Answer:
Explanation:
Given
Initial Thickness=45 mm
Final thickness=20 mm
Roll diameter=600 mm
Radius(R)=300 mm
coefficient of friction between rolls and strip ()=0.15
maximum draft
Minimum no of passes
(b)draft per each pass
Ohm's law states that the current (I) in amps equals the voltage (E) in volts decided by the resistance (R) in ohm's. If you con
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Answer: 1.98 × 10^4 N
Explanation:
Form similar triangle ADE and ABC
a/x= 2/3, a=2/3x
Width of the strip w= 2(4+a) = 8+2a
W= 8 +2 (2/3x)= 8+4/3x
Area of the strip = w Δx
(8 +4/3x) Δx
Pressure on the strip p= pgx= 10^3 ×9.81x= 9810x
But,
Force= Pressure × area= 9810x × (8+4/3x)Δx
Adding the forces and taking lim n to infinity
F total= lim n--> infinity E 9810x × (8+4/3x)Δx
Ftotal= Integral 2,0 9810x × (8+4/3x)Δx
F total= 9810 integral 2, 0 (8+4/3x)dx
= 9810(8+x^2/2 + 4/3x^3/3)2,0
=9810(4x^2 + 4/9x^3)
=9810(4x2^2 + 4/9×2^3-0)
=9810(16 + 32/9)
Hydrostatic force as an integral
Ft= 19.18 ×10^4N
Answer:
<em>Heat rejected to cold body = 3.81 kJ</em>
Explanation:
Temperature of hot thermal reservoir Th = 1600 K
Temperature of cold thermal reservoir Tc = 400 K
<em>efficiency of the Carnot's engine = 1 - </em><em> </em>
eff. of the Carnot's engine = 1 -
eff = 1 - 0.25 = 0.75
<em>efficiency of the heat engine = 70% of 0.75 = 0.525</em>
work done by heat engine = 2 kJ
<em>eff. of heat engine is gotten as = W/Q</em>
where W = work done by heat engine
Q = heat rejected by heat engine to lower temperature reservoir
from the equation, we can derive that
heat rejected Q = W/eff = 2/0.525 = <em>3.81 kJ</em>