Answer:
0.0297M^3/s
W=68.48kW
Explanation:
Hello! To solve this problem, we must first find all the thermodynamic properties at the input (state 1) and the compressor output (state 2), using the thermodynamic tables
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
state 1
X=quality=1
T=-26C
density 1=α1=5.27kg/m^3
entalpy1=h1=234.7KJ/kg
state 2
T2=70
P2=8bar=800kPa
density 2=α2=31.91kg/m^3
entalpy2=h2=306.9KJ/kg
Now to find the flow at the outlet of the compressor, we remember the continuity equation that states that the mass flow is equal to the input and output.
m1=m2
(Q1)(α1)=(Q2)(α2)
the volumetric flow rate at the exit is 0.0297M^3/s
To find the power of the compressor we use the first law of thermodynamics that says that the energy that enters must be equal to the energy that comes out, in this order of ideas we have the following equation
W=m(h2-h1)
m=Qα
W=(0.18)(5.27)(306.9-234.7)
W=68.48kW
the compressor power is 68.48kW
Answer:
The distance from the entrance at which the boundary layers meet is 0.516m
The distance from the entrance at which the thermal boundary layers meet is 1.89m
Explanation:
For explanation, look at the attached file
Answer:
The answer is
C. Split phase motor
Explanation:
Clamp meters rely on the principle of magnetic induction to make non contact AC current measurements. Electric current flowing through a wire produces a magnetic field.
Which is similar to basic mode of operation of electric motor and split phase motor is a type of electric motor.
What is a a clamp on meter?
Clamp meters are electrical testers which have wide jaws that are able to clamp around an electrical conductor. Originally designed as a single purpose tool for measuring AC current, clamp meters now include inputs for accepting test leads and other probes that support a wide range of electrical measurements, the jaws of a clamp meter permit work in tight spaces and permits current measurements on live conductors without circuit interruption.
Answer:
835,175.68W
Explanation:
Calculation to determine the required power input to the pump
First step is to calculate the power needed
Using this formula
P=V*p*g*h
Where,
P represent power
V represent Volume flow rate =0.3 m³/s
p represent brine density=1050 kg/m³
g represent gravity=9.81m/s²
h represent height=200m
Let plug in the formula
P=0.3 m³/s *1050 kg/m³*9.81m/s² *200m
P=618,030 W
Now let calculate the required power input to the pump
Using this formula
Required power input=P/μ
Where,
P represent power=618,030 W
μ represent pump efficiency=74%
Let plug in the formula
Required power input=618,030W/0.74
Required power input=835,175.68W
Therefore the required power input to the pump will be 835,175.68W
Answer:
Explanation:
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