Answer a) roof equilibrium temperature is 400K
Explanation:
The roof surface is sorrounded by ambient air whose temperature is 300K and which has a convective coefficient h of 10W/m2-K.
This means that heat will be conducted to the roof from the air around at an heat Flux of 300K x 10W/m2-K = 3000w/m2
For a clear solar day with solar heat Flux of 500W/m2, total heat Flux on roof will be
Q = 500 + 3500 = 3500W/m2
Q = h(Tr-Ta)
Ts is temperature of roof,
Ta is temperature of air
3500 = 10(Tr - 50)
350 = Tr - 50
Tr = 400k
Answer b): roof equilibrium temperature will be 350K
Explanation:
At night total heat Flux is only due to hot ambient air sorrounding the surface of the roof.
Q = 3000W/m2
Q = h(Tr-Ta)
3000 = 10(Tr-50)
300 = Tr - 50
Tr = 350K
Approximation of PI: [50 marks] Write a C program that computes an approximation of pi (the mathematical constant used in many trigonometric and calculus applications) to four decimal places. A summation that can be used to compute pi is the following (from calculus).
Answer:
#include <iostream>
#include <cmath>
#include <iomanip>
using namespace std;
double piValue() {
double p = 1;
int i = 1;
while(true) {
double prev = p;
p += pow(-1, i) * (1.0 / (1 + 2*i));
if(abs(prev - p) < 0.00005) {
break;
}
i++;
}
return 4*p;
}
int main() {
cout << piValue() << endl;
}
**************************************************
Consider the economies of Hermes and Tralfamadore, both of which produce glops of gloop using only tools and workers. Suppose that, during the course of 30 years, the level of physical capital per worker rises by 5 tools per worker in each economy, but the size of each labor force remains the same.
Missing Part of Question
Complete the following tables by entering productivity (in terms of output per worker) for each economy in 2012 and 2042.
Hermes
Year (2002)
Physical Capital: 11 tools per worker
Labour Force: 30 workers
Output: 1,800 Gobs of goo
Productivity:__________
Year (2042)
Physical Capital: 16 tools per worker
Labour Force: 30 workers
Output: 2,160 Gobs of goo
Productivity:__________
Tralfamadore
Year (2002)
Physical Capital: 8 tools per worker
Labour Force: 30 workers
Output: 900 Gobs of goo
Productivity:__________
Year (2042)
Physical Capital: 13 tools per worker
Labour Force: 30 workers
Output: 1,620 Gobs of goo
Productivity:__________
Answer:
The Complete Table is as follows
Hermes
Year (2002)
Physical Capital: 11 tools per worker
Labour Force: 30 workers
Output: 1,800 Gobs of goo
Productivity:60 workers
Year (2042)
Physical Capital: 16 tools per worker
Labour Force: 30 workers
Output: 2,160 Gobs of goo
Productivity:72 workers
Tralfamadore
Year (2002)
Physical Capital: 8 tools per worker
Labour Force: 30 workers
Output: 900 Gobs of goo
Productivity:30 workers
Year (2042)
Physical Capital: 13 tools per worker
Labour Force: 30 workers
Output: 1,620 Gobs of goo
Productivity:54 workers
Explanation:
We calculate the productivity of both Hermes and Tralfamadore by dividing Output by Labor Force. This is given by:
Productivity = Output/Labour Force
For Hermes
In 2002
Productivity = 1800/30 = 60 workers
In 2042,
Productivity = 2160/30 = 72 workers
For Tralfamadore
In 2002
Productivity = 900/30 = 30 workers
In 2042,
Productivity = 1620/30 = 54 workers
A zener diode exhibits a constant voltage of 5.6 V for currents greater than five times the knee current. IZK is specified to be 1 mA. The zener is to be used in the design of a shunt regulator fed from a 15-V supply. The load current varies over the range of 0 mA to 15 mA. Find a suitable value for the resistor R. What is the maximum power dissipation of the zener diode?
Answer:
The maximum power dissipation of the zener diode 112mV.
Explanation:
The minimum zener current should be:
5 * Iza= 5 * 1= 5 mA.
Since the load current can be at maximum 15 mA, we should select R so that, IL= 15 mA.
A zener current of 5 mA is available, Thus the current should be 20 mA, which leads to,
R = [tex]\frac{15 - 5.6}{20 mA}[/tex] = 470 Ω.
Maximum power dissipated in the diode occours when, IL=0 is
Pmax = 20 * [tex]10^{3}[/tex] * 5.6 = 112mV.
A 1.5-m-long aluminum rod must not stretch more than 1 mm and the normal stress must not exceed 40 MPa when the rod is subjected to a 3.0-kN axial load. Knowing that E = 70 GPa, determine the required diameter of the rod.
Answer:
the required diameter of the rod is 9.77 mm
Explanation:
Given:
Length = 1.5 m
Tension(P) = 3 kN = 3 × 10³ N
Maximum allowable stress(S) = 40 MPa = 40 × 10⁶ Pa
E = 70 GPa = 70 × 10⁹ Pa
δ = 1 mm = 1 × 10⁻³ m
The required diameter(d) = ?
a) for stress
The stress equation is given by:
[tex]S = \frac{P}{A}[/tex]
A is the area = πd²/4 = (3.14 × d²)/4
[tex]S = \frac{P}{(\frac{3.14*d^{2} }{4}) }[/tex]
[tex]S = \frac{4P}{{3.14*d^{2} } }[/tex]
[tex]3.14*S*{d^{2}} = {4P}[/tex]
[tex]{d^{2}} =\frac{4P}{3.14*S}[/tex]
[tex]d= \sqrt{\frac{4P}{3.14*S} }[/tex]
Substituting the values, we get
[tex]d= \sqrt{\frac{4*3*10^{3} }{3.14*40*10^{6} } }[/tex]
[tex]d= \sqrt{\frac{12000 }{125600000 } }[/tex]
[tex]d= \sqrt{9.55*10^{-5} }[/tex]
d = (9.77 × 10⁻³) m
d = 9.77 mm
b) for deformation
δ = (P×L) / (A×E)
A = (P×L) / (E×δ) = (3000 × 1.5) / (1 × 10⁻³ × 70 × 10⁹) = 0.000063
d² = (4 × A) / π = (0.000063 × 4) / 3.14
d² = 0.0000819
d = 9.05 × 10⁻³ m = 9.05 mm
We use the larger value of diameter = 9.77 mm
an existing highway-railway at-grade crossing is being redesigned as grade separated to improve traffic operations. The railway must remain at the same elevation. The highway is being reconstructed to travel under the railway. The underpass will be a sag curve that has an initial grade of -2% and a final grade of 2%. The PVI of the sag curve will be centered under the railway (a symmetrical alignment). The sag curve design speed is 45 mi/h. How many feet below the railway should the curve PVI be located
Answer:
Please find attached solution
Explanation:
Write a SELECT statement that returns the same result set as this SELECT statement. Substitute a subquery in a WHERE clause for the inner join. SELECT DISTINCT VendorName FROM Vendors JOIN Invoices ON Vendors.VendorID = Invoices.VendorID ORDER BY VendorName
Answer:
SELECT distinct VendorName FROM Vendors
WHERE VendorID IN (
SELECT VendorID FROM Invoices
)
Explanation:
Please use in-memory join to answer this question. Given any two Users (they are friend) as input, output the list of the names and the date of birth (mm/dd/yyyy) of their mutual friends. Note: use the userdata.txt to get the extra user information. Output format: UserA id, UserB id, list of [names: date of birth (mm/dd/yyyy)] of their mutual Friends.
Answer:
See the attached java files for code.
Explanation:
See the attached files.
One cycle of the power dissipated by a resistor (R = 700 Ω) is given by P(t) 55 W, 0 < t < 18.0 s P(t) = 35 W, 18.0 < t < 30 s This periodic signal repeats in both directions of time. What is the average power dissipated by the 700-Ω resistor? Pau to within three significant digits)
Answer:
Average power dissipated by the 700-Ω resistor=47W
Explanation:
average power dissipated by the 700-Ω resistor
[tex]\frac{1}{t} \int\limits^t_0 {P(t)} \, dx \\=\frac{1}{30} (\int\limits^18_0 {55} \, dx + \int\limits^30_18 {35} \, dx )\\\\=\frac{1}{30} (55*(18-0) + 35(30-18))\\\\=\frac{1410}{30}\\[/tex]
=47W
Answer:
[tex]P=47[/tex] [tex]W[/tex]
Explanation:
The average power dissipated in a resistor is given by
[tex]P=\frac{1}{T} \int\limits^T_0 {} \, p(t)dt[/tex]
Where T is the time taken by the sine wave to complete one cycle.
We have instantaneous power P(t) = 55 W for 0 < t < 18s and P(t) = 35 W for 18 < t < 30s
So the time period is T = 30 seconds
Now we will integrate both of the instantaneous powers
[tex]P=\frac{1}{30} (\int\limits^b_a {} \, 55dt + \int\limits^b_a {35} \, dt )[/tex]
[tex]P=\frac{1}{30} (55t +{35} t )[/tex]
[tex]P=\frac{1}{30} (55(18-0) +{35(35-18)} )[/tex]
[tex]P=\frac{1}{30} (55(18) +{35(12}))[/tex]
[tex]P=\frac{1}{30} (990 +420)[/tex]
[tex]P=\frac{1410}{30}[/tex]
[tex]P=47[/tex] [tex]W[/tex]
A positive electric charge is moved at a constant speed between two locations in an electric field, with no work done by or against the field at any time during the motion This situation can occur only if the
(A) charge is moved in the direction of the field
(B) charge is moved opposite to the direction of the field
(C) charge is moved perpendicular to an equipotential line
(D) charge is moved along an equipotential line
(E) electric field is uniform
Answer:
D. Charge is moved along an equipotential line
Explanation:
This means that the potential will be the same sown each equipotential line, which implies that the charge does not require any work before it moves anywhere along the line. Although work is required for a charge to move from an equipotential line to another, in all situations equipotential lines are vertical to electric field lines.
A horizontal curve was designed for a four-lane highway for adequate SSD. Lane widths are 12 ft, and the superelevation is 0.06 and was set assuming maximum fs. If the necessary sight distance required 52 ft of lateral clearance from the roadway centerline, what design speed was used for the curve
Answer and Explanation:
The answer is attached below
Answer:
80 mi/h
Explanation:
To get the adequate design speed the assumed design speed ( Fs ) for the Horizontal curve has to produce the necessary Middle ordinate distance that is equal to the provided Middle ordinate
The provided middle ordinate is gotten form the formula
Ms = Available clearance - lane width - lane width / 2 ( equation 1 )
available clearance = 52 ft
lane width = 12 ft
hence equation 1 becomes
Ms = 52 - 12 - 12/2
= 40 - 6 = 34 ft
To check if the assumed design speed will produce the necessary middle ordinate distance that is equal to the provided middle ordinate the formula below is used
Ms₍necessary₎ = Rv ( 1 - cos[tex]\frac{90SSD}{\pi Rv }[/tex] ) (equation 2 )
Assuming a design speed ( Fs ) of 80 mi/h and referring to the engineering table and also considering the super elevation of 0.06 ( 6% ) the values of
Rv = 3050 ft
SSD = 910 ft
substituting these values into (equation 2) Ms ( necessary ) becomes
= 3050 ( 1 - cos [tex]\frac{90*910}{\pi * 3050 }[/tex] )
= 3050 ( 1 - cos [tex]\frac{81900}{\pi * 3050 }[/tex] )
= 33.876 ft
33.876 ft is approximately equal to the provided middle ordinate of 34 ft hence the design speed used would be 80mi/h
Oil with a specific gravity of 0.72 is used as the indicating fluid in a manometer. If the differential pressure across the ends of the manometer is 6kPa, what will be the difference in oil levels in the manometer?
Answer:
the difference in oil levels is 0.850 m
Explanation:
given data
specific gravity ρ = 0.72
pressure across P = 6 kPa = 6000 Pa
solution
we get here difference in oil levels h is
P = ρ × g × h .................1
here ρ = 0.72 × 1000 = 720 kg/m³
and g is 9.8
put here value in equation 1 and we get h
6000 = 720 × 9.8 × h
h = [tex]\frac{6000}{720\times 9.8}[/tex]
h = 0.850 m
so the difference in oil levels is 0.850 m
Using the given specific gravity of oil and the differential pressure, the difference in oil levels in the manometer is calculated as 0.84 meters.
Explanation:The difference in oil levels in a manometer when the differential pressure is 6kPa, and the specific gravity of the oil is 0.72, can be calculated using the formula: Δh = ΔP/(ρ*g), where Δh is the height difference, ΔP is the differential pressure, ρ is the density of the oil, and g is the acceleration due to gravity. This calculation assumes standard gravity (g = 9.81 m/s^2). The density of oil can be obtained from its specific gravity and the density of water (1000 kg/m^3). Thus:
ρ = 0.72 * 1000 kg/m^3 = 720 kg/m^3
Substituting the values into the formula, we get:
Δh = 6000 Pascals / (720 kg/m^3*9.81 m/s^2) = 0.84 meters
Hence, the difference in the oil levels in the manometer is approximately 0.84 meters.
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A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 19 kN are applied at the centers of the ends of the bar. Knowing that a = 28 mm and sigma_all = 130 MPa, determine the smallest allowable depth d of the milled portion of the bar.
Answer:
Complete question for your reference is below.
A milling operation was used to remove a portion of a solid bar of square cross section. Forces of magnitude P = 19 kN are applied at the centers of the ends of the bar. Knowing that a = 28 mm and sigma_all = 130 MPa, determine the smallest allowable depth d of the milled portion of the bar
Explanation:
Please find attached file for complete answer solution and explanation.
To find the smallest allowable depth 'd', we use the stress formula and the data provided in the question. By substituting these values into the formula, we can solve for 'd'.
Explanation:The question asks for the smallest allowable depth 'd' of the milled portion in an Engineering situation involving a solid square bar, forces, and a given stress limit. First, we need to realize this is a problem of Stress Analysis. According to the formula for stress in a bar under axial load, Stress (σ) = Force (P) / Area. The cross-sectional area of the reduced portion of the bar is now a*d (width*depth), hence σ = P / (a*d). Here, 'a' is 28 mm and 'P' is 19 kN. As per the problem, the maximum allowable stress (σ_all) is 130 MPa. By substitifying known values into formula, we get 130 = 19,000 / (28 * d), and solving this equation will give the minimum value for 'd'.
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The aluminum rod AB (G 5 27 GPa) is bonded to the brass rod BD (G 5 39 GPa). Knowing that portion CD of the brass rod is hollow and has an inner diameter of 40 mm, determine the angle of twist at A.
Answer:
Qcd=0.01507rad
QT= 0.10509rad
Explanation:
The full details of the procedure and answer is attached.
