Areciprocating engine of 750mm stroke runs at 240 rpm. if the length of the connecting rod is 1500mm find the piston speed and acceleration when the crank is 45 past the top dead center position.

speed = 16.44 m/s

Acceleration = 71.36 m/s²

Explanation:

Given data

Speed ( N) = 240 rpm

angle  = 45°

stoke length(L)  = 750 mm

length of rod ( l )  = 1500 mm

To find out

the piston speed and acceleration

Solution

we find speed by this formula

speed = r ω (sin(θ) + (sin2(θ)/ 2n))  ...................1

here we have find  r and ω

ω = 2 N / 60

so ω = 2 × 240 / 60

ω =  25.132 rad/s

n = l/r =  1500/750 = 2

we know  L = 2r

so r = L/2 = 750/2 = 375 mm

put these value in equation 1

speed = 375 × 25.132 (sin(45) + (sin2(45)/ 2×2))  

speed = 16444.811823 mm/s = 16.44 m/s

Acceleration = r ω² (cos(θ) + (cos2(θ)/ n))  ...................2

put the value  r, ω and n in equation 2

Acceleration = r ω² (cos(θ) + (cos2(θ)/ n))

Acceleration  = 375 × (25.132)² (cos(45) + (cos2(45)/2))  

Acceleration = 71361.363659 = 71.36 m/sec²

yes indeed

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answer:

yes

explanation: