Solution :
Given :
External diameter of the hemispherical shell, D = 500 mm
Thickness, t = 20 mm
Internal diameter, d = D - 2t
= 500 - 2(20)
= 460 mm
So, internal radius, r = 230 mm
= 0.23 m
Density of molten metal, ρ = [tex]$7.2 \ g/cm^3$[/tex]
= [tex]$7200 \ kg/m^3$[/tex]
The height of pouring cavity above parting surface is h = 300 mm
= 0.3 m
So, the metallostatic thrust on the upper mold at the end of casting is :
[tex]$F=\rho g A h$[/tex]
Area, A [tex]$=2 \pi r^2$[/tex]
[tex]$=2 \pi (0.23)^2$[/tex]
[tex]$=0.3324 \ m^2$[/tex]
[tex]$F=\rho g A h$[/tex]
[tex]$=7200 \times 9.81 \times 0.3324 \times 0.3$[/tex]
= 7043.42 N
In a residence, the control of a room's baseboard electric resistance heating system would be accomplished automatically using a?
A. wall-mounted pressure sensor.
B. duct-mounted thermostat.
C. wall-mounted humidity sensor.
D. wall-mounted thermostat.
Answer:
I think D is the correct answer
Explanation:
hope it is
HELP! It’s for an architecture class on PLATO
Select the correct answer.
Which association maintains the International Building Code?
A. NFPA
B. ICC
C. EPA
D. DOJ
Answer:
ICC
Explanation:
The International Building Code (IBC) is a model building code developed by the International Code Council (ICC). It has been adopted for use as a base code standard by most jurisdictions in the United States.
The following true stresses produce the corresponding true plastic strains for a brass alloy: True Stress (psi) True Strain 48400 0.11 60400 0.19 What true stress is necessary to produce a true plastic strain of 0.26
Answer:
70,900
Explanation:
Given :
True stress (psi) _____ True strain (psi)
48400 ______________ 0.11
60400 ______________ 0.19
Using ratio simplification :
Let :
s = True stress ; t = true strain
s1 = 48400
s2 = 60400
t1 = 0.11
t2 = 0.19
True stress, s0 ; needed to produce a True plastic strain, tp = 0.26
(s0 - s1) / (s2 - s1) = (tp - t1) / (t2 - t1)
(s0 - 48400)/(60400 - 48400) = (0.26 - 0.11)/(0.19 - 0.11)
(s0 - 48400)/12000 = 0.15/0.08
Cross multiply :
0.08(s0 - 48400) = 0.15 * 12000
0.08s0 - 3872 = 1800
0.08s0 = 1800 + 3872
0.08s0 = 5672
s0 = 5762 / 0.08
s0 = 70,900
The true stress required to produce a true plastic strain of 0.26 is 70,900
_____ is the set of processes used to get members of an organization to work together to further the interests of the organization.
Answer:
organizational goals
Explanation:
The mass fraction of eutectoid cementite in a hypoeutectoid iron-carbon alloy (just below its eutectoid temperature) is 0.109. On the basis of this information, determine the composition of the alloy. If it is not possible to determine the composition from the information provided, enter 0.
Answer:
The composition of an alloy is 0.75%wt
Explanation:
Let alloy is a hypoeutectoid alloy.
So, we can apply lever rule which is shown below.
[tex]W_{a}=[/tex][tex]\frac{C_{0} -C_{a} }{C_{b}-C_{a} }[/tex]
We know that [tex]C_{a}=0.022[/tex] and [tex]C_{b}=6.7[/tex]
Given that [tex]W_{a}=0.109[/tex], we have to find [tex]C_{0}[/tex]
Thus,
[tex]0.109=\frac{C_{0}-0.022 }{6.7-0.022}[/tex]
Hence
[tex]C_{0}=0.75[/tex]%wt
Explain in your own words how standard form can be used to better communicate information in certain scientific or engineering contexts.
Answer:
Technical communication is a means to convey scientific, engineering, or other technical information.[1] Individuals in a variety of contexts and with varied professional credentials engage in technical communication. Some individuals are designated as technical communicators or technical writers. These individuals use a set of methods to research, document, and present technical processes or products. Technical communicators may put the information they capture into paper documents, web pages, computer-based training, digitally stored text, audio, video, and other media. The Society for Technical Communication defines the field as any form of communication that focuses on technical or specialized topics, communicates specifically by using technology or provides instructions on how to do something.[2][3] More succinctly, the Institute of Scientific and Technical Communicators defines technical communication as factual communication, usually about products and services.[4] The European Association for Technical Communication briefly defines technical communication as "the process of defining, creating and delivering information products for the safe, efficient and effective use of products (technical systems, software, services)".[5]
Whatever the definition of technical communication, the overarching goal of the practice is to create easily accessible information for a specific audience.[6]
A distillation column at 101 kPa is used to separate 350 kmol/h of a bubble point mixture of toluene and benzene into an overhead product with 3 mole % toluene and a bottoms product with 98 mole % toluene. The feed contains 0.45 mole fraction benzene. The overhead reflux occurs at saturation temperature (implies it is total condenser – if it was not, you would have a vapor liquid mixture or a partial condenser akin to a partial reboiler). VLE data for the mixture is given below. An average alpha can be found from the VLE data if solving analytically.
