Question #25633

a hoisting engine lifts a 200 kg bucket containing 1200 Lof clay of density 2*10^3 kg m^3 to a height of 7 m in 20 sec. the power supplied to the engine is 12kW. calculate: work output in kJ, power output in kW and efficiency of the engine.

Expert's answer

The work done by a constant force of magnitude F on a point that moves a distance d in the direction of the force is the product:

A = F*d

F = (m+M)g

m -& mass of& bucket

M - mass of clay

M = density*V,

V& - volume of clay

A = ( m+density*V)*g*d = (200 kg + 2100kg/m^3 *1.2m^3) 9.8m/s^2 7m = (200+2100*1.2)*9.8*7 J = 186592 J = 187 kJ

A = 187 kJ

P = F*v,

P - power, v - velocity.

v = d/t,

d - distance ( 7 m)

t - time (20sec)

P = F*d/t = A/t = 186592 J /20sec =& 9329.6 W = 9.33 kW

P = 9.33 kW

efficiency of the engine = P/P_in,

P_in - power supplied to the engine

efficiency of the engine = P/P_in = 9.33/12 = 0.7775 = 77.75 %

efficiency of the engine = 77.75 %

A = F*d

F = (m+M)g

m -& mass of& bucket

M - mass of clay

M = density*V,

V& - volume of clay

A = ( m+density*V)*g*d = (200 kg + 2100kg/m^3 *1.2m^3) 9.8m/s^2 7m = (200+2100*1.2)*9.8*7 J = 186592 J = 187 kJ

A = 187 kJ

P = F*v,

P - power, v - velocity.

v = d/t,

d - distance ( 7 m)

t - time (20sec)

P = F*d/t = A/t = 186592 J /20sec =& 9329.6 W = 9.33 kW

P = 9.33 kW

efficiency of the engine = P/P_in,

P_in - power supplied to the engine

efficiency of the engine = P/P_in = 9.33/12 = 0.7775 = 77.75 %

efficiency of the engine = 77.75 %

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