Answer:
BOD5 = 200 mg/L
Explanation:
given data
diluted = 1% = 0.01
time = 5 day
oxygen consumption = 2.00 mg · L−1
solution
we get here BOD5 that is BOD after 5 day
and here total volume is 100% = 1
so dilution factor is = 100
so BOD5 is
BOD5 = oxygen consumption × dilution factor
BOD5 = 2 × 100
BOD5 = 200 mg/L
Answer:
Explanation:
Given that
R=8 ft
Width= 10 ft
We know that hydro statics force given as
F=ρ g A X
ρ is the density of fluid
A projected area on vertical plane
X is distance of center mass of projected plane from free surface of water.
Here
X=8/2 ⇒X=4 ft
A=8 x 10=80
So now putting the values
F=ρ g A X
F=62.4(32.14)(80)(4)
Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
Answer:
A) energy loss E = pgQtH
Where p = density in kg/m3
g = gravity acceleration in m/s2
Q = flow rate in m3/s
t = time taken for flow in sec
H = height of flow in m
B) power required to run pump;
P = pgQH
Explanation:
Detailed explanation and calculation is shown in the image below
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.
