5) performed at 1.9 ° C below (T1) and 2.9 ° C above (T2) the ambient temperature and taken respectively at 8:43:31 and at 09:42:34, that is, 3543 seconds away, we have that the power transferred from the pump to the water is equal to 4.6W.

As can be seen the power is slightly higher than that measured in the first test.

In Figure 6 the motor temperature is visible (measured by a pyrometer): it is slightly higher than that of the first test;

Figure 6 – Motor temperature measured by the pyrometer

Figure 6 – Motor temperature measured by the pyrometer

this is normal, being the delivered power slightly greater; however, it is also possible that at least part of the difference is due to a slightly higher supply voltage or reading at a warmer point.

We therefore confirm our measurement of the heat transferred to the water from the pump.

As Jed Rothwell continues to maintain that our value is impossible, we invite him to run again (he himself or together with Mizuno) measurement according to our procedure, since, as we have shown, the test takes no more than 2-3 hours and to publish photos and diagrams as we did.

At this point we must ask ourselves how it is possible that the pump delivers more than 4W to the water, if the theoretical mechanical work exerted by the pump on the water, as we have seen, is 20 times smaller.

One explanation is that these small magnetic drive centrifugal pumps have a very low efficiency, rarely exceeding 0.3.

But this would explain a dissipated power in the order of 1 – 2W, but does not appear sufficient to explain 4.6W.

The explanation (that we already communicated in private to Rothwell) is that a not negligible amount of heat flows by conduction from the motor (which, as seen in Figure 6 is at 45 ° C) to the water through the wall of the magnetic coupling.

For clarity in Figure 7 and 8 one can see the disassembled pump.

Figure 7 - Disassembled pump and magnetic driver temperature magnetico

Figure 7 – Disassembled pump and magnetic driver temperature

Figure 8 - Disassembled pump

Figure 8 – Disassembled pump

As it is clearly seen the water surrounding the magnet is in direct contact with the thin plastic wall that, externally, is in contact with the magnetic driver, whose temperature, after disassembly, at the end of the test was still 42.5 °C, as shown in Fig 7.

These photos contradict what Rothwell states, that these pumps are constructed so as to prevent any flow of heat from the motor to the fluid.

Equipment used:
Iso-Tech IDM91E Tester
Hanna Hi935005 Thermometer
RS 1319A Thermometer
Fluke 80T-IR IR Meter
8A Variac
850cc Dewar vessel
Radio Controlled Clock
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