(a) A drilling operation is required to produce a hole with a diameter of 25 mm...

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(a) A drilling operation is required to produce a hole with a diameter of 25 mm and depth of 20 mm at one end of the Aluminum alloy block. If the feed of a 25-mm drill is 0.1 mm/rev and the spindle of that drilling machine is rotating at 800 rpm. (i) Calculate the material removal rate and time required to drill that hole on the block. Given: Feed rate fr = Nxf Time to machine Tm = (h+D/2)/fr Material removal rate RMR = TD²f/4 Where N=Rotational speed in rpm, f=feed motion in mm/rev, D-diameter of the hole in mm and h-depth of the hole in mm. (7 Marks) (ii) Discuss and explain whether the hole of the same quality cannot be produced by other machining process. (4 Marks) (b) A 150 mm long having 12.5 mm diameter 304 stainless-steel solid rod is being turned to a diameter of 10 mm by a lathe. The spindle rotates at N = 400 rpm, and the tool is traveling at an axial speed of 200 mm/min. Calculate the cutting speed, material removal rate and machining time. (7 Marks) Given: Rotational speed N = v/Do Feed rate fr= Nxf Time to machine Tm = πDoL/fv Material removal rate RMR = vfd where N=Rotational speed; f-feed motion; D.-outer diameter of the workpiece; v=cutting speed and d = depth of cut. (c) A slab-milling operation is being carried out on a 300 mm long having 100 mm wide mild-steel block at a feed f=0.25 mm/tooth and a depth of cut d=3.13 mm. The cutter is D=50 mm in diameter and has 20 straight teeth, which rotates at N=100 rpm and is wider than the block to be machined. Calculate the material removal rate and the machining time. Given: MRR = wxdxfr fr=Nxnt xf (7 Marks) Tm=(L+A)/fr A = Vd (D-d) where w and L are the width and length of the workpiece respectively. D, n, and N are the diameter, number of teeth and rotational speed of the cutter respectively. d is the depth of cut; f, is the feed rate; A is the vertical distance between the cutter centre and workpiece edge; Tm is the machining time. (a) A drilling operation is required to produce a hole with a diameter of 25 mm and depth of 20 mm at one end of the Aluminum alloy block. If the feed of a 25-mm drill is 0.1 mm/rev and the spindle of that drilling machine is rotating at 800 rpm. (i) Calculate the material removal rate and time required to drill that hole on the block. Given: Feed rate fr = Nxf Time to machine Tm = (h+D/2)/fr Material removal rate RMR = TD²f/4 Where N=Rotational speed in rpm, f=feed motion in mm/rev, D-diameter of the hole in mm and h-depth of the hole in mm. (7 Marks) (ii) Discuss and explain whether the hole of the same quality cannot be produced by other machining process. (4 Marks) (b) A 150 mm long having 12.5 mm diameter 304 stainless-steel solid rod is being turned to a diameter of 10 mm by a lathe. The spindle rotates at N = 400 rpm, and the tool is traveling at an axial speed of 200 mm/min. Calculate the cutting speed, material removal rate and machining time. (7 Marks) Given: Rotational speed N = v/Do Feed rate fr= Nxf Time to machine Tm = πDoL/fv Material removal rate RMR = vfd where N=Rotational speed; f-feed motion; D.-outer diameter of the workpiece; v=cutting speed and d = depth of cut. (c) A slab-milling operation is being carried out on a 300 mm long having 100 mm wide mild-steel block at a feed f=0.25 mm/tooth and a depth of cut d=3.13 mm. The cutter is D=50 mm in diameter and has 20 straight teeth, which rotates at N=100 rpm and is wider than the block to be machined. Calculate the material removal rate and the machining time. Given: MRR = wxdxfr fr=Nxnt xf (7 Marks) Tm=(L+A)/fr A = Vd (D-d) where w and L are the width and length of the workpiece respectively. D, n, and N are the diameter, number of teeth and rotational speed of the cutter respectively. d is the depth of cut; f, is the feed rate; A is the vertical distance between the cutter centre and workpiece edge; Tm is the machining time.

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a i Calculate the material removal rate and time required to drill that hole on the block Given N 800 rpm f 01 mmrev D 25 mm h 20 mm Material Removal ... View the full answer