The velocity of a particle which moves along the s-axis is given by v = 2-4t+5t^(3/2), where t is in seconds and v is in meters per second. Evaluate the position s, velocity v, and acceleration a when t=4s. The particle is at the position s0 =2m when t=0
Answer:
a) s = 42 m
b) V = 26 m/s
c) a = 11 m/s²
Explanation:
The velocity(V) = 2 - 4t + 5t^3/2 where t is in seconds and V is in m/s
a) To get the position, we have to integrate the velocity with respect to time.
We get the position(s) from the equation:
[tex]V=\frac{ds}{dt}[/tex]
[tex]ds=Vdt[/tex]
Integrating both sides,
[tex]s=\int\ {V} \, dt[/tex]
[tex]s=\int\ {(2-4t+5t^{\frac{3}{2} }) \, dt[/tex]
Integrating, we get
[tex]s=(2t-2t^{2} +2t^{\frac{5}{2} }+c)m[/tex]
But at t=0, s(0) = 2 m
Therefore,
[tex]s(0)=2(0)-2(0)^{2} +2(0)^{\frac{5}{2} }+c[/tex]
[tex]2=0+0+0+c[/tex]
c = 2
Therefore
[tex]s=(2t-2t^{2} +\frac{10}{5}t^{\frac{5}{2} }+2)m[/tex]
When t = 4,
[tex]s=2(4)-2(4)^{2} +2(4)^{\frac{5}{2} }+2[/tex]
[tex]s=8-32+64+2[/tex]
s = 42 m
At t= 4s, the position s = 42m
b) The velocity equation is given by:
[tex]V=(2-4t+5t^{\frac{3}{2} })m/s[/tex]
At t = 4,
[tex]V=2-4(4)+5(4)^{\frac{3}{2} }[/tex]
V = 2 - 16 + 40 = 26 m/s
The velocity(V)at t = 4 s is 26 m/s
c) The acceleration(a) is given by:
[tex]a=\frac{dv}{dt}[/tex]
[tex]a=\frac{d}{dt}(2-4t+5t^{ \frac{3}{2}})[/tex]
Differentiating with respect to t,
[tex]a=-4+\frac{15}{2}t^{\frac{1}{2} }[/tex]
[tex]a=(-4+\frac{15}{2}t^{\frac{1}{2} })m/s^{2}[/tex]
At t = 4 s,
[tex]a=-4+\frac{15}{2}(4)^{\frac{1}{2} }[/tex]
[tex]a=-4+15[/tex]
a = 11 m/s²
The position s, velocity v, and acceleration a when t = 4s are 40m, 26m/s and 11m/s² respectively.
Explanation:The velocity, v, of the particle at any time, t, is given by;
v = 2 - 4t + 5[tex]t^{\frac{3}{2} }[/tex] ----------(i)
Analysis 1: To get the position, s, of the particle at any time, t, we integrate equation (i) with respect to t as follows;
s = ∫ v dt
Substitute the value of v into the above as follows;
s = ∫ (2 - 4t + 5[tex]t^{\frac{3}{2} }[/tex]) dt
s = 2t - [tex]\frac{4t^2}{2}[/tex] + [tex]\frac{5t^{5/2}}{5/2}[/tex]
s = 2t -2t² + 2[tex]t^{\frac{5}{2} }[/tex] + c [c is the constant of integration] ------------(ii)
According to the question;
when t = 0, s = 2m
Substitute these values into equation (ii) as follows;
2 = 2(0) -2(0)² + 2[tex](0)^{\frac{5}{2} }[/tex] + c
2 = 0 - 0 - 0 + c
c = 2
Substitute the value of c = 2 back into equation (ii) as follows;
s = 2t -2t² + 2[tex]t^{\frac{5}{2} }[/tex] + 2 --------------------------------(iii)
Analysis 2: To get the acceleration, a, of the particle at any time, t, we differentiate equation (i) with respect to t as follows;
a = [tex]\frac{dv}{dt}[/tex]
Substitute the value of v into the above as follows;
a = [tex]\frac{d(2 - 4t + 5t^{3/2})}{dt}[/tex]
a = -4 + [tex]\frac{15}{2}[/tex][tex]t^{1/2}[/tex] ----------------------------------(iv)
Now;
(a) When t = 4s, the position s, of the particle is calculated by substituting t=4 into equation (iii) as follows;
s = 2(4) -2(4)² + 2([tex]4^{\frac{5}{2} }[/tex]) + 2
s = 8 - 32 + 2(4)²°⁵
s = 8 - 32 + 2(32)
s = 8 - 32 + 64
s = 40m
(b) When t = 4s, the velocity v, of the particle is calculated by substituting t=4 into equation (i) as follows;
v = 2 - 4(4) + 5([tex]4^{\frac{3}{2} }[/tex])
v = 2 - 16 + 5(4)¹°⁵
v = 2 - 16 + 5(8)
v = 2 - 16 + 40
v = 26m/s
(c) When t = 4s, the acceleration a, of the particle is calculated by substituting t = 4 into equation (iv) as follows;
a = -4 + [tex]\frac{15}{2}[/tex]([tex]4^{1/2}[/tex])
a = -4 + [tex]\frac{15}{2}[/tex](2)
a = -4 + 15
a = 11m/s²
Therefore, the position s, velocity v, and acceleration a when t=4s are 40m, 26m/s and 11m/s² respectively.
