Considering only (110), (1 1 0), (101), and (10 1 ) as the possible slip planes, calculate the stress at which a BCC single crystal will yield if the critical resolved shear stress is 50 MPa and the load is applied in the [100] direction.
Answers
Solution :
i. Slip plane (1 1 0)
Slip direction -- [1 1 1]
Applied stress direction = ( 1 0 0 ]
τ = 50 MPa ( Here slip direction must be perpendicular to slip plane)
τ = σ cos Φ cos λ
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
ii. Slip plane --- (1 1 0)
Slip direction -- [1 1 1]
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
iii. Slip plane --- (1 0 1)
Slip direction --- [1 1 1]
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
iv. Slip plane -- (1 0 1)
Slip direction ---- [1 1 1]
τ = σ cos Φ cos λ
∴
σ = 122.47 MPa
∴ (1, 0, -1). (1, -1, 1) = 1 + 0 - 1 = 0
Related Questions
Air enters a compressor operating at steady state at 1.05 bar, 300 K, with a volumetric flow rate of 39 m3/min and exits at 12 bar, 400 K. Heat transfer occurs at a rate of 6.5 kW from the compressor to its surroundings. Assuming the ideal gas model for air and neglecting kinetic and potential energy effects, determine the power input, in kW.
A furnace is shaped like a long equilateral triangular duct where the width of each side is 2 m. Heat is supplied from the base surface, whose emissivity is ε1 = 0.8, at a rate of 800 W/m2 while the side surfaces, whose emissivities are 0.5, are maintained at 500 K. Neglecting the end effects, determine the temperature of the base surface. Can you treat this geometry as a two-surface enclosure?
Q/For the circuit showm bellow:a) find the mathematical expression for the transient behavior of ve and ic after closing the switeh,b) sketch vc and ic.
How much heat is lost through a 3’× 5' single-pane window with a storm that is exposed to a 60°F temperature differential?A. 450 Btu/hB. 900 Btu/hC. 1350 Btu/hD. 1800 Btu/h
Answers
Answer:
y=0.12 lbmol(water)/lbmol(products)
Explanation:
First we find the humidity of air. Using humidity tables and the temperature we find that is 0.01 g water/L air.
Now we set the equation assuming dry air:
With this equation we have almost all moles that exit the reactor, we are just missing the initial moles of water due to the humidity. So we proceed to calculate it with the ideal gas law:
PV=nRT
Vair=867.7L
With the volume and the fraction of water, we can calculate the mass of water:
0.01 * 867.7=8.677 g of water
Now we calculate the moles of water:
8.677 g / 18 g/mol = 0.48 moles of water
Now we can calculate the total moles of water in the exit of the reactor:
0.48 + 4 = 4.48 moles of water
And finally we just need to sum all moles at the exit of the reactor and divide:
3 moles of CO2 + 2.5 moles of O2+ 4.48 moles of H2O + 28.2 moles of N2
And we have 38.18 moles in total, then:
4.48/38.18=y=0.12 moles of water/moles of products
As the relation moles/moles is equal to lb-moles/lb-moles, we have our fina result:
y=0.12 lbmol(water)/lbmol(products)
Answers
Answer:
558.1918 kilocalories = 558191.8 calories
Explanation:
Data provided in the question:
Atmospheric pressure = 84.6 KPa
Mass of water, m = 900 g = 0.90 kg
Temperature = 15°C
Now,
Temperature at 84.6 KPa = 94.77°C
Therefore,
Heat energy required = m(CΔT + L)
here,
C is the specific heat of the water = 4.2 KJ/kg.°C
L = Latent heat of water = 2260 KJ/kg
Thus,
Heat energy required = 0.90[ 4.2 × (94.77 - 15) + 2260 ]
= 2335.53 KJ
also,
1 KJ = 0.239 Kilocalories
Therefore,
2335.53 KJ = 0.239 × 2335.53 Kilocalories
= 558.1918 kilocalories = 558191.8 calories
Answers
Answer:
robotic technology
Explanation:
Innovation is nothing but the use of various things such as ideas, products, people to build up a solution for the benefit of the human. It can be any product or any solution which is new and can solve people's problems.
Innovation solution makes use of technology to provide and dispatch new solutions or services which is a combination of both technology and ideas.
One such example of an innovative solution we can see is the use of "Robots" in medical science or in any military operations or rescue operation.
Sometimes it is difficult for humans to do everything or go to everywhere. Thus scientist and engineers have developed many advance robots or machines using new ideas and technology to find solutions to these problems.
Using innovations and technologies, one can find solutions to many problems which is difficult for the peoples. Robots can be used in any surveillance operation or in places of radioactive surrounding where there is a danger of humans to get exposed to such threats. They are also used in medical sciences to operate and support the patient.
b. the back work ratio
c. the net power developed, in kW
d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.
Answers
Answer:
a. = 77.65%
b. bwr = 6.5%
c. 3538.986 kW
d. -163.169 kJ
Explanation:
a. The given property are;
P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa
p₄/p₁ = 10
P₂/P₁ = p₄/p₃ = √10
p₂ = 100·√10
= T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K
T₂ = T₁ + ( - T₁)/
= 300 + (416.85 - 300)/0.8 = 446.0625 K
p₄ = 10×p₁ = 10×100 = 1000 kPa
p₄/p₃ = √10 =
p₃ = 100·√10
T₃ = 300 K
T₃/ = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)
= T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K
T₄ = T₃ + ( - T₃)/
= 300 + (215.905- 300)/0.8 = 194.881 K
The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28
T₄ = 446.0625 K
T₆ = 1400 K
/T₆ = (1/√10)^(0.4/1.4)
= 1400×(1/√10)^(0.4/1.4) = 1007.6 K
T₇ = T₆ - (T₆ -
) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K
T₈ = 1400 K
T₉ = 1086.08 K
T₅ = T₄ + (T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K
=(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))
(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765
= 77.65%
b. Back work ratio, bwr =
((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))
40.9435/627.84 = 6.5%
c.
Power developed is given by the relation;
= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW
d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)
-163.169 kJ
Answers
Answer: the thermal conductivity of the second material is 125.9 W/m.k
Explanation:
Given that;
The two rods could be approximated as a fins of infinite length.
TA = 75°C, θA = (TA - T∞) = 75 - 25 = 50°C
TB = 70°C θB = (TB - T∞) = 70 - 25 = 45°C
Tb = 100°C θb = (Tb - T∞) = (100 - 25) = 75°C
T∞ = 25°C
KA = 200 W/m · K, KB = ?
Now
The temperature distribution for the infinite fins are given by
θ/θb = e^(-mx)
θA/θb= e^-√(hp/A.kA) x 1 --------------1
θB/θb = e^-√(hp/A.kB) x 1---------------2
next we take the natural logof both sides,
ln(θA/θb) = -√(hp/A.kA) x 1 ------------3
In(θB/θb) = -√(hp/A.kB) x 1 ------------4
now we divide 3 by 4
[ ln(θA/θb) /in(θB/θb)] = √(KB/KA)
we substitute
[ In(50/75) /In(45/75)] = √(KB/200)
In(0.6666) / In(0.6) = √KB / √200
-0.4055/-0.5108 = √KB / √200
0.7938 = √KB / 14.14
√KB = 11.22
KB = 125.9 W/m.k
So the thermal conductivity of the second material is 125.9 W/m.k
Answers
Explanation:
Step1
Absolute pressure is the pressure above zero level of the pressure. Absolute pressure is considering atmospheric pressure in it. Absolute pressure is always positive. There is no negative absolute pressure.
The expression for absolute pressure is given as follows:
Here, is absolute pressure,
is gauge pressure and
is atmospheric pressure.
Step2
Gauge pressure is the pressure that measure above atmospheric pressure. It is not considering atmospheric pressure. It can be negative called vacuum or negative gauge pressure. Gauge pressure used to simplify the pressure equation for fluid analysis.