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2 years ago

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Consider a single crystal of some hypothetical metal that has the FCC crystal structure and is oriented such that a tensile stre

ss is applied along a [112] direction. If slip occurs on a (111) plane and in a [011] direction, and the crystal yields at a stress of 5.12 MPa, compute the critical resolved shear stress.

1 answer:

Naddik [55]2 years ago

5 0

Answer:

imma leabe

Explanation:

leabe leave*

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A window‐mounted air‐conditioning unit (AC) removes energy by heat transfer from a room, and rejects energy by heat transfer to

Arada [10]

Solution :

Given :

The power of the air‐conditioning (AC) unit is , W = 0.434 kW

The coefficient of performance or the COP of the air‐conditioning (AC) unit is given by = 6.22

Therefore he heat removed is given by , $Q_H = 6.22 \times 0.434$

$Q_H = 2.7 \ kW$

Now if the electricity is valued at 0.10 dollar per kW hour, then the operating cost of the air conditioning unit in 24 hours is given by = 0.10 x 2.7 x 24

= 6.48

Therefore the operating cost = $ 6.48 for 24 hours.

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2 years ago

A pump is used to transport water from a reservoir at one elevation to another reservoir at a higher elevation. If the elevation

erastova [34]

Answer:b

Explanation:

We know power delivered by Pump is

$P=\rho \times Q\times g\times \Delta H$

where $P\rho$ =Density of fluid

$Q$ =Flow rate

$g$ =acceleration due to gravity

$\Delta H$ =Change in Elevation

If $\Delta H$ is increased by 4 time then

$P'=\rho \times Q\times g\times (4\Delta H)$

$P'=4 P$

So power increases by four times.

4 0

3 years ago

The components of an electronic system dissipating 180 W are located in a 1-m-long horizontal duct whose cross section is 16 cm

oee [108]

Answer:

a) The exit temperature is 39.25°C

b) The highest component surface is 132.22°C

c) The average temperature for air equal to 35°C is a good assumption because the air temperature at the inlet will increase due to the result in the heat gain produced by the duct and whose surface is exposed to a flow of hot.

Explanation:

a) The properties of the air at 35°C:

p = density = 1.145 kg/m³

v = 1.655x10⁻⁵m²/s

k = 0.02625 W/m°C

Pr = 0.7268

cp = 1007 J/kg°C

a) The mass flow rate of air is equal to:

$m=\rho *V = 1.145*0.65=0.7443kg/min=0.0124kg/s$

The exit temperature is:

$T=T_{i} +\frac{Q}{m*c_{p} } =27+\frac{0.85*180}{0.0124*1007} =39.25$ °C

b) The mean fluid velocity is:

$V_{m} =\frac{V}{A} =\frac{0.65}{0.16*0.16} =25.4m/min=0.4232m/s$

The hydraulic diameter is:

$D_{h} =\frac{4A}{p} =\frac{4*0.16*0.16}{4*0.16} =0.16m$

The Reynold´s number is:

$Re=\frac{VD_{h} }{v} =\frac{0.4232*0.16}{1.655x10^{-5} } =4091.36$

Assuming fully developed turbulent flow, the Nusselt number is:

$Nu=0.023Re^{0.8} *Pr^{0.4} =0.023*4091.36^{0.8} *0.7268^{0.4} =15.69$

$h=\frac{k*Nu}{D_{h} } =\frac{0.02625*15.69}{0.16} =2.57W/m^{2} C$

The highest component surface temperature is:

$T=T_{e} +\frac{\frac{Q}{A} }{h} =39.2+\frac{0.85*\frac{180}{4*0.16*1} }{2.57} =132.22$ °C

6 0

3 years ago

For a bolted assembly with eight bolts, the stiffness of each bolt is kb = 1.0 MN/mm and the stiffness of the members is km = 2.

rjkz [21]

Answer:

a) 0.978

b) 0.9191

c) 1.056

d) 0.849

Explanation:

Given data :

Stiffness of each bolt = 1.0 MN/mm

Stiffness of the members = 2.6 MN/mm per bolt

Bolts are preloaded to 75% of proof strength

The bolts are M6 × 1 class 5.8 with rolled threads

Pmax =60 kN, Pmin = 20kN

<u>a) Determine the yielding factor of safety</u>

$n_{p} = \frac{S_{p}A_{t} }{CP_{max}+ F_{i} }$ ------ ( 1 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

Input the given values into the equation above

equation 1 becomes ( np ) = $\frac{380*20.1}{0.277*7500*5728.5}$ = 0.978

note : values above are derived values whose solution are not basically part of the required solution hence they are not included

<u>b) Determine the overload factor of safety</u>

$n_{L} = \frac{S_{p}A_{t}-F_{i} }{C(P_{max} )}$ ------- ( 2 )

Sp = 380 MPa, At = 20.1 mm^2, C = 0.277, Pmax = 7500 N, Fi = 5728.5 N

input values into equation 2 above

hence : $n_{L}$ = 0.9191 $n_{L} = 0.9191$

<u>C) Determine the factor of safety based on joint separation</u>

$n_{0} = \frac{F_{i} }{P_{max}(1 - C ) }$

Fi = 5728.5 N, Pmax = 7500 N, C = 0.277,

input values into equation above

Hence $n_{0}$ = 1.056

<u>D) Determine the fatigue factor of safety using the Goodman criterion.</u>

nf = 0.849

attached below is the detailed solution .

4 0

2 years ago

Subcooled liquid water flows adiabatically in a constant diameter pipe past a throttling valve that is partially open. The liqui

Llana [10]

Answer:

hi-he = 0

pi-pe = positive

ui-ue = negative

ti-te = negative

Explanation:

we know that fir the sub cool liquid water is

dQ = Tds = du + pdv ............1

and Tds = dh - v dP .............2

so now for process of throhling is irreversible when v is constant

then heat transfer is = 0 in irreversible process

so ds > 0

so here by equation 1 we can say

ds > 0

dv = 0 as v is constant

so that Tds = du .................3

and du > 0

ue - ui > 0

and

now by the equation 2 throttling process

here enthalpy is constant

so dh = 0

and Tds = -vdP

so ds > 0

so that -vdP > 0

as here v is constant

so -dP =P1- P2

so P1-P2 > 0

so pressure is decrease here

5 0

3 years ago