Tech A says that a clutch must engage completely, all at once. Tech B says that the clutch must slip slightly during engagement. Who is correct?
A. Tech A
B. Tech B
C. Both A and B
D. Neither A or B
Answer: Tech B
Explanation:
Tech A is wrong by saying that a clutch must engage completely, all at once. Engaging the clutch completely will affect the clutch and may get it spoilt.
Tech B says that the clutch must slip slightly during engagement. That is the right way to engage the clutch.
Ridif bar ABC is supported with a pin at A and an elastic steel rod at C. The elastic rod has a diameter of 25mm and modulus of elasticity E = 200 GPA. The bar is subjected to a uniform load q on span AC and a pointload at B.
Calculate the change in length of the elastic rod.
What is the vertical displacement at point B?
Answer:
the change in length of the elastic rod is 0.147 mm and the vertical displacement of the point b is 0.1911 mm.
Explanation:
As the complete question is not visible , the complete question along with the diagram is found online and is attached herewith.
Part a
Take moments at point A to calculate the tensile force on the steel
rod at point C
[tex]\sum M_A=0\\T_c*2.5-P(2.5+0.75)-q*2.5*2.5/2=0\\T_c*2.5-10(3.25)-5*3.125=0\\T_c=19.25 kN\\[/tex]
Calculate the change in length of the elastic rod
[tex]\Delta l=\dfrac{T_c*L}{A*E}\\\Delta l=\dfrac{19.25\times 10^3*0.75}{\pi/4\times 0.025^2*200\times 10^9}\\\Delta l=0.000147 m \approx 0.147 mm[/tex]
So the change in length of the elastic rod is 0.147 mm.
As by the relation of the similar angles the vertical displacement of the point B is given as
[tex]\Delta B=\Delta l\dfrac{AC+BC}{AC}[/tex]
Here AC=2.5 m
BC=0.75 m
Δl=0.147 mm so the value is as
[tex]\Delta B=\Delta l\dfrac{AC+BC}{AC}\\\Delta B=0.147 mm\dfrac{2.5+0.75}{2.5}\\\Delta B=0.147 mm\dfrac{3.25}{2.5}\\\Delta B=0.1911 mm[/tex]
So the vertical displacement of the point b is 0.1911 mm.
A steam pipe has an outside diameter of 0.12 m. It is insulated with calcium silicate. The insulation is 20 mm thick. The temperature between the outer edge of the pipe and the inner radius of the insulation is maintained at 600 K. Convection and radiation are driving by a temperature of 25 degree C on the outside of the insulation. The convection heat transfer coefficient is 25 W/m^2 -K. The radiation heat transfer coefficient is 30 W/m^2-K. What is the rate of heat loss from the pipe on a per length basis? What is the temperature on the outside surface of the calcium silicate?
Answer:
[tex]\dot Q = 524.957 W[/tex], [tex]T_{out} = 317.048 K[/tex].
Explanation:
a) The rate of heat loss is determined by following expression:
[tex]\dot Q = \frac{T_{ins, in} - T_{air}}{R_{th}}[/tex]
Where [tex]R_{th}[/tex] is the thermal resistance throughout the system:
[tex]R_{th} = R_{cond} + R_{conv || rad}[/tex]
Since convection and radiation phenomena are occuring simultaneously, the equivalent thermal resistance should determined:
[tex]\frac{1}{R_{conv||rad}} = \frac{1}{R_{conv}} + \frac{1}{R_{rad}}[/tex]
[tex]R_{conv||rad} = \frac{R_{conv}\cdot R_{rad}}{R_{conv} + R_{rad}}[/tex]
Where:
[tex]R_{conv} = \frac{1}{h_{conv} \cdot A} \\R_{rad} = \frac{1}{h_{rad} \cdot A}[/tex]
Outer surface area is given by:
[tex]A = 2 \cdot \pi \cdot r_{out} \cdot L[/tex]
Heat resistance associated to conduction through a hollow cylinder is:
[tex]R_{cond} = \frac{\ln \frac{r_{out}}{r_{in}} }{2 \cdot \pi L \cdot k}[/tex]
According to an engineering database, calcium silicate has a thermal conductivity k = [tex]0.085 \frac{W}{m \cdot K}[/tex]. Then, needed variables are calculated (L = 1 m):
[tex]r_{out} = 0.08 m, r_{in} = 0.06 m[/tex]
[tex]A \approx 0.503 m^2[/tex]
[tex]h_{conv} = 25 \frac{W}{m^{2}\cdot K}\\h_{rad} = 30 \frac{W}{m^2 \cdot K}[/tex]
[tex]R_{conv} = 0.080 \frac{K}{W}\\R_{rad} = 0.066 \frac{K}{W}[/tex]
[tex]R_{conv||rad} = 0.036 \frac{K}{W}[/tex]
[tex]R_{cond} = 0.539 \frac{K}{W}[/tex]
[tex]R_{th} = 0.575 \frac{K}{W}[/tex]
[tex]\dot Q = 524.957 W[/tex]
The temperature on the outside surface of the calcium silicate can be determined from the following expression:
[tex]\dot Q = \frac{T_{in}-T_{out}}{R_{cond}}[/tex]
Then,
[tex]T_{out}=T_{in}-\dot Q \cdot R_{cond}\\T_{out} = 317.048 K[/tex]
An insulated piston-cylinder device initially contains 0.16 m2 of CO2 at 150 kPa and 41 °C. Electric resistance heater supplied heat for 10 mins. During this procedure the volume has doubled while pressure stayed the same. Considering electric resistance heater running on 110 V, calculate the needed current in A ((Give your answer with three decimals, and do NOT enter units!!!).
