11.A heat engine operates between two reservoirs at 800 and 20°C. One-half of the work output of the heat engine is used to driv
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Answer:
9500 kJ; 9000 Btu
Explanation:
Data:
m = 100 lb
T₁ = 25 °C
T₂ = 75 °C
Calculations:
1. Energy in kilojoules
ΔT = 75 °C - 25 °C = 50 °C = 50 K
2. Energy in British thermal units
The omitted part of the question shown in bold format
The lead of the threaded shaft of the C-clamp is 0.05 in., and the mean radius of the thread is r = 0.15 in. Assume μs = 0.18 and μk = 0.16. What couple must be applied to the shaft to exert a 30-lb force on the clamped object?
Answer:
C = 0.967 in. lb
Explanation:
Given that:
The lead of the threaded shaft of the C-clamp is 0.05 in.
∴ the pitch of the screw = 0.05 in
the mean radius of the thread is r = 0.15 in.
Assuming:
(μs)= 0.18 which implies the coefficient of the static friction
(μk) = 0.16 (coefficient of kinetic friction)
Force = 30-lb
What couple must be applied to the shaft to exert a 30-lb force on the clamped object?
To determine the Couple (C) that must be applied; we use the expression:
C = Fr tan ( +
)
where; F = force
r = mean radius
= angle of kinetic friction
= pitch angle
NOW, let's take then one after the other.
From the coefficient of the static friction and the kinetic friction; we can solve for their respective angles, so we have:
Angle of static friction ()
() =
() = 10.204°
Angle of kinetic friction ()
= 9.0903°
To determine the pitch angle(); we apply the expression:
() =
() =
() =
() = 3.0368°
Have gotten our parameters to solve for Couple (C); then we have:
C = Fr tan ( +
)
substituting our values; we have:
C = (30 × 0.15) tan ( 9.0903 + 3.0368)
C = 4.5 × tan ( 12.1271)
C = 4.5 × 0.2148761968
C = 0.96669428854 in.lb
C = 0.967 in. lb
Therefore, 0.967 in. lb couple must be applied to the shaft to exert a 30-lb force on the clamped object.