Answer:
There are six conditions
1. Poles should contain some residual flux.
2. Field and armature winding must be correctly connected so that initial mmm adds residual flux.
3. Resistance of field winding must be less than critical resistance.
4. Speed of prime mover of generator must be above critical speed.
5. Generator must be on load.
6. Brushes must have proper contact with commutators.
Explanation:
Answer:
The diameter is 50mm
Explanation:
The answer is in two stages. At first the torque (or twisting moment) acting on the shaft and needed to transmit the power needs to be calculated. Then the diameter of the shaft can be obtained using another equation that involves the torque obtained above.
T=(P×60)/(2×pi×N)
T is the Torque
P is the the power to be transmitted by the shaft; 40kW or 40×10³W
pi=3.142
N is the speed of the shaft; 250rpm
T=(40×10³×60)/(2×3.142×250)
T=1527.689Nm
Diameter of a shaft can be obtained from the formula
T=(pi × SS ×d³)/16
Where
SS is the allowable shear stress; 70MPa or 70×10⁶Pa
d is the diameter of the shaft
Making d the subject of the formula
d= cubroot[(T×16)/(pi×SS)]
d=cubroot[(1527.689×16)/(3.142×70×10⁶)]
d=0.04808m or 48.1mm approx 50mm
Answer:
The amperage draw of the condensing unit will be low.
Explanation:
A condensing unit is made up of a compressor and condenser, while an evaporating unit is made up of an evaporator coil.
A split AC system is a type of air conditioner system that has a condensing unit which is placed separately from the evaporative coil unit. Then the two units are connected to each other via a copper tube containing refrigerants.
The liquid line connects the condenser to the evaporator, and if this liquid line is restricted, the amp consumed by the condensing unit will be low.
Architects must have a professional bachelor's or master's degree in architecture from a program that has been accredited by the National Architectural Accrediting Board, and a state license.
Answer:
Explanation:
(c). looking for the radiation of the collector is given thus
C = 0.095 + 0.04 sin [360/365(n-100)] = 0.095 + 0.04 sin [360/365(1-100)]
C = 0.05535
∴ Diffuse radiation of the collector Idc = C*Ib + (1+cosσ/2)
Idc = 0.5535 * 908.7 (1+cos40/ 2) = 44.41 W/m²
Idc = 44.41 W/m²
