Question 26.60: The chocolate crumb mystery. This story begins with Problem ......
The chocolate crumb mystery. This story begins with Problem 60 in Chapter 23 and continues through Chapters 24 and 25. The chocolate crumb powder moved to the silo through a pipe of radius R with uniform speed ν and uniform charge density ρ. (a) Find an expression for the current i (the rate at which charge on the powder moved) through a perpendicular cross section of the pipe. (b) Evaluate i for the conditions at the factory: pipe radius R = 5.0 cm, speed ν = 2.0 m/s, and charge density ρ = 1.1 × 10^{-3} C/m³.
If the powder were to flow through a change V in electric potential, its energy could be transferred to a spark at the rate P = iV. (c) Could there be such a transfer within the pipe due to the radial potential difference discussed in Problem 70 of Chapter 24? As the powder flowed from the pipe into the silo, the electric potential of the powder changed. The magnitude of that change
was at least equal to the radial potential difference within the pipe (as evaluated in Problem 70 of Chapter 24). (d) Assuming that value for the potential difference and using the current found in (b) above, find the rate at which energy could have been transferred from the powder to a spark as the powder exited the pipe. (e) If a spark did occur at the exit and lasted for 0.20 s (a reasonable expectation), how much energy would have been transferred to the spark?
Recall from Problem 60 in Chapter 23 that a minimum energy transfer of 150 mJ is needed to cause an explosion. (f) Where did the powder explosion most likely occur: in the powder cloud at the unloading bin (Problem 60 of Chapter 25), within the pipe, or at the exit of the pipe into the silo?
(a) Current is the transport of charge; here it is being transported “in bulk” due to the volume rate of flow of the powder. From Chapter 14, we recall that the volume rate of flow is the product of the cross-sectional area (of the stream) and the (average) stream velocity. Thus, i = ρAν where ρ is the charge per unit volume. If the cross section is that of a circle, then i = ρπR²ν.
(b) Recalling that a coulomb per second is an ampere, we obtain
i=\left(1.1\times10^{-3}\,\mathrm{C/m}^{3}\right)\pi\left(0.050\,\mathrm{m}\right)^{2}\left(2.0\ \mathrm{m}/s\right)=1.7\times10^{-5}\,\mathrm{A}.
(c) The motion of charge is not in the same direction as the potential difference computed in problem 70 of Chapter 24. It might be useful to think of (by analogy) Eq. 7-48; there, the scalar (dot) product in P= \vec{F}\cdot{\vec{\nu}} makes it clear that P = 0 if \vec{F}\perp{\vec{\nu}}. This suggests that a radial potential difference and an axial flow of charge will not together produce the needed transfer of energy (into the form of a spark).
P={\vec{F}}.{\vec{\nu}} (instantaneous power). (7-48)
(d) With the assumption that there is (at least) a voltage equal to that computed in problem 70 of Chapter 24, in the proper direction to enable the transference of energy (into a spark), then we use our result from that problem in Eq. 26-26:
P= iV (rate of electrical energy transfer). (26-26)
P=i V=\left(1.7\times10^{-5}\,{A}\right)\left(7.8\times10^{4}\,{V}\right)=1.3\,{W}.
(e) Recalling that a joule per second is a watt, we obtain (1.3 W)(0.20 s) = 0.27 J for the energy that can be transferred at the exit of the pipe.
(f) This result is greater than the 0.15 J needed for a spark, so we conclude that the spark was likely to have occurred at the exit of the pipe, going into the silo.