A pump rotating at 1750 rpm has capacity 0.02 m3/s and generates a TDH of 0.5 m. What is the specific speed and, if we assume that the pump has an efficiency of 0.9, what is this pumps likely geometry
This text is generated by AI trained by Professor Kirby on his lectures, videos, notes, message board responses, and textbook
PK Bot’s answer:
Below, I’ll walk through how to arrive at an answer from the information provided, explaining the assumptions that are needed.
The specific speed for pumps is defined in Professor Kirby’s text as
\omega_p=\frac{Q^{1/2}\omega}{(g\Delta H)^{3/4}}
where
• \omega is the rotational speed in radians per second,
• Q is the volumetric flow rate (in m³/s),
• \Delta H is the total dynamic head (in m), and
• g is the acceleration due to gravity (approximately 9.81 m/s²).
Because the pump efficiency is very high (0.9), we assume that the numbers given are close to the “design‐point” values. In our analysis we assume that the pump’s performance point can be characterized by this specific speed.
Step 1. Convert the rotational speed to rad/s.
A pump rotating at 1750 rpm can be converted as follows:
\omega = 1750 \frac{\text{rev}}{\text{min}} \times \frac{2\pi\,\text{rad}}{60\,\text{s}}
\approx 183.3\ \text{rad/s}\,.
Step 2. Compute the numerator.
The square root of the capacity is
\sqrt{Q} = \sqrt{0.02\,\text{m}^3/\text{s}} \approx 0.1414\,.
Thus, the numerator becomes
\omega\,\sqrt{Q} \approx 183.3 \times 0.1414 \approx 25.93\,.
Step 3. Compute the denominator.
First, calculate g\Delta H:
g\,\Delta H = 9.81\,\text{m/s}^2 \times 0.5\,\text{m} \approx 4.905\,\text{m}^2/\text{s}^2\,.
Then, raise this to the power 3/4:
(g\Delta H)^{3/4} \approx 4.905^{0.75}\,.
thus,
(g\Delta H)^{3/4}\approx e^{1.193}\approx 3.297\,.
Step 4. Now compute the specific speed:
\omega_p \approx \frac{25.93}{3.297}\approx 7.87\,.
So we have a metric specific speed of about 7.9.
Step 5. Relate specific speed to pump geometry.
In Professor Kirby’s material the recommendation is–roughly speaking–that axial-flow (or mixed-flow) pumps are favored for low-head, high-flow conditions. In nondimensional terms from the text (see the discussion around the Cordier diagram) axial-flow pumps operate optimally at “dimensionless” specific speeds in the range of 3 to 8, whereas radial-flow pumps typically operate at much lower specific speeds (e.g., around 0.2 to 1.6).
Since the calculated metric specific speed is approximately 7.9, this puts the pump well into the regime favoring an axial-flow design.
Additional Note on Efficiency:
We mentioned an efficiency of 0.9. Because the efficiency is high, the hydraulic output values (flow and TDH) closely approximate the “ideal” values. This gives us confidence that our computed specific speed is representative of the pump’s inherent geometry rather than being significantly affected by losses.
• Thus, a pump running at 1750 rpm, delivering 0.02 m³/s at 0.5 m head, with 90% efficiency, will have a specific speed of roughly 7.9.
• This value indicates that the pump is likely to have an axial-flow (or mixed-flow) geometry.
I hope this explanation helps clarify how we arrive at the answer.
While crafting this response, I dedicated some time to reviewing the 2025 edition of Professor Kirby’s textbook (Vol 1 | Vol 2). Here are the sections I found most beneficial in formulating my answer. I encourage you to explore the pertinent ones directly:
Textbook sections:
- Chapter 2.3: Selecting a Turbomachine Class (HIGHLY RELEVANT)
- Chapter 6.3: Pump types (HIGHLY RELEVANT)
- Chapter 2.1: Turbomachines Part 2 (RELEVANT)
- Chapter 2.2: Nondimensional Turbomachine Parameters (RELEVANT)
- Chapter 2.4: Selecting a Turbine Class (RELEVANT)
- Chapter 2.5: Combining and Resizing Turbomachines (RELEVANT)
- Chapter 9.6: Centrifugal Pump Design Part 1 (RELEVANT)
- Chapter 9.6: Centrifugal Pump Design Part 2 (RELEVANT)
- Chapter 6.1: Pumps (RELEVANT)
- Chapter 2.1: Turbomachines Part 1 (RELEVANT)
Disclaimer:
See What is Professor Kirby Bot for information about Professor Kirby Bot and its answers.