Answer:
Updated question
The big ben clock tower in London has clocks on all four sides. If each clock has a minute hand that is 11.5 feet in length, how far does the tip of each hand travel in 52 minutes?
The distance traveled by the tip of the minute hand of the clock would be 62.59 ft
Explanation:
Let us assume the shape of the clock is circular.
the minute hand is equal to the radius = 11.5 ft
Diameter = radius x 2
Diameter = 11.5 x 2 = 23 ft
The distance traveled by the tip of the minute hand can be calculated thus;
the fraction of the circumference traveled by the minute hand would be;
52/60 = 0.8667
Circumference of the clock would be;
C = pi x d
where C is the circumference
pi is a constant
d is the diameter
C = 3.14 x 23
C = 72.22 ft
Therefore the fraction of the circumference covered by the minute hand would be;
72.22 ft x 0.8667 = 62.59 ft
Therefore the distance traveled by the tip of the minute hand of the clock would be 62.59 ft
A hair dryer is basically a duct in which a few layers of electric resistors are placed. A small fan pulls the air in and forces it to flow over the resistors, where it is heated. Air enters a 1400-W hair dryer at 100 kPa and 22°C and leaves at 47°C. The cross-sectional area of the hair dryer at the exit is 60 cm2. Neglect the power consumed by the fan and the heat losses through the walls of the hair dryer. The gas constant of air is R = 0.287 kPa·m3/kg·K. Also, cp = 1.007 kJ/kg·K for air at room temperature.
determine
(a) the volume flow rate of air at the inlet and
(b) the velocity of the air at the exit.
Answer:
a) volume flow rate of air at the inlet is 0.0471 m³/s
b) the velocity of the air at the exit is 8.517 m/s
Explanation:
Given that;
The electrical power Input W_elec = -1400 W = -1.4 kW
Inlet temperature of air T_in = 22°C
Inlet pressure of air p_in = 100 kPa
Exit temperature T_out = 47°C
Exit area of the dyer is A_out = 60 cm²= 0.006 m²
cp = 1.007 kJ/kg·K
R = 0.287 kPa·m3/kg·K
Using mass balance
m_in = m_out = m_air
W _elec = m_air ( h_in - h_out)
we know that h = CpT
so
W _elec = m_air.Cp ( T_in - T_out)
we substitute
-1.4 = m_air.1.007 ( 22 - 47 )
-1.4 = - m_air.25.175
m_air = -1.4 / - 25.175
m_ air = 0.0556 kg/s
a) volume flow rate of air at the inlet
we know that
m_air = P_in × V_in
now from the ideal gas equation
P_in = p_in / RT_in
we substitute our values
= (100×10³) / ((0.287×10³)(22+273))
= 100000 / 84665
P_in = 1.18 kg/m³
therefore inlet volume flowrate will be;
V_in = m_air / P_in
= 0.0556 / 1.18
= 0.0471 m³/s
the volume flow rate of air at the inlet is 0.0471 m³/s
b) velocity of the air at the exit
the mass flow rate remains unchanged across the duct
m_ air = P_in.A_in.V_in = P_out.A_out.V_out
still from the ideal gas equation
P_out = p_out/ RT_out ( assume p_in = p_out)
P_out = (100×10³) / ((0.287×10³)(47+273))
P_out = 1.088 kg/m³
so the exit velocity will be;
V_out = m_air / P_out.A_out
we substitute our values
V_out = 0.0556 / ( 1.088 × 0.006)
= 0.0556 / 0.006528
= 8.517 m/s
Therefore the velocity of the air at the exit is 8.517 m/s
An equal-tangent sag vertical curve is designed with a PVC as station 109+00 and elevation 950ft, the PVI has a station of 110+77 and elevation of 947.34ft, and the low point at station 110+50. Determine the design speed of the cure.
