Consider the rigid pressure vessel containing air in the diagram at the

Describe the greatest power in design according to Aravena?

Ann [662]

Answer: Describe the greatest power in design according to Aravena? The subject of Aravena’s recent Futuna Lecture Series in New Zealand was ‘the power of design,’ which he described as ultimately being “the power of synthesis” because, increasingly, architects are dealing with complex issues and problems.

What are the three problems with global urbanization? 1. Degraded Environmental Quality ...

2. Overcrowding ...

3. Housing Problems ...

4. Unemployment ...

5. Development of Slums...

How could you use synthesis in your life to solve problems? Hence, synthesis is often not a one-time process of solution design but is used in combination with problem understanding and solution analysis to progress towards a more complete understanding of problems and solutions over time (see Applying the Systems Approach topic for a more complete discussion of the dynamics of this aspect of the approach).

I got all three answers

4 0

2 years ago

The total solids production rate in an activated sludge aeration tank is 7240 kg/d on a dry mass basis. It is necessary to maint

snow_lady [41]

Answer:

volume of biological sludge = 28.566 m³ per day

Explanation:

given data

mass of solid = 7240 kg/day

initial moisture content = 78%

solution

here percentage of solid will be

% of solid = 100 - initial moisture content

% of solid = 100 - 78 = 22 %

so that

mass of sludge produced = $\frac{100}{100 - P} M$ kg per day

put her value

mass of sludge produced = $\frac{100}{100 - 78} 7240$ kg

mass of sludge produced = 32909.09 kg

so

specific gravity of sludge = $\frac{\rho sludge}{\rho water }$

and as we know that

$\frac{100}{S sludge} = \frac{solid percentage}{S solid} = \frac{water percentage}{S water}$

$\frac{100}{S sludge} = \frac{22}{2.5} = \frac{78}{1}$

$S sludge$ = 1.152

so that

density of sludge = S sludge × density of water

density of sludge = 1.152 × 1000

density of sludge = 1152 kg/m³

so that

volume of biological sludge = $\frac{mass sludge produce}{\rho sludge}$

volume of biological sludge = $\frac{32909.09}{1152}$

volume of biological sludge = 28.566 m³ per day

6 0

3 years ago

A 50 Hz, four pole turbo-generator rated 100 MVA, 11 kV has an inertia constant of 8.0 MJ/MVA. (a) Find the stored energy in the

raketka [301]

Given Information:

Frequency = f = 60 Hz

Complex rated power = G = 100 MVA

Intertia constant = H = 8 MJ/MVA

Mechanical power = Pmech = 80 MW

Electrical power = Pelec = 50 MW

Number of poles = P = 4

No. of cycles = 10

Required Information:

(a) stored energy = ?

(b) rotor acceleration = ?

(c) change in torque angle = ?

(c) rotor speed = ?

Answer:

(a) stored energy = 800 Mj

(b) rotor acceleration = 337.46 elec deg/s²

(c) change in torque angle (in elec deg) = 6.75 elec deg

(c) change in torque angle (in rmp/s) = 28.12 rpm/s

(c) rotor speed = 1505.62 rpm

Explanation:

(a) Find the stored energy in the rotor at synchronous speed.

The stored energy is given by

$E = G \times H$

Where G represents complex rated power and H is the inertia constant of turbo-generator.

$E = 100 \times 8 \\\\E = 800 \: MJ$

(b) If the mechanical input is suddenly raised to 80 MW for an electrical load of 50 MW, find rotor acceleration, neglecting mechanical and electrical losses.

The rotor acceleration is given by

$P_a = P_{mech} - P_{elec} = M \frac{d^2 \delta}{dt^2}$

Where M is given by

$M = \frac{E}{180 \times f}$

$M = \frac{800}{180 \times 50}$

$M = 0.0889 \: MJ \cdot s/ elec \: \: deg$

So, the rotor acceleration is

$P_a = 80 - 50 = 0.0889 \frac{d^2 \delta}{dt^2}$

$30 = 0.0889 \frac{d^2 \delta}{dt^2}$

$\frac{d^2 \delta}{dt^2} = \frac{30}{0.0889}$

$\frac{d^2 \delta}{dt^2} = 337.46 \:\: elec \: deg/s^2$

(c) If the acceleration calculated in part(b) is maintained for 10 cycles, find the change in torque angle and rotor speed in revolutions per minute at the end of this period.

The change in torque angle is given by

$\Delta \delta = \frac{1}{2} \cdot \frac{d^2 \delta}{dt^2}\cdot (t)^2$