Question

I need someone to double check my math. So, here we go. I need to know how much energy it would take to move (or displace in a 4" tube) 13.6 gallons of water (or 114.6 lbs of water) vertically 6 inches in 1 second. Energy required will need to be in joules.

Answer 1

I need someone to double check my math. So, here we go. I need to know how much energy it would take to move (or displace in a 4" tube) 13.6 gallons of water (or 114.6 lbs of water) vertically 6 inches in 1 second. Energy required will need to be in joules.

Solution:

The law of conservation of total mechanical energy:

$$W _ {1} = W _ {2} (1)$$ $$W _ {1} = E _ {\text {kinetic1}} + E _ {\text {potential1}} + E _ {\text {work}}$$ $$E _ {\text {kinetic1}} = 0 \text{ (the initial velocity of zero)}$$

$E_{\text{potential1}} = 0$ (relative to the initial level, potential energy is zero)

$E_{\text{work}}$ – this energy we need to find

$$W _ {1} = E _ {\text {work}} (2)$$ $$W _ {1} = E _ {\text {kinetic2}} + E _ {\text {potential2}}$$ $$E _ {\text {kinetic2}} = \frac {m V ^ {2}}{2}; m = 114.6 \text{ lbs} = 52 \text{ kg}, V = \frac {6 \text{ inches}}{1 \text{ s}} = 0.1524 \frac {m}{\text{s}}$$ $$E _ {\text {potential2}} = mgh, h = 6 \text{ inches} = 0.1524m$$ $$W _ {1} = \frac {m V ^ {2}}{2} + mgh (3)$$

(3) and (2) in (1):

$$\begin{array}{l} E _ {work} = \frac {m V ^ {2}}{2} + mgh = m \left(\frac {V ^ {2}}{2} + gh\right) = 52 \text{ kg} \left(\frac {\left(0.1524 \frac {m}{s}\right) ^ {2}}{2} + 9.8 \frac {N}{kg} \cdot 0.1524m\right) = \\ = 78.26 J \\ \end{array}$$

Answer: energy is 78.26 J