1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Rainbow [258]
3 years ago
13

A large particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of

tungsten and copper are 0.60 and 0.40, respectively, estimate the upper and lower bounds of the modulus of this composite. Modulus of Elasticity (GPa) Tungsten 407 Copper 110
Engineering
1 answer:
OverLord2011 [107]3 years ago
7 0

Answer:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

Explanation:

Calculation to estimate the upper and lower bounds of the modulus of this composite.

First step is to calculate the maximum modulus for the combined material using this formula

Modulus of Elasticity for mixture

E= EcuVcu+EwVw

Let pug in the formula

E =( 110 x 0.40)+ (407 x 0.60)

E=44+244.2 GPa

E=288.2GPa

Second step is to calculate the combined specific gravity using this formula

p= pcuVcu+pwTw

Let plug in the formula

p = (19.3 x 0.40) + (8.9 x 0.60)

p=7.72+5.34

p=13.06

Now let calculate the UPPER BOUNDS and the LOWER BOUNDS of the Specific stiffness

UPPER BOUNDS

Using this formula

Upper bounds=E/p

Let plug in the formula

Upper bounds=288.2/13.06

Upper bounds=22.07 GPa

LOWER BOUNDS

Using this formula

Lower bounds=EcuVcu/pcu+EwVw/pw

Let plug in the formula

Lower bounds =( 110 x 0.40)/8.9+ (407 x 0.60)/19.3

Lower bounds=(44/8.9)+(244.2/19.3)

Lower bounds=4.94+12.65

Lower bounds=17.59 GPa

Therefore the Estimated upper and lower bounds of the modulus of this composite will be:

Upper bounds 22.07 GPa

Lower bounds 17.59 GPa

You might be interested in
A 3-in-thick slab is 10 in wide and 15 ft long. The thickness of the slab is reduced by 20% and width increases by 3% in a hot-r
xxMikexx [17]

Answer: l = 2142.8575 ft

v = 193.99 ft/min.

Explanation:

Given data:

Thickness of the slab = 3in

Length of the slab = 15ft

Width of the slab = 10in

Speed of the slab = 40ft/min

Solution:

a. After three phase

three phase = (0.2)(0.2)(0.2)(3.0)

= 0.024in.

wf = (1.03)(1.03)(1.03)(10.0)

= 10.927 in.

Using constant volume formula

= (3.0)(10.0)(15 x 15) = (0.024)(10.927)Lf

Lf = (3.0)(10.0)(15 x 15)/(0.024)(10.927)

= 6750 /0.2625

= 25714.28in = 2142.8575 ft

b.

vf = (0.2 x 0.2 x 3.0)(1.03 x 1.03 x 10.0)(40)/(0.024)(10.927)

= (0.12)(424.36)/0.2625

= 50.9232/0.2625

= 193.99 ft/min.

4 0
3 years ago
One - tenth kilogram of air as an ideal gas with k= 1.4 executes a carnot refrigeration cycle as shown i fig. 5,16, the isotherm
maria [59]

Answer:

Hello your question is incomplete attached below is the missing part

a) p1 = 454.83 kPa,  p2 = 283.359 Kpa , p3 = 536.423 kpa , p4 = 860.959kPa

b) W12 = 3.4 kJ, W23 = -3.5875 KJ, W34 = -4.0735 KJ, W41 = 3.5875 KJ

c) 5

Explanation:

Given data:

mass of air ( m ) = 1/10 kg

adiabatic index ( k ) = 1.4

temperature for isothermal expansion = 250K

rate of heat transfer ( Q12 ) = 3.4 KJ

temperature for Isothermal compression ( T4 ) = 300k

final volume ( V4 ) = 0.01m ^3

a)  Calculate the pressure, in Kpa, at each of the four principal states

from an ideal gas equation

P4V4 = mRT4 ( input values above )

hence P4 = 860.959kPa

attached below is the detailed solution

b) Calculate work done for each processes

attached below is the detailed solution

C) Calculate the coefficient of performance

attached below is detailed solution

6 0
3 years ago
A very long pipe of 0.05 m (r1) radius and 0.03 m thickness (r2 - r1) is buried at a depth of 2m (z) to transport liquid nitroge
Elenna [48]

Answer:

The surface temperature of the ground is = 296.946K

Explanation:

Solution

Given

r₁= 0.05m

r₂= 0.08m

Tn =Ti = 77K

Ki = 0.0035 Wm-1K-1

Kg =  1 Wm-1K-1

Z= 2m

Now,

The outer type temperature (Skin temperature pipe)

Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi  T0 -T1/ln (r2/r1)

Thus,

10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05

⇒ T₀ -77 = 231.72

    T₀= 290.72K

The shape factor between the cylinder and he ground

S = 2πL/ln 4z/D

where L = length of pipe

D = outer layer of pipe

S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m

The heat gained in the pipe is = S  * Kg * (Tg- T₀)

(10* 1) = 1.606 * 1* (Tg- 290.72)

Tg - 290.72 = 6.2266

Tg = 296.946K

Therefore the surface temperature to the ground is 296.946K

6 0
3 years ago
For a continuous and oriented fiber-reinforced composite, the moduli of elasticity in the longitudinal and transverse directions
goldfiish [28.3K]

Answer:

attached below

Explanation:

6 0
3 years ago
Flue gases from an incinerator are released to atmosphere using a stack that is 0.6 m in diameter and 10.0 m high. The outer sur
Goshia [24]

Answer:

(a) Rate of heat transfer assuming there is no wind is 152.70J/s

(b) Rate of heat transfer assuming the stack is exposed to 20km/h winds is 151.49J/s

Explanation:

Q/t = KA(T2 - T1)/d

Range of heat transfer coefficient (K) for flue gases is 60-180W/mk

Assuming K=180W/mk, diameter (d) =0.6m, A=3.142×0.6^2/4 = 028278m^2, distance (d) = 10m, T2=40°C, T1=10°C

(a) Q/t = 180×0.28278×(40-10)/10 = 180×0.28278×30/10 = 152.70J/s

(b) V= 20km/h=20×1000/3600m/s= 5.6m/s, t=d/v=10/5.6=1.8s

Q = KA(T2 - T1)/V = 180×0.28278×(40 - 10)/5.6 = 180×0.28278×30/5.6 = 272.68J

Q/t = 272.68J/1.8s = 151.48J/s

7 0
3 years ago
Other questions:
  • What is the power of a parallel circuit with a resistance of 1,000 omh and current of 0.03a
    11·1 answer
  • What is the braks mean effictive pressure?
    9·1 answer
  • Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet temperature
    13·1 answer
  • Plot the following trig functions using subplots, choosing an appropriate layout for the number of functions displayed. The subp
    8·1 answer
  • Write a program to input 6 numbers. After each number is input, print the biggest of the numbers entered so far.
    12·1 answer
  • A 10-mm steel drill rod was heat-treated and ground. The measured hardness was found to be 290 Brinell. Estimate the endurance s
    14·1 answer
  • In a flow over a flat plate, the Stanton number is 0.005: What is the approximate friction factor for this flow a)- 0.01 b)- 0.0
    8·1 answer
  • In a much smaller model of the Gizmo apparatus, a 5 kg mass drops 86 mm (0.086 m) and raises the temperature of 1 gram of water
    10·1 answer
  • T/f
    10·1 answer
  • The controller determines if a(n) _________ exists by calculating the difference between the SP and the PV.
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!