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2 years ago

Compute the mass fractions of α ferrite and cementite in pearlite. Assume T=726⁢ C∘.

Engineering

1 answer:

kotykmax [81]2 years ago

4 0

Answer:

The mass composition of ∝ ferrite is 88.94%.

The mass composition of cementite is 11.05%.

Explanation:

Given that

T=726 °C

We have to find the mass fractions of ferrite nad cementite in pearlite.

Lets take data from ideal Iron -carbon diagram at 726 °C

Composition of ∝ ferrite=0.022 5

Composition of pearlite =0.76 %

Composition of cementite =7.6%

We know that if we want to find the mass fraction the we use Lever rule .So now by using lever rule

$The\ mass\ composition\ of\alpha \ ferrite =\dfrac{6.7-0.76}{6.7-0.022}$

So the mass composition of ∝ ferrite is 88.94%.

$The\ mass\ composition\ of\ cementite =\dfrac{0.76-0.022}{6.7-0.022}$

So the mass composition of cementite is 11.05%.

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A pipe is insulated such that the outer radius of the insulation is less than the critical radius. Now the insulation is taken o

irina [24]

Answer:

the heat transfer from the pipe will decrease when the insulation is taken off for r₂< $r_{cr}$

where;

r₂ = outer radius

$r_{cr}$ = critical radius

Explanation:

Note that the critical radius of insulation depends on the thermal conductivity of the insulation k and the external convection heat transfer coefficient h .

$r_{cr} =\frac{k}{h}$

The rate of heat transfer from the cylinder increases with the addition of insulation for outer radius less than critical radius (r₂< $r_{cr}$ ) 0, and reaches a maximum when r₂ = $r_{cr}$ , and starts to decrease for r₂< $r_{cr}$ . Thus, insulating the pipe may actually increase the rate of heat transfer from the pipe instead of decreasing it when r₂< $r_{cr}$ .

7 0

2 years ago

Consider a steady developing laminar flow of water in a constant-diameter horizontal discharge pipe attached to a tank. The flui

dalvyx [7]

Answer:

hello attached is the free body diagram of the missing figure

Fr = $\frac{\pi }{4} D^2 [ ( P1 - P2) - pV^2 ]$

Explanation:

Average velocity is constant i.e V1 = V2 = V

The momentum equation for the flow in the Z - direction can be expressed as

-Fr + P1 Ac - P2 Ac = mB2V2 - mB1V1 ------- equation 1

Fr = horizontal force on the bolts

P1 = pressure of fluid at entrance

V1 = velocity of fluid at entrance

Ac = cross section area of the pipe

P2 and V2 = pressure and velocity of fluid at some distance

m = mass flow rate of fluid

B1 = momentum flux at entrance , B2 = momentum flux correction factor

Note; average velocity is constant hence substitute V for V1 and V2

equation 1 becomes

Fr = ( P1 - P2 ) Ac + mV ( 1 - 2 )

Fr = ( P1 - P2 ) Ac - mV ---------------- equation 2

equation for mass flow rate

m = pAcV

p = density of the fluid

insert this into equation 2 EQUATION 2 BECOMES

Fr = ( P1 - P2) Ac - pAcV^2

= Ac [ (P1 - P2) - pV^2 ] ---------- equation 3

Note Ac = $\frac{\pi }{4} D^2$

Equation 3 becomes

Fr = $\frac{\pi }{4} D^2$ [ (P1 -P2 ) - pV^2 ] ------- relation for the horizontal force acting on the bolts

6 0

2 years ago

A sign erected on uneven ground is guyed by cables EF and EG. If the force exerted by cable EF at E is 46 lb, determine the mome

Vikki [24]

Answer:

M_AD = 1359.17 lb-in

Explanation:

Given:

- T_ef = 46 lb

Find:

- Moment of that force T_ef about the line joining points A and D.

Solution:

- Find the position of point E:

mag(BC) = sqrt ( 48^2 + 36^2) = 60 in

BE / BC = 45 / 60 = 0.75

Hence, E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in

- Find unit vector EF:

mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in

vec(EF) = < -15 , -110 , 30 >

unit(EF) = (1/115) * < -15 , -110 , 30 >

- Tension T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb

- Find unit vector AD:

mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in

vec(AD) = < 48 , -12 , 36 >

unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >

unit (AD) = <0.7845 , -0.19612 , 0.58835 >

M_AD = unit(AD) . ( E x T_EF)

$M_d = \left[\begin{array}{ccc}0.7845&-0.19612&0.58835\\36&96&27\\-6&-44&12\end{array}\right]$

M_AD = 1835.73 + 116.49528 - 593.0568

M_AD = 1359.17 lb-in

3 0

2 years ago

The hot water needs of an office are met by heating tab water by a heat pump from 16 C to 50 C at an average rate of 0.2 kg/min.

Alex777 [14]

Answer:

option B

Explanation:

given,

heating tap water from 16° C to 50° C

at the average rate of 0.2 kg/min

the COP of this heat pump is 2.8

power output = ?

$COP = \dfrac{Q_H}{W_{in}}\\W_{in} = \dfrac{Q_H}{COP}\\W_{in} = \dfrac{\dfrac{0.2}{60}\times 4.18\times (50-16)}{2.8}\\W_{in} = 0.169$

the required power input is 0.169 kW or 0.17 kW

hence, the correct answer is option B

7 0

2 years ago

A 5-cm-diameter shaft rotates at 4500 rpm in a 15-cmlong, 8-cm-outer-diameter cast iron bearing (k = 70 W/m·K) with a uniform cl

-BARSIC- [3]

Answer:

(a) the rate of heat transfer to the coolant is Q = 139.71W

(b) the surface temperature of the shaft T = 40.97°C

Explanation:

See explanation in the attached files

5 0

2 years ago