DC shunt motor hoist: raising and rheostatic lowering

Raising the load

Power needed to lift the load:

$$P_{load} = mgv = 400 \times 9.81 \times 2.5 = 9810\ \text{W}$$

Motor output (hoist efficiency 80 %) and input (motor efficiency 85 %):

$$\begin{aligned} P_{out} &= \frac{9810}{0.80} = 12262.5\ \text{W} \\ P_{in} &= \frac{12262.5}{0.85} = 14426.5\ \text{W} \\ I &= \frac{P_{in}}{V} = \frac{14426.5}{220} = 65.57\ \text{A} \end{aligned}$$

Lowering at 2.5 m/s by rheostatic braking

When lowering, the falling load drives the motor, which runs as a generator disconnected from the supply with a resistor across its armature (field stays excited). Power now flows the other way, so each efficiency reduces it:

$$\begin{aligned} \text{Shaft power to machine} &= 9810 \times 0.80 = 7848\ \text{W} \\ \text{Electrical power generated} &= 7848 \times 0.85 = 6670.8\ \text{W} \end{aligned}$$

The speed is the same as when raising and the field is unchanged, so the generated emf is taken as about the supply voltage, $E \approx 220$ V (armature resistance not given; its drop is neglected). All this power is dissipated in the braking resistance:

$$\begin{aligned} I_b &= \frac{6670.8}{220} = 30.32\ \text{A} \\ R &= \frac{E^2}{P} = \frac{220^2}{6670.8} = 7.26\ \Omega \end{aligned}$$

Answer: Supply current while raising = 65.6 A; braking resistance for lowering at 2.5 m/s ≈ 7.26 Ω.

(If the armature resistance $R_a$ were known, the external resistance to add would be $7.26 - R_a$.)