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Romashka [77]
3 years ago
14

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s)

with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.
Required:

a. Find the initial DO deficit (Do) after mixing.
b. What is the reaeration constant after mixing?
c. What is the critical time for this system?
d. What is the downstream distance fitting the critical time?
Engineering
1 answer:
4vir4ik [10]3 years ago
8 0

A wastewater plant discharges a treated effluent (w) with a flow rate of 1.1 m^3/s, 50 mg/L BOD5 and 2 mg/L DO into a river (s) with a flow rate of 8.7 m^3/s, 6 mg/L BOD5 and 8.3 mg/L DO. Both streams are at 20°C. After mixing, the river is 3 meters deep and flowing at a velocity of 0.50 m/s. DOsat for this river is 9.0 mg/L. The deoxygenation constant is kd= 0.20 d^-1 and The reaction rate constant k at 20 °C is 0.27 d^-1.

The answer therefore would be the number 0.27 divided by two and then square while getting the square you would make it a binomial.

I wont give the answer but the steps

Your Welcome

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An AX ceramic compound has the rock salt crystal structure. If the radii of the A and X ions are 0.137 and 0.241 nm, respectivel
Tju [1.3M]

Answer:

c) 1.75 g/cm³

Explanation:

Given that

Radii of the A ion, r(c) = 0.137 nm

Radii of the X ion, r(a) = 0.241 nm

Atomic weight of the A ion, A(c) = 22.7 g/mol

Atomic weight of the X ion, A(a) = 91.4 g/mol

Avogadro's number, N = 6.02*10^23 per mol

Solution is attached below

3 0
2 years ago
For some metal alloy, a true stress of 345 MPa (50040 psi) produces a plastic true strain of 0.02. How much will a specimen of t
Strike441 [17]

Answer:

the elongation of the metal alloy is 21.998 mm

Explanation:

Given the data in the question;

K = σT/ (εT)ⁿ

given that metal alloy true stress σT = 345 Mpa, plastic true strain εT = 0.02,

strain-hardening exponent n = 0.22

we substitute

K = 345 / 0.02^{0.22

K = 815.8165 Mpa

next, we determine the true strain

(εT) = (σT/ K)^1/n

given that σT = 412 MPa

we substitute

(εT) = (412 / 815.8165 )^(1/0.22)

(εT) = 0.04481 mm

Now, we calculate the instantaneous length

l_i = l_0e^{ET

given that l_0 = 480 mm

we substitute

l_i =480mm × e^{0.04481

l_i =  501.998 mm

Now we find the elongation;

Elongation = l_i - l_0

we substitute

Elongation = 501.998 mm - 480 mm

Elongation = 21.998 mm

Therefore, the elongation of the metal alloy is 21.998 mm

6 0
2 years ago
A heat pump cycle is used to maintain the interior of a building at 15°C. At steady state, the heat pump receives energy by heat
Hoochie [10]

Answer:

a) Ql=33120000 kJ

b) COP = 5.6

c) COPreversible= 29.3

Explanation:

a) of the attached figure we have:

HP is heat pump, W is the work supplied, Th is the higher temperature, Tl is the low temperature, Ql is heat supplied and Qh is the heat rejected. The worj is:

W=Qh-Ql

Ql=Qh-W

where W=2000 kWh

Qh=120000 kJ/h

Q_{l}=14days(\frac{24 h}{1 day})(\frac{120000 kJ}{1 h})-2000 kWh(\frac{3600 s}{1 h})=33120000 kJ

b) The coefficient of performance is:

COP=\frac{Q_{h} }{W}=\frac{120000 kJ/h*14(\frac{24 h}{1 day}) }{2000 kWh(\frac{3600 s}{1 h}) } = 5.6

c) The coefficient of performance of a reversible heat pump is:

COP_{reversible}=\frac{T_{h} }{T_{h}-T_{l}  }

Th=20+273=293 K

Tl=10+273=283K

Replacing:

COP_{reversible}=\frac{293}{293-283}=29.3

4 0
3 years ago
A 15-ft beam weighing 570 lb is lowered by means of two cables unwinding from overhead cranes. As the beam approaches the ground
7nadin3 [17]

Answer:

I. Tension (cable A) ≈ 6939 lbf

II. Tension (cable B) ≈ 17199 lbf

Explanation:

Let's begin by listing out the data that we were given:

mass of beam (m) = 570 lb, deceleration (cable A) = -20 ft/s², deceleration (cable B) = -2 ft/s²,

g = 32.17405 ft/s²

The tension on an object is given by the product of mass of the object by gravitational force plus/minus the product of mass by acceleration.

Mathematically represented thus:

T = mg + ma

where:

T = tension, m = mass, g = gravitational force,

a = acceleration

I. For Cable A, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-20)]

T = 18339.2085 - 11400 = 6939.2085

T ≈ 6939 lbf

II. For Cable B, we have:

T = mg + ma = (570 * 32.17405) + [570 * (-2)]

T = 18339.2085 - 1140 = 17199.2085

T ≈ 17199 lbf

4 0
3 years ago
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Only put ciilant into ur radiator when the engine is cool (D)

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