Know the principle of the conservation of energy and apply this principle to complex examples involving multiple stages, including the interpretation of Sankey diagrams

1.7.1 Energy – Conservation of Energy

Learning Objective

Know the principle of the conservation of energy and apply this principle to complex examples involving multiple stages, including the interpretation of Sankey diagrams.

Key Concepts

• Energy is the capacity to do work or produce heat.
• Energy can be transferred or transformed but cannot be created or destroyed – the principle of conservation of energy.
• Common forms of energy: kinetic, gravitational potential, elastic, chemical, electrical, thermal, nuclear.
• Sankey diagrams are graphical representations that show the flow of energy between forms, with the width of arrows proportional to the amount of energy.

Mathematical Formulation

The total energy of an isolated system remains constant:

\$\sum E{\text{initial}} = \sum E{\text{final}}\$

For a single object moving under gravity:

\$E{\text{total}} = KE + PE{\text{g}} = \frac{1}{2}mv^{2} + mgh\$

When energy is transferred between several stages, the principle is applied sequentially:

\$E{1} \rightarrow E{2} \rightarrow E{3} \rightarrow \dots \rightarrow E{n}\$

Energy lost as heat or sound is accounted for as a reduction in the useful energy transferred.

Forms of Energy and Their Equations

Understanding Sankey Diagrams

A Sankey diagram shows:

1. The total input energy (usually at the left).
2. Each stage of energy conversion, represented by arrows whose width is proportional to the amount of energy transferred.
3. Energy losses (often shown as thin arrows or shaded sections) that represent energy that is not useful for the intended purpose.

Applying Conservation of Energy to Multi‑Stage Problems

When solving a problem with several stages, follow these steps:

1. Identify all forms of energy at the start of each stage.
2. Write the conservation equation for that stage, including any known losses.
3. Solve for the unknown quantity (usually a speed, height, or amount of energy transferred).
4. Carry the result forward to the next stage, treating it as the initial energy for that stage.

Worked Example – Roller Coaster with Multiple Drops

A roller coaster car of mass \$m = 500\ \text{kg}\$ starts from rest at a height of \$30\ \text{m}\$. The track has three successive drops: 30 m, 20 m, and 10 m. At each drop, \$5\%\$ of the mechanical energy is lost as heat and sound. Determine the speed of the car at the bottom of the final drop.

Solution

1. Initial mechanical energy (all gravitational potential):

   \$E{0}=mgh{1}=500\times9.8\times30=147\,000\ \text{J}\$

2. After the first drop (30 m) the car reaches the bottom with kinetic energy \$KE_{1}\$, but \$5\%\$ is lost:

   • Energy available for conversion: \$0.95E_{0}=139\,650\ \text{J}\$
   • All of this becomes kinetic energy: \$KE_{1}=139\,650\ \text{J}\$
   • Speed after first drop:

     \$v{1}= \sqrt{\frac{2KE{1}}{m}} = \sqrt{\frac{2\times139\,650}{500}} \approx 23.6\ \text{m s}^{-1}\$

3. Car ascends to the next height (20 m). Its kinetic energy is partially converted back to gravitational potential:

   • Potential energy at 20 m: \$PE{2}=mgh{2}=500\times9.8\times20=98\,000\ \text{J}\$
   • Remaining kinetic energy at the top of the second drop:

     \$KE{\text{top}} = KE{1} - PE_{2}=139\,650-98\,000=41\,650\ \text{J}\$

   • Again \$5\%\$ of the total mechanical energy at this stage is lost:

     \$E{\text{available}} = 0.95\,(KE{\text{top}}+PE_{2}) = 0.95\times139\,650 = 132\,667.5\ \text{J}\$

   • At the bottom of the second drop the total mechanical energy is \$E_{\text{available}}\$, all kinetic:

     \$KE_{2}=132\,667.5\ \text{J}\$

     \$v{2}= \sqrt{\frac{2KE{2}}{m}} \approx 23.0\ \text{m s}^{-1}\$

4. Repeat for the third drop (10 m):

   • Potential energy at 10 m: \$PE_{3}=500\times9.8\times10=49\,000\ \text{J}\$
   • Mechanical energy before loss: \$KE{2}+PE{3}=132\,667.5+49\,000=181\,667.5\ \text{J}\$
   • After \$5\%\$ loss:

     \$E_{\text{final}} = 0.95\times181\,667.5 = 172\,584.1\ \text{J}\$

   • All of \$E_{\text{final}}\$ is kinetic at the bottom:

     \$v{3}= \sqrt{\frac{2E{\text{final}}}{m}} = \sqrt{\frac{2\times172\,584.1}{500}} \approx 26.3\ \text{m s}^{-1}\$

Thus the speed of the car at the bottom of the final drop is approximately \$26\ \text{m s}^{-1}\$.

Interpreting a Sankey Diagram – Example

Consider a Sankey diagram for a simple electric heater:

• Input electrical energy: \$2000\ \text{J}\$ (shown as the widest arrow).
• Thermal energy delivered to water: \$1800\ \text{J}\$ (arrow width 90% of input).
• Heat lost to surroundings: \$200\ \text{J}\$ (arrow width 10% of input).

From the diagram we can state:

1. The efficiency of the heater is \$\displaystyle \frac{1800}{2000}=0.90\$ or \$90\%\$.
2. Energy is conserved because \$1800\ \text{J}+200\ \text{J}=2000\ \text{J}\$.

Practice Questions

1. A 2 kg ball is dropped from a height of 5 m. Air resistance removes \$10\%\$ of the mechanical energy during the fall. Calculate the speed of the ball just before it hits the ground.
2. In a hydroelectric plant, water of mass \$1.5\times10^{5}\ \text{kg}\$ falls 50 m. If the turbine and generator together are \$85\%\$ efficient, how much electrical energy is produced? (Use \$g=9.8\ \text{m s}^{-2}\$.)
3. Interpret the following Sankey diagram (described in words): Input chemical energy \$5000\ \text{kJ}\$ → Thermal energy \$3500\ \text{kJ}\$ → Mechanical energy \$1200\ \text{kJ}\$ → Electrical energy \$300\ \text{kJ}\$. Identify the total energy loss and the overall efficiency of converting chemical to electrical energy.

Summary

• The total energy of an isolated system never changes; it only changes form.
• When solving multi‑stage problems, apply the conservation equation at each stage, accounting for any losses.
• Sankey diagrams provide a visual check of energy conservation and help quantify efficiencies and losses.