Answer:
I=0.3636
Explanation:
See the attached picture for explanation.
Yanay gets a lot of spam calls. An area code is defined to be a three digit number from 200-999 inclusive. In reality, many of these area codes are not in use, but for this question we'll simplify things and assume they all are. **Throughout these questions, you should assume that Yanay's area code is 781\.\**
Answer:
Determine the usable area codes.
Explanation:
The best way will be the installation of an antispyware software in the mobile phone to identify the source of the call.
Design a digital integrator using the impulse invariance method. Find and give a rough sketch of the amplitude response, and compare it with that of the ideal integrator. If this integrator is used primarily for integrating audio signals (whose bandwidth is 20 kHz), determine a suitable value for T.
Answer:
50 μsec
Explanation:
See the attached pictures for detailed answer.
A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at −30°C by rejecting its waste heat to cooling water that enters the condenser at 18°C at a rate of 0.25 kg/s and leaves at 26°C. The refrigerant enters the condenser at 1.2 MPa and 65°C and leaves at 42°C. The inlet state of the compressor is 60 kPa and −34°C and the compressor is estimated to gain a net heat of 450 W from the surroundings. Determine (a) the quality of the refrigerant at the evaporator inlet, (b) the refrigeration load, (c) the COP of the refrigerator
Answer:
A) The quality of the refrigerant at the evaporator inlet = 0.48
B) The refrigeration load = 5.39 kW
C) COP = 2.14
Explanation:
A) From the refrigerant R-144 table I attached,
At P=60kpa and interpolating at - 34°C,we obtain enthalpy;
h1 = 230.03 Kj/kg
Also at P= 1.2MPa which is 1200kpa and interpolating at 65°C,we obtain enthalpy ;
h2 = 295.16 Kj/Kg
Also at P= 1.2MPa which is 1200kpa and interpolating at 42°C,we obtain enthalpy ;
h3 = 111.23 Kj/Kg
h4 is equal to h3 and thus h4 = 111.23 Kj/kg
We want to find the refrigerant quality at the evaporation inlet which is state 4 and P= 60 Kpa.
Thus, from the table attached, we see that hf = 3.84 at that pressure and hg = 227.8
Now, to find the quality of the refrigerant, we'll use the formula,
x4 = (h4 - hg) /(hf - hg)
Where x4 is the quality of the refrigerant. Thus;
x4 = (111.23 - 3.84)/(227.8 - 3.84) = 0.48
B) The mass flow rate of the refrigerant can be determined by applying a 1st law energy balance across the
condenser. Thus, the water properties can be obtained by using a saturated liquid at the given temperatures;
So using the first table in the image i attached; interpolating at 18°C; hw1 = hf = 75.54 kJ/kg
Also interpolating at 26°C; hw2 = hf = 109.01 kJ/kg
Now;
(mr) (h2 − h3)= (mw) (hw2 − hw1)
mr is mass flow rate
Making mr the subject, we get;
mr = [(mw) (hw2 − hw1)] /(h2 − h3)
mr = [(0.25 kg/s)(109.01 − 75.54) kJ/kg
] /(295.13 − 111.37) kJ/kg
mr = 8.3675/183.76
mr = 0.0455 kg/s
Formula for refrigeration load is;
QL = mr(h1 − h4)
Thus,
QL = (0.0455 kg/s)(230.03 − 111.37) kJ/kg = 5.39 kW
C) The formula for specific work into the compressor is;
W(in) = [(h2 − h1)] − (Q(in
)/mr)
= (295.13 − 230.03) kJ/kg − (0.450kJ/s
/0.0455 kg/s)
= 65.10 Kj/kg - 9.89 Kj/kg
55.21 kJ/kg
Formula for COP is;
COP = qL
/W(in)
Thus; COP = (h1 − h4)/W(in
)
= [(230.03 − 111.37) kJ/kg
] /55.24 kJ/kg
= 2.14
Thirty-six grams of air in a piston–cylinder assembly undergo a Stirling cycle with a compression ratio of 7.5. At the beginning of the isothermal compression, the pressure and volume are 1 bar and 0.03 m3, respectively. The temperature during the isothermal expansion is 1200 K. Assuming the ideal gas model and ignoring kinetic and potential energy effects, determine the net work, in kJ.
The net work done during the Stirling cycle is 2.7 kJ.
Explanation:In order to determine the net work done during the Stirling cycle, we first need to calculate the initial and final volumes of the air in the piston-cylinder assembly.
