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Answer:
the power that must be supplied to the motor is 101.45 hp
Explanation:
Given that,
The weight of elevator = 1000lb
The motor exerts a constant force of 500 lb on the cable
The load has been hoisted s = 15 ft starting from rest
The motor has an efficiency of e = 0.65
According to the equation of motion:
F = ma
To calculate the output power:
P(out) = F. v
P(out) = 3 (500) * 21.98
P(out) = 32970 lb.ft/s
As efficiency is given and output power is known, we can calculate the input power.
ε = P(out) / P(in)
0.65 = 32970 / P(in )
P(in) = 50,723 lb.ft/s
P(in) = 50,723 / 500 hp
= 101.45 hp
the power that must be supplied to the motor is 101.45 hp
Answer:
a) W=85.225 kW
b)
Explanation:
First, consider the energy balance for the compressor: The energy that enters to the system (W and enthalpy of the feed flow) is equal to the energy that goes out from it (Heat Q and enthalpy of the exit flow):
Consider the enthalpy data from van Wylen 6th edition, Table B.2.2. According to that, ,
So, the power input to the compressor is:
b) The differential entropy change dS for a reversible heat transfer dQ at a temperature T is:
This equation can be integrated if the heat transfer surface temperature remains constant, which is the case, giving as a result: