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Ber [7]
3 years ago
15

An electrical current of 700 A flows through a stainlesssteel cable having a diameter of 5 mm and an electricalresistance of 610

4/m (i.e., per meter of cablelength). The cable is in an environment having a tem-perature of 30C, and the total coefficient associatedwith convection and radiation between the cable andthe environment is approximately 25 W/m2K.(a) If the cable is bare, what is its surface temperature
Engineering
1 answer:
KatRina [158]3 years ago
6 0

Answer:

778.4°C

Explanation:

I = 700

R = 6x10⁻⁴

we first calculate the rate of heat that is being transferred by the current

q = I²R

q = 700²(6x10⁻⁴)

= 490000x0.0006

= 294 W/M

we calculate the surface temperature

Ts = T∞ + \frac{q}{h\pi Di}

Ts = 30+\frac{294}{25*\frac{22}{7}*\frac{5}{1000}  }

Ts=30+\frac{294}{0.3928} \\

Ts =30+748.4\\Ts = 778.4

The surface temperature is therefore 778.4°C if the cable is bare

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The entire system of components that produces power and transmits it to the road is called the vehicle's _____.
IrinaK [193]

Answer:

Powertrain

Explanation:

6 0
3 years ago
A cold air standard gas turbine engine with a thermal efficiency of 56.9 % has a minimum pressure of 100 kP
Aleks04 [339]

Answer:

a) 5.2 kPa

b) 49.3%

Explanation:

Given data:

Thermal efficiency ( л ) = 56.9% = 0.569

minimum pressure ( P1 ) = 100 kpa

<u>a) Determine the pressure at inlet to expansion process</u>

P2 = ?

r = 1.4

efficiency = 1 - [ 1 / (rp)\frac{r-1}{r} ]

   0.569   = 1 - [ 1 / (rp)^0.4/1.4

1 - 0.569  = 1 / (rp)^0.285

∴ (rp)^0.285 = 0.431

rp = 0.0522

note : rp = P2 / P1

therefore P2 = rp * P1 = 0.0522 * 100 kpa

                                   = 5.2 kPa  

b) Thermal efficiency

Л = 1 - [ 1 / ( 10.9 )^0.285 ]

   = 0.493 = 49.3%

4 0
3 years ago
A cylinder with a 6.0 in. diameter and 12.0 in. length is put under a compres-sive load of 150 kips. The modulus of elasticity f
jeka94

Answer:

Final Length = 11.992 in

Final Diameter = 6.001 in

Explanation:

First we calculate the cross-sectional area:

Area = A = πr² = π(3 in)² = 28.3 in²

Now, we calculate the stress:

Stress = Compressive Load/Area

Stress = - 150 kips/28.3 in²

Stress = -5.3 ksi

Now,

Modulus of Elasticity = Stress/Longitudinal Strain

8000 ksi = -5.3 ksi/Longitudinal Strain

Longitudinal Strain = -6.63 x 10⁻⁴

but,

Longitudinal Strain = (Final Length - Initial Length)/Initial Length

-6.63 x 10⁻⁴ = (Final Length - 12 in)/12 in

Final Length = (-6.63 x 10⁻⁴)(12 in) + 12 in

<u>Final Length = 11.992 in</u>

we know that:

Poisson's Ratio = - Lateral Strain/Longitudinal Strain

0.35 = - Lateral Strain/(- 6.63 x 10⁻⁴)

Lateral Strain = (0.35)(6.63 x 10⁻⁴)

Lateral Strain = 2.32 x 10⁻⁴

but,

Lateral Strain = (Final Diameter - Initial Diameter)/Initial Diameter

2.32 x 10⁻⁴ = (Final Diameter - 6 in)/6 in

Final Diameter = (2.32 x 10⁻⁴)(6 in) + 6 in

<u>Final Diameter = 6.001 in</u>

8 0
3 years ago
The one end of a hollow square bar whose side is (10+N/100) in with (1+N/100) in thickness is under a tensile stress 102,500 psi
netineya [11]

Answer:

The one end of a hollow square bar whose side is (10+N/100) in wit

Explanation:

3 0
3 years ago
Name the seven physical qualities for which standards have been developed.
SIZIF [17.4K]

Answer:

HUMAN DEVELOPMENT

MOTOR BEHAVIOR

EXERCISE SCIENCE

MEASUREMENT AND EVALUATION

HISTORY AND PHILOSOPHY

UNIQUE ATTRIBUTES OF LEARNERS

CURRICULUM THEORY AND DEVELOPMENT

Explanation:

6 0
4 years ago
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