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.
Answer:
You need a 120V to 24V commercial transformer (transformer 1:5), a 100 ohms resistance, a 1.5 K ohms resistance and a diode with a minimum forward current of 20 mA (could be 1N4148)
Step by step design:
- Because you have a 120V AC voltage supply you need an efficient way to reduce that voltage as much as possible before passing to the rectifier, for that I recommend a standard 120V to 24V transformer. 120 Vrms = 85 V and 24 Vrms = 17V = Vin
- Because 17V is not 15V you still need a voltage divider to step down that voltage, for that we use R1 = 100Ω and R2 = 1.3KΩ. You need to remember that more than 1 V is going to be in the diode, so for our calculation we need to consider it. Vf = (V*R2)/(R1+R2), V = Vin - 1 = 17-1 = 16V and Vf = 15, Choosing a fix resistance R1 = 100Ω and solving the equation we find R2 = 1.5KΩ
- Finally to select the diode you need to calculate two times the maximum current and that would be the forward current (If) of your diode. Imax = Vf/R2 = 10mA and If = 2*Imax = 20mA
Our circuit meet the average voltage (Va) specification:
Va = (15)/(pi) = 4.77V considering the diode voltage or 3.77V without considering it
Answer:
a. The very first liquid process, when heated from 1250 degree Celsius, is expected to form at the temperature by which the vertical line crosses the phase boundary (a -(a + L)) which is about <em>1310 degree Celsius. </em>
b. The structure of that first liquid is identified by the intersection with ((a+ L)-L) phase boundary; <em>47wt %of Ni</em> is of a tie line formed across the (a+ L) phase area <em>at 1310 degrees.</em>
c. To find the alloy's full melting, it is determined that the intersection of the same vertical line at 60 wt percent Ni with (a -(a+L)) phase boundary is around <em>1350 degrees.</em>
c. The structure of the last remaining solid before full melting correlates to the intersection with the phase boundary (a -(a + L), of the tie line built at 1350 degrees across the (a + L) phase area, <em>being 72wt % of Ni.</em>
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