Rubbing hands together is a dissipative process that we would like to model and quantify.
a) Determine the mechanical power P_W dissipated by friction during this process, in terms of the friction force F^{fr} and the the mean relative velocity v of a hand with respect to the other.
b) At room temperature T, determine the entropy production rate Π_S of this process.
Numerical Application:
\left\|F^{fr}\right\|=1 N, \left\|ν\right\|= 0.1m/s and T = 25^◦C.
a) With respect to the centre of mass frame of reference of the hands, each hand is moving with a velocity of norm \left\|ν\right\|/2. The friction force F^{fr} is opposed to the relative motion of the hands, i.e. F^{fr} · v < 0. The mechanical power is written,
P_W = \bigl(-F^{fr}\bigr) .\Bigl(\frac{\nu }{2} \Bigr)+\bigl(F^{fr}\bigr).\Bigl(-\frac{\nu }{2} \Bigr) = −F^{fr} · v = 0.1 W.
b) It can be presumed here that no heat is transferred to the environment, so that the evolution equation (2.18) for S implies \dot{S} = Π_S. The evolution equation (1.29) for the internal energy implies \dot{U} = P_W. As the internal energy U is a function of the entropy S only, \dot{U} = T\dot{S} . Consequently, the entropy production rate is given by,
\dot{S} = Π_S + \frac{P_Q}{T}. (2.18)
\dot{U} = P_W + P_Q. (closed system) (1.29)
Π_S = \frac{P_W }{T} = – \frac{F^{fr} \cdot ν }{t}= 3.36 · 10^{−4} W/K.