Briefly discuss if it would be better to operate with pumps in parallel or series and how your answer would change as the steepn
1 answer:
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.
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Answer:
110 m or 11,000 cm
Explanation:
- let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
- let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
- let heat capacity rate for cold and hot fluid = C<em>c</em> and C<em>h </em>respectively
M<em>c</em> = 1.2 kg/s and M<em>h = </em>2 kg/s
C<em>pc</em> = 4.18 kj/kg °c and C<em>ph</em> = 4.31 kj/kg °c
<u>Using effectiveness-NUT method</u>
- <em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>
C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c
C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c
From the result above cold fluid heat capacity rate is smaller
Dimensionless heat capacity rate, C = minimum capacity/maximum capacity
C= C<em>min</em>/C<em>max</em>
C = 5.016/8.62 = 0.582
.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>
Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)
Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW
.<em>3 Third, we determine the actual heat transfer rate, Q</em>
Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)
Q = (5.016 kW/°c)(80 - 20) °c
Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW
.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε
ε<em> </em>= Q/Qmax
ε <em>= </em>303.66 kW/702.24 kW
ε = 0.432
.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>
NTU =
NTU =
NTU = 0.661
<em>.6 sixth, we determine Heat Exchanger surface area, As</em>
From the question, the overall heat transfer coefficient U = 640 W/m²
As =
As =
As = 5.18 m²
<em>.7 Finally, we determine the length of the heat exchanger, L</em>
L =
L =
L= 109.91 m
L ≅ 110 m = 11,000 cm
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