Answer:
K = 96 and the design speed of the curve = 50mph
Explanation:
109+00 = 10900
Elevation = 950ft
110+77 = 11077
Elevation = 947.34ft
Station of low point = 110+50 = 11050
To get grade of curve
Gi = 947.34-950/11077-10900
= -2.66/177
= -0.015x100
= -1.5%
Locate of low point (XL)
= 11050-10900
= 150
To get the value of K
XL = |GL| x K
When we substitute values
150 = 1.5 x K
150 = 1.5K
K = 150/1.5
K = 100
The suitable and most nearest value is K = 96
Then we use the standard chart to get the design speed for K = 96
On this chart, the design speed for the curve = 50mph
Therefore K = 96 and speed = 50mph
Consider the string length equal to 7. This string is distorted by a function f (x) = 2 sin(2x) - 10sin(10x). What is the wave formed in this string? a. In=12cos (nit ) sin(max) b. 2cos(2t)sin (2x) - 10cos(10t ) sin(10x) c. n 2 sin 2x e' – 10sin 10x e
Answer:
hello your question has a missing part below is the missing part
Consider the string length equal to [tex]\pi[/tex]
answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)
Explanation:
Given string length = [tex]\pi[/tex]
distorted function f(x) = 2sin(2x) - 10sin(10x)
Determine the wave formed in the string
attached below is a detailed solution of the problem
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Answer:
The Answer is C!
Explanation:
Tubular centrifuge is used for recovering cells 60% of the cells or recover data flow rate of 12 l/min with a rotational speed of 4000 RPM what is the RPM to increase the recovery rate of the cells to 95% at the same flow rate
Answer:
The RPM to increase the recovery rate of the cells to 95% at the same flow rate is 6,333.3 RPM.
Explanation:
If the tubular centrifuge rotates at about 4,000 revolutions per minute to recover 60% of the cells, in case of wanting to recover 95% of the cells, the following calculation must be carried out to determine the required number of revolutions per minute:
60 = 4,000
95 = X
((95 x 4,000) / 60)) = X
(380,000 / 60) = X
6,333.3 = X
Therefore, as the calculation emerges, the tubular centrifuge will need to rotate at about 6,333.3 revolutions per minute to recover 95% of the cells in the same time.
Instead of running blood through a single straight vessel for a distance of 2 mm, one mammalian species uses an array of 100 tiny parallel pipes of the same total cross-sectional area, 4.0 mm2. Volume flow is 1000 mm3/s. 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
Solution:
Given that :
Volume flow is, [tex]$Q_1 = 1000 \ mm^3/s$[/tex]
So, [tex]$Q_2= \frac{1000}{100}=10 \ mm^3/s$[/tex]
Therefore, the equation of a single straight vessel is given by
[tex]$F_{f_1}=\frac{8flQ_1^2}{\pi^2gd_1^5}$[/tex] ......................(i)
So there are 100 similar parallel pipes of the same cross section. Therefore, the equation for the area is
[tex]$\frac{\pi d_1^2}{4}=1000 \times\frac{\pi d_2^2}{4} $[/tex]
or [tex]$d_1=10 \ d_2$[/tex]
Now for parallel pipes
[tex]$H_{f_2}= (H_{f_2})_1= (H_{f_2})_2= .... = = (H_{f_2})_{10}=\frac{8flQ_2^2}{\pi^2 gd_2^5}$[/tex] ...........(ii)
Solving the equations (i) and (ii),
[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{\frac{8flQ_1^2}{\pi^2 gd_1^5}}{\frac{8flQ_2^2}{\pi^2 gd_2^5}}$[/tex]
[tex]$=\frac{Q_1^2}{Q_2^2}\times \frac{d_2^5}{d_1^5}$[/tex]
[tex]$=\frac{(1000)^2}{(10)^2}\times \frac{d_2^5}{(10d_2)^5}$[/tex]
[tex]$=\frac{10^6}{10^7}$[/tex]
Therefore,
[tex]$\frac{H_{f_1}}{H_{f_2}}=\frac{1}{10}$[/tex]
or [tex]$H_{f_2}=10 \ H_{f_1}$[/tex]
Thus the answer is option A). 10
please help i have no xlue
A system samples a sinusoid of frequency 230 Hz at a rate of 175 Hz and writes the sampled signal to its output without further modification. Determine the frequency that the sampling system will generate in its output.
a. 120
b. 55
c. 175
d. 230
Compute the discharge observed at a v-notch weir. The weir has an angle of 90-degrees. The height above the weir is 3 inches.
