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2 years ago

15

) In a disk test performed on a specimen 32-mm in diameter and 7 mm thick, the specimen fractures at a stress of 680 MPa. What w

as the load on the disk at fracture

1 answer:

Radda [10]2 years ago

6 0

Answer:

Sorry it doesnt tall me anythikng

Explanation:

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What are the laws that apply to one vehicle towing another?

Sergeu [11.5K]

The drawbar or other connections must be strong enough to pull all the weight of the vehicle being towed. The drawbar or other connection may not exceed 15 feet from one vehicle to the other.

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3 years ago

A certain power-supply filter produces an output with a ripple of 100 mV peak-to-peak and a dc value of 20 V. The ripple factor

Oksanka [162]

Answer: the ripple factor is 0.005

Explanation:

Given the data in the question;

we know that expression of ripple factor is;

r = Vr(pp) / Vdc

where Vr(pp) the peak to peak is ripple voltage ( 100mv = 0.1 V)

and Vdc is the dc value of the filter output voltage ( 20 V)

so we substitute our given values;

r = 0.1 / 20

r = 0.005

Therefore; the ripple factor is 0.005

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3 years ago

An insulated mixing chamber receives 0.5 kg/s of steam at 3 MPa and 300°C through one inlet, and saturated liquid water at 3 MPa

maxonik [38]

Answer:

$\dot m_{2} = 0.199\,\frac{kg}{s}$

Explanation:

The mixing chamber can be modelled by applying the First Law of Thermodynamics:

$\dot W_{in}+\dot m_{1}\cdot h_{1} +\dot m_{2} \cdot h_{2} - \dot m_{3}\cdot h_{3} = 0$

Since that mass flow rate of water at inlet 1 is the only known variable, the expression has to be simplified like this:

$\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - z\cdot h_{3} = 0$

Besides, the following expression derived from the Principle of Mass Conservation is presented below:

$1 + y = z$

Then, the expression is simplified afterwards:

$\frac{\dot W_{in}}{\dot m_{1}} + h_{1}+ y\cdot h_{2} - (1+y)\cdot h_{3} = 0$

$\frac{\dot W_{in}}{\dot m_{1}} +h_{1} - h_{3} + y\cdot (h_{2}-h_{3}) = 0$

Specific enthalpies are obtained from steam tables and described as follows:

State 1 (Superheated vapor)

$h = 2994.3\,\frac{kJ}{kg}$

State 2 (Saturated liquid)

$h = 1008.3\,\frac{kJ}{kg}$

State 3 (Liquid-Vapor mixture)

$h = 2444.22\,\frac{kJ}{kg}$

The ratio of the stream at state 2 to the stream at state 1 is:

$y = \frac{\frac{\dot W_{in}}{\dot m_{1}}+h_{1}-h_{3}}{h_{3}-h_{2}}$

$y = \frac{\frac{10\,kW}{0.5\,\frac{kg}{s} }+2994.3\,\frac{kJ}{kg}-2444.22\,\frac{kJ}{kg} }{2444.22\,\frac{kJ}{kg}-1008.3\,\frac{kJ}{kg} }$

$y = 0.397$

The mass flow rate of the saturated liquid is:

$\dot m_{2} = y\cdot \dot m_{1}$

$\dot m_{2} = 0.397\cdot (0.5\,\frac{kg}{s} )$

$\dot m_{2} = 0.199\,\frac{kg}{s}$

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3 years ago

Why do automotive technicians need to have good computer skills?

Elden [556K]

Answer:

Just think about the number of computerized functions there are in modern automobiles.

Explanation:

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3 years ago

H. Blasius correlated data on turbulent friction factor in smooth pipes. His equation f s m o o t h ≈ 0.3164 Re − 1 / 4 fsmooth≈

tiny-mole [99]

Answer:

Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

Explanation:

The first step to take is to calculate the the velocity of flow through a pipe

Q =Av

Where Q = is the discharge through pipe

A = Area of the pipe

v = the flow of velocity

We substitute 0.001 m^3/s for Q and 0.03 m for D

Q= Av

0.001=Av

Substitute π/4 D² for A

0.001 = π/4 D² (v)

v = 0.004/πD²

D = he diameter of the pipe

substitute 3 cm for D

v= 0.004/π * [3 cm * 1 m/100 cm]²

v =1.414 m/s

Obtain fluid properties from the table Kinematic viscosity and Dynamic of water

p =1000 kg /m³

μ= 1.002 * 10^ ⁻³ N.s/m³

Thus,

we write the expression to determine the Reynolds number of flow

Re = pvD/μ

Re = is the Reynolds number

p =density

μ = dynamic viscosity at 20⁰C

We then substitute 1000 kg /m³ in place of p, 1.002 * 10^ ⁻³ N.s/m³ for μ,

1.414 m/s for v and 0.03 m for D

Thus,

Re = 1000 * 1.414 * 0.03/ 1.002 * 10^ ⁻³ = 42335

The next step is to calculate the friction factor form the Blasius equation

f = 0.3164 (Re)^1/4

f = friction factor

We substitute 42335 for Re

f = 0.3164 (42335)1/4

=0.022

The next step is to write the expression to determine the friction head loss

hl = flv²/2gD

hl = head loss

l = length of pipe

g= acceleration due to gravity

We then again substitute 0.022 for f, 1.414 m/s for v, 0.03 m for D, and 9.8 m/s² for g.

so,

hl = flv²/2gD

hl/L = 0.022 * 1.414²/2 * 9.81 * 0.03

sinθ = 0.07473

θ = 4° 16'

Therefore the angle the pipe needed to make the static pressure constant along the pipe is θ = 4° 16'

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3 years ago