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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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Does a train have good sorce of power

Alexeev081 [22]

Answer: here you go sir

How trains work. On an electric locomotive, the wheels are moved by electric motors. ... On a diesel locomotive, a diesel engine drives the wheels via a mechanical transmission. Cutaway illustration of an electric power carHigh-speed trains are powered by electric current, collected from an overhead cable by a pantograph.

Explanation:

Chugging across short distances or entire continents, trains act as a major form of transportation worldwide. Also called railroads or railways, trains carry within their cars passengers or freight -- such as raw materials, supplies or finished goods -- and sometimes both.

7 0

3 years ago

Helium gas is compressed from 90 kPa and 30oC to 450 kPa in a reversible, adiabatic process. Determine the final temperature and

ra1l [238]

Answer:

T2 ( final temperature ) = 576.9 K

a) 853.4 kJ/kg

b) 1422.3 kJ / kg

Explanation:

given data :

pressure ( P1 ) = 90 kPa

Temperature ( T1 ) = 30°c + 273 = 303 k

P2 = 450 kPa

Determine final temperature for an Isentropic process

$T2 = T1 (\frac{p2}{p1} )^{(k-1)/k}$ ----------- ( 1 )

T2 = 303 $( \frac{450}{90})^{(1.667- 1)/1.667}$ = 576.9K

Work done in a piston-cylinder device can be calculated using this formula

$w_{in} = c_{v} ( T2 - T1 )$ ------- ( 2 )

where : cv = 3.1156 kJ/kg.k for helium gas

T2 = 576.9K , T1 = 303 K

substitute given values Back to equation 2

$w_{in}$ = 853.4 kJ/kg

work done in a steady flow compressor can be calculated using this

$w_{in} = c_{p} ( T2 - T1 )$

where : cp ( constant pressure of helium gas ) = 5.1926 kJ/kg.K

T2 = 576.9 k , T1 = 303 K

substitute values back to equation 3

$w_{in}$ = 1422.3 kJ / kg

4 0

3 years ago

A circular copper bar with diameter d 5 3 in. is subjected to torques T 5 30 kip-in. at its ends. Find the maximum shear, tensil

Ivenika [448]

Answer:

Maximum shear stress= 5.66 ksi

Maximum tensile stress= 5.66 ksi

Maximum compressive stress=-5.66 ksi

Maximum shear strain=0.000943

Maximum tensile strain= 0.0004715

Maximum compressive strain= -0.0004715

Explanation:

For acircular bar, the maximum shear stress will be given by

$\frac {16T}{\pi d^{3}}$ where d is the diameter and T is torque.

By substituting 30 kip-in for torque and 3 in for d then

Maximum shear stress= $\frac {16*30}{\pi *3^{3}}\approx 5.66 ksi$

Also, the maximum tensile and compressive stresses will be 5.66 ksi and -5.66 ksi respectively.

The maximum shear strain will be given by stress divided by modulus of elasticity, in this case 6000 G

Maximum shear strain will be $\frac {5.66}{6000}\approx 0.000943$

The maximum tensile strain will be the above divided by 2 whereas the maximum compressive strain will be negative of tensile strain hence $\frac {0.000943}{2}=0.0004715$

Maximum compressive strain will be $\frac {-0.000943}{2}=-0.0004715$

4 0

4 years ago

Read 2 more answers

A hydraulic lift is to be used to lift a 2500 kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Deter

Lilit [14]

Answer:here in Hydraulic system the pressure on both sides will remain the same

P_1 = P_2

\frac{F_1}{A_1} = \frac{F_2}{A_2}

\frac{mg}{\pi*\frac{10^2}{4}} = \frac{Mg}{\pi \frac{d^2}{4}}

\frac{25}{10^2} = \frac{1900}{d^2}

d = 87.2 cm

so its diameter is 87.2 cm

Explanation:

8 0

3 years ago

Obtain a relation for the logarithmic mean temperature difference for use in the LMTD method?

kolezko [41]

Answer:

The log mean temperature difference is:

ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)

Explanation:

To evaluate the equivalent average temperature difference between two fluids we consider a parallel-flow double-pipe heat exchanger (see attached diagram). The temperature of the hot and cold fluids is large at the inlet of the heat exchanger and decreases exponentially toward the outlet.

We can assume that the outer surface of the heat exchanger is well insulated and that heat transfer only occurs between the two fluids. We can also assume negligible kinetic and potential. The energy balance on each fluid can be written as the rate of heat loss from the hot fluid is equal to the rate of heat gained by the cold fluid in any section of the heat exchanger:

Q = -m,h×c,ph×dT,h (1)

where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the hot fluid, dT,h= differential temperature of the hot fluid

Q = m,c×c,pc×T.c (2)

where Q=rate of heat loss, m=mass flow rate, c,ph=heat capacity of the cold fluid, dT,h= differential temperature of the cold fluid

The temperature of the hot fluid change is negative and is added to make Q positive. Solving equations 1 and 2 in terms of dT:

dT.h = - Q/(m,h×c,ph)

dT.c = Q/(m,c×c,pc)

and taking the difference:

dT,h-dT,c= d(T,h - T,c) = -Q(1/(m,h×c,ph) + 1/(m,c×c,pc)) (3)

The heat transfer rate in the differential section of the heat exchanger can be expressed as:

Q = U(T,h-T,c)×dA,s (4)

where U=overall heat transfer coefficients, dA,s = differential sectional area. Substitute equation 4 into 3:

d(T,h - T,c)/(T,h - T,c) = -U×dA,s×(1/(m,h×c,ph) + 1/(m,c×c,pc)) (5)

Integrating equation 5:

㏑((T,h out - T,c out)/(T,h in - T,c in)) = -U×A,s×(1/(m,h×c,ph) + 1/(m,c×c,pc)) (6)

The first law of thermodynamics requires the rate of heat transfer from hot and cold fluid to be equal.

Q= m×c, pc×(T, c out-T, c in) (7)

Q= m×c, ph×(T,h out-T, h in) (8)

Solve equations 7 and 8 for m,c×c, pc and m,h×c, ph and substituting into equation 6:

Q = U×A,s×ΔT,lm

Where the log mean temperature difference is:

ΔT,lm=(ΔT1-ΔT2)/㏑(ΔT1/ΔT2)

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8 0

3 years ago