The answer is b, I hope this helps you
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:
Eutectic product in Fe-C system is called Ledeburite-C.
Answer:
<em> - 14.943 W/m^2K ( negative sign indicates cooling ) </em>
Explanation:
Given data:
Area of FPC = 4 m^2
temp of water = 60°C
flow rate = 0.06 l/s
ambient temperature = 8°C
exit temperature = 49°C
<u>Calculate the overall heat loss coefficient </u>
Note : heat lost by water = heat loss through convection
m*Cp*dT = h*A * ( T - To )
∴ dT / T - To = h*A / m*Cp ( integrate the relation )
In ( 
) = h* 4 / ( 0.06 * 10^-3 * 1000 * 4180 )
In ( 41 / 52 ) = 0.0159*h
hence h = - 0.2376 / 0.0159
= - 14.943 W/m^2K ( heat loss coefficient )
