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:
The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.
<h3>What is altitude?</h3>
Altitude or height exists as distance measurement, usually in the vertical or "up" approach, between a reference datum and a point or object. The exact meaning and reference datum change according to the context.
The MOCA exists in the lower published altitude in effect between fixes on VOR airways, off-airway routes, or route segments that satisfy obstacle support conditions for the whole route segment. This altitude also ensures acceptable navigational signal coverage only within 22 NM of a VOR.
The altitude ensures acceptable navigational signal coverage only within 22 NM of a VOR.
Therefore, the correct answer is 22 NM of a VOR.
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Answer:
I would say that it is forming.
Explanation:
Give brainliest if u can. :S
Answer:
flow ( m ) = 4.852 kg/s
Explanation:
Given:
- Inlet of Turbine
P_1 = 10 MPa
T_1 = 500 C
- Outlet of Turbine
P_2 = 10 KPa
x = 0.9
- Power output of Turbine W_out = 5 MW
Find:
Determine the mass ow rate required
Solution:
- Use steam Table A.4 to determine specific enthalpy for inlet conditions:
P_1 = 10 MPa
T_1 = 500 C ---------- > h_1 = 3375.1 KJ/kg
- Use steam Table A.6 to determine specific enthalpy for outlet conditions:
P_2 = 10 KPa -------------> h_f = 191.81 KJ/kg
x = 0.9 -------------> h_fg = 2392.1 KJ/kg
h_2 = h_f + x*h_fg
h_2 = 191.81 + 0.9*2392.1 = 2344.7 KJ/kg
- The work produced by the turbine W_out is given by first Law of thermodynamics:
W_out = flow(m) * ( h_1 - h_2 )
flow ( m ) = W_out / ( h_1 - h_2 )
- Plug in values:
flow ( m ) = 5*10^3 / ( 3375.1 - 2344.7 )
flow ( m ) = 4.852 kg/s
Answer:
a) 4.1 kw
b) 4.68 tons
c) 4.02
Explanation:
Saturated vapor enters compressor at ( p1 ) = 2.6 bar
Saturated liquid exits the condenser at ( p2 ) = 12 bar
Isentropic compressor efficiency = 80%
Mass flow rate = 7 kg/min
A) Determine compressor power in KW
compressor power = m ( h2 - h1 )
= 7 / 60 ( 283.71 - 248.545 )
= 4.1 kw
B) Determine refrigeration capacity in tons = m ( h1 - h4 )
= 7/60 ( 248.545 - 107.34 )
= 16.47 kw = 4.68 tons
C) coefficient of performance ( COP )
= Refrigeration capacity / compressor power
= 16.47 / 4.1 = 4.02
Attached below is the beginning part of the solution
