A centrifugal irrigation pump operates at a speed of $N_1 = 1450.0\text{ rpm}$ developing a total head of $H_1 = 25.0\text{ m}$ at a discharge of $Q_1 = 40.0\text{ L/s}$. If the pump rotational speed is increased to $N_2 = 1740.0\text{ rpm}$ without altering the impeller diameter, according to pump affinity laws, the new operating head $H_2$ in meters is ________ (round off to one decimal place).
Answer Key:
36.0
Step-by-Step Solution & Explanation
**Method 1: Centrifugal Pump Affinity Laws**
For constant impeller diameter $D$:
1. Discharge varies linearly with speed:
$\frac{Q_2}{Q_1} = \frac{N_2}{N_1}$
2. Total head varies with the square of speed:
$\frac{H_2}{H_1} = \left( \frac{N_2}{N_1} \right)^2$
3. Given speed ratio:
$\frac{N_2}{N_1} = \frac{1740.0}{1450.0} = 1.20$
4. Calculating new head $H_2$:
$H_2 = H_1 \times (1.20)^2 = 25.0 \times 1.44 = 36.0\text{ m}$
**Method 2: Head Coefficient Conservation**
$C_H = \frac{g H}{N^2 D^2} = \text{constant} \implies \frac{H_1}{N_1^2} = \frac{H_2}{N_2^2}$
$H_2 = 25.0 \times \left( \frac{1740}{1450} \right)^2 = 25.0 \times 1.44 = 36.0\text{ m}$