Answer:
- the capacity of the pump reduces by 35%.
- the head gets reduced by 57%.
the power consumption by the pump is reduced by 72%
Explanation:
the pump capacity is related to the speed as speed is reduces by 35%
so new speed is (100 - 35) = 65% of orginal speed
speed Q ∝ N ⇒ Q1/Q2 = N1/N2
Q2 = (N2/N1)Q1
Q2 = (65/100)Q1
which means that the capacity of the pump is also reduces by 35%.
the head in a pump is related by
H ∝ N² ⇒ H1/H2 = N1²/N2²
H2 = (N2N1)²H1
H2 = (65/100)²H1 = 0.4225H1
so the head gets reduced by 1 - 0.4225 = 0.5775 which is 57%.
Now The power requirement of a pump is related as
P ∝ N³ ⇒ P1/P2 = N1³/N2³
P2 = (N2/N1)³P1
H2 = (65/100)²P1 = 0.274P1
So the reduction in power is 1 - 0.274 = 0.725 which is 72%
Therefore for a reduction of 35% of speed there is a reduction of 72% of the power consumption by the pump.
The power that must be supplied to the motor is 136 hp
<u>Explanation:</u>
Given-
weight of the elevator, m = 1000 lb
Force on the table, F = 500 lb
Distance, s = 27 ft
Efficiency, ε = 0.65
Power = ?
According to the equation of motion:
F = ma

a = 16.1 ft/s²
We know,

To calculate the output power:
Pout = F. v
Pout = 3 (500) * 29.48
Pout = 44220 lb.ft/s
As efficiency is given and output power is known, we can calculate the input power.
ε = Pout / Pin
0.65 = 44220 / Pin
Pin = 68030.8 lb.ft/s
Pin = 68030.8 / 500 hp
= 136 hp
Therefore, the power that must be supplied to the motor is 136 hp
Answer:
A wheelbarrow, a bottle opener, and an oar are examples of second class levers