Please help describe the impact that a toy robot has had or could have on its intended audience
2 answers:
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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:
The graphs are attached below
Explanation:
Ans) We know,
F(t) = K(t - to) + zΔ∅ ln[(1 + F(t)/zΔ∅]
where K = hydraulic conductivity =0.1 cm/hr
z = capillary suction =20cm
Δ∅ = ∅ (1 - Se)
∅ = effective porosity , Se = initial moisture content
Δ∅ = 0.40(1 - 0.15)
Δ∅ = 0.34
Now, F(t) = 0.1(t - to) + 20(0.34) ln[ 1 + F(t)/6.8]
F(t) = 0.1(t - to) + 6.8 ln [1+ 0.147 F(t)]
Also, infiltration rate (f),
f = K [(zΔ∅ + F)/F]
f = 0.40 [6.8 + F]/F
Condition of ponding,
Ponding time tp = K zΔ∅ i(i-k)
where, i = rainfall rate (1cm/hr)
tp = 0.40(6.8) / 0.60
tp= 4.53 hours
Now, cumulative infiltraton at ponding Fp = i tp
Fp = 1 x 4.53 or 4.53 cm
For infiltration at time less then ponding time , infiltration rate = rainfall rate
For t = 0.25 hr , f = 1cm/hr ; F = 0.25 x 1 = 0.25 cm
For t = 0.50 hr , f = 1cm/hr ; F = 0.50 cm
For t = 0.75 hr , f = 1cm/hr ; F = 0.75 cm
For t = 1 hr , f = 1cm/hr ; F = 1cm/hr
For t > tp ,
Equivalent time origin(to),
to = tp - 1/K [ Fp - z Δ∅ ln (1+ Fp/zΔ∅)
to = 4.53 - 1/0.40[ 4.53 - 6.8 ln(1 + 0.66)
to = 4.53 - 2.5 ( 4.53 - 3.44)
to = 1.82 hr
Hence,
F = 0.10( t - 1.82) + 6.8 ln[ 1+ 0.147F(t)]
Solving above equation for t by assuming F, and further solving equation for infiltration rate
Ans (a) Following is required curve :
O.S 1-5 2 time Chr)ホー Lt 2 time Chr)
Ans (c) Following is required table :
Ans (d) For time step t = 0.25 hrs
At t < tp ; i = f = 1 cm/hr
F = i x t
F = 1 x 0.25
F = 0.25 cm
Answer:
The total load carried by the fiber will be "98%".
Explanation:
The given values are:
,
As we know,
⇒
On putting the estimated values, we get
⇒
⇒
Now,
⇒
On putting the estimated values, we get
⇒
⇒
Therefore,
The load carried by fiber,
i.e., 98%