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3 years ago

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A. Calculate the concentration of a toxic (Cout) that is added to a lake with an input concentration of 100 μg/l. The flow rate

is constant at 10 m3/s. Assume that the system is at the steady state and that the toxic has a degradation rate (reactions term) of 50kg/day.
What is the removed fraction of toxic (in %)?

1 answer:

Mazyrski [523]3 years ago

5 0

Answer:

Calt =42.129 mg/eit

Explanation:

See attached image

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A water contains 50.40 mg/L as CaCO3 of carbon dioxide, 190.00 mg/L as CaCO3 of Ca2 and 55.00 mg/L as CaCO3 of Mg2 . All of the

Galina-37 [17]

Answer:

Total sludge = 123426kg/d

Explanation:

The reaction is given as;

H2Co3 + Ca(OH)2 ⇆ CaCo3 + 2H20

1 1 1 2 moles

Calculating the concentration of C02, we have

Concentration of C02 = concentration of CaCo3/Molecular weight of Caco3

= 50.4/100.09

= 0.5035mol/L

Sludge of Co2 = Conc. of Co2 * Q * MW of CaCo3 *10^-6

= 0.5035 * 253.6 *10^6 * 100.09 * 10^-6

= 12780kg/d

From the equation Ca2+ + 2HCo3- + Ca(OH)2 ⇄ 2CaCo3 + 2H2O

1 mole of calcium yields 2 moles of CaCo3

Therefore, Concentration of Ca2+ = Conc. of CaCo3/Mw of CaCO3

= 190-30/100.09

=1.599mol/L

Calculating sludge of calcium:

Sludge of Ca = 2 * Conc. of ca * Q * mw of CaCO3 * 10^-6

= 2 * 1.599 *253.6*10^6* 100.09 * 10^-6

= 811742kg/d

From the equation,

Mg2+ +2HCO3- + Ca(OH)2 ⇄ MgCO3 + 2CaCO3 + 2H2O

1 mole of mg yields 2 moles CaCO3 and 1 mole of Mg(OH)2

Concentration of Mg2+ = Conc, of CaCO3 /Mw of CaCo3

= 55- 10/100.09

= 0.4496mol/L

Sludge of Mg = 2 * Conc. of Mg * Q * mw of CaCO3 * 10^-6 +* Conc. of Mg * Q * mw of Mg(OH)2 * 10^-6

= 2 * 0.4496 * 253.5*10^6 * 100.09 * 10^-6 + 0.4996* 253.5*10^6 58.3 * 10^-6

= 29472kg/d

Total Sludge = Sludge of CO2 + Sludge of Ca + Sludge of Mg

12780+ 81174 + 29472

= 123426kg/d

6 0

3 years ago

Read 2 more answers

The mechanical properties of a metal may be improved by incorporating fine particles of its oxide. Given that the moduli of elas

Sedaia [141]

Answer:

a) 254.6 GPa

b) 140.86 GPa

Explanation:

a) Considering the expression of rule of mixtures for upper-bound and calculating the modulus of elasticity for upper bound;

Ec(u) = EmVm + EpVp

To calculate the volume fraction of matrix, 0.63 is substituted for Vp in the equation below,

Vm + Vp = 1

Vm = 1 - 0.63

Vm = 0.37

In the first equation,

Where

Em = 68 GPa, Ep = 380 GPa, Vm = 0.37 and Vp = 0.63,

The modulus of elasticity upper-bound is,

Ec(u) = EmVm + EpVp

Ec(u) = (68 x 0.37) + (380 x 0.63)

Ec(u) = 254.6 GPa.

b) Considering the express of rule of mixtures for lower bound;

Ec(l) = (EmEp)/(VmEp + VpEm)

Substituting values into the equation,

Ec(l) = (68 x 380)/(0.37 x 380) + (0.63 x 68)

Ec(l) = 25840/183.44

Ec(l) = 140.86 GPa

6 0

3 years ago

Consider two different types of motors. Motor A has a characteristic life of 4100 hours (based on a MTTF of 4650 hours) and a sh

Daniel [21]

Answer:B

Explanation:

Given

For motor A

Characteristic life(r)=4100 hr

MTTF=4650 hrs

shape factor(B )=0.8

For motor B

Characteristic life(r)=336 hr

MTTF=300 hr

Shape Factor (B)=3

Reliability for 100 hours

$R_a=e^{-\left ( \frac{T-r}{n}\right )B}$

$R_a=e^{-\left ( \frac{4650-4100}{100}\right )0.8}$

$R_a=e^{-4.4}=0.01227$

For B

$R_b=e^{-\left ( \frac{300-336}{100}\right )3}$

$R_b=e^{1.08}=2.944$

B is better for 100 hours

(b)For 750 hours

$R_a=e^{-0.5866}=0.55621$

$R_b=e^{0.144}=1.154$

So here B is more Reliable.

3 0

3 years ago

P9.28 A large vacuum tank, held at 60 kPa absolute, sucks sea- level standard air through a converging nozzle whose throat diame

eimsori [14]

Answer:

a) $m=0.17kg/s$

b) $Ma=0.89$

Explanation:

From the question we are told that:

Pressure $P=60kPa$

Diameter $d=3cm$

Generally at sea level

$T_0=288k\\\\\rho_0=1.225kg/m^3\\\\P_0=101350Pa\\\\r=1.4$

Generally the Power series equation for Mach number is mathematically given by

$\frac{p_0}{p}=(1+\frac{r-1}{2}Ma^2)^{\frac{r}{r-1}}$

$\frac{101350}{60*10^3}=(1+\frac{1.4-1}{2}Ma^2)^{\frac{1.4}{1.4-1}}$

$Ma=0.89$

Therefore

Mass flow rate

$\frac{\rho_0}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\frac{1.225}{\rho}=(1+\frac{1.4-1}{2}(0.89)^2)^{\frac{1.4}{1.4-1}}$

$\rho=0.848kg/m^3$

Generally the equation for Velocity at throat is mathematically given by

$V=Ma(r*T_0\sqrt{T_e}$ )

Where

$T_e=\frac{P_e}{R\rho}\\\\T_e=\frac{60*10^6}{288*0.842\rho}$

$T_e=248$

Therefore

$V=0.89(1.4*288\sqrt{248})\\\\V=284$

Generally the equation for Mass flow rate is mathematically given by

$m=\rho*A*V$

$m=0.84*\frac{\pi}{4}*3*10^{-2}*284$

$m=0.17kg/s$

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3 years ago

Build a 32-bit accumulator circuit. The circuit features a control signal inc and enable input en. If en is 1 and inc is 1, the

Alex_Xolod [135]

Sorry need.points I'm new

7 0

3 years ago