Answer:
The time necessary to purge 95% of the NaOH is 0.38 h
Explanation:
Given:
vfpure water(i) = 3 m³/h
vNaOH = 4 m³
xNaOH = 0.2
vfpure water(f) = 2 m³/h
pwater = 1000 kg/m³
pNaOH = 1220 kg/m³
The mass flow rate of the water is = 3 * 1000 = 3000 kg/h
The mass of NaOH in the solution is = 0.2 * 4 * 1220 = 976 kg
When the 95% of the NaOH is purged, thus the NaOH in outlet is = 0.95 * 976 = 927.2 kg
The volume of NaOH in outlet after time is = 927.2/1220 = 0.76 m³
The time required to purge the 95% of the NaOH is = 0.76/2 = 0.38 h
78950W the answer
Explanation:
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
A 75- kw, 3-, Y- connected, 50-Hz 440- V cylindrical synchronous motor operates at rated condition with 0.8 p.f leading. the motor efficiency excluding field and stator losses, is 95%and X=2.5ohms. calculate the mechanical power developed, the Armature current, back e.m.f, power angle and maximum or pull out torque of the motor
Answer:
Explanation:
In the diagram there three gears in which gear 1 is input gear ,gear 2 is idle gear and gear 3 is out put gear.
Lets take
All external matting gears will rotates in opposite direction with respect to each other.
So the speed of gear third can be given as follows
