Might a synchronous motor equally well be called a synchronous inductor
1 answer:
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Answer:
1791 secs ≈ 29.85 minutes
Explanation:
( Initial temperature of slab ) T1 = 300° C
temperature of water ( Ts ) = 25°C
T2 ( final temp of slab ) = 50°C
distance between slab and water jet = 25 mm
<u>Determine how long it will take to reach T2</u>
First calculate the thermal diffusivity
∝ = 50 / ( 7800 * 480 ) = 1.34 * 10^-5 m^2/s
<u>next express Temp as a function of time </u>
T( 25 mm , t ) = 50°C
next calculate the time required for the slab to reach 50°C at a distance of 25mm
attached below is the remaining part of the detailed solution
Answer:
Go to explaination for the details of the answer.
Explanation:
In order to determine the lifetime (75 years) chronic daily exposure for each individual, we have to first state the terms of our equation:
CDI = Chronic Daily Intake
C= Chemical concentration
CR= Contact Rate
EFD= Exposure Frequency and Distribution
BW= Body Weight
AT = Average Time.
Having names our variables lets create the equations that will be used to derive our answers.
Please kindly check attachment for details of the answer.
Answer:
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
Explanation:
given data
pressure p1 = 1.4 MPa = 14 bar
temperature t1 = 32°C
exit pressure = 0.08 MPa = 0.8 bar
to find out
the quality of the refrigerant exiting the expansion valve
solution
we know here refrigerant undergoes at throtting process so
h1 = h2
so by table A 14 at p1 = 14 bar
t1 ≤ Tsat
so we use equation here that is
h1 = hf(t1) = 332.17 kJ/kg
this value we get from table A13
so as h1 = h2
h1 = h(f2) + x(2) * h(fg2)
so
exit quality =
exit quality =
so exit quality = 0.2337 = 23.37 %
the quality of the refrigerant exiting the expansion valve is 0.2337 = 23.37 %
