Chapter 6, Problem 67P

To determine

The minimum output rating $\left(\text{watts}\right)$ must the pump motor have.

Answer to Problem 67P

Solution: $15.44\text{ watt}$

Explanation of Solution

Given:

Mass of water $=27.0\text{kg}$

Time $=6\text{s}$

Height $=3.50\text{m}$

Formula Used: Work done to move m $\left(\text{kg}\right)$ mass to height $\left(h\right)\text{ against gravity}$

  $=mgh$

Power $=\frac{work\left(W\right)}{time\left(t\right)}$

Calculation:

W (work) done by pump to lift 27.0kg water up to height $3.50$ m.

  $=mgh$

  $=\left(27.0\text{kg}\right)\times \left(9.80\text{m}/{\text{s}}^2\right)\times \left(3.50\text{m}\right)$

  $=926.1\text{J}$

Power of pump $=\frac{W}{t}$

  $=\frac{926.1\text{J}}{60\text{ s}}$

  $=15.44\text{ watt}$

Conclusion: Minimum output to lift $27.0\text{ kg}$ of water per minute up to height $3.50\text{m}$ is equal to. $15.44\text{W}$ .