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5.4: Examples of Using the Steady-State Energy Density Model

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    • Contributed by Wendell Potter and David Webb et al.
    • Physics at UC Davis

    We now apply our energy density model to sections of pipe and electrical circuits to illustrate several general points regarding the use of the model.

    Uniform Section of Pipe or Conductor with No Pumps or Batteries

    In these first examples we are concerned with changes in the total head as we move along a section of pipe, or the change in voltage as we move along a conductor when there are no pumps or sources of emf along the section in question. In this case, the complete energy-density equations simplify to:

    \[\Delta(total~ head) = – I R \]

    \[\Delta(V) = – IR\]

    We will look at the electrical case first, because it is less complicated.


    We imagine a wire carrying a steady current. The wire must be attached to batteries or other source of emf, and be part of a larger circuit, but we won’t worry about those details at this point. Our question, is simply, how does the energy-density system, i.e., the voltage, change as we move along the wire. (The shape of section of wire or hose in the figure is meant to represent any general section of wire or hose. It continues in both directions.)

    Figure 5.4.1

    If we move from A to B, we are moving in the direction of the current. Energy is being transferred from the electric potential energy system to the thermal system. (The temperature of the wire will increase.) So we can write:

    \[\Delta V_{A ~to ~B} = V_B - V_A = - IR_{AB} \]


    \[V_B = V_A- IR_{AB}\]

    The voltage has dropped by an amount IR in moving from A to B. We speak of there being an “IR" drop along the wire.

    Suppose the current in the wire is 10 amps and the resistance per meter of wire length is 0.01 ohm/meter. If the length of wire between points A and B is 200 m, then the resistance of the section of wire between A and B is 2\(\Omega\). And

    \[ \Delta V_{A~ to~ B} = - 10A \times 2 \Omega = - 20 V\]

    If we attached the two leads of a voltmeter at the points A and B , we would get a reading of |20 V|. If the black (negative) lead were attached at point A and the red (positive) lead attached at point B, the reading would be negative: -20 V (the minus sign will actually be there on a digital voltmeter). The standard color coding of voltmeter leads is that the reading gives the voltage of the red lead with respect to the voltage of the black lead. Since there is a transfer of energy to the thermal system in going from A to B, there is a drop in the voltage. The change in voltage is negative.

    If we measured the voltage drop from B to C, we might find it was -16 V. And the drop from C to D might be \(-12 V\). What is the voltage drop in going from A to D? The charge energy system continuously transfers energy to the thermal system of the wire as we move along. The total transfer from A to D is simply the sum of the transfer along each section. Thus, we can immediately say that the voltage decrease in going from A to D is 48 V, or \(\Delta V_{A~ to~ D}= ~- 48 V\)


    Let’s consider the analogous fluid example. Imagine that we now have a section of constant diameter fire hose carrying a current of \(1.0 \times 10^{-3} ~m^3/s\). Perhaps the resistance per meter of length of the hose is \(1.0~ MJs/m^6\). Then the drop in total head in going from A to B is

    \[ \Delta (total~ head)_{A to B} = - 10^{-3} \dfrac{m^3}{s} \times 200 \frac{MJs}{m^6} \]

    \[= - 2.0 \times10^5 \frac{J}{m^3}\]

    \[= - 2.0 \times 10^5 Pa\]

    \[= - 2.0~~ atm~~ change\]

    Can we say that the “pressure drop” is -2 atm? Only if points A and B of the hose are at the same elevation, for only in this case will there be no change in the gravity head (gravitational potential energy density system). (Because the hose is uniform, we know there will be no change in the velocity head.) Thus, from what we have been given so far, all we can say is that there is a change in the combination of pressure and gravity head of -2.0~ atm.

    If point B were 10 m lower in elevation than point A, then

    \[ \Delta(hydraulic~ head)_{A ~to ~B} = \Delta (pressure~ head)_{A ~to ~B} + \Delta(gravity~ head)_{A ~to ~B} \]

    \[- 2.0~ atm = \Delta P_{A ~to~ B} - 1.0 ~atm\]


    \[\Delta P_{A~ to ~B} = - 1.0~ atm. \]

    In this example, half of the total head loss was in pressure and half in the gravity head. What would the drop in pressure be if point B were 20 m lower than point A? Answer: \(zero\)!

    Uniform Section of Pipe or Conductor Containing a Pump or Battery

    We will consider the same example as previously, but put a pump or battery at the location of point C.

    Figure 5.4.2

    Suppose we have exactly the same currents in the wire or hose as before. Would the voltage drop or the head loss from A to B be any different because we now have a battery or pump at point C? Neither the current in the section from A to B nor the resistance between A and B is different. Thus, the loss in energy density (voltage or head) must be exactly the same as before between points A and B .

    What about the loss between points A and D? If the currents are the same as before, then the IR drops from A to D will be exactly the same as before, because the resistance also will be the same as before. But, the pump or battery will add energy to the fluid or charge system (if they are installed in a manner that would tend to make the fluid or charge flow in the direction it is already going). Now we need to use the complete energy-density equations:

    \[\Delta(V) = \varepsilon – I R\]

    \[\Delta (total~ head) = \dfrac{E_{pump}}{vol} – IR \]

    So, the emf of the battery and the energy per volume added by the pump will certainly decrease the voltage drop and head loss, and could even cause gains in going from A to D instead of losses, if the energy they add is larger than the IR losses.

    We can also calculate the power losses and gains in the different sections of these wires and hoses. Let’s focus on the electrical example:

    The power loss of the electric charge system from A to B is simply the product of the voltage drop and the current: P = \(\Delta\)V I. (You can remember this relation by recalling that power is energy per time. Current is charge per time and voltage is energy per charge, so the product is energy per time.) So with \(\Delta\)V = -20 V and I = 10 A, the power loss in this section of wire is 200 VA = 200 W (watts). We could also have obtained this value from the relation P = I2R. If the battery were a 12 V car battery, the power supplied by the battery would be given by P = \(\varepsilon\)I and would be 120 W.