Answer: the discharge observed at a v-notch weir is 66.7 in³/s
Explanation:
Given that;
Notch angle ∅ = 90°
height above the weir is 3 inches { head + head correction factor) h + k = 3 in
Discharge Q = ?
To determine the discharge observed, we us the following expression
Q = 4.28Ctan(∅/2) ( h + k )^5/2
where Q is discharge, C is discharge coefficient, ∅ is notch angle, h is head and k is head correction factor
now we substitute
Q = 4.28 × 1 × tan(90/2) ( 3 )^5/2
Q = 4.28 × 1 × 1 × 15.5884
Q = 66.7 in³/s
Therefore the discharge observed at a v-notch weir is 66.7 in³/s
The structure of a house is such that it loses heat at a rate of 3800 kJ/h per C di erence between the indoors and outdoors. A heat pump that requires a power input of 4 kW is used to maintain this house at 24C. Determine the lowest outdoor temperature for which the heat pump can meet the heating requirements of this house.
Answer:
-9.5° C
Explanation:
See attachment for calculations.
On the concluding parts, from the attachment, we have that
√[(297000 * 4)/(1056)] = 297 - T(l), and solving further, we get
297 - T(l) = √(1188000/1056)
297 - T(l) = √1125
297 - T(l) = 33.5
T(l) = 297 - 33.5
T(l) = 263.5
When you convert back to °C, we have
263.5 - 273 = -9.5° C
The drone hovers for 29 seconds at a height of 53 meters above the ground while taking a picture of a crowd. What is the vertical impulse by the lift force (combined force of the four rotors) during this time (in Newton-seconds)?
This question is incomplete, the complete question is;
All of the questions for this quiz involve a drone with a mass of 1.2 kg.
The drone hovers for 29 seconds at a height of 53 meters above the ground while taking a picture of a crowd. What is the vertical impulse by the lift force (combined force of the four rotors) during this time (in Newton-seconds)?
Answer: the vertical impulse is 341.388 Ns
Explanation:
Given that;
mass m = 1.2 kg
t = 29 sec
h = 53 m
vertical impulse = ?
Vertical impulse by lift force is given as;
Vertical Impulse = F × t
= mg × t
so we substitute
Vertical impulse = 1.2 ×9.81 × 29
= 341.388 Ns
therefore the vertical impulse is 341.388 Ns
Which method of freezing preserves the quality and taste of food?
Answer:
commercial freezing
Explanation:
smaller ice crystals are formed this causes less damage to cell membranes so the quality is less effected
A spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons. If the spring is stretched 1 meters beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.
Answer:
0.75 kg
Explanation:
c = Damping coefficient = 3 kg/s
x = Displacement of spring = 0.5 m
F = Force = 1.5 N
From Hooke's law we get
[tex]F=kx\\\Rightarrow k=\dfrac{F}{x}\\\Rightarrow k=\dfrac{1.5}{0.5}\\\Rightarrow k=3\ \text{N/m}[/tex]
In the case of critical damping we have the relation
[tex]c^2-4mk=0\\\Rightarrow m=\dfrac{c^2}{4k}\\\Rightarrow m=\dfrac{3^2}{4\times 3}\\\Rightarrow m=0.75\ \text{kg}[/tex]
Mass that would produce critical damping is 0.75 kg.
0.75 kg is the mass that would produce critical damping. As spring with an m-kg mass and a damping constant 3 (kg/s) can be held stretched 0.5 meters beyond its natural length by a force of 1.5 newtons.
What is zero velocity?A change in time and position is referred to as an object's velocity. When there is no movement of the object, the velocity of the object is said to be 0.
For any body in planar motion, the velocity is always instantaneously 0 at some point in the plane of motion (if it were rigidly connected to the body). This place is known as the instantaneous center of zero velocity, or IC.
Example: The gravitational pull of the earth pushes the ball away from the thrower when a ball is thrown upwards on Earth at a constant speed. The speed of the ball increases until it reaches its maximum, at which point it starts to plummet.
