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4 years ago

What is the most advanced engineering school?

Engineering

2 answers:

MA_775_DIABLO [31]4 years ago

5 0

I don't really now.... I think its the Massachusetts Institute of Technology (MIT)

IamSugarBee

Elden [556K]4 years ago

4 0

Answer:

Massachusetts Institute of Technology (MIT) , Stanford University, University of California, Berkeley (UCB)

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For a bronze alloy, the stress at which plastic deformation begins is 297 MPa and the modulus of elasticity is 113 GPa. (a) What

Alenkinab [10]

Answer:

a) 93.852 kN

b) 128.043 mm

Explanation:

Stress is load over section:

σ = P / A

If plastic deformation begins with a stress of 297 MPa, the maximum load before plastic deformation will be:

P = σ * A

316 mm^2 = 3.16*10^-4

P = 297*10^6 * 3.16*10^-4 = 93852 N = 93.852 kN

The stiffness of the specimen is:

k = E * A / l

k = 113*10^9 * 3.16*10^-4 / 0.128 = 279 MN/m

Hooke's law:

x' = x0 * (1 + P/k)

x' = 0.128 * (1 + 93.852*10^3 / 279*10^6) = 0.128043 m = 128.043 mm

5 0

3 years ago

A heat engine receives 6 kW from a 250oC source and rejects heat at 30oC. Examine each of three cases with respect to the inequa

taurus [48]

Answer:

Explanation:

Given

$T_h=250^{\circ}C\approx 523\ K$

$T_L=30^{\circ}C\approx 303\ K$

$Q_1=6 kW$

From Clausius inequality

$\oint \frac{dQ}{T}=0$ =Reversible cycle

$\oint \frac{dQ}{T}$ =Irreversible cycle

$\oint \frac{dQ}{T}>0$ =Impossible

(a)For $P_{out}=3 kW$

Rejected heat $Q_2=6-3=3\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3}{303}=1.57\times 10^{-3} kW/K$

thus it is Impossible cycle

(b) $P_{out}=2 kW$

$Q_2=6-2=4 kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{4}{303}=-1.73\times 10^{-3} kW/K$

Possible

(c)Carnot cycle

$\frac{Q_2}{Q_1}=\frac{T_1}{T_2}$

$Q_2=3.47\ kW$

$\oint \frac{dQ}{T}= \frac{Q_1}{T_1}-\frac{Q_2}{T_2}$

$=\frac{6}{523}-\frac{3.47}{303}=0$

and maximum Work is obtained for reversible cycle when operate between same temperature limits

$P_{out}=Q_1-Q_2=6-3.47=2.53\ kW$

Thus it is possible

6 0

4 years ago

Moist air enters a duct at 10 oC, 70% relative humidity, and a volumetric flow rate of 150 m3/min. The mixture is heated as it f

AleksAgata [21]

70% of the chances you a car be a because if you see or 40° 51 5030 minutes and to 70% can be ignored to kinetic energy

7 0

3 years ago

Serratia marcescens bacteria are used for the production of threonine. The maximum specific oxygen uptake rate of S. marcescens

Sati [7]

Answer:

The rate of cell metabolism is limited by mass transfer since the value of maximum cell concentration obtained (38 g/l) is lower than 50 g l-1, the value planed.

Explanation:

Data

kLa = 0.17/s

Solubility of oxygen = 8 × 10^-3 kg / m^3

The maximum specific oxygen uptake rate = 4 mmol O2 / g h.

Concentration of oxygen = 0.5 × 10^-3 kg/ m^3

**The maximum cell density = 50 g/l

___________________

The calculated maximum cell concentration:

xmax= kLa · CAL*/ qo

CAL* is the solubility of oxygen in the broth and qo is the specific oxygen uptake rate

Replacing the data given

xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 4 mmol O2 / g h

4 mmol O2 / g h to kg O2/ g s

$4 \frac{mmol}{gh} \frac{1 gmol}{1000mmol}\frac{1h}{3600s}\frac{32 g}{gmol} \frac{1Kg}{1000g}$

= 3.56 x 10^-3 kg O2/ g s

So then,

xmax= ( 0.17/s ) · (8 × 10^-3 kg / m^3) / 3.56 x 10^-3 o kg O2/ g s

xmax= 3. 8 x 10^4 g/ m^3 = 38 g/l

_____________________

5 0

3 years ago

You should be extra careful during the hours of sunrise, sunset, and nighttime because _____. A. of increased law enforcement ac

torisob [31]

Answer:

B, its the only valid answer

5 0

4 years ago