Answer:
Connect the test light in series with the negative post, and start pulling feed wires. The first to check is the heavy charging wire from the alternator. A bad or leaky diode in an alternator is a very common source of overnight battery drain. Connect wires one at a time to see what lead is drawing current.
Solution :
Given :
Water have quality x = 0.7 (dryness fraction) at around pressure of 200 kPa
The phase diagram is provided below.
a). The phase is a standard mixture.
b). At pressure, p = 200 kPa, T =
Temperature = 120.21°C
c). Specific volume
d). Specific energy ()
e). Specific enthalpy
At
f). Enthalpy at m = 0.5 kg
= 1022.91 kJ
R01= 14.1 Ω
R02= 0.03525Ω
<h3>Calculations and Parameters</h3>
Given:
K= E2/E1 = 120/2400
= 0.5
R1= 0.1 Ω, X1= 0.22Ω
R2= 0.035Ω, X2= 0.012Ω
The equivalence resistance as referred to both primary and secondary,
R01= R1 + R2
= R1 + R2/K2
= 0.1 + (0.035/9(0.05)^2)
= 14.1 Ω
R02= R2 + R1
=R2 + K^2.R1
= 0.035 + (0.05)^2 * 0.1
= 0.03525Ω
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Answer:
Explanation:
The pictures below shows the whole explanation for the problem