Given:
Mass of air, m = 36 g = 0.036 kgInitial pressure, P1 = 1 bar = 100,000 PaInitial volume, V1 = 0.03 m3Compression ratio, r = 7.5Temperature during isothermal expansion, TH = 1200 KThe compression ratio is given by:
r = V1/V2
where V2 is the final volume. Solving for V2, we get:
V2 = V1/r = 0.03/7.5 = 0.004 m3
Next, we can use the ideal gas law to relate the initial and final pressures and volumes:
P1V1/T1 = P2V2/T2
Since the process is isothermal, T1 = T2 = TH = 1200 K.
Solving for P2, we get:
P2 = (P1V1T2)/V2T1 = (100,000 x 0.03 x 1200)/(0.004 x 1200) = 100,000 Pa
Finally, we can calculate the net work done using the formula:
Wnet = (P1V1) - (P2V2)
Substituting the values, we get:
Wnet = (100,000 x 0.03) - (100,000 x 0.004) = 2700 J = 2.7 kJ
The two aluminum rods AB and AC with diameters of 10 mm and 8 mm respectively, have a pin joint at an angle of 45.°
Determine the largest vertical force P that can be supported at the joint. The allowable tensile stress for the aluminum is 150 MPa.
Answer:
attached below
Explanation:
A rope having a weight per unit length of 0.4 lb/ft is wound 2 1/2 times around a horizontal rod. Knowing that the coefficient of static friction between the rope and the rod is 0.30, determine the minimum length x of rope that should be left hanging if a 100-lb load is to be supported
Explanation:
Tension in the rope
[tex]\begin{aligned}T_{1} &=0.4 \times x \\T_{2} &=100+0.4 \times 10 \\&=104\end{aligned}[/tex]
[tex]M s=0.3[/tex] ∅ [tex]=2 \cdot 5(2 \pi)=5 \pi[/tex]
[tex]\begin{aligned}\frac{T_{1}}{T_{2}}=e^{\mu \theta} & \Rightarrow \frac{104}{0.4 x}=e^{0.3(5 \pi)} \\& \Rightarrow x=2.34 \mathrm{H}\end{aligned}[/tex]
NOTE : Refer the image
An refrigerator with an average COP 2.8 is used to cool a well insulated container whose contents are equivalent to 12 kg of water from 40 C to 10 C. when the running , the refrigerator consumes 400 W of electric power. Calculate the time required for the refrigerator to accomplish this cooling.
Answer:
It is going to take about 22.43 minutes
Explanation:
The Coefficient of Performance of a refrigerator is the ratio of useful cooling to work required to achieve it. The formula would be:
COP = Cooling Effect/Power Input
The COP is given as 2.8 and the power rating is 400 Watts. We can find the cooling effect as follows:
2.8 = Cooling Effect/400
Cooling Effect = 2.8 * 400 = 1120 Watts
Now,
The cooling effect can be done using the formula:
[tex]Q=mc \Delta T[/tex]
Where
Q is the thermal energy (or work)
m is the mass
c is the specific heat capacity
[tex]\Delta T[/tex] is the temperature change
We know
m = 12 kg
c is the specific heat of water, 4187 J/kg*C
[tex]\Delta T[/tex] is 30, from 40C to 10C
Substituting, we get:
[tex]Q=mc \Delta T\\Q=12*4187*30\\Q=1507320[/tex]
This thermal energy is the Work, which is Power * Time required
Thus, we can say:
Time Required = Q/Power
So,
Time Required = 1507320/1120 = 1346 Seconds
To minutes, we can say:
1346/60 = 22.43 minutes
An inventor has developed a machine that extracts energy from water and other effluents drained from apartments in a building. The machine essentially consists of a hydraulic turbine placed in the basement (level zero), through which the effluents pass before being dumped into the sewage system.
(a) If the average height of an apartment relative to the location of the turbine is 30 meters and the average flow from each apartment (assume it to be water) is 100 liters per day, calculate the average power produced by the contraption if there are 100 apartments in the building. Assume ambient state to be 300 K, 1 bar.
Answer:
34.06 W.
Explanation:
Assumptions : ideal turbine " no loss of work", no pipe or friction losses, all the available energy of water in converted to useful power.
(A) total volume of water per day in cubic meters:
1 cubic meters=1000 L
average flow from each apartment*total apartments/1000= (100/1000)*100
=10 m^3
total mass of water m= density*volume=1000*10=10000 kg
total energy of water at 30 m height= m*g*h= 10000*9.81*30=2943000 J
if all the available energy is converted to power.
