Problem
Rectangular slabs cast from 1020 annealed steel are heated and flat-rolled to improve their metallurgical structure and reduce their thickness. The slabs are 1.5m×9m×18cm with a density of 7590kg/m3. After passing through a 5-stand roughing mill, the rolled product will have a thickness of 3mm. Vertical rolls are used to prevent widening of the slabs so the width of the finished product does not change; however, the density of the finished product increases by 0.1%.
a) What is the length of the final rolled product? We want the first stand in the 5-stand roughing mill to reduce the thickness of the slab from 18cm to 16cm. The roll’s diameter is 600mm and it rotates at 75rpm. The motor driving the roll has 8,000kW of power with an efficiency of 75%. For 1020 annealed steel, K=745MPa and n=0.20.
b) Will you be able to perform this rolling job in one pass (i.e., reduce from 18cm to 16cm in one pass through the rolls)?
c) Will you be able to perform this rolling job in two passes (with equal reduction in every pass)? Justify your answer with calculations.
Submission Instructions
Final answers are not accepted – show the steps of your work
The correct answer and explanation is:
Let’s go step by step to solve the problem.
Given Data:
- Initial slab dimensions:
- Width (WW) = 1.5 m
- Length (LL) = 9 m
- Thickness (hih_i) = 18 cm = 0.18 m
- Density of 1020 annealed steel (ρi\rho_i) = 7590 kg/m³
- Final thickness (hfh_f) = 3 mm = 0.003 m
- No change in width (Wf=Wi=1.5W_f = W_i = 1.5 m)
- Final density (ρf\rho_f) increases by 0.1%, meaning:
- ρf=1.001×ρi=1.001×7590\rho_f = 1.001 \times \rho_i = 1.001 \times 7590 kg/m³
- Roll diameter (DD) = 600 mm = 0.6 m
- Roll rotational speed (NN) = 75 rpm
- Motor power = 8,000 kW
- Motor efficiency = 75%
- Flow stress equation: K=745K = 745 MPa, n=0.20n = 0.20
Step 1: Find the Final Length of the Slab
Since the volume remains constant: Vi=VfV_i = V_f Wi×Li×hi=Wf×Lf×hfW_i \times L_i \times h_i = W_f \times L_f \times h_f (1.5)(9)(0.18)=(1.5)(Lf)(0.003)(1.5)(9)(0.18) = (1.5)(L_f)(0.003)
Solving for LfL_f: Lf=(1.5)(9)(0.18)(1.5)(0.003)L_f = \frac{(1.5)(9)(0.18)}{(1.5)(0.003)} Lf=2.430.0045=540 mL_f = \frac{2.43}{0.0045} = 540 \text{ m}
So, the final length of the slab is 540 meters.
Step 2: Checking If the First Pass is Feasible
We first determine the maximum draft (dd), which is the reduction in thickness per pass: d=hi−hf=0.18−0.16=0.02 md = h_i – h_f = 0.18 – 0.16 = 0.02 \text{ m}
Using the equation for maximum draft: dmax=μ2Rd_{\max} = \mu^2 R
where RR is the roll radius: R=D2=0.62=0.3 mR = \frac{D}{2} = \frac{0.6}{2} = 0.3 \text{ m}
To find the coefficient of friction (μ\mu), we approximate: μ=0.3\mu = 0.3
Thus, the maximum draft is: dmax=(0.3)2×0.3=0.027 md_{\max} = (0.3)^2 \times 0.3 = 0.027 \text{ m}
Since our required draft (0.02 m) is less than the maximum draft (0.027 m), the pass is feasible.
Step 3: Checking If Two Passes Are Feasible
Each pass would need to reduce the thickness by: 0.18−0.162=0.01 m per pass\frac{0.18 – 0.16}{2} = 0.01 \text{ m} \text{ per pass}
Since 0.010.01 m is significantly less than the max draft of 0.0270.027, the reduction can be done easily in two passes.
Conclusion
- Part (a): The final length of the rolled product is 540 meters.
- Part (b): The rolling job can be performed in one pass since the required draft is within the permissible limit.
- Part (c): The rolling job can also be performed in two passes, and this would actually be easier since each pass would require a smaller reduction.
Now, I’ll generate an image to represent the rolling process.
I’ve generated a technical illustration showing the rolling process of the steel slab, including its transformation through the mill. Let me know if you need any modifications or additional explanations!
