Problem Statement
Find the number of pumps required to take water from a deep well under a total head of 156 m. Also the pumps are identical and are running at 1000 r.p.m. The specific speed of each pump is given as 20 while the rated capacity of each pump is 150 litre/s.
Given Data & Constants
- Total Head, \(H_{\text{total}} = 156 \, \text{m}\)
- Speed, \(N = 1000 \, \text{r.p.m.}\)
- Specific Speed, \(N_s = 20\)
- Discharge per pump, \(Q = 150 \, \text{L/s} = 0.15 \, \text{m}^3/\text{s}\)
Solution
1. Calculate the Head Generated by a Single Pump (\(H\))
The specific speed formula relates the speed, discharge, and head for a single pump. We need to rearrange it to solve for the head (\(H\)).
2. Calculate the Number of Pumps Required
To achieve the total required head, the pumps must be connected in series. The total head is the sum of the heads generated by each pump.
The number of pumps required is 3.
Explanation of Specific Speed and Pump Staging
Specific Speed (\(N_s\)): This is a dimensionless parameter used to classify pumps based on their performance and impeller type. A low specific speed (like 20 in this problem) indicates a pump designed to produce high head and low flow rate (a radial flow impeller). The formula \(N_s = \frac{N \sqrt{Q}}{H^{3/4}}\) allows us to predict one performance variable if the other two are known.
Pump Staging (Pumps in Series): When the head required by a system is greater than what a single pump can deliver, multiple identical pumps are connected in series. The outlet of the first pump feeds into the inlet of the second, and so on. This arrangement has the following effect:
- The total head is the sum of the individual pump heads.
- The flow rate remains the same as the flow rate through a single pump.
In this case, a single pump can generate about 52 m of head. To reach the required 156 m, we need to stack the heads of three pumps (3 x 52 m = 156 m).
