What is the most important task of a spark plug?
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) Tբ = 151.8°C
b) ΔV = - 0.194 m³
c) The T-V diagram is sketched in the image attached.
Explanation:
Using steam tables,
At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.
Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C
b) The volume change
Using data from A-5 and A-6 of the steam tables,
The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume
(αբ) (which is calculated from the final quality and the consituents of the specific volumes).
ΔV = m(αբ - αᵢ)
αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)
q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg
αբ = 0.00109 + 0.5(0.3748 - 0.00109)
αբ = 0.187945 m³/kg
αᵢ = 0.5226 m³/kg
ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³
c) The T-V diagram is sketched in the image attached
Answer:
a)Δs = 834 mm
b)V=1122 mm/s
Explanation:
Given that
a)
When t= 2 s
s= 114 mm
At t= 4 s
s= 948 mm
So the displacement between 2 s to 4 s
Δs = 948 - 114 mm
Δs = 834 mm
b)
We know that velocity V
At t= 5 s
V=1122 mm/s
We know that acceleration a
a= 90 t
a = 90 x 5