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Zina [86]
3 years ago
14

A town is designing a rectangular, 4m deep settling tank for treating surface water intake. The tank will have a flow velocity o

f 5 cm/s. The untreated water has particles with the following properties: d=0.025 mm and density of 2,650 kg m-3 The settler is placed following a coagulation-flocculation step which is able to do one of the following: a) Triple the diameter of the particles but have no impact on the particle density. b) Triple the density of the particles but have no impact on their diameter. 1. Which of the options would you select (a or b)? (Please explain your selection in the next question) 2. According to your selection, calculate the length of the settler (in m) required to remove the particles assuming they reach terminal velocity.
Engineering
1 answer:
Oksana_A [137]3 years ago
7 0

Answer:

333335790089

Explanation:

blh

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Which is the correct order for handwashing
cupoosta [38]

Answer:

Follow these five steps every time.

1.Wet your hands with clean, running water (warm or cold), turn off the tap, and apply soap.

2.Lather your hands by rubbing them together with the soap. Lather the backs of your hands, between your fingers, and under your nails.

3.Scrub your hands for at least 20 seconds. Need a timer? Hum the “Happy Birthday” song from beginning to end twice.

4.Rinse your hands well under clean, running water.

5.Dry your hands using a clean towel or air dry them.

8 0
3 years ago
Read 2 more answers
Steam enters a well-insulated turbine operating at steady state at 4 MPa with a specific enthalpy of 3015.4 kJ/kg and a velocity
Sedbober [7]

Answer:

attached below

Explanation:

6 0
4 years ago
Technician A says that 18 gauge AWG wire can carry more current flow that 12 gauge AWG wire. Technician B says that metric wire
denpristay [2]

Answer:

Technician B

Explanation:

Both AWG and metric are sized by cross-sectional area.

Technician A is wrong:  12 gauge wire is larger diameter rated for 20 amps in free air.  18 awg is smaller diameter and typically used for speaker wiring, Class II or low voltage and sub-circuits within appliances.

6 0
4 years ago
An insulated piston-cylinder device contains 0.15 of saturated refrigerant-134a vapor at 0.8 MPa pressure. The refrigerant is no
Stells [14]

Answer:

Assumption:

1. The kinetic and potential energy changes are negligible

2. The cylinder is well insulated and thus heat transfer is negligible.

3. The thermal energy stored in the cylinder itself is negligible.

4. The process is stated to be reversible

Analysis:

a. This is reversible adiabatic(i.e isentropic) process and thus s_{1} =s_{2}

From the refrigerant table A11-A13

P_{1} =0.8MPa   \left \{ {{ {{v_{1}=v_{g}  @0.8MPa =0.025645 m^{3/}/kg } } \atop { {{u_{1}=u_{g}  @0.8MPa =246.82 kJ/kg } -   also  {{s_{1}=s_{g}  @0.8MPa =0.91853 kJ/kgK } } \right.

sat vapor

m=\frac{V}{v_{1} } =\frac{0.15}{0.025645} =5.8491 kg\\and \\\\P_{2} =0.2MPa  \left \{ {{x_{2} =\frac{s_{2} -s_{f} }{s_{fg }}=\frac{0.91853-0.15449}{0.78339}   = 0.9753 \atop {u_{2} =u_{f} +x_{2} }(u_{fg}) =  38.26+0.9753(186.25)= 38.26+181.65 =219.9kJ/kg \right. \\s_{1} = s_{2}

T_{2} =T_{sat @ 0.2MPa} = -10.09^{o}  C

b.) We take the content of the cylinder as the sysytem.

This is a closed system since no mass leaves or enters.

Hence, the energy balance for adiabatic closed system can be expressed as:

E_{in} - E_{out}  =ΔE

w_{b, out}  =ΔU

w_{b, out} =m([tex]u_{1} -u_{2)

w_{b, out}  = workdone during the isentropic process

=5.8491(246.82-219.9)

=5.8491(26.91)

=157.3993

=157.4kJ

4 0
3 years ago
Risks are Not Perceived Differently from What is Happening<br> False<br> True
Gnoma [55]

Answer:

False......................

8 0
3 years ago
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