Answer:
The average thickness of the blubber is 0.077 m
Explanation:
Here, we want to calculate the average thickness of the Walrus blubber.
We employ a mathematical formula to calculate this;
The rate of heat transfer(H) through the Walrus blubber = dQ/dT = KA(T2-T1)/L
Where dQ is the change in amount of heat transferred
dT is the temperature gradient(change in temperature) i.e T2-T1
dQ/dT = 220 W
K is the conductivity of fatty tissue without blood = 0.20 (J/s · m · °C)
A is the surface area which is 2.23 m^2
T2 = 37.0 °C
T1 = -1.0 °C
L is ?
We can rewrite the equation in terms of L as follows;
L × dQ/dT = KA(T2-T1)
L = KA(T2-T1) ÷ dQ/dT
Imputing the values listed above;
L = (0.2 * 2.23)(37-(-1))/220
L = (0.2 * 2.23 * 38)/220 = 16.948/220 = 0.077 m
You are tasked with designing a thin-walled vessel to contain a pressurized gas. You are given the parameters that the inner diameter of the tank will be 60 inches and the tank wall thickness will be 5/8" (0.625 inches). The allowable circumferential (hoop) stress and longitudinal stresses cannot exceed 30 ksi.
(1) What is the maximum pressure that can be applied within the tank before failure? = psi(2) If you had the opportunity to construct a spherical tank having an inside diameter of 60 inches and a wall thickness of 5/8" (instead of the thin-walled cylindrical tank as described above), what is the maximum pressure that can be applied to the spherical tank? = psi
Answer:
Explanation:
For cylinder
Diameter d = 60 inches
thickness t = 0.625 inches
circumferential (hoop) stress = 30 ksi
[tex]hoop \ \ stress =\sigma_1=\frac{P_1d}{2t}\\\\\sigma_1=30ksi\\\\30000=\frac{P_1\times 60}{2\times0.625}\\\\P_1=624psi[/tex]
[tex]longitudinal \ \ stress =\sigma_2=\frac{P_2d}{2t}\\\\\sigma_2=30ksi\\\\30000=\frac{P_2\times 60}{4\times0.625}\\\\30000=\frac{P_2\times 60}{2.5}\\\\75000=P_2\times60\\\\P_2=\frac{75000}{60} \\\\P_1=1250psi[/tex]
Therefore maximum pressure without failure is P₁ = 625 psi
ii) For Sphere
[tex]\sigma_1=\sigma_2=\frac{Pd}{4t} \\\\P=\frac{30000\times 4 \times 0.625}{60} \\\\=\frac{75000}{60}\\\\=1250\ \ psi[/tex]
While having a discussion about O-rings at the bottom of filters, Technician A says that the Automotive Filter Manufacturers Council recommends that the filter O-ring be lubricated with oil after installing the filter. Technician B says that the filter O-ring should be lubricated before installation. Who is correct
Answer:
Technician B is correct
Explanation:
O- rings are used with oil transmission filters to avoid transmission failures but some people use lip seals as well. either of them is inserted onto the outer part of the transmission system i.e it is inserted/found in-between Transmission filters and the transmission systems and it main purpose is to avoid leaks and transmission failure in the short and long term.
0-rings should be lubricated before installation this is because the o-rings are usually super tight when installing and would require lubrication to ease the installation process else the rubber might get ruptured and this would lead to instant transmission failure.
Choose two consumer services careers and research online to determine what kinds of professional organizations exist for these professions. Write a paragraph describing the purpose of the organization, the requirements for joining, and the benefits of membership.
Bank loan facilitator, and hospital emergency care specialist are the two consumer or customer services careers.