EQUILIBRIUM DATA IN MOLE- FRACTION BENZENE, 101 kPA
y 0.21 0.37 0.51 0.64 0.72 0.79 0.86 0.91 0.96 0.98
x 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 0.95
Required:
a. What are the distillate and bottoms flow rates (D and B)?
b. What is the minimum reflux ratio, Rmin (corresponding to N= infinity)?
Answer:
A)
D = 158.42 kmol/h
B = 191.578 kmol/h
B) Rmin = 1.3095
Explanation:
a) Determine the distillate and bottoms flow rates ( D and B )
F = D + B ----- ( 1 )
Given data :
F = 350 kmol/j
Xf = 0.45 mole
yD ( distillate comp ) = 0.97
yB ( bottom comp ) = 0.02
back to equation 1
350(0.45) = 0.97 D + 0.02 B ----- ( 2 )
where; B = F - D
Equation 2 becomes
350( 0.45 ) = 0.97 D + 0.02 ( 350 - D ) ------ 3
solving equation 3
D = 158.42 kmol/h
resolving equation 2
B = 191.578 kmol/h
B) Determine the minimum reflux ratio Rmin
The minimum reflux ratio occurs when the enriching line meets the q line in the VLE curve
first we calculate the value of the enriching line
Y =( Rm / R + 1 m ) x + ( 0.97 / Rm + 1 )
q - line ; y = ( 9 / 9-1 ) x - xf/9-1
therefore ; x = 0.45
Finally Rmin
= (( 0.97 / (Rm + 1 )) = 0.42
0.42 ( Rm + 1 ) = 0.97
∴ Rmin = 1.3095
A square plate of titanium is 12cm along the top, 12cm on the right side, and 5mm thick. A normal tensile force of 15kN is applied to the top side of the plate. A normal tensile force of 20kN is applied to the right side of the plate. The elastic modulus, E, is 115 GPa for titanium. If the left and bottom edges of the plate are xed, calculate the normal strain and elongation of both the TOP and RIGHT side of the plate. Report your answer with proper units and signicant digits.
Answer:
For the Top Side
- Strain ε = 0.00021739
- Elongation is 0.00260868 cm
For The Right side
- Strain ε = 0.00021739
-Elongation is 0.00347826 cm
Explanation:
Given the data in the question;
Length of the squared titanium plate = 12 cm by 12 cm = 0.12 m by 0.12 m
Thickness = 5 mm = 0.005 m
Force to the Top F[tex]_t[/tex] = 15 kN = 15000 Newton
Force to the right F[tex]_r[/tex] = 20 kN = 20000 Newton
elastic modulus, E = 115 GPa = 115 × 10⁹ pascal
Now, For the Top Side;
- Strain = σ/E = F[tex]_t[/tex] / AE
we substitute
= 15000 / ( 0.12 × 0.005 × (115 × 10⁹) )
= 15000 / 69000000
Strain ε = 0.00021739
- Elongation
Δl = ε × l
we substitute
Δl = 0.00021739 × 12 cm
Δl = 0.00260868 cm
Hence, Elongation is 0.00260868 cm
For The Right side
- Strain = σ/E = F[tex]_r[/tex] / AE
we substitute
Strain = 20000 / ( 0.12 × 0.005 × (115 × 10⁹) )
= 20000 / 69000000
Strain ε = 0.000289855
- Elongation
Δl = ε × l
we substitute
Δl = 0.000289855× 12 cm
Δl = 0.00347826 cm
Hence, Elongation is 0.00347826 cm
4. Which of the following is a mechanical property of a material? Conductivity o Strength Heat resistivity Weight
Answer:
The Mechanical Properties include Elasticity, Plasticity, Ductility, Malleability, Hardness, Toughness, Brittleness, Tenacity, Fatigue, Fatigue resistance, Impact Resistance property, Machineability, Strength, Strain Energy, Resilience, Proof Resilience, Modulus of Resilience, Creep, Rupture, and Modulus of Toughness.
A small hot surface at temperature Ti-430K having an emissivity 0.8 dissipates heat by radiation into a surrounding area at T2-400K. If this radiation transfer process is characterized by a radiation heat transfer coefficient h, calculate the value of h (a) 14.4 W/m2.C (b) 114.4 W/m2C (c) 314.4 W/m2.C ( 514.4 W/m2.c
Answer:
389.6 W/m²
Explanation:
The power radiated to the surroundings by the small hot surface, P = σεA(T₁⁴ - T₂⁴) where σ = Stefan-Boltzmann constant = 5.67 × 10⁻⁸ W/m²-K⁴, ε = emissivity = 0.8. T₁ = temperature of small hot surface = 430 K and T₂ = temperature of surroundings = 400 K
So, P = σεA(T₁⁴ - T₂⁴)
h = P/A = σε(T₁⁴ - T₂⁴)
Substituting the values of the variables into the equation, we have
h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 ((430 K )⁴ - (400 K)⁴)
h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 (34188010000 K⁴ - 25600000000 K⁴)
h = 5.67 × 10⁻⁸ W/m²-K⁴ × 0.8 × 8588010000K⁴
h = 38955213360 × 10⁻⁸ W/m²
h = 389.55213360 W/m²
h ≅ 389.6 W/m²
Select the correct answer.
