Lesson 9. Power Requirements

The power is the rate of doing work. It can be written as

L 9 eq.1

1 hp           = 33000 ft-lb/min

                  = 550 ft-lb/s

                  = 75 m-kgf/s

                  = 4500 m-kgf/min


L 9 eq.2


hp        = horse power

D         = distance travelled in ft

F          = force exerted in lb

T          = time taken in min

It can also be written as

L 9 eq.3


hp        = horse power

D         = distance travelled in m

F          = force exerted in kgf

T          = time taken in min

It can be further written as

L 9 eq.4

Lesson 9 : Solved examples

Example 1 : An implement requires 300 kg force to pull it through a distance of 100 m in 5 min. Find the power required to pull the implement.


L 9 ex1 eq.1

Given        F = 300 kgf

D = 100 m

T = 5 min


L 9 ex1 eq.2

Example 2 : A tractor pulls a draught load of 1000 kg while travelling at a speed of 60 m/min. Find the horse power (hp) developed by the tractor.


L 9 ex2 eq.1

Drawbar Power: Power requirement of a tractor at its drawbar can also be estimated by the following formulae:

L 9 ex2 eq.2


P  = drawbar power, kW

F  = draught, kN

S  = speed, km/h


L 9 ex2 eq.3


P  = drawbar power, hp

F  = draught, kgf

S  = speed, km/h

Power requirement for some of the implements is given in Table 3.1. The drawbar power output is always less than Power-Take-Off (PTO) power output because of drive wheel slippage, tractor rolling resistance and friction losses in the drive train between the engine and the wheels. The sum of these losses may be represented by a tractive & transmission (T&T) coefficient. The tractive & transmission coefficient is defined as

L 9 ex2 eq.4

T&T coefficients for four-wheel drive tractors are somewhat higher than those for two wheels drive tractors.  Track type tractors seldom have more than 5% slips even on soft soil. The value of tractive and transmission coefficient for track type tractors varies between 0.80 and 0.85 on firm soil and 0.70 and 0.75 on soft soil. The PTO power corresponding to drawbar power is determined by applying an appropriate T&T coefficient.

Rotary power: It is defined as

L 9 ex2 eq.5 

Example: If a belt produces a tangential force of 1000 N on a pulley having radius of 0.30 m at a speed of 200 rpm, the power it will be

L 9 ex2 eq.6

Given F = 1000 N                 

          D = distance covered = 2 p r x rpm

              = 2 \[\pi \] x 0.30 x 200 m/min

          T = 1 min

L 9 ex2 eq.7

1 W     = 1 N-m/s

P          = 6283.18 W               =          6.28 kW

Fluid power: It is defined as the product of a weight rate of flow and the resistance to that flow called the head of the flow. Head describes the height of a column of fluid whose mass creates at the bottom a pressure equivalent to that in the flowing system. Thus,

L 9 ex2 eq.8

Example: If 100 kg of water is to be pumped to the height of 30 m in 100 s, the power required would be

L 9 ex2 eq.9

Given        F = 100 kg       = 100 x 9.8 N  = 980 N

                      D = 30 m

                      T = 100 s


L 9 ex2 eq.10

The pressure of the fluid and not its head is used in most calculations and expressed in Pascal (Pa).

1 Pa     = 1 N/m2

1 kPa   = 1000 Pa        = 0.145 psi

1 MPa = 1000 kPa

1 bar    = 100 kPa        = 1 atm

Hydraulic power (Ph): It is expressed as

L 9 ex2 eq.11       


p          = gauge pressure, k Pa

Q         = flow rate, l/s

C         = constant, 1000

It can also be expressed as

      Ph        = 0.01667 Q Dp, kW


Q         = fluid flow rate, l/min

Dp       = pressure change, M Pa

Power-Take-Off (PTO) power: It is expressed as

L 9 ex2 eq.12


P          = PTO power, kW

F          = tangential force, kN

R         = radius of force rotation, m

N         = revolutions per minute, rpm

T          = torque, N-m = F R

Last modified: Monday, 24 March 2014, 5:02 AM