Answer:
overflow rate 20.53 m^3/d/m^2
Detention time 2.34 hr
weir loading 114.06 m^3/d/m
Explanation:
calculation for single clarifier
![sewag\ flow Q = \frac{12900}{2} = 6450 m^2/d](https://tex.z-dn.net/?f=sewag%5C%20%20flow%20Q%20%3D%20%5Cfrac%7B12900%7D%7B2%7D%20%3D%206450%20m%5E2%2Fd)
![surface\ area =\frac{pi}{4}\times diameter ^2 = \frac{pi}{4}\times 20^2](https://tex.z-dn.net/?f=surface%5C%20%20area%20%3D%5Cfrac%7Bpi%7D%7B4%7D%5Ctimes%20diameter%20%5E2%20%3D%20%5Cfrac%7Bpi%7D%7B4%7D%5Ctimes%2020%5E2)
![surface area = 314.16 m^2](https://tex.z-dn.net/?f=surface%20area%20%3D%20314.16%20m%5E2)
volume of tank![V = A\times side\ water\ depth](https://tex.z-dn.net/?f=%20V%20%20%3D%20A%5Ctimes%20side%5C%20water%5C%20depth)
![=314.16\times 2 = 628.32m^3](https://tex.z-dn.net/?f=%3D314.16%5Ctimes%202%20%3D%20628.32m%5E3)
![Length\ of\ weir = \pi \times diameter of weir](https://tex.z-dn.net/?f=Length%5C%20of%5C%20%20weir%20%3D%20%5Cpi%20%5Ctimes%20diameter%20of%20weir)
![= \pi \times 18 = 56.549 m](https://tex.z-dn.net/?f=%20%3D%20%5Cpi%20%5Ctimes%2018%20%3D%2056.549%20m)
overflow rate =![v_o = \frac{flow}{surface\ area} = \frac{6450}{314.16} = 20.53 m^3/d/m^2](https://tex.z-dn.net/?f=%20v_o%20%3D%20%5Cfrac%7Bflow%7D%7Bsurface%5C%20area%7D%20%3D%20%5Cfrac%7B6450%7D%7B314.16%7D%20%3D%2020.53%20m%5E3%2Fd%2Fm%5E2)
Detention time![t_d = \frac{volume}{flow} = \frac{628.32}{6450} \times 24 = 2.34 hr](https://tex.z-dn.net/?f=%20t_d%20%3D%20%5Cfrac%7Bvolume%7D%7Bflow%7D%20%3D%20%5Cfrac%7B628.32%7D%7B6450%7D%20%5Ctimes%2024%20%3D%202.34%20hr)
weir loading![= \frac{flow}{weir\ length} = \frac{6450}{56.549} = 114.06 m^3/d/m](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bflow%7D%7Bweir%5C%20length%7D%20%3D%20%5Cfrac%7B6450%7D%7B56.549%7D%20%3D%20114.06%20m%5E3%2Fd%2Fm)
Answer:
hi-he = 0
pi-pe = positive
ui-ue = negative
ti-te = negative
Explanation:
we know that fir the sub cool liquid water is
dQ = Tds = du + pdv ............1
and Tds = dh - v dP .............2
so now for process of throhling is irreversible when v is constant
then heat transfer is = 0 in irreversible process
so ds > 0
so here by equation 1 we can say
ds > 0
dv = 0 as v is constant
so that Tds = du .................3
and du > 0
ue - ui > 0
and
now by the equation 2 throttling process
here enthalpy is constant
so dh = 0
and Tds = -vdP
so ds > 0
so that -vdP > 0
as here v is constant
so -dP =P1- P2
so P1-P2 > 0
so pressure is decrease here
is the volume of the sample when the water content is 10%.
