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:
increases by a factor of 6.
Explanation:
Let us assume that the initial cross sectional area of the pipe is A m² while the initial velocity of the water is V m/s², hence the flow rate of the water is:
Initial flow rate = area * velocity = A * V = AV m³/s
The water speed doubles (2V m/s) and the cross-sectional area of the pipe triples (3A m²), hence the volume flow rate becomes:
Final flow rate = 2V * 3A = 6AV m³/s = 6 * initial flow rate
Hence, the volume flow rate of the water passing through it increases by a factor of 6.
Answer:
The maximum theoretical height that the pump can be placed above liquid level is 
Explanation:
To pump the water, we need to avoid cavitation. Cavitation is a phenomenon in which liquid experiences a phase transition into the vapour phase because pressure drops below the liquid's vapour pressure at that temperature. As a liquid is pumped upwards, it's pressure drops. to see why, let's look at Bernoulli's equation:
(
stands here for density, 
for height)
Now, we are assuming that there aren't friction losses here. If we assume further that the fluid is pumped out at a very small rate, the velocity term would be negligible, and we get:
This means that pressure drop is proportional to the suction lift's height.
We want the pressure drop to be small enough for the fluid's pressure to be always above vapour pressure, in the extreme the fluid's pressure will be almost equal to vapour pressure.
That means:
We insert that into our last equation and get:
And that is the absolute highest height that the pump could bear. This, assuming that there isn't friction on the suction pipe's walls, in reality the height might be much less, depending on the system's pipes and pump.
Answer:
4. A series of steps engineers use to solve problems.
Explanation:
The process of engineering design is a sequence of procedures that engineers pursue to arrive at a solution to a specific problem. Most times the solution includes creating a product such as a computer code, which fulfills certain conditions or performs a function. If the project in-hand includes designing, constructing, and testing it, then engineers probably adopt the design process. Steps of the process include defining the problem, doing background research, specifying requirements, brainstorming solutions, etc.
