An elevator is designed to deliver books for the bookshop located on the fourth floor of the library. The power required to run the elevator when it is fully loaded and empty needs to be determined.

Consider the Car and the
Counterweight as a System

Assumptions:

• The guide rails and the pulleys are frictionless
• The ropes can not be stretched

Consider the car and the counterweight as a system. It is a closed system because the car (with or without a load) and the counterweight are solids, hence the mass remains constant during the process.

The energy balance for a closed system is:

      Q - W = (ΔU + ΔKE + ΔPE)_system

It can be simplified according to the following conditions:

• No heat is transferred in or out, Q = 0
• No internal energy change, ΔU = 0
• The velocity remains constant during the process, ΔKE = 0

The energy balance equation becomes:

      - W =(ΔPE)_system

Locations of the Elevator at
Time t₁and t₂ when the
Elevator Moves Upwards

(1) Determine the power needed when the elevator is fully loaded and moving upwards

From the analysis above, the work needed to run the fully loaded elevator upwards is:

      W = - (ΔPE)_system

The mass of the counterweight is:

      m_counter = (m_car + m_load )(50%)
                   = (150 + 650)(50%) = 400 kg

At time t₁, the total potential energy of the system is:

      (PE)₁ = (m_car + m_load )g x₁ + m_counterg y₁

At time t₂, the total potential energy of the system changes to:

      (PE)₂ =(m_car + m_load )g x₂ + m_counterg y₂

      (ΔPE)_system = (PE)₂ - (PE)₁
      = (m_car + m_load )g ( x₂ - x₁) + m_counterg ( y₂ - y₁)

Since the ropes cannot be stretched, the distance that the car moves upwards is the same as the distance that the counterweight moves downwards.

      (x₂ - x₁) = - (y₂ - y₁) = (t₂ - t₁)v

Hence,

      (ΔPE)_system= (m_car + m_load- m_counter)g (t₂ - t₁)v

      W = - (m_car + m_load- m_counter)g (t₂ - t₁)v

Power is work per unit time, and it is given by

       = W/(t₂ - t₁) = - (m_car + m_load- m_counter)gv
             = - (150 + 650 - 400)(9.81)(2.0)
             = - 7,840 W

The negative sign means that work needs to be applied to the system when the elevator runs upwards with full load.

(2) Determine the power needed when the elevator is empty and moving downwards

At time t₁, the total potential energy of the system is:

      (PE)₁ = (m_car)g x₁ + m_counterg y₁

At time t₂, the total potential energy of the system is:

      (PE)₂ =(m_car)g x₂ + m_counterg y₂

      (ΔPE)_system = (PE)₂ - (PE)₁
                        = (m_car )g (x₂ - x₁) +
                           m_counterg (y₂ - y₁)

The counterweight moves upwards while the car moves downwards.

      y₂ - y₁ = - (x₂ - x₁) = (t₂ - t₁)v

Hence,

      (ΔPE)_system= (m_counter - m_car )g(t₂ - t₁)v

Power can be determined by dividing the work by time.

       = W/(t₂ - t₁) = - (m_counter- m_car)gv
            = - (400 - 150)(9.81)(2.0)
            = - 4,900 W

The result is negative which means that input power is required.

(3) Determine the energy saved by using counterweight

For a single delivery, the elevator runs upwards with full load and downwards with no load. Set m_counter equals to 0 in the above analysis. The energy needed for the elevator moving upwards and downwards are:

      _up = W/(t₂ - t₁) = - (m_car + m_load)gv
            = - (150 + 650)(9.8)(2.0)
            = - 15,680 W

      _down = W/(t₂ - t₁) = - (- m_car)gv
            = - ( - 150)(9.8)(2.0)
            = 2,940 W

The work is positive which means that no energy is needed to be applied to the system.

The percent of energy saved by using the counterweight is

      (15,680 - 7,840 - 2,940)/15,680 = 31.3%