Respuesta :
Answer:
roll force = 1.38 x 10 ∧ 6 N
power = 409 kw
Explanation:
The pictures below can be seen for full explanation
The frictional force acting on the roller is 2517.3kN and the power is 1.49*10^9W
To solve this problem, we would need some rolling formula
Data given are;
- k = 530mPa
- η = 0.26
- width (w) = 200mm
- roll radius (r) = 200mm
- speed(N) = 200 rpm
- coefficient of friction (μ) = 0.1
- Initial thickness (x[tex]_o[/tex]) = 10mm
- final thickness (x[tex]_f[/tex]) = 6mm
Rolling Force
The rolling force is a product of stress and projected area
The rolling force (F[tex]_r[/tex]) = r[tex]_o[/tex] * (L[tex]_p[/tex] * w) …equation (i)
Stress
The stress (r[tex]_o[/tex]) = k(E[tex]_T[/tex])^n
E[tex]_[/tex][tex]_T[/tex] = True strain.
[tex](E_T) = In(\frac{x_o}{x_f})= In(\frac{10}{6})\\E_T= 0.51082[/tex]
Projected Length (L[tex]_p[/tex])
The formula for the projected length is given as
[tex]L_p=\sqrt{R{\delta}x\\\\[/tex]
[tex]L_p=\sqrt{R(x_o-x_f}\\L_p=\sqrt{200(10-6)} \\L_p=28.28mm[/tex]
From equation (i)
F[tex]_r[/tex] = r[tex]_o[/tex]* (L[tex]_p[/tex] * w)
[tex]F_r = k(E_T)^n*(l_p*w)\\F_r=530*(0.51082)^0^.^2^6*(28.28*200)\\F_r=2517299.69\\F_r= 2517.3kN[/tex]
But the method of rolling wasn't given and we can assume the roller is a hot roller.
Torque (τ)
This is can be calculated as
[tex]F_r * a\\a = 0.5L_p[/tex]
This is the value for a hot rolled roller.
hence,
Torque (T) =
[tex]0.5*F_r*L_p\\T= 3.559*10^7N.m[/tex]
Power
Power is the rate at which work is done with respect to time.
[tex]p = 2Tw[/tex] since we are working with two rollers here.
[tex]p = 2*T*\frac{2\pi N}{60} \\p = 2 * 3.559*10^7*\frac{2*3.14*200}{60}\\p = 1.49*10^9w[/tex]
From the calculations above, we have the force of acting on the roller or force on roller (F[tex]_r[/tex]) as 2317.3kN and the power acting on the roller as[tex]1.49*10^9W[/tex]
Learn more on strain, stress, torque and frictional force here;
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