Anyone know anything about Drivers Ed?
1 answer:
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Answer:
1) In series, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.
2) In parallel, the combined head and volume flow will move along the system curve from point 1 to point 3.
Explanation:
1) Pump in series:
When two or more pumps are connected in series, their resulting pump performance curve will be obtained by adding their respective heads at the same flow rate as shown in the first diagram attached.
In the first diagram, we have 3 curves namely:
- system curve
- single pump curve
- 2 pump in series curve
Also, we have points labeled 1, 2 and 3.
- Point 1 represents the point that the system operates with one pump running.
- Point 2 represents the point where the head of two identical pumps connected in series is twice the head of a single pump flowing at the same rate.
- Point 3 is the point where the system is operating when both pumps are running.
Now, since the flowrate is constant, the combined head will move from point 1 to point 2 in theory. However, practically speaking, the combined head and flow rate will move along the system curve to point 3.
2) Pump in parallel:
When two or more pumps are connected in parallel, their resulting pump performance curve will be obtained by adding their respective flow rates at same head as shown in the second diagram attached.
In the second diagram, we have 3 curves namely:
- system curve
- single pump curve
- 2 pump in series curve
Also, we have points labeled 1, 2 and 3
- Point 1 represents the point that the system operates with one pump running.
- Point 2 represents the point where the flow rate of two identical pumps connected in series is twice the flow rate of a single pump.
- Point 3 is the point where the system is operating when both pumps are running.
In this case, the combined head and volume flow will move along the system curve from point 1 to point 3.
Answer:
A) attached below
B) 743 KJ
C) 1.8983 KJ/K
Explanation:
A) Diagram of system schematic and set up states
attached below
<u>B) Calculate the amount of work received from the paddle wheel </u>
assuming ideal gas situation
v1 = v2 ( for a constant volume process )
work generated by paddle wheel = system internal energy
dw = mCv dT . where ; Cv = 0.743 KJ/kgk
= 5 * 0.743 * ( 500 - 300 )
= 3.715 * 200 = 743 KJ
<u>C) calculate the amount of entropy generated ( KJ/K )</u>
S2 - S1 = 1.8983 KJ/K
attached below is the detailed solution