Thus, it is 0.75 kg.
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The pascal is actually a very small unit of pressure. To show this, convert 1 Pa = 1 N/m² to lb/ft². Atmosphere pressure at sea level is 14.7 lb/in². How many pascals is this?
Answer:
pascals is this = 101352.972 Pa
Explanation:
given data
Atmosphere pressure at sea level = 14.7 lb/in²
we convert 1 Pa = 1 N/m² to lb/ft²
so we convert here 14.7 lb/in² to pascals
we know that 1 lb/ft² = 47.990172 N/m²
so
1 lb/ft² × ft²/(12in)² = 47.990172 × 144 N/m²
it will be simplyfy
1 lb/ft² = 6894.76 N/m²
so
14.7 lb/in² = 14.7 × 6894.76 N/m²
14.7 lb/in² = 101352.972 Pa
The structure of PF3(C6H5)2 is trigonal bipyramidal, with one equatorial and two axial F atoms which interchange positions when heated. Describe the low- and high- temperature 31P and 19F NMR spectra.
Answer:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
Explanation:
For 31 P NMR spectra
low temperature
there is two types of 19f seen in low temperature and they are
one at equitorial one at axialtherefore in low temperature the 31p couples with the two types of 19F seen ( [tex]b_{f} and c_{f}[/tex]to form a triplet and this couples more with [tex]a_{f}[/tex] to form a doublet. i.e. one (1) peak
High temperature
At High temperature The exchange is fast here therefore the 31p spectra sees all 19p at once and in the same environment leading to the formation of one (1) peak
For 19 P NMR spectra
low temperature
In low temperature [tex]a_{f}, b_{f} , c_{f}[/tex] is fixed and the environment where [tex]b_{f} and c _{f}[/tex] is the same hence a peak is formed and another peak is formed by [tex]a_{f}[/tex] that makes the number of peaks = 2 peaks
High temperature
In high temperature [tex]a_{f}, b_{f} , c_{f}[/tex] exchange very fast therefore one peak is formed for all, since the fast exchanges makes NMR machine to take an average and produce just one peak for all
What is the Bernoulli formula?
Answer:
P1+1/2pv2/1+pgh1=P2+1/2pv2/2+pgh2
What is computer programming
Answer:
Computer programming is where you learn and see how computers work. People do this for a living as a job, if you get really good at it you will soon be able to program/ create a computer.
Explanation:
Hope dis helps! :)
While at a concert you notice five people in the crowd headed in the same direction. Your tendency to group them is due to? *
Answer:
common fate
Explanation:
The gestalt effect may be defined as the ability of our brain to generate the whole forms from the groupings of lines, points, curves and shapes. Gestalt theory lays emphasis on the fact that whole of anything is much greater than the parts.
Some of the principles of Gestalt theory are proximity, similarity, closure, symmetry & order, figure or ground and common fate.
Common fate : According to this principle, people will tend to group things together which are pointed towards or moving in a same direction. It is the perception of the people that objects moving together belongs together.
A uniform edge load of 500 lb/in. and 350 lb/in. is applied to the polystyrene specimen. If it is originally square and has dimensions of a = 2 in., b = 2 in., and a thickness of t = 0.25 in., determine its new dimensions a, b, and t after the load is applied. Ep = 597(10)3 psi, vp = 0.25.
The image of the load applied to the polystyrene is missing, so i have attached it.
Answer:
a_new = 2.00302 in
b_new = 2.00552
Explanation:
From the image attached, we can see that the load of 500 lb/in is applied in the x-direction while the load of 350 lb/in acts in the y-direction.
Now, formula for stress is;
Stress(σ) = Force/Area
We are not given force and area but the load and plate thickness.
Thus, stress = load/thickness
We are given;
Load in x - direction = 500 lb/in.
Load in y - direction = 350 lb/in.
Thickness; t = 0.25 in
Thus;
σ_x = 500/0.25
σ_x = 2000 ksi
σ_y = 350/0.25
σ_y = 1400 ksi
From Hooke's law for 2 dimensions, strain is given by the formula;
ε_x = (1/E)(σ_x - vσ_y)
ε_y = (1/E)(σ_y - vσ_x)
We are given v_p = 0.25 and Ep = 597 × 10³ psi
Thus;
ε_x = (1/(597 × 10^(3)))(2000 - (0.25 × 1400)
ε_x = 0.00276
ε_y = (1/(597 × 10^(3)))(1400 - (0.25 × 2000)
ε_y = 0.00151
From elongation formula, we know that;
Startin is: ε = ΔL/L
Thus; ΔL = Lε
We are given a = 2 and b = 2
Thus;
ΔL_x = 2 × 0.00276
ΔL_x = 0.00552
ΔL_y = 2 × 0.00151
ΔL_y = 0.00302
New dimensions are;
a_new = 2 + 0.00302
a_new = 2.00302 in
b_new = 2 + 0.00552
b_new = 2.00552
Calculate the normal stress along the x-direction.
[tex]\to \sigma_x =\frac{500}{t}=\frac{500}{0.25}= 2000 \ \frac{lb}{in^2}\\\\[/tex]
Calculate the normal stress along the y-direction.
[tex]\to \sigma_y =\frac{350}{t}=\frac{350}{0.25}= 1400 \ \frac{lb}{in^2}[/tex]
Calculate the strain along the x-direction.
[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_y-\sigma_z)]\\\\[/tex]
[tex]=\frac{1}{597\times 10^3} [2000-0.25(1400+0)] \\\\= 2.764\times 10^{-3}\\\\[/tex]
Calculate the strain along the y-direction.
[tex]\to \varepsilon_y=\frac{1}{E}[\sigma_x-v(\sigma_x+\sigma_z)][/tex]
[tex]=\frac{1}{597\times 10^3}[1400 -0.25 (2000+0)]\\\\=1508\times 10^{-3}[/tex]
Calculate the strain along the z-direction.
[tex]\to \varepsilon_x= \frac{1}{E}[\sigma_z-v(\sigma_x +\sigma_y)][/tex]
[tex]=\frac{1}{597\times 10^3} [0-0.25(2000+1400)] \\\\=-1.424\times 10^{-3}[/tex]
Calculate the new dimensions.
[tex]\to b' =b+ \varepsilon_xb\\\\[/tex]
[tex]= 2+2.764\times 10^{-3} \times 2\\\\= 2.005528 \ in\\\\[/tex]
[tex]\to a' = a + \varepsilon_y a\\\\[/tex]
[tex]= 2+1.508\times 10^{-3} \times 2\\\\= 2.003016\ in\\\\[/tex]
[tex]\to c'= c + \varepsilon_x c\\\\[/tex]
[tex]= 0.25 +(-1.424\times 10^{-3}) \times 0.25\\\\= 0.249644\ in\\\\[/tex]
Find out more about the dimensions here:
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Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3
/s and at a
velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Figure.
Determine the force acting on the shaft (which is
also the force acting on the bearing of the shaft) in
the axial direction.
Answer:
Water enters a centrifugal pump axially at atmospheric pressure at a rate of 0.12 m3/s and at a velocity of 7 m/s, and leaves in the normal direction along the pump casing, as shown in Fig. PI3-39. Determine the force acting on the shaft (which is also the force acting on the bearing of the shaft) in the axial direction.
Step-by-step solution:
Step 1 of 5
Given data:-
The velocity of water is .
The water flow rate is.
A smooth ceramic sphere (SG 5 2.6) is immersed in a fl ow of water at 208C and 25 cm/s. What is the sphere diameter if it is encountering (a) creeping motion, Red 5 1 or (b) transition to turbulence, Red 5 250,000
Answer:
a. 4[tex]\mu m[/tex]
b. 1 m
Explanation:
According to the question, the data is as follows
The Density of water at 20 degrees celcius is 1000 kg/m^3
Viscosity is 0.001kg/m/.s
Velocity V = 25 cm/s
V = 0.25 m/s
Now
a. The creeping motion is
As we know that
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
1 = (1,000 × 0.25 × d) ÷ 0.0001
d = (1 × 0.001) ÷ (1,000 × 0.25)
= 4E - 06^m
= 4[tex]\mu m[/tex]
b. Now the sphere diameter is
Reynold Number = (Density of water × V × d) ÷ (Viscosity)
250,000 = (1,000 × 0.25 × d) ÷ 0.0001
d = (250,000 × 0.001) ÷ (1,000 × 0.25)
= 1 m
Which of the following is an example of a tax
Answer:
A tax is a monetary payment without the right to individual consideration, which a public law imposes on all taxable persons - including both natural and legal persons - in order to generate income. This means that taxes are public-law levies that everyone must pay to cover general financial needs who meet the criteria of tax liability, whereby the generation of income should at least be an auxiliary purpose. Taxes are usually the main source of income of a modern state. Due to the financial implications for all citizens and the complex tax legislation, taxes and other charges are an ongoing political and social issue.
I dont know I asked this to
Explanation:
Consider diodes in a rectifier circuit. Input voltage is sinusoidal with a peak of +/-10 V. Diode drop is 0.7 V. What is the PIV for each type rectifier 1. 0.7 V 2. 1.4 V 3. 10.7 V 4. 11.4 V Bridge rectifier 5. 19.3 V Full-wave rectifier 6. 8.6 V 7. 9.3 V Half-wave rectifier 8. 7.2 V 9. 12.1 V 10. 12.8 V 11. 10 V
Answer is given below:
Explanation:
Peak inverse voltage (PIV) can be defined as the maximum value of the reverse voltage of the diode, which is the maximum value of the input cycle when the diode is on. In reverse bias. Happens. 9.3V for braid rectifiers cut at 0.7The center tapered rectifier has 2 diodes in parallel so the maximum voltage is 2Vm so the answer to cut off the 0.7 voltage is19.3V. For a half wave rectifier it is Vm i.e. 10 V.A layer of viscous fluid of constant thickness (no velocity perpenducilar to plate) flows steadily down an infinite, inclined plane. Determine, with the Navier Stokes equations, the flowrate per unit width as a function of flow height.
Answer:
q = (ρg/μ)(sin θ)(h³/3)
Explanation:
I've attached an image of a figure showing the coordinate system.
In this system: the velocity components v and w are equal to zero.
From continuity equation, we know that δu/δx = 0
Now,from the x-component of the navier stokes equation, we have;
-δp/δx + ρg(sin θ) + μ(δ²u/δy²) = 0 - - - - - (eq1)
Due to the fact that we have a free surface, it means we will not have a pressure gradient in the x-component and so δp/δx = 0
Then our eq 1 is now;
ρg(sin θ) + μ(δ²u/δy²) = 0
μ(δ²u/δy²) = -ρg(sin θ)
Divide both sides by μ to get;
(δ²u/δy²) = -(ρg/μ)(sin θ)
Integrating both sides gives;
δu/δy = -(ρg/μ)(sin θ)y + b1 - - - - (eq2)
Now, the shear stress is given by the formula;
τ_yx = μ[δu/δy + δv/δx]
From the diagram, at the free surface,τ_yx = 0 and y = h
This means that δu/δy = 0
Thus, putting 0 for δu/δy in eq 2, we have;
0 = -(ρg/μ)(sin θ)h + b1
b1 = h(ρg/μ)(sin θ)
So, eq 2 is now;
δu/δy = -(ρg/μ)(sin θ)y + h(ρg/μ)(sin θ)
Integrating both sides gives;
u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y + b2 - - - eq3
Because u = 0 when y = 0, it means that b2 = 0 also because when we plug 0 for u and y into eq3, we will get b2 = 0.
Thus, we now have:
u = -[(y²/2) × (ρg/μ)(sin θ)] + h(ρg/μ)(sin θ)y
Factorizing like terms, we have;
u = (ρg/μ)(sin θ)[hy - y²/2] - - - (eq 4)
The flow rate per unit width is gotten by Integrating eq 4 between the boundaries of h and 0 to give;
∫u = (h,0)∫(ρg/μ)(sin θ)[hy - y²/2]
q = (ρg/μ)(sin θ)[hy²/2 - y³/6] between h and 0
q = (ρg/μ)(sin θ)[h³/2 - h³/6]
q = (ρg/μ)(sin θ)(h³/3)
Find the magnitude of the steady-state response of the system whose system model is given by dx(t)/dt+ x(t)-f(t), where f(t) 2cos8t. Keep 3 significant figures
This question is incomplete, the complete question is;
Find the magnitude of the steady-state response of the system whose system model is given by
dx(t)/dt + x(t) = f(t)
where f(t) = 2cos8t. Keep 3 significant figures
Answer: The steady state output x(t) = 0.2481 cos( 8t - 45° )
Explanation:
Given that;
dx(t)/dt + x(t) = f(t) where f(t) = 2cos8t
dx(t)/dt + x(t) = f(t)
we apply Laplace transformation on both sides
SX(s) + x(s) = f(s)
(S + 1)x(s) = f(s)
f(s) / x(s) = S + 1
x(s) / f(s) = 1 / (S + 1)
Therefore
transfer function = H(s) = x(s)/f(s) = 1/(S+1)
f(t) = 2cos8t → [ 1 / ( S + 1 ) ] → x(t) = Acos(8t - ∅ )
A = Magnitude of steady state output
S = jw
S = j8
so
A = 2 × 1 / √( 8² + 1 ) = 2 / √ (64 + 1 )
A = 2/√65 = 0.2481
∅ = tan⁻¹( 1/1) = 45°
therefore The steady state output x(t) = 0.2481 cos( 8t - 45° )
Which kind of fracture (ductile or brittle) is associated with each of the two crack propagation mechanisms?
dutile is the correct answer
I need help please thank for the help on the last one <3
g A smooth pipeline with a diameter of 5 cm carries glycerin at 20 degrees Celsius. The flow rate in the pipe is 0.01 m3/s. What is the friction factor
Answer:
The friction factor is 0.303.
Explanation:
The flow velocity ([tex]v[/tex]), measured in meters per second, is determined by the following expression:
[tex]v = \frac{4\cdot \dot V}{\pi \cdot D^{2}}[/tex] (1)
Where:
[tex]\dot V[/tex] - Flow rate, measured in cubic meters per second.
[tex]D[/tex] - Diameter, measured in meters.
If we know that [tex]\dot V = 0.01\,\frac{m^{3}}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the flow velocity is:
[tex]v = \frac{4\cdot \left(0.01\,\frac{m^{3}}{s} \right)}{\pi\cdot (0.05\,m)^{2}}[/tex]
[tex]v \approx 5.093\,\frac{m}{s}[/tex]
The density and dinamic viscosity of the glycerin at 20 ºC are [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex] and [tex]\mu = 1.5\,\frac{kg}{m\cdot s}[/tex], then the Reynolds number ([tex]Re[/tex]), dimensionless, which is used to define the flow regime of the fluid, is used:
[tex]Re = \frac{\rho\cdot v \cdot D}{\mu}[/tex] (2)
If we know that [tex]\rho = 1260\,\frac{kg}{m^{3}}[/tex], [tex]\mu = 1.519\,\frac{kg}{m\cdot s}[/tex], [tex]v \approx 5.093\,\frac{m}{s}[/tex] and [tex]D = 0.05\,m[/tex], then the Reynolds number is:
[tex]Re = \frac{\left(1260\,\frac{kg}{m^{3}} \right)\cdot \left(5.093\,\frac{m}{s} \right)\cdot (0.05\,m)}{1.519 \frac{kg}{m\cdot s} }[/tex]
[tex]Re = 211.230[/tex]
A pipeline is in turbulent flow when [tex]Re > 4000[/tex], otherwise it is in laminar flow. Given that flow has a laminar regime, the friction factor ([tex]f[/tex]), dimensionless, is determined by the following expression:
[tex]f = \frac{64}{Re}[/tex]
If we get that [tex]Re = 211.230[/tex], then the friction factor is:
[tex]f = \frac{64}{211.230}[/tex]
[tex]f = 0.303[/tex]
The friction factor is 0.303.
Refrigerant-22 absorbs heat from a cooled space at 50°F as it flows through an evaporator of a refrigeration system. R-22 enters the evaporator at 10°F at a rate of 0.08 lbm/s with a quality of 0.3 and leaves as a saturated vapor at the same pressure. Determine:
a. The rate of cooling provided, in Btu/h.
b. The rate of exergy destruction in the evaporator.
c. The second-law efficiency of the evaporator.
Take T0 = 77°F. The properties of R-22 at the inlet and exit of the evaporator are: h1 = 107.5 Btu/lbm, s1 = 0.2851 Btu/lbm·R, h2 = 172.1 Btu/ lbm, s^2 = 0.4225 Btu/lbm·R.
Answer:
a) the rate of cooling provided is 18604.8 Btu/h
b) the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm
c) the second-law efficiency of the evaporator is 37.39%
Explanation:
Given that;
Temperature of sink TL = 50°F = 510 R
Temperature at evaporator inlet TI = 10°F = 470 R
mass flow rate m" = 0.08 lbm/s
quality of refrigerant at evaporator inlet x1 = 0.3
quality of refrigerant at evaporator exit x2 = 1.0
T₀ = 77°F = 537 R
h1 = 107.5 Btu/lbm
s1 = 0.2851 Btu/lbm·R,
h2 = 172.1 Btu/ lbm,
s2 = 0.4225 Btu/lbm·R.
a) rate of cooling provided, in Btu/h.
QL = m"( h2 - h1)
we substitute
QL = 0.08( 172.1 - 107.5
= 0.08 × 64.6
= 5.168 Btu/s
we convert to Btu/h
5.168 × 60 × 60
QL = 18604.8 Btu/h
Therefore the rate of cooling provided is 18604.8 Btu/h
b) The rate of exergy destruction in the evaporator
Entropy generation can be expressed as;
S_gen = m"(s2 - s1) - QL/TL
so we substitute
S_gen = 0.08( 0.4225 - 0.2851 ) - 5.168 / 510
= 0.010992 - 0.01013
S_gen = 0.00086 Btu/ibm.R
now the energy destroyed expressed as;
X_dest = T₀ × S_gen
so
X_dest = 537 × 0.00086
X_dest = 0.46 Btu/Ibm
Therefore the rate of exergy destruction in the evaporator is 0.46 Btu/Ibm
c) The second-law efficiency of the evaporator.
Energy expended is expressed as;
X_exp = m"(h1 - h2) - m"T₀(s1 - s2)
we substitute
= 0.08( 107.5 - 172.1 ) - [0.08 × 537 ( 0.2851 - 0.4225 )
= -5.168 - [ - 5.9027)
= -5.168 + 5.9027
= 0.7347 Btu/s
Now second law efficiency is expressed as;
nH = 1 - (X_dest / X_esp)
= 1 - ( 0.46 / 0.7347 )
= 1 - 0.6261
= 0.3739
nH = 37.39%
Therefore the second-law efficiency of the evaporator is 37.39%
A tube of diameter 3 cm and length 3 m has a water flow of 100 cm3/s. If the pollutant concentration in the water is constant at 2 mg/L, find the mass flux (mg/cm2-s) of pollutant through the tube due to advection.
Answer: the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s
Explanation:
Given that;
Diameter of tube = 3 cm, radius r = 1.5 cm
water flow is 100 cm³/s
pollutant concentration = 2 mg/L
first we find the rate of flow of pollutant
we know that
1 L = 1000 cm³
xL = 100 cm³
100Lcm³ = xL1000cm³
xL = 100/1000
xL = 1/10 L
so 100cm³ = 1/10 L
now pollutant concentration in 100 cm³ = 1/10L × 2mg/L = 0.2 mg
Rate of flow of pollutant = 0.2 mg/s
Mass flux density is the pollutant mass per unit time per unit area
so Area of tube = πr² = 3.14 × 1.5² = 7.065 cm²
So
Mass flux = 0.2 / 7.065
Mass flux = 0.0283 mg/cm².s
Therefore, the mass flux of pollutant through the tube due to advection is 0.0283 mg/cm².s