<u>Explanation:</u>
Given Data:
![V_{1}=100\ \mathrm{ft}^{3}](https://tex.z-dn.net/?f=V_%7B1%7D%3D100%5C%20%5Cmathrm%7Bft%7D%5E%7B3%7D)
First has a natural water content of 25% =
= 0.25
Shrinkage limit, ![w_{1}=12 \%=\frac{12}{100}=0.12](https://tex.z-dn.net/?f=w_%7B1%7D%3D12%20%5C%25%3D%5Cfrac%7B12%7D%7B100%7D%3D0.12)
![G_{s}=2.70](https://tex.z-dn.net/?f=G_%7Bs%7D%3D2.70)
We need to determine the volume of the sample when the water content is 10% (0.10). As we know,
![V \propto[1+e]](https://tex.z-dn.net/?f=V%20%5Cpropto%5B1%2Be%5D)
------> eq 1
![e_{1}=\frac{w_{1} \times G_{s}}{S_{r}}](https://tex.z-dn.net/?f=e_%7B1%7D%3D%5Cfrac%7Bw_%7B1%7D%20%5Ctimes%20G_%7Bs%7D%7D%7BS_%7Br%7D%7D)
The above equation is at
,
![e_{1}=w_{1} \times G_{s}](https://tex.z-dn.net/?f=e_%7B1%7D%3Dw_%7B1%7D%20%5Ctimes%20G_%7Bs%7D)
Applying the given values, we get
![e_{1}=0.25 \times 2.70=0.675](https://tex.z-dn.net/?f=e_%7B1%7D%3D0.25%20%5Ctimes%202.70%3D0.675)
Shrinkage limit is lowest water content
![e_{2}=w_{2} \times G_{s}](https://tex.z-dn.net/?f=e_%7B2%7D%3Dw_%7B2%7D%20%5Ctimes%20G_%7Bs%7D)
Applying the given values, we get
![e_{2}=0.12 \times 2.70=0.324](https://tex.z-dn.net/?f=e_%7B2%7D%3D0.12%20%5Ctimes%202.70%3D0.324)
Applying the found values in eq 1, we get
![\frac{V_{2}}{100}=\frac{1+0.324}{1+0.675}=\frac{1.324}{1.675}=0.7904](https://tex.z-dn.net/?f=%5Cfrac%7BV_%7B2%7D%7D%7B100%7D%3D%5Cfrac%7B1%2B0.324%7D%7B1%2B0.675%7D%3D%5Cfrac%7B1.324%7D%7B1.675%7D%3D0.7904)
![V_{2}=0.7904 \times 100=79\ \mathrm{ft}^{3}](https://tex.z-dn.net/?f=V_%7B2%7D%3D0.7904%20%5Ctimes%20100%3D79%5C%20%5Cmathrm%7Bft%7D%5E%7B3%7D)
Answer:
b. 10A
Explanation:
Using the formula, E= k × r×I
200= 0.5 ×2000×0.02×I
200=20×I
Dividing with 20
I = 200/20= 10A
Assumptions:
- Steady state.
- Air as working fluid.
- Ideal gas.
- Reversible process.
- Ideal Otto Cycle.
Explanation:
Otto cycle is a thermodynamic cycle widely used in automobile engines, in which an amount of gas (air) experiences changes of pressure, temperature, volume, addition of heat, and removal of heat. The cycle is composed by (following the P-V diagram):
- Intake <em>0-1</em>: the mass of working fluid is drawn into the piston at a constant pressure.
- Adiabatic compression <em>1-2</em>: the mass of working fluid is compressed isentropically from State 1 to State 2 through compression ratio (r).
![r =\frac{V_1}{V_2}](https://tex.z-dn.net/?f=r%20%3D%5Cfrac%7BV_1%7D%7BV_2%7D)
- Ignition 2-3: the volume remains constant while heat is added to the mass of gas.
- Expansion 3-4: the working fluid does work on the piston due to the high pressure within it, thus the working fluid reaches the maximum volume through the compression ratio.
![r = \frac{V_4}{V_3} = \frac{V_1}{V_2}](https://tex.z-dn.net/?f=r%20%3D%20%5Cfrac%7BV_4%7D%7BV_3%7D%20%3D%20%5Cfrac%7BV_1%7D%7BV_2%7D)
- Heat Rejection 4-1: heat is removed from the working fluid as the pressure drops instantaneously.
- Exhaust 1-0: the working fluid is vented to the atmosphere.
If the system produces enough work, the automobile and its occupants will propel. On the other hand, the efficiency of the Otto Cycle is defined as follows:
![\eta = 1-(\frac{1}{r^{\gamma - 1} } )](https://tex.z-dn.net/?f=%5Ceta%20%3D%201-%28%5Cfrac%7B1%7D%7Br%5E%7B%5Cgamma%20-%201%7D%20%7D%20%29)
where:
![\gamma = \frac{C_{p} }{C_{v}} : specific heat ratio](https://tex.z-dn.net/?f=%5Cgamma%20%3D%20%5Cfrac%7BC_%7Bp%7D%20%7D%7BC_%7Bv%7D%7D%20%3A%20specific%20heat%20ratio)
Ideal air is the working fluid, as stated before, for which its specific heat ratio can be considered constant.
![\gamma = 1.4](https://tex.z-dn.net/?f=%5Cgamma%20%3D%201.4)
Answer:
See image attached.