Which type of material functions as heat shields in space shuttles?
ОА. .
elastomers
OB.
metals
O c.
semiconductors
D.
plastics
E.
composites
C. Semiconductors.
They are made up of what is called a porous silicon material that is very light and extremely heat resistant.
Answer:
composites
Explanation:
When you're running a test bench, you would like to include a printout to the screen to inform yourself on the status of the current simulation. Which command you could use to monitor specific variables or signals in a simulation every time one of the signals changes value?
Answer:
$Monitor
Explanation:
The command that would be used when running a test bench to monitor variables or signals ( i.e. changes in the values of specific variables and signa)
is the $Monitor command
This command is also used to monitor the varying values of signals during simulation.
Which type of container traps the most heat a shoe box covered in aluminum foil plastic wrap or wax paper?
The container that traps the most heat is a shoebox covered in aluminum foil.
What is the use of aluminum foil?An aluminum foil is used for food storage, and to wrap items, such as meats, to minimize moisture loss while cooking.
Aluminum foil is also an excellent insulator. This is due to the fact that it reduces heat emission by reflecting it back.
Therefore, in a shoebox container, the aluminum foil helps to trap most heat better than a plastic wrap or wax paper.
Learn more about aluminum foil here:
https://brainly.com/question/277354
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g Given a state of stress and , where is a positive constant with units of , what is the maximum allowable value of to avoid failure according to the Tresca/von Mises criterion if the material has yield strength .
Answer: Hello your question is poorly written attached below is the complete question
answer :
: max value to avoid failure = 59 MPa
; max value to avoid failure = 34.064 Mpa
Explanation:
Attached below is the detailed solution of the given problem
For Tresca criterion : max value to avoid failure = 59 MPa
For Von-Nissen criterion ; max value to avoid failure = 34.064 Mpa
what happens to the recievng end voltage as the load on a transmission line is increased if the load has a lagging power facotr
Answer:
The receiving end Voltage ( VR ) will reduce significantly
Explanation:
Determine what happens to the receiving end voltage
Given that ; Load on transmission line is increased and Load has a lagging power factor
Increase in load = Increase in reactive/real power drawn from the transmission line and this will cause an increase in the current drawn as well.
Vs = VR + j XL I
Vs = sending end voltage
VR = receiving end voltage
jXL I = voltage drop across reactance
The rotating and shifting of Vs to a new power angle ( lagging power factor ) without a change in magnitude will cause the value of the VR receiving end voltage to drop significantly
Pie charts should have no more than eight segments. True or False?
Answer:
Explanation:
Pie charts generally should have no more than eight segments.
IV. An annealed copper strip 9 inches wide and 2.2 inches thick, is rolled to its maximum possible draft in one pass. The following properties of annealed copper are given: strength coefficient is 90,000 psi; true strain at the onset of non-uniform deformation is 0.45; and, engineering strain at yield is 0.11. The coefficient of friction between strip and roll is 0.2. The roll radius is 14inches and the rolls rotate at 150 rpm. Calculate the roll-strip contact length. Calculate the absolute value of thetrue strain that the strip undergoes in this operation. Determine the average true stress of the strip in theroll gap. Calculate the roll force. Calculate the horsepower required.
Answer:
13.9357 horse power
Explanation:
Annealed copper
Given :
Width, b = 9 inches
Thickness, [tex]$h_0=2.2$[/tex] inches
K= 90,000 Psi
μ = 0.2, R = 14 inches, N = 150 rpm
For the maximum possible draft in one pass,
[tex]$\Delta h = H_0-h_f=\mu^2R$[/tex]
[tex]$=0.2^2 \times 14 = 0.56$[/tex] inches
[tex]$h_f = 2.2 - 0.56$[/tex]
= 1.64 inches
Roll strip contact length (L) = [tex]$\sqrt{R(h_0-h_f)}$[/tex]
[tex]$=\sqrt{14 \times 0.56}$[/tex]
= 2.8 inches
Absolute value of true strain, [tex]$\epsilon_T$[/tex]
[tex]$\epsilon_T=\ln \left(\frac{2.2}{1.64}\right) = 0.2937$[/tex]
Average true stress, [tex]$\overline{\gamma}=\frac{K\sum_f}{1+n}= 31305.56$[/tex] Psi
Roll force, [tex]$L \times b \times \overline{\gamma} = 2.8 \times 9 \times 31305.56$[/tex]
= 788,900 lb
For SI units,
Power = [tex]$\frac{2 \pi FLN}{60}$[/tex]
[tex]$=\frac{2 \pi 788900\times 2.8\times 150}{60\times 44.25\times 12}$[/tex]
= 10399.81168 W
Horse power = 13.9357