Answer:
The question is incomplete, below is the complete question
"The real power delivered by a source to two impedance, Z1=4+j5Ω and Z2=10Ω connected in parallel, is 1000 W. Determine (a) the real power absorbed by each of the impedances and (b) the source current."
answer:
a. 615W, 384.4W
b. 17.4A
Explanation:
To determine the real power absorbed by the impedance, we need to find first the equivalent admittance for each impedance.
recall that the symbol for admittance is Y and express as
Hence for each we have,
for the second impedance we have
we also determine the voltage cross the impedance,
P=V^2(Y1 +Y2)
The real power in the impedance is calculated as
for the second impedance
b. We determine the equivalent admittance
We convert the equivalent admittance back into the polar form
the source current flows is
Answer:
mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Explanation:
The properties of water at 1 atm and thus at the saturation temperature of 100C are hfg =2256.4 kJ/kg (Table A-4). The net rate of heat transfer to the water is ˙Q=0 . 60×3 kW=1 . 8 kWNoting that it takes 2256.4 kJ of energy to vaporize 1 kg of saturated liquid water, therate of evaporation of water is determined to be mevaporation=˙Qhfg=1. 8 kJ /s2269. 6 kJ /kg=0 . 793×10−3kg/ s=2. 855 kg /h
Answer:
Calculate the individual residential lifetime cancer risk.
Risk = Potential factor x CDI …… (1)
Calculate the value of C
C = 
t = d / v
t = 150 miles / 2 mile per hrs.
t = 75 hrs
t = 75/24
t = 3.13 days
Substitute the obtained value in (2).
C =
C = 0.72 x e^(-0.1*3.13)
C = 0.72 x 0.7313
C = 0.526 mg/L
Substitute the obtained value in (1).
Risk = Potential factor x CDI
Risk = 0.30kg.d/mg x 0.526mg/L x 2L/d x 350day/365days
Risk = 0.3026
Answer:
In engineering design, failure is expected. It helps you find the best solutions before implementing them in the “real world”. Having a prototype fail is a GOOD thing, because that means you have learned something new about the problem and potential solutions.
Explanation:
