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kondaur [170]
3 years ago
8

A series of experiments is conducted in which a thin plate is subjected to biaxial tension/compression, σ1, σ2 , the plane surfa

ces of the plate being traction-free (i.e. σ 3-O) Unbeknown to the experimenter, the material contains microscopic defects which can be idealized as a sparse distribution of small circular holes through the thickness of the plate. The hoop stress around the circumference of one of these holes when the plate is loaded in uniaxial tension σ is known to be σ 0-o(1-cos 2θ), where the angle θ is measured from the direction of the applied stress. Show graphically the relation that will hold at yield between the stresses ' σ2 applied to the defective plate if the Tresca criterion applies for the undamaged material (Hint: The hoop stress due to biaxial stress can be constructed by superposition. The maximum must occur at either θ = 0 or θ-m2, depending on the relative magnitude of σ, σ2 .)
Engineering
1 answer:
Andrew [12]3 years ago
3 0

Answer:

                                               

Explanation:

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Calculate the energy in kJ added to 1 ft3 of water by heating it from 70° to 200°F.
Serga [27]

Answer:

The energy in kJ is 8558.16 kJ.

Explanation:

Data presented in the problem:

Water is heated from 70 (T1) to 200 °F (T2).

Volume (V) of the water is 1 ft3.

It is required for the specific heat of water(HW), which is 1 BTU/lb°F.

First, we need to calculate the mass of water (M) presented in the process. Water density (D) is 62. 4 lb/ft3.

M = V*D = (1ft3)*(62.4 lb/ft3) = 62.4 lb Water.

.After that, we can calculate the heat required (Q).

Q = M*HW*(T2 - T1) = (62.4 lb)*(1 BTU/lb°F)(200 °F - 70°F)

Q = 62.4 * 130 BTU = 8112 BTU.

Q is converted to kJ units using the conversion factor 1 BTU = 1.055 kJ

Q = 8112 BTU * (1.055 kJ/1BTU) = 8558.16 kJ.

Finally, the energy required is 8558.16 kJ.

3 0
3 years ago
The ampere draw of a 5000 watt electric heater used on 120 volts is
Salsk061 [2.6K]

Answer:

41.67amps

Explanation:

v*a=w

120*a=5000

5000/12=41.67

8 0
3 years ago
Convert the following pairs of voltage and current waveforms to phasor form. Each pair of waveforms corresponds to an unknown el
exis [7]

Answer:

a) V = 20 ∠30⁰    ,    I = 4 ∠-210⁰    Z inductive    L = 0,0125 H

b) V = 9∠-60⁰      ,    I = 4 ∠ 190⁰    Z capacitive C = 4,94 *10⁻⁴ F

c) V = 13 ∠240⁰   ,    I = 7 ∠ 150⁰    Z Inductive  L = 0,0074 H

Explanation:

a) v(t) = 20 cos (400*t + 30 )

Phasor form    V = 20 ∠30⁰

i(t) = 4 sin (400*t - 120)

First we need to transform 4sin( 400t - 120 ) as  function cosine

we know that  sin ( x + 90 )  =  cos x

Then  sin ( 400*t -120 )  = cos ( 400*t  - 120 -90 )  = cos ( 400t - 120 - 90)

Phasor form  I = 4 ∠-210⁰

To have the impedance nature we compute

Z = V / I      ⇒  Z = 20 ∠30⁰ / 4  ∠-210⁰    Z = 5 ∠-180⁰

We notice that  voltage advances the current then we are in presence of an inductive impedance

5 = wl      ⇒  5  = 400 *L       ⇒  L  =    0,0125 H        

b) v(t) = 9 cos ( 900t - 60 )

V = 9∠-60⁰

i(t)  = 4 sin ( 900t + 280 )    ⇒  i(t) = 4 cos ( 900t + 280 - 90)

i(t) = 4 cos (900t + 190 )    ⇒  I = 4 ∠ 190⁰

Z = V/I    ⇒  Z = 9∠-60⁰ / 4  ∠ 190⁰    Z = 2,25 ∠-250

In this case the current advances the voltage. Impedance capacitive

1/wc  = 1/ 900*C       1/wc = Z   ⇒ 2,25 = 1/ 900*C

2,25*900 = 1/C     ⇒  2025 =1/C     ⇒  C = 4,94 *10⁻⁴ F

c) v(t) = - 13 cos ( 250t + 60 )

v(t) = 13 cos ( 250t + 60 +180 )    ⇒ v(t) = 13 cos ( 250t +240)

Phasor Form

V = 13 ∠240⁰

i(t) = 7 sin (250t + 240 - 90)  ⇒  i(t) = 7 sin (250t + 150)

Phasor Form  I = 7  ∠150⁰

Z = 13∠240⁰ / 7 ∠150⁰    ⇒  Z = 1,86 ∠ 90⁰

Voltage advances the current then the impedance is inductive

wl = 250L     250 L = 1,86     L  = 1,86/250     L = 0,0074 H

7 0
3 years ago
Three parallel three-phase loads are supplied from a 480V (line-line RMS), 60 Hz three-phase supply. The loads are as follows: L
Travka [436]

Answer:

The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Explanation:

Load 1: Active power P_1 = 20 HP = 14.91kW;

Reactive power Q1 = P tan(\phi)

                               = 14.91\times tan(cos^{-}0.8) = 11.18 kvar


Load 2: Active power P_2 = 20 kW;

Reactive power Q2 = 0 since the load is purely resistive.

Load 3: Active power P_3 = 0 due to purely capacitiveload

           Reactive power Q_3 = -20 Var

a) since all three loads are connected in parallel therefore

    The total system active power P = P_1 + P_2 + P_3 = 34.91 KW

Total system reactive power Q = Q_1 + Q_2 + Q_3 = 11.18 + 0 -20 = -8.82 kVar

Since Q = 0, the power factor is unity.

Supply current per phase is given by

I = \frac{P}{\sqrt{3}V_{L}}

= \frac{34910}{\sqrt{3}\times 480} = 41.99 A

5 0
3 years ago
A particular motor rotates at 3000 revolutions per minute (rpm). What is its speed in rad/sec, and how many seconds does it take
raketka [301]

Answer:

ω=314.15 rad/s.

0.02 s.

Explanation:

Given that

Motor speed ,N= 3000 revolutions per minute

N= 3000 RPM

The speed of the motor in rad/s given as

Now by putting the values in the above equation

ω=314.15 rad/s

Therefore the speed in rad/s will be 314.15 rad/s.

The speed in rev/sec given as

ω= 50 rev/s

It take 1 sec to cover 50 revolutions

That is why to cover 1 revolution it take

\dfrac{1}{50}=0.02\ s

5 0
4 years ago
Read 2 more answers
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