A linear frequency-modulated signal makes a good test for aliasing, because the frequency moves over a range. This signal is

A storage tank, used in a fermentation process, is to be rotationally molded from polyethylene plastic. This tank will have a co

NNADVOKAT [17]

Answer:

The volume up to cylindrical portion is approx 32355 liters.

Explanation:

The tank is shown in the attached figure below

The volume of the whole tank is is sum of the following volumes

1) Hemisphere top

Volume of hemispherical top of radius 'r' is

$V_{hem}=\frac{2}{3}\pi r^3$

2) Cylindrical Middle section

Volume of cylindrical middle portion of radius 'r' and height 'h'

$V_{cyl}=\pi r^2\cdot h$

3) Conical bottom

Volume of conical bottom of radius'r' and angle $\theta$ is

$V_{cone}=\frac{1}{3}\pi r^3\times \frac{1}{tan(\frac{\theta }{2})}$

Applying the given values we obtain the volume of the container up to cylinder is

$V=\pi 1.5^2\times 4.0+\frac{1}{3}\times \frac{\pi 1.5^{3}}{tan30}=32.355m^{3}$

Hence the capacity in liters is $V=32.355\times 1000=32355Liters$

3 0

3 years ago

A liquid stream containing 52.0 mole% benzene and the balance toluene at 20.0°C is fed to a continuous single-stage evaporator a

OLga [1]

Answer:

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

Explanation:

Given that;

liquid phase leaving the evaporator = 32.5 mole%

Equi Temp T = 99.0°C = 99 + 273.15 = 372.15 K

Now let 1 and 2 represent Benzene and Toluene respectively.

Antoine's Constant for these components are;

COMPONENETS A B C

Benzene 1 4.72583 1660.652 -1.461

Toluene 2 4.07827 1343.943 -53.773

Antoine's equation is expressed as;

Ps = 10^(A - (B/(T+C)))

Ps is in Bar and T is in Kelvin

so

P1s = 10^( 4.72583 - (1660.652/(372.15 - (-1.461)))) = 1.7617 Bar

P2s = 10^( 4.07827 - (1343.943/(372.15 - (-53.773)))) = 0.7195 Bar

now here, liquid leaving and vapor are both in equilibrium

composition of liquid leaving are;

X1 = 32.5% = 0.325

X2 = 1 - X1 = 1 - 0.325 = 0.675

Now

Raoult's Law is expressed as;

p × y1=x1 × pis for all components

So for Benzene ; p × y1=x1 × p1s ------let this be equation 1

for Toluene ; p × y2=x2 × p2s ------let this be equation 2

lets add equ 1 and 2

p × y1=x1 × p1s + p × y2=x2 × p2s

p(y1 + y2) = x1 × p1s + x2 × p2s

buy y1 + y2 = 1

therefore we substitute

p(1) = 0.325 × 1.7617 + 0.675 × 0.7195 = 1.0582 Bar

we know that 1 Bar = 750.062 mmHg

so p = 1.0582 × 750.062

p = 793.716 mmHg

Also from equation 1

p × y1=x1 × p1s

y1 = (x1 × p1s) / p

y1 = (0.325 × 1.7617) / 1.0582

y1 = 0.541

Therefore;

Operating Pressure P = 793.716 mmHg

Y_Benzene y1 = 0.541

5 0

2 years ago

Find the derivative of y = sin(ln(5x2 − 2x))

pickupchik [31]

Answer:

$y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$

Explanation:

Let $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ and we proceed to find the derivative by the following steps:

1) $y = \sin[\ln(5\cdot x^{2}-2\cdot x)]$ Given

2) $y = \sin [\ln[x\cdot (5\cdot x - 2)]]$ Distributive property

3) $y = \sin[\ln x + \ln (5\cdot x - 2 )]$ $\ln (a\cdot b) = \ln a + \ln b$

4) $y = \cos[\ln x + \ln (5\cdot x - 2)]\cdot \left(\frac{1}{x} + \frac{5}{5\cdot x-2} \right)$ $\frac{d}{dx} (\sin x) = \cos x$ / $\frac{d}{dx}(\ln x) = \frac{1}{x}$ / $\frac{d}{dx}(c\cdot x^{n}) = n\cdot c\cdot x^{n-1}$ /Rule of chain/Result

3 0

3 years ago

Answers:

frutty [35]

Answer:

the chassis ground wire

an inverter

is connected to one side of the power line

failure of a component such as the hash filter

is the hot wire

the common wire

6 0

3 years ago

A rapid sand filter has a loading rate of 8.00 m/h, surface dimensions of 10 m ´ 8 m, an effective filtration rate of 7.70 m/h.

galben [10]

Answer:

Explanation:

given data

loading rate = 8.00 m/h

filtration rate = 7.70 m/h

dimensions = 10 m × 8 m

filter cycle duration = 52 h

time = 20 min

to find out

flow rate and volume of water is used for back washing plus rinsing the filter

solution

we consider here production efficiency is 96%

so here flow rate will be

flow rate = area × rate of filtration

flow rate = 10 × 8 × 7.7

flow rate = 616 m³/h

and

we know back washing generally 3 to 5 % of total volume of water per cycle so

volume of water is = 616 × 52

volume of water is 32032 m³

and

volume of water of back washing is = 4% of 32032

volume of water of back washing is 1281.2 m³

8 0

3 years ago