Answer:
Explanation:
Hello,
Let's get the data for this question before proceeding to solve the problems.
Mass of flywheel = 40kg
Speed of flywheel = 590rpm
Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.
Time = 30s = 0.5 min
During the power off, the flywheel made 230 complete revolutions.
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + ω₂) / 2] × 0.5
But ∇θ = 230 revolutions
∇θ/t = (530 + ω₂) / 2
230 / 0.5 = (530 + ω₂) / 2
Solve for ω₂
460 = 295 + 0.5ω₂
ω₂ = 330rpm
a)
ω₂ = ω₁ + αt
but α = ?
α = (ω₂ - ω₁) / t
α = (330 - 590) / 0.5
α = -260 / 0.5
α = -520rev/min
b)
ω₂ = ω₁ + αt
0 = 590 +(-520)t
520t = 590
solve for t
t = 590 / 520
t = 1.13min
60 seconds = 1min
X seconds = 1.13min
x = (60 × 1.13) / 1
x = 68seconds
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + 0) / 2] × 1.13
∇θ = 333.35 rev/min
Answer:
ωB = 300 rad/s
ωC = 600 rad/s
Explanation:
The linear velocity of the belt is the same at pulley A as it is at pulley D.
vA = vD
ωA rA = ωD rD
ωD = (rA / rD) ωA
Pulley B has the same angular velocity as pulley D.
ωB = ωD
The linear velocity of the belt is the same at pulley B as it is at pulley C.
vB = vC
ωB rB = ωC rC
ωC = (rB / rC) ωB
Given:
ω₀A = 40 rad/s
αA = 20 rad/s²
t = 3 s
Find: ωA
ω = αt + ω₀
ωA = (20 rad/s²) (3 s) + 40 rad/s
ωA = 100 rad/s
ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s
ωB = ωD = 300 rad/s
ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s
There are two different processes here:
1) we must add heat in order to bring the temperature of the water from

to

(the temperature at which the water evaporates)
2) other heat must be added to make the water evaporates
1) The heat needed for process 1) is

where

is the water mass

is the water specific heat

is the variation of temperature of the water
If we plug the numbers into the equation, we find

2) The heat needed for process 2) is

where

is the water mass

is the latent heat of evaporation of water
If we plug the numbers into the equation, we find

So, the total heat needed for the whole process is