Bank loan facilitator is a consumer service facilitator who ask and provide people loan in emergency, for the purpose of education, treatment, family events, and for other reasons. For bank loan facilitator the professional organizations should be banking and finance sector. The purpose of these organizations is to help people in financial matter seeking benefit by getting interest from customers. The requirements for joining of the employee must include strong convincing power for the employee, time management, strong and tactful communication skills. Benefits of membership of the customers can help them to seek loans on need basis on lower interest. Hospital emergency care specialist provides help to the staff and the customers in medical emergency. These professionals are necessary for the hospital, clinics, and rehabilitation centers. Purpose of the organization is to provide medical care to the patients. The requirements for joining of the employee includes ability to give information to patients and staff during emergency conditions, facilitating ambulance to rescue patients from their homes, and from other areas, providing medicine, medical equipment, and other facilities to the patients and other medical staff necessary for treatment. Benefits of membership in clinical or hospital settings can help the patient in frequent visits for treatment, concession in laboratory tests, and medication.Learn more about customer:
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What's the "most common" concern with using variable frequency drives (VFDs)? 1) carrier frequency 2) harmonic distortion 3) hertz modulation
The common" concern with using variable frequency drives (VFDs) is C. hertz modulation.
What is variable frequency drive?It should be noted that a variable frequency drive simply means a type of motor drive that us used in mechanical drive system.
In this case, common" concern with using variable frequency drives (VFDs) is hertz modulation
Learn more about frequency on:
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An isentropic steam turbine processes 5.5 kg/s of steam at 3 MPa, which is exhausted at 50 kPa and 100°C. Five percent of this flow is diverted for feedwater heating at 500 kPa. Determine the power produced by this turbine. Use steam tables.
Answer:
The answer is 1823.9
Explanation:
Solution
Given that:
m = 5.5 kg/s
= m₁ = m₂ = m₃
The work carried out by the energy balance is given as follows:
m₁h₁ = m₂h₂ +m₃h₃ + w
Now,
By applying the steam table we have that<
p₃ = 50 kPa
T₃ = 100°C
Which is
h₃ = 2682.4 kJ/KJ
s₃ = 7.6953 kJ/kgK
Since it is an isentropic process:
Then,
p₂ = 500 kPa
s₂=s₃ = 7.6953 kJ/kgK
which is
h₂ =3207.21 kJ/KgK
p₁ = 3HP0
s₁ = s₂=s₃ = 7.6953 kJ/kgK
h₁ =3854.85 kJ/kg
Thus,
Since 5 % of this flow diverted to p₂ = 500 kPa
Then
w =m (h₁-0.05 h₂ -0.95 )h₃
5.5(3854.85 - 0.05 * 3207.21 - 0.95 * 2682.4)
5.5( 3854.83 * 3207.21 - 0.95 * 2682.4)
5.5 ( 123363249.32 -0.95 * 2682.4)
w=1823.9
The Rappahannock River near Warrenton, VA, has a flow rate of 3.00 m3/s. Tin Pot Run (a pristine stream) discharges into the Rappahannock at a flow rate of 0.05 m3/s. To study mixing of the stream and river, a conserva- tive tracer is to be added to Tin Pot Run. If the instruments that can mea- sure the tracer can detect a concentration of 1.0 mg/L, what minimum concentration must be achieved in Tin Pot Run so that 1.0 mg/L of tracer can be measured after the river and stream mix? Assume that the 1.0 mg/L of tracer is to be measured after complete mixing of the stream and Rappa- hannock has been achieved and that no tracer is in Tin Pot Run or the Rap- pahannock above the point where the two streams mix. What mass rate (kg/d) of tracer must be added to Tin Pot Run?
Find the given attachments for complete explanation
The basic behind equal driving is to
Follow traffic signs , Keep distance between cars , Be patient in traffic.
An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,500 kg/hr. The feed is known to contain174 kg of aluminum and 2,326 kg of reject. After operating for 1 hour, a total of 256 kg ofmaterials is collected in the product stream. On close inspection, it is found that 140 kg ofproduct is aluminum. Estimate the % recovery of aluminum product and the % purity of thealuminum produc
Answer:
the % recovery of aluminum product is 80.5%
the % purity of the aluminum product is 54.7%
Explanation:
feed rate to separator = 2500 kg/hr
in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine
of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.
