How would you achieve the linear convolution of a 100 sample time series and a 20 tap filter in the frequency domain?

Evaporation in Double-Effect Reverse-Feed Evaporators. A feed containing 2 wt % dissolved organic solids in water is fed to a do

Ne4ueva [31]

Answer:

472,826 lb/hr

Explanation:

As per the given data:

Solids in feed= 2%

Solids in concentrate= 25%

HTA1 = HTA2 = 1000 ft^2

U1 = 500 Btu/ h ft^2 F & U2 = 700 Btu/ h ft^2 F

Overall material balance: Feed= Distillate + concentrate ----> Eq-1

Component balance: Feed * 0.02 = Distillate * 0 + concentrate * 0.25

Feed = 12.5 * concentrate ---> Eq-2

Boiling point rise = negligible, so solution & solvent vapor temperature will be same.

Assumed that the 1st effect is operating under atmospheric pressure (Boiling point - 212F).

As per the data:

Latent heat 212F = 300 Btu/lb

Latent heat 100F = 320 Btu/lb

As per material balance:

Vapor flowrate * latent heat = Overall HT coefficient * HTA * DT

1st effect: M-1= (500 * 1000 * (326-212)) / 300 = 190,000 lb/hr

2nd effect: M-2= (700*1000 * (212-100)) / 320 = 245,000 lb/hr

Distillate = M-1 + M-2 = 190,000+245,000 = 435,000 lb/hr

Substituting the above in Eq-1

Feed = 435,000 + concentrate

Substitute Eq-2 in the above

12.5 * concentrate = 435,000 + concentrate

Concentrate, L1 = 435,000/11.5 = 37826 lb /hr

Feed, F = 435,000 + 37826 = 472,826 lb/hr

1st effect operating pressure is not given, That may be the reason we are not getting the given answer. But procedure is right.

[ Find the figure in the attachment ]

5 0

3 years ago

Air is compressed steadily from 100kPa and 20oC to 1MPa by an adiabatic compressor. If the mass flow rate of the air is 1kg/s an

igomit [66]

Answer:

(a). 575 kJ/kg.

(b). 290kw.

Explanation:

We have the following set of information or parameters from the question above;

Pressure(1) = 100kPa, Pressure (2) = 1MPa, temperature(1) = b1= 12°C = 285K = 285kJ/kg, efficiency = 80% and the mass flow rate of the air = 1kg/s.

At a temperature of 12°C, we have the value from steam table; gx= 1.2, thus gx22 = 1.2 × (1000/100) = 12.

We have that the value for b12 = 517.

Therefore, the value for h2a can be calculated as;

80/100 = (517 - 285)/ (tp at exist) - 285.

0.8 = 232/ (tp at exist) - 285.

232 = 0.8 × (tp at exist) - 285).

232 = 0.8 (tp at exist) - 228 .

(tp at exist) = 575.

Therefore, the temperature 575 kJ/kg.

Thus, the required power input of the compressor = 1kg/s × ( 575 - 285) = 290kw.

6 0

3 years ago

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

Two technicians are discussing solder wire repair. Technician A says that electrical tape can be used to cover the joint. Techni

Oksanka [162]

Answer:

8282777727e7e87e7e8e88282883

8 0

3 years ago

As shown, a load of mass 10 kg is situated on a piston of diameter D1 = 140 mm. The piston rides on a reservoir of oil of depth

telo118 [61]

Answer:

165 mm

Explanation:

The mass on the piston will apply a pressure on the oil. This is:

p = f / A

The force is the weight of the mass

f = m * a

Where a in the acceleration of gravity

A is the area of the piston

A = π/4 * D1^2

Then:

p = m * a / (π/4 * D1^2)

The height the oil will raise is the heignt of a colum that would create that same pressure at its base:

p = f / A

The weight of the column is:

f = m * a

The mass of the column is its volume multiplied by its specific gravity

m = V * S

The volume is the base are by the height

V = A * h

Then:

p = A * h * S * a / A

We cancel the areas:

p = h * S * a

Now we equate the pressures form the piston and the pil column:

m * a / (π/4 * D1^2) = h * S * a

We simplify the acceleration of gravity

m / (π/4 * D1^2) = h * S

Rearranging:

h = m / (π/4 * D1^2 * S)

Now, h is the heigth above the interface between the piston and the oil, this is at h1 = 42 mm. The total height is

h2 = h + h1

h2 = h1 + m / (π/4 * D1^2 * S)

h2 = 0.042 + 10 / (π/4 * 0.14^2 * 0.8) = 0.165 m = 165 mm

7 0

3 years ago