A steam power plant with a power output of 230 MW consumes coal at a rate of 60 tons/h. If the heating value of the coal is 30,0
1 answer:
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Complete Question:
A metal plate of 400 mm in length, 200mm in width and 30 mm in depth is to be machined by orthogonal cutting using a tool of width 5mm and a depth of cut of 0.5 mm. Estimate the minimum time required to reduce the depth of the plate by 20 mm if the tool moves at 400 mm per second.
Answer:
26 mins 40 secs
Explanation:
Reduction in depth, Δd = 20 mm
Depth of cut,
Number of passes necessary for this reduction,
n = 20/0.5
n = 40 passes
Tool width, w = 5 mm
Width of metal plate, W = 200 mm
For a reduction in the depth per pass, tool will travel W/w = 200/5 = 40 times
Speed of tool, v = 100 mm/s
minimum time required to reduce the depth of the plate by 20 mm:
number of passes * Time/pass
n * Time/pass
40 * 40
1600 = 26 mins 40 secs
The question is incomplete. We have to calculate :
a). the cutting force
b). volumetric metal removal rate, MRR
c). the horsepower required at the cut
d). if the power efficiency of the machine tool is 90%, determine the motor horsepower
Solution :
Given :
Cutting velocity (v) = 500 ft/min
= 500 x 12 in/min
= 6000 in/min
Feed , f = 0.025 in/rev
Depth of cut, d = 0.2 in
b). Volumetric material removal rate, MRR = v.f.d
= 6000 x 0.025 x 0.2
= 30
c). Horsepower required = MRR x unit horsepower
= 30 x 1.6
= 48 hp
a). Cutting force,
(1 hp = 550 ft lbf /sec)
= 3168 lbf
d). Machine HP required
= 53.33 HP