After the separation, 256 kg is collected in the product stream.
of this 256 kg, 140 kg is aluminium.
% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material
% recovery of aluminium = 140kg/174kg x 100% = 80.5%
% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream
% purity of the aluminium product = 140kg/256kg
x 100% = 54.7%
WHAT IS A VACUOMETER?
Solid spherical particles having a diameter of 0.090 mm and a density of 2002 kg/m3 are settling in a solution of water at 26.7C. The volume fraction of the solids in the water is 0.45. Calculate the settling velocity and the Reynolds number.
Answer:
Settling Velocity (Up)= 2.048*10^-5 m/s
Reynolds number Re = 2.159*10^-3
Explanation:
We proceed as follows;
Diameter of Particle = 0.09 mm = 0.09*10^-3 m
Solid Particle Density = 2002 kg/m3
Solid Fraction, θ= 0.45
Temperature = 26.7°C
Viscosity of water = 0.8509*10^-3 kg/ms
Density of water at 26.7 °C = 996.67 kg/m3
The velocity between the interface, i.e between the suspension and clear water is given by,
U = [ ((nf/ρf)/d)D^3] [18+(1/3)D^3)(1/2)]
D = d[(ρp/ρf)-1)g*(ρf/nf)^2]^(1/3)
D = 2.147
U = 0.0003m/s (n = 4.49)
Up = 0.0003 * (1-0.45)^4.49 = 2.048*10^-5 m/s
Re=0.09*10^-3*2.048*10^-5*996.67/0.0008509 = 2.159*10^-3
cubical tank 1 meter on each edge is filled with water at 20 degrees C. A cubical pure copper block 0.46 meters on each edge with an initial temperature of 100 degrees C is quickly submerged in the water, causing an amount of water equal to the volume of the smaller cube to spill from the tank. An insulated cover is placed on the tank. The tank is adiabatic. Estimate the equilibrium temperature of the system (block + water). Be sure to state all applicable assumptions.
Answer:
final temperature = 26.5°
Explanation:
Initial volume of water is 1 x 1 x 1 = 1 [tex]m^{3}[/tex]
Initial temperature of water = 20° C
Density of water = 1000 kg/[tex]m^{3}[/tex]
volume of copper block = 0.46 x 0.46 x 0.46 = 0.097 [tex]m^{3}[/tex]
Initial temperature of copper block = 100° C
Density of copper = 8960 kg/[tex]m^{3}[/tex]
Final volume of water = 1 - 0.097 = 0.903 [tex]m^{3}[/tex]
Assumptions:
since tank is adiabatic, there's no heat gain or loss through the wallsthe tank is perfectly full, leaving no room for cooling airtotal heat energy within the tank will be the summation of the heat energy of the copper and the water remaining in the tank.mass of water remaining in the tank will be density x volume = 1000 x 0.903 = 903 kg
specific heat capacity of water c = 4186 J/K-kg
heat content of water left Hw = mcT = 903 x 4186 x 20 = 75.59 Mega-joules
mass of copper will be density x volume = 8960 x 0.097 = 869.12 kg
specific heat capacity of copper is 385 J/K-kg
heat content of copper Hc = mcT = 869.12 x 385 x 100 = 33.46 Mega-joules
total heat in the system = 75.59 + 33.46 = 109.05 Mega-joules
this heat will be distributed in the entire system
heat energy of water within the system = mcT
where T is the final temperature
= 903 x 4186 x T = 3779958T
for copper, heat will be
mcT = 869.12 x 385 = 334611.2T
these component heats will sum up to the final heat of the system, i.e
3779958T + 334611.2T = 109.05 x [tex]10^{6}[/tex]
4114569.2T = 109.05 x [tex]10^{6}[/tex]
final temperature T = (109.05 x [tex]10^{6}[/tex])/4114569.2 = 26.5°
