Determine the constant speed at which the cable at A must be drawn in by the motor in order to hoist the load 6 m in 1.5s
1 answer:
Answer:
4m/s
Explanation:
We know that power supplied by the motor should be equal to the rate at which energy is increased of the mass that is to be hoisted
Mathematically
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We also know that Power = force x velocity ..................(i)
The force supplied by the motor should be equal to the weight (mg) of the block since we lift the against a force equal to weight of load
=> power = mg x Velocity........(ii)
While hoisting the load at at constant speed only the potential energy of the mass increases
Thus Potential energy = Mass x g x H...................(iii)
where
g = accleration due to gravity (9.81m/s2)
H = Height to which the load is hoisted
Equating equations (ii) and (iii) we get
m x g x v =
thus we get v = H/t
Applying values we get
v = 6/1.5 = 4m/s
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Answer:
a) Tբ = 151.8°C
b) ΔV = - 0.194 m³
c) The T-V diagram is sketched in the image attached.
Explanation:
Using steam tables,
At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.
Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C
b) The volume change
Using data from A-5 and A-6 of the steam tables,
The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume
(αբ) (which is calculated from the final quality and the consituents of the specific volumes).
ΔV = m(αբ - αᵢ)
αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)
q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg
αբ = 0.00109 + 0.5(0.3748 - 0.00109)
αբ = 0.187945 m³/kg
αᵢ = 0.5226 m³/kg
ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³
c) The T-V diagram is sketched in the image attached
