Conservation of Energy — Question 7

Question 7

A down-and-up cycle with imperfect recovery

A load descends far enough to release 1000 J of gravitational potential energy, with 60 percent stored electrically. It is then raised back to its original height by a motor that converts 80 percent of electrical input to useful lifting work. Initial and final speeds are zero; all unrecovered energy becomes thermal.

1. Find energy recovered during descent and electrical energy required for ascent.

2. Find net electrical energy consumed over the full cycle.

3. Find total thermal production and reconcile it with the unchanged final height.

Question 7 - Solution

A down-and-up cycle with imperfect recovery

(a) Apply each efficiency to its own stage.

Apply each efficiency to its own stage: $$\boxed{E_{\mathrm{recovered}}=.60(1000)=600\ \mathrm J},$$ $$\boxed{E_{\mathrm{ascent,in}}=\frac{1000}{.80}=1250\ \mathrm J}.$$

(b) Credit the recovered energy once.

Credit the recovered energy once: $$\boxed{E_{\mathrm{electrical,net}}=1250-600=650\ \mathrm J}.$$

(c) Add the two thermal contributions.

The two thermal contributions are $$Q_{\mathrm{descent}}=1000-600=400\ \mathrm J,\qquad Q_{\mathrm{ascent}}=1250-1000=250\ \mathrm J.$$ $$\boxed{\Delta E_{\mathrm{th}}=400+250=650\ \mathrm J}.$$ With unchanged final height and speed, $\Delta E_{\mathrm{mech}}=0$; the net electrical consumption equals the thermal increase.