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3 years ago

Which of the two materials (brittle vs. ductile) usually obtains the highest ultimate strength and why?

Engineering

1 answer:

sveta [45]3 years ago

5 0

Answer:

Explanation:

Ductile materials typically have a higher ultimate strength because they stretch absorbing more energy before breaking. While fragile materials snap in half before larger deformations due to larger loads occur.

It should be noted that when ductile materials stretch their section becomes smaller, and in that reduced section the stresses concentrate.

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1. If a Gear with a 3 inch Diameter is being turned by a Gear with a 6 inch Diameter, which Gear will rotate at a higher Rate?

pshichka [43]

Answer:

The smaller gear will rotate faster.

Explanation:

If a larger gear is driven by a smaller gear, the large gear will rotate slower than the smaller gear but will have a greater moment. For example, a low gear on a bike or car. If a smaller gear is driven by a larger gear, the smaller gear will rotate quicker than the larger gear but will have a smaller moment.

I hope this helps! :)

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2 years ago

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Lilit [14]

Answer:

Explanation:

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3 years ago

A thick steel slab ( 7800 kg/m3 , c 480 J/kg K, k 50 W/m K) is initially at 300 C and is cooled by water jets impinging on one o

Nutka1998 [239]

Answer:

1791 secs ≈ 29.85 minutes

Explanation:

( Initial temperature of slab ) T1 = 300° C

temperature of water ( Ts ) = 25°C

T2 ( final temp of slab ) = 50°C

distance between slab and water jet = 25 mm

<u>Determine how long it will take to reach T2</u>

First calculate the thermal diffusivity

∝ = 50 / ( 7800 * 480 ) = 1.34 * 10^-5 m^2/s

<u>next express Temp as a function of time </u>

T( 25 mm , t ) = 50°C

next calculate the time required for the slab to reach 50°C at a distance of 25mm

attached below is the remaining part of the detailed solution

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3 years ago

Air modeled as an ideal gas enters a turbine operating at steady state at 1040 K, 278 kPa and exits at 120 kPa. The mass flow ra

gladu [14]

Answer:

a) $T_{2}=837.2K$

b) $e=91.3$ %

Explanation:

A) First, let's write the energy balance:

$W=m*(h_{2}-h_{1})\\W=m*Cp*(T_{2}-T_{1})$ (The enthalpy of an ideal gas is just function of the temperature, not the pressure).

The Cp of air is: 1.004 $\frac{kJ}{kgK}$ And its specific R constant is 0.287 $\frac{kJ}{kgK}$ .

The only unknown from the energy balance is $T_{2}$ , so it is possible to calculate it. The power must be negative because the work is done by the fluid, so the energy is going out from it.

$T_{2}=T_{1}+\frac{W}{mCp}=1040K-\frac{1120kW}{5.5\frac{kg}{s}*1.004\frac{kJ}{kgk}} \\T_{2}=837.2K$

B) The isentropic efficiency (e) is defined as:

$e=\frac{h_{2}-h_{1}}{h_{2s}-h_{1}}$

Where ${h_{2s}$ is the isentropic enthalpy at the exit of the turbine for the isentropic process. The only missing in the last equation is that variable, because $h_{2}-h_{1}$ can be obtained from the energy balance $\frac{W}{m}=h_{2}-h_{1}$

$h_{2}-h_{1}=\frac{-1120kW}{5.5\frac{kg}{s}}=-203.64\frac{kJ}{kg}$

An entropy change for an ideal gas with constant Cp is given by:

$s_{2}-s_{1}=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})$

You can review its deduction on van Wylen 6 Edition, section 8.10.

For the isentropic process the equation is:

$0=Cpln(\frac{T_{2}}{T_{1}})-Rln(\frac{P_{2}}{P_{1}})\\Rln(\frac{P_{2}}{P_{1}})=Cpln(\frac{T_{2}}{T_{1}})$

Applying logarithm properties:

$ln((\frac{P_{2}}{P_{1}})^{R} )=ln((\frac{T_{2}}{T_{1}})^{Cp} )\\(\frac{P_{2}}{P_{1}})^{R}=(\frac{T_{2}}{T_{1}})^{Cp}\\(\frac{P_{2}}{P_{1}})^{R/Cp}=(\frac{T_{2}}{T_{1}})\\T_{2}=T_{1}(\frac{P_{2}}{P_{1}})^{R/Cp}$