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2 months ago

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A 2.0-in-thick slab is 10.0 in wide and 12.0 ft long. Thickness is to be reduced in three steps in a hot rolling operation. Each

step will reduce the slab to 75% of its previous thickness. It is expected that for this metal and reduction, the slab will widen by 3% in each step. If the entry speed of the slab in the first step is 40 ft/min, and roll speed is the same for the three steps.
Calculate:

a) lenghtb) exit velocity of the final slab

1 answer:

Mrrafil [318]2 months ago

8 0

Answer:

L_f = 26.025 ft

v_f = 51.77 ft/min

Explanation:

Given:

- Initial slab thickness, t_o = 2 in

- Initial slab width, w_o = 10 in

- Initial slab length, L_o = 12.0 ft

- Thickness reduction each step, r = 75%

- Width increase each step, m = 3%

- Initial entry speed vi = 40 ft/min

- Roll speed remains constant

Required:

a) Length

b) Exit velocity

For the final slab

Solution:

- After three passes, the final thickness (t_f) is determined as:

t_f = ( r / 100 )^n * t_o

Where n signifies the number of passes.

t_f = ( 75 / 100 ) ^3 * ( 2.0 )

t_f = 0.844 in

- The final width after three passes is calculated as:

w_f = ( m / 100 + 1 )^n * w_o

Where n denotes the number of passes.

w_f = ( 3 / 100 + 1 ) ^3 * ( 10.0 )

w_f = 10.927 in

- Assuming there is no loss in material, the final length of the slab can be obtained:

t_o*w_o*L_o = t_f*w_f*L_f

L_f = ( 2 * 10 * 12 ) / ( 0.844 * 10.927 )

L_f = 26.025 ft

- The equation for volume rate can be applied since the roll speed stays the same. Thus, we equate the conditions before and after the third step:

t_i*w_i*v_i = t_f*w_f*v_f

Where v_i is the initial speed at entry:

v_f signifies the exit speed after the third step.

(0.75)^2 * 2 * (1.03)^2 * 10 * 40 = (0.844)*(10.927)*v_f

v_f = 51.77 ft/min

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Provided data

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Final temperature T2 = 80°C

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We aim to determine percentage change in volume and the rise in water level.

Solution

We will apply the bulk modulus equation, which relates the change in pressure to the change in volume.

It can similarly relate to density changes.

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E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

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Initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

Final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

From equations 3 and 4, subtract v1 from v2.

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

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Substituting all values yields

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Thus, dl = 0.004138 m.

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