Answer:
The maximum water pressure at the discharge of the pump (exit) = 496 kPa
Explanation:
The equation expressing the relationship of the power input of a pump can be computed as:
[tex]E _{pump,u} = \dfrac{m(P_2-P_1)}{\rho}[/tex]
where;
m = mass flow rate = 120 kg/min
the pressure at the inlet [tex]P_1[/tex] = 96 kPa
the pressure at the exit [tex]P_2[/tex] = ???
the pressure [tex]\rho[/tex] = 1000 kg/m³
∴
[tex]0.8 \times 10^{3} \ W = \dfrac{(120 \ kg/min * 1min/60 s)(P_2-96000)}{1000}[/tex]
[tex]0.8 \times 10^{3}\times 1000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}[/tex]
[tex]800000 = {(120 \ kg/min * 1min/60 s)(P_2-96000)}[/tex]
[tex]\dfrac{800000}{2} = P_2-96000[/tex]
400000 = P₂ - 96000
400000 + 96000 = P₂
P₂ = 496000 Pa
P₂ = 496 kPa
Thus, the maximum water pressure at the discharge of the pump (exit) = 496 kPa
You have a 12-in. diameter pile that is embedded in the ground 50-ft. The soil is a clay and has a cohesion of 1,000-psf. Determine the Ultimate Pile Capacity, Qult.
Answer:
hello your question is incomplete attached below is the complete question
answer : 0.75 ( A )
Explanation:
Given data:
12 - in diameter pile
embedded 50-ft
type of soil ; clay
Cohesion = 1000 psf
Determine the Ultimate pile Capacity
cohesion = 1000 psf
hence 1000 psf = 1 ksf (where ; 1 psf = 0.0.001ksf )
form the given table the value of α corresponding to 1 ksf = 0.75
Link BD consists of a single bar 36 mm wide and 18 mm thick. Knowing that each pin has a 12-mm diameter, determine the maximum value of the average normal stress in link BD if (a) θ 5 0, (b) θ 5 90°.
Answer:
hello the diagram attached to your question is missing attached below is the missing diagram
answer :
a) 48.11 MPa
b) - 55.55 MPa
Explanation:
First we consider the equilibrium moments about point A
∑ Ma = 0
( Fbd * 300cos30° ) + ( 24sin∅ * 450cos30° ) - ( 24cos∅ * 450sin30° ) = 0
therefore ; Fbd = 36 ( cos ∅tan30° - sin∅ ) kN ----- ( 1 )
A ) when ∅ = 0
Fbd = 20.7846 kN
link BD will be under tension when ∅ = 0, hence we will calculate the loading area using this equation
A = ( b - d ) t
b = 12 mm
d = 36 mm
t = 18
therefore loading area ( A ) = 432 mm^2
determine the maximum value of average normal stress in link BD using the relation below
бbd = [tex]\frac{Fbd}{A}[/tex] = 20.7846 kN / 432 mm^2 = 48.11 MPa
b) when ∅ = 90°
Fbd = -36 kN
the negativity indicate that the loading direction is in contrast to the assumed direction of loading
There is compression in link BD
next we have to calculate the loading area using this equation ;
A = b * t
b = 36mm
t = 18mm
hence loading area = 36 * 18 = 648 mm^2
determine the maximum value of average normal stress in link BD using the relation below
бbd = [tex]\frac{Fbd}{A}[/tex] = -36 kN / 648mm^2 = -55.55 MPa
The maximum value of the average normal stress in link BD at the given angles are;
At θ = 0°; 64.15 MPa
At θ = 90°; 66.66 MPa
Average Normal Stress
The image of the link and the single bar is missing and so i have attached it.
From the image of the link and single bar attached, i have drawn a free body diagram of link ABC that will help us to solve this question.
Taking Moments about point A and summing to zero, we can solve for F_bd at the given angles as;
A) At θ = 0°;
From the diagram, AC = 450 mm = 0.45 m and force acting at point C is 24 kN or 24000 N. Thus;
(0.45 * sin 30)(24000) - F_bd(0.3 * cos 30) = 0
Thus;
(0.45 * sin 30)(24000) = F_bd(0.3 * cos 30)
⇒ 5400 = 0.2598F_bd
F_bd = 5400/0.2598
F_bd = 20785.22 N
Area at tension Loading is;
A = (0.03 - 0.012)0.018
A = 324 × 10⁻⁶ m²
Thus;
Average Normal stress is;
σ = 20785.22/(324 × 10⁻⁶)
σ = 64.15 × 10⁶ Pa = 64.15 MPa
B) At θ = 90°;
(0.45 * cos 30)(24000) + F_bd(0.3 * cos 30) = 0
Thus;
-(0.45 * cos 30)(24000) = F_bd(0.3 * cos 30)
F_bd = -36000 N
Area at compression Loading is;
A = 0.03 * 0.018
A = 540 × 10⁻⁶ m²
Thus;
Average Normal stress is;
σ = -36000/(540 × 10⁻⁶)
σ = 66.66 × 10⁶ Pa = 66.66 MPa
Read more about Average Normal Stress at; https://brainly.com/question/14468674
Pin supports, such as that at A, may have horizontal and vertical components to the support reaction. Roller supports, such as that at B, have only a vertical component. What are the support reactions for this beam
Answer:
hello your question is incomplete below is the missing part of the question and attached is the missing diagram
In the simply-supported beam shown in the figure below, d1=14 ft, d2=7 ft, and F=15 kips. so find Az, Ay, By.(in kips)
answer :
Reaction force at B = 10 kips
Reaction force in the y axis = 5 kips
Reaction force in the Z direction = 0 kips
Explanation:
Taking moment about point A
∑ Ma = 0
By + ( d1 + d2 ) - F*d1 = 0
By ( reaction force at B ) = ( 15 * 14 ) / ( 21 ) = 10 kips
Applying equilibrium forces in the Y-axis
∑ Fy = 0
Ay - F + By = 0
where : F = 15, By = 10
hence ; Ay = 5 kips
applying equilibrium forces in the Z-direction
∑ Fz = 0
Az = 0 kips
Compute the theoretical density of ZnS given that the Zn-S distance and bond angle are 0.234 nm and 109.5o, respectively. The atomic weights of zinc and sulfur are 65.41 g/mol and 32.06 g/mol.
Answer: the theoretical density is 4.1109 g/cm³
Explanation:
first the image of one set of ZnS bonding in the crystal structure, we calculate the value of angle θ
θ + ∅ + 90° = 180°
θ = 90° - ∅
θ = 90° - ( 109.5° / 2 )
θ = 35.25°
next we calculate the value of x from the geometry
given that; distance angle d = 0.234
x = dsinθ
= 0.234 × sin35.25°)
= 0.135 nm = 0.135 × 10⁻⁷ cm
next we calculate the length of the unit cell
a = 4x
a = 4(0.135)
a = 0.54 nm = 0.54 × 10⁻⁷ cm
next we calculate number of formula units
n' = (no of corner atoms in unit ell × contribution of each corner atom in unit cell) + ( no of face center atom in a unit cell × contribution of each face center atom in a unit cell)
n' = 8 × 1/8) + ( 6 × 1/2)
= 1 + 3
= 4
next we calculate the theoretical density using this equation
P = [n'∑(Ac + AA)] / [Vc.NA]
= [n'∑(Ac + AA)] / [(a)³NA]
where the ∑Ac is sum of atomic weights of all cations in the formula unit( 65.41 g/mol)
∑AA is the sum of weights of all anions in the formula unit( 32.06 g/mol)
Na is the Avogadro’s number( 6.023 × 10²³ units/mole)
so we substitute
P = [4( 65.41 + 32.06)] / [ ( 0.54 × 10⁻⁷ )³ × (6.023 × 10²³)]
= 389.88 / 94.84
= 4.1109 g/cm³
therefore the theoretical density is 4.1109 g/cm³
The average age of engineering students at graduation is a little over 23 years. This means that the working career of most engineers is almost exactly 500 months. How much would an engineer need to save each month to accrue $5 million by the end of her working career? Assume a 9% interest rate, compounded monthly.
Answer:
$916
Explanation:
To solve this, we use the formula
FV = P/i * [(1+i)^n - 1], where
FV = future value of the all the money invested, $5 million
n = time span, = 500 months
P = payment per month
I = interest rate, 9% by 12 months, = 0.0075
Considering that we have been given all in our question, then we substitute directly and solve. So we have,
5000000 = P/0.0075 * [(1+0.0075)^500 -1]
5000000 * 0.0075 = P * [1.0075^500 - 1]
37500 = P * [41.93 - 1]
37500 = P * 40.93
P = 37500/40.93
P = $916.20
Therefore, the engineer needs to save $916 in a month which is the accrued
Why is the drawdown cone for a well completed in a low permeability aquifer narrower and deeper than a drawdown cone for a well in a high permeability'aquifer? a) Low permeability aquifers typically produce water at a higher discharge rate b) The pump in a low permeability well casing typically has a higher capacity c) It takes a greater hydraulic head to drive the groundwater laterally to the well casing in the lower permeability aquifer d) The radius of influence of a high permeability well is typically shorter than that of a low permeability well
Answer:
c) It takes a greater hydraulic head to drive the groundwater laterally to the well casing in the lower permeability aquifer
Explanation:
The groundwater are contains under the rock and in the open spaces within the rocks and the unconsolidated sediments. Aquifer refers to the underground layers of the permeable sand or rocks that transmits the groundwater below water table which provides a sufficient supply of water to the well. Groundwater is present everywhere where there is porosity in the rocks and it depends on the permeability of the rocks to allow them flow.
A drawdown cone is completed in the lower permeable aquifer deeper and narrower than the high permeable aquifer as it takes more amount hydraulic head or energy to drive groundwater to the well casing which is in the lower permeable aquifer.
Think of an employee object. What are several of the possible states that the object may have over time?
Hi, your question is unclear. However, I inferred you meant 'object' in computer programming.
Explanation:
Remember, the term 'object' used in programming refers to stored data that can take different forms or states.
For example, a company's employee database may have several object states. Which includes;
New Employee (meaning the database can contain newly employed employees)Former Employee (meaning the database can contain past/formerly employed employee) Current Employee (meaning the database can contain present/current employees)Suspended Employee (meaning the database could contain employees on suspension)
how to change a fuel fiter
In flowing from section (1) to section (2) along an open channel, the water depth decreases by a factor of two and the Froude number changes from a subcritical value of 0.4 to a supercritical value of 2.5. Determine the channel width at (2) if it is 13 ft wide at (1).
Answer:
channel width = 2.621 ft
Explanation:
Given data :
Decreasing Factor = 2
subcritical value = 0.4
super critical value = 2.5
width = 13 ft
attached below is a detailed solution and
PLS HELPPP!!!!!!!!!!!!!!!
What is the sense of a vector that is 20° CW from the negative x-axis?
A. Up and right
B. Down and right
C. Up and left
D. Down and left
Answer:
C. Up and left
Explanation:
The negative x-axis is on the left side of a graph.
If you turn 20° clockwise, then it would still be in the second quadrant pointing up a little bit.
Just loook at the picture i linked below
Answer:
c
Explanation:
I need a thesis statement about Engineers as Leaders.
Answer:
Engineers are a very beneficial contribution in which offers great solutions to national problems.
A 2 m3 insulated rigid tank contains 3 kg of nitrogen at 90 kPa. Now work is done on the system until the pressure in the tank rises to 175 kPa. Determine the entropy change of nitrogen in kJ/K during this process assuming constant specific heats.
Answer: [tex]\Delta S[/tex] = 1.47kJ/K
Explanation: Entropy is the measure of a system's molecular disorder, i.e, the unuseful work a system does.
The nitrogen gas in the insulated tank can be described as an ideal gas, so it can be used the related formulas.
For the entropy, the ratio of initial and final temperatures is needed and as volume is constant, we use:
[tex]\frac{P_{1}}{T_{1}} =\frac{P_{2}}{T_{2}}[/tex]
[tex]\frac{P_{2}}{P_{1}} =\frac{T_{2}}{T_{1}}[/tex]
[tex]\frac{T_{2}}{T_{1}} =\frac{175}{90}[/tex]
[tex]\frac{T_{2}}{T_{1}} =1.94[/tex]
Specific Heat is the quantity of heat required to increase the temperature 1 degree of a unit mass of a substance. Specific heat of nitrogen at constant volume is [tex]c_{v}=[/tex] 0.743kJ/kg.K
The change in entropy is calculated by
[tex]\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})-Rln(\frac{V_{2}}{V_{1}} )][/tex]
For the nitrogen insulated in a rigid tank:
[tex]\Delta S= m[c_{v}ln(\frac{T_{2}}{T_{1}})][/tex]
Substituing:
[tex]\Delta S= 3[0.743ln(1.94)][/tex]
[tex]\Delta S=[/tex] 1.47
The entropy change of nitrogen in an insulated rigid tank is 1.47kJ/K
A bearing is to be used as shaft support to carry both radial and thrust forces. Calculations show that a 02 series ball bearing with a bore size of 20 mm would work well for this application. What are the static and dynamic load ratings of this bearing in kN
Answer:
The correct answer is "20.8 kN" and "31 kN". A further explanation is given below.
Explanation:
The angular touch bearing seems to be a fine replacement while accommodating radial and even some displacement pressures. You may receive static as well as dynamic scores from either the manufacturer's collections.The load ratings should be for the SKF bearing including its predetermined distance:
Static load
= 20.8 kN
Dynamic load
= 31 kN
- What will happen if high voltage from the HV battery or motor-generator is shorting to frame ground?
Answer:
Unlike a low voltage battery such as 12V, high voltage from a High Voltage battery should not be grounded to the chassis for several numbers of reason which are;
- HV up to 350V have a corresponding high current which generate unwanted magnetic field and causes magnetic interference. This can be reduced by using a twisted conductor so that the interference can be cancelled.
-HV can result to surges which result to spark over and flash over between phase and ground.
A skier wears a jacket filled with goose down that is 15 mm thick. Another skier wears a wool sweater that is 4.0 mm thick. Both have the same surface area. Assuming the temperature difference between the inner and outer surfaces of each garment is the same, calculate the ratio (wool/goose down) of the heat lost due to conduction during the same time interval. Isn't heat flow directly proportional to the distance?
Answer:
Why the f do you have a dumb goose on your clothes?
Explanation:
Chickens are so much better than gooses, first of all, and second of all, drop out of school. Its so much easier. Next year eligible to drop out by law and ima f school and live in da hood with JB Money. Care to join me in Racine? I need some more "employees" to help be a look-out for cops
The ratio of the heat lost due to conduction during the same time interval is Qw / Qs = 6.
What is heat loss?The deliberate or accidental transfer of heat from one material to another is known as heat loss. Radiation, convection, and conduction can all contribute to this.
When a component comes in direct touch with another component, whether it is insulated or not, conduction frequently happens. U value x Wall area x Delta T is the formula used to calculate the amount of heat loss via a wall in BTUs.
Q = KA ΔTФ / L
QL /K = A ΔTФ
Where, A T and is constant
so, QL/K = constant
Qg Lg / Kg = Qw Lw /Kw
0.04 x 15 x 10-3 / 0.025 x 4 x 10-3
Qw / Qs = 6
Therefore, the heat loss is Qw / Qs = 6.
To learn more about heat loss, refer to the link:
https://brainly.com/question/14228650
#SPJ5
If the phase shift is π/2 rads and T is the period, then the voltage at (T/2) is _____
a. zero
b. +peak/2
c. +peak
d. -peak
Answer:
d. - peak
Explanation:
In alternating current, the voltage is represented by the following formula:
[tex]V=V_{max}sin(\omega t+\phi)[/tex]
where,
[tex]V_{max}[/tex]=Maximum voltage
[tex]\omega[/tex]=Angular frequency
[tex]\phi[/tex]=phase shift
t=time
The angular frequency can be written in terms of the period (T), so:
[tex]\omega=\frac{2\pi}{T}[/tex]
So the equation will now lok like this:
[tex]V=V_{max}sin(\frac{2\pi}{T} t+\phi)[/tex]
we know that [tex]\phi=\frac{\pi}{2}[/tex] and that [tex]t=\frac{T}{2}[/tex] so the equation will now look like this:
[tex]V=V_{max}sin(\frac{2\pi}{T} (\frac{T}{2})+\frac{\pi}{2})[/tex]
which can be simplified to:
[tex]V=V_{max}sin(\pi+\frac{\pi}{2})[/tex]
[tex]V=V_{max}sin(\frac{3\pi}{2})[/tex]
Which solves to:
[tex]V=-V_{max}[/tex]
so the answer is d. -peak
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 2.5 × 10^-4 mm (0.9843 × 10^-5 in.) and a crack length of 5 × 10^-2 mm (1.969 × 10^-3 in.) when a tensile stress of 130 MPa (18860 psi) is applied?
Answer:
2600 MPa
Explanation:
The formula to be used for the question is
σ(m) = 2 * σ(o) * [α/ρ(t)]^0.5, where
σ(m) = maximum stress
σ(o) = maximum applied tensile stress
α = length of surface crack
ρ(t) = radius of curvature of the crack
It's an easy one, as we have all the values given from the question, and all we do is plug them in directly
σ(m) = 2 * 130 * [(0.05/2)/0.00025]^0.5
σ(m) = 260 * [0.025/0.00025]^0.5
σ(m) = 260 * 100^0.5
σ(m) = 260 * 10
σ(m) = 2600 MPa
Saturated humid air at 1 atm and 10°C is to be mixed with atmospheric air at 1 atm, 32°C, and 80 percent relative humidity to form air of 70 percent relative humidity. Determine the proportion at which these two streams are to be mixed and the temperature of the resulting air.
Answer:
Explanation:
From the information given in the question:
The pressure = 1 atm
The saturated humid air temperature [tex]T_1 = 10^0 \ C[/tex]
The saturated humid air relative humidity [tex]\phi_1[/tex] = 100%
The atmospheric air temperature [tex]T_2[/tex] = 32°C; &
The atmospheric relative humidity [tex]\phi_2[/tex] = 80%
The data obtained at 1 atm pressure from property psychometric chart at [tex]T_1[/tex] = 10°C
[tex]h_1 = 29.4 \ kJ/kg[/tex] of air ; [tex]\omega _1[/tex] = 0.0077 kg/kg of air
At [tex]T_2= 12^0 \ C[/tex]
[tex]h_2 = 94.6 \ kJ/kg[/tex] of air; [tex]\omega _2 = 0.024 \ kg/kg \ of \ air[/tex]
If we take a look at the expression used in combining the conservation of energy and mass for adiabatic mixing of two streams; we have:
[tex]\dfrac{m_1}{m_2}= \dfrac{\omega_2 -\omega _3}{\omega _3-\omega _1}= \dfrac{h_2-h_3}{h_3-h_1}[/tex]
[tex]\dfrac{m_1}{m_2}= \dfrac{0.024 -\omega _3}{\omega _3-0.0077}= \dfrac{94.6-h_3}{h_3-29.4}[/tex]
The mixture temperature [tex]T_3[/tex] is determined through a trial and error method.
At trial and error method [tex]T_3[/tex] = 24°C
From the relative humidity of 70%;
From the psychometric chart;
The specific humidity [tex]\omega _3[/tex] = 0.0143 kg/kg of air
The enthalpy [tex]h_3[/tex] = 57.6 kJ/kg of air
Then;
[tex]\dfrac{m_1}{m_2}=1.3[/tex]
Thus, 1.3 is the proportion at which the two streams are being mixed.
A water treatment plant is designed to process 100 ML/d (mega liter per day). The flocculator is 30 m long, 15 m wide, and 5 m deep. Revolving paddles are attached to four horizontal shafts that rotate at 1.5 rpm. Each shaft supports four paddles that are 200 mm wide, 15 m long and centered 2 m from the shaft. Assume the mean water velocity to be 70% less than paddle velocity and CD = 1.8. all paddles remain submerged all the time.
Calculate the following:
a. Difference in velocity between paddles and water
b. Value of G
c. Retention time
d. Camp number.
Answer:
A) 0.22 m/sec
B) 10.717 sec^-1
C) 32.4 min
D) 20833.848
Explanation:
A) calculate the difference in velocity between paddles and water
Vr = Vp - Vw
Vp = paddle velocity
Vw = water velocity
Vw = 0.3 Vp therefore Vr = 0.7vp
also ; Vp = ωr = [tex]\frac{2\pi N}{60} r[/tex] = [tex]\frac{2*3.14*1.5 *2}{60}[/tex] = 0.314 m/sec
therefore
Vr = 0.7 * 0.314 = 0.22 m/sec
B) Value of G
attached below is the detailed solution
C) Retention time
Td = V / Q = Volume / Discharge = [tex]\frac{30* 15*5*24}{100*10^6*10^-3} * 60 min[/tex]
= 32.4 min
D) Camp number
camp number = G * t
= 10.717 sec^-1 * 32.4 * 60
= 20833.848
System reliability improves by using redundant systems. The reliability of the system can be improved by using two such systems in parallel. Again, if the probability of failure of any one subsystem is 0.01, what is the reliability of this system?
Answer:
Reliability is 0.99
Explanation:
Reliability is complementary to probability of failure, i.e. R(t) = 1 –F(t)
F(t) = 0.01
R(t) = 1 - 0.01 = 0.99
Reliability is 0.99
Its means that the probability of failure has dropped 10 times.
Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 smaller parallel pipes of the same total cross-sectional area, 4.0 mm2. Total volume flow is 1000 mm3/is. The pressure drop for fluid passing through the single pipe is lower than that through the 100 vessel array by a factor of:_____.A. 10.B. 100.
C. 1000.
Answer:
A. 10
Explanation:
For a single straight vessel; we can express the equation as;
[tex]H_{f_1} = \dfrac{8 \ fl \ Q_1^2}{\pi ^2 gd_1^5} \ \ \ \ \ ... (1)[/tex]
Given that:
The total volume Q₁ = 1000 m/s²
Then the Q₂ = 1000/100 = 10 mm/s₂
However, the question proceeds by stating that 100 pipes of the same cross-section is being used.
Therefore, the formula for the area can be written as:
[tex]\dfrac{\pi}{4}d_1^2 = 100 \bigg ( \dfrac{\pi}{4} d_2^2\bigg)[/tex]
Divide both sides by [tex]\dfrac{\pi}{4}[/tex]
[tex]d_1^2 = 100 \ d_2^2[/tex]
Making [tex]d_1[/tex] the subject of the formula;
[tex]d_1 = 10d_2[/tex]
However, considering a pipe in parallel
[tex]H_{f_2} = (H_f_2)_1 = (H_f_2)_2=...= (H_f_2)_{10}= \dfrac{8 \ fl Q_2^2}{\pi^2 \ gd _2^5} \ \ \ \ \ \ \ ...(2)[/tex]
Relating equation (1) by (2); then solving; we have;
[tex]\dfrac{H_{f_1}}{H_{f_2}} = \dfrac{\dfrac{8flQ_1^2}{\pi^2 \ gd _1^5} }{\dfrac{8\ fl Q_2^2 }{\pi^2 gd_2^5} }[/tex]
[tex]\dfrac{H_{f_1}}{H_{f_2}} =\dfrac{Q_1^2}{Q_2^2} \times \dfrac{d_2^5}{d_1^5}[/tex]
[tex]\dfrac{H_{f_1}}{H_{f_2}} =\dfrac{(1000)^2}{(10)^2} \times \dfrac{d_2^5}{(10 \ d_2)^5}[/tex]
[tex]\dfrac{H_{f_1}}{H_{f_2}} =\dfrac{1}{10}[/tex]
[tex]H_{f_2} =10H_{f_1}[/tex]
(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. The student's skin temperature is 94.4° F. Determine the net energy transfer from the student's body during the 20.00 min ride to school due to electromagnetic radiation. Note: Skin emissivity is 0.90, and the surface area of the student is 1.50m2.
Answer:
The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.
Explanation:
From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation ([tex]\dot Q[/tex]), measured in BTU per hour, is represented by this formula:
[tex]\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4})[/tex] (1)
Where:
[tex]\epsilon[/tex] - Emissivity, dimensionless.
[tex]A[/tex] - Surface area of the student, measured in square feet.
[tex]\sigma[/tex] - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.
[tex]T_{s}[/tex] - Temperature of the student, measured in Rankine.
[tex]T_{b}[/tex] - Temperature of the bus, measured in Rankine.
If we know that [tex]\epsilon = 0.90[/tex], [tex]A = 16.188\,ft^{2}[/tex], [tex]\sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}[/tex], [tex]T_{s} = 554.07\,R[/tex] and [tex]T_{b} = 527.67\,R[/tex], then the heat transfer rate due to electromagnetic radiation is:
[tex]\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}][/tex]
[tex]\dot Q = 417.492\,\frac{BTU}{h}[/tex]
Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:
[tex]Q = \dot Q \cdot \Delta t[/tex] (2)
Where [tex]\Delta t[/tex] is the heat transfer time, measured in hours.
If we know that [tex]\dot Q = 417.492\,\frac{BTU}{h}[/tex] and [tex]\Delta t = \frac{1}{3}\,h[/tex], then the net energy transfer is:
[tex]Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)[/tex]
[tex]Q = 139.164\,BTU[/tex]
The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.
A thin-walled sphere of 2m mean diameter with a wall thickness of100mm is subjected to an internal pressure of 10MPa. Biaxialcircumferential stresses are developed and calculated by σ 1 = σ 2 =pd/(4t), where d is the mean diameter and t the thickness. Neglectingthe radial stress, calculate the ratio of the von-Mises stress over themaximum shear stress.
Answer:
ratio = 1
Explanation:
Given data :
mean diameter ( D ) = 2m
wall thickness( t ) = 100 mm
internal pressure ( P ) = 10 MPa
where : σ1 = pd/4t = ( 10*2000 ) / ( 4 * 100 ) = 50 MPa
also ; σ2 = 50 MPa
next calculate the Von-mises stress
attached below is the remaining part of the solution
next calculate the maximum stress
attached below
hence ratio of Von-mises stress over maximum shear stress =
= 50 / (2*25 ) = 1
In a paragraph explain the tradeoffs an engineer would make in selecting a wood with a rectangle shape versus manufactured beams with other stronger but lighter weight shapes.
Answer:
Wood is strong
Explanation:
Determine the force in members FD and DB of the frame. Also, find the horizontal and vertical components of reaction the pin at C exerts on member ABC and member EDC.
Answer:
A.
Explanation:
____ emotions can influence your driving. A. Only some B. All of your C. Only negative D. Only positive
The task of framing a building has been estimated to take anaverage of 25 days with astandard deviation of 4 days. What duration should be used if there is to be a 90%confidence that the duration will not be exceeded?
Answer:
30.128 days
Explanation:
Given that:
Mean = 25
Standard deviation = 4
Confidence interval = 90% = 0.9
Since the confidence interval should not exceed 90%
Then using z test table
P(z) = 0.9
For 0.8997 , we get = 1.28
For 0.9015, we get = 1.29
∴
[tex]\dfrac{0.9 - 0.8897}{0.9015 - 0.8997 }=\dfrac{ z - 1.28}{1.29 -1.28}[/tex]
By solving
Z = 1.282
Thus, the duration to be used so that it will not exceed 90% C.I is:
Z = (x - μ)/σ
1.282= ( x - 25)/4
1.282 * 4 = x - 25
(1.282*4)+25 = x
x = 30.128 days
A wet electrode can cause arc blow ?
Answer:
yes it can
Explanation:
The transition zone in which the ocean's density increases rapidly with depth is called the:___________
Answer: Pycnocline.
Explanation:
The Pycnocline is the layer, where there is a significant change in density of water with respect to depth.
In freshwater environments such as lakes this density change is primarily as a result of change in water temperature, while in seawater environments e.g oceans the cause of change in density change are changes in water temperature or salinity.
35 points and brainiest A, B, C, D
Which option identifies the terms that correctly complete the following statement?
A hammer is to a nail as a _____ is to a _____.
bolt, nut
mallet, screw
socket wrench, bolt
posthole digger, earth
Answer:
Mallet, Screw is the correct answer.
Explanation:
Hope this will help you.
By,
Zeeshan Khan