power produced per day=total energy of water / time
time in seconds=24*3600=86400 s
power produced in a day=2943000/86400= 34.06 W
Answer:
34.06 W
Explanation:
Assuming ideal conditions which are assuming no fiction or pipe loss is made along the line of extraction of water
Energy of water at ideal condition ( Eₐ ) = 1000 kg/m³
height given = 30 meters
quantity of water = 100 Liters
to calculate the quantity of water in M³/s ( cubic per second )
= [tex]\frac{100*10^{-3} }{24*3600}[/tex] = 1.157 * [tex]10^{-6}[/tex]
power produced by water ( Pw) = energy of water * quantity of water in m^3/s
= 1.157 * [tex]10^{-6}[/tex] * [tex]10^{3}[/tex] = 1.157 *[tex]10^{-3}[/tex]
since it is an ideal condition all the power produced by water is converted to power produced by the contraception
Pc = ( Pw * g * h ) * n
H = height = 30
g = 9.81
n = number of apartments = 100
Pc =( 1.157 * [tex]10^{-3}[/tex] * 9.81 * 30) *100 = 34.06 W
A long circular cylinder of diameter 2a meters is set horizontally in a steady stream (perpendicular to the cylinder axis) of velocity U m/s. The cylinder is caused to rotate at ω rad/s around its axis. Obtain an expression in terms of ω and U for the ratio of the pressure difference between the top and bottom of the cylinder divided by the dynamic pressure of the stream (i.e., the pressure coefficient difference).
Answer:
The ratio of the difference of the pressure at the top and bottom of the cylinder to dynamic pressure is given as
[tex]\dfrac{4a (U\omega- g)}{U^2}[/tex]
Explanation:
As the value of the diameter is given as d=2a
The velocity is given as v=U
The rotational velocity is given as ω rad/s
Point A is at the top of the cylinder and point B is at the bottom of the cylinder
Such that the point A is at the highest point on the circumference and point B is at the bottom of the cylinder
Now the velocity at point A is given as
[tex]v_A=U-\dfrac{d}{2}\omega\\v_A=U-\dfrac{2a}{2}\omega\\v_A=U-a\omega\\[/tex]
Now the velocity at point B is given as
[tex]v_B=U+\dfrac{d}{2}\omega\\v_B=U+\dfrac{2a}{2}\omega\\v_B=U+a\omega\\[/tex]
Considering point B as datum and applying the Bernoulli's equation between the point A and B gives
[tex]\dfrac{P_A}{\rho g}+\dfrac{v_A^2}{2 g}+z_A=\dfrac{P_B}{\rho g}+\dfrac{v_B^2}{2 g}+z_B[/tex]
Here P_A and P_B are the local pressures at the point A and point B.
v_A and v_B are the velocities at the point A and B
z_A and z_B is the height of point A which is 2a and that of point B is 0
Now rearranging the equation of Bernoulli gives
[tex]\dfrac{P_A-P_B}{\rho g}=\dfrac{v_B^2-v_A^2}{2 g}+z_B-z_A[/tex]
Putting the values
[tex]\dfrac{P_A-P_B}{\rho g}=\dfrac{v_B^2-v_A^2}{2 g}+z_B-z_A\\\dfrac{P_A-P_B}{\rho g}=\dfrac{(U+a\omega)^2-(U-a\omega)^2}{2 g}+0-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{(U^2+a^2\omega^2+2Ua\omega)-(U^2+a^2\omega^2-2Ua\omega)}{2g}-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{U^2+a^2\omega^2+2Ua\omega-U^2-a^2\omega^2+2Ua\omega)}{2g}-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{4Ua\omega}{2g}-2a\\\dfrac{P_A-P_B}{\rho g}=\dfrac{2Ua\omega}{g}-2a\\P_A-P_B=\dfrac{2Ua\omega}{g}*\rho g-2a*\rho g\\[/tex]
[tex]P_A-P_B=2Ua\omega\rho-2a\rho g[/tex]
Now the dynamic pressure is given as
[tex]P_D=\dfrac{1}{2}\rho U^2[/tex]
[tex]\dfrac{P_A-P_B}{P_D}=\dfrac{2Ua\omega\rho-2a\rho g}{1/2 \rho U^2}\\\dfrac{P_A-P_B}{P_D}=\dfrac{2a\rho (U\omega- g)}{1/2 \rho U^2}\\\dfrac{P_A-P_B}{P_D}=\dfrac{4a\rho (U\omega- g)}{\rho U^2}\\\dfrac{P_A-P_B}{P_D}=\dfrac{4a (U\omega- g)}{U^2}[/tex]
So the ratio of the difference of the pressure at the top and bottom of the cylinder to dynamic pressure is given as
[tex]\dfrac{4a (U\omega- g)}{U^2}[/tex]
During the run-up at a high-elevation airport, a pilot notes a slight engine roughness that is not affected by the magneto check but grows worse during the carburetor heat check.
Under these circumstances, what would be the most logical initial action?
A. Check the results obtained with a leaner setting of the mixture.
B. Taxi back to the flight line for a maintenance check.
C. Reduce manifold pressure to control detonation.
Answer:
A). Check the results obtained with a leaner setting of the mixture.
Explanation:
Answer:
Under the explained circumstances, the most logical initial action would be:
A. Check the results obtained with a leaner setting of the mixture.
Explanation:
Check the results obtained with a leaner setting of the mixture, as it has been enriched by the carburetor-heated air causing engine roughness at that high altitude.
Answer b is wrong because the pilot should first try the runup with a leaner mixture and answer c is not correct because detonation is the result of a mixture that is to lean.
Suppose an op amp has a midband voltage gain of 500,000. If the upper cutoff frequency is 15 Hz, what does the frequency response look like? (Malvino, 20150123) Malvino, A. (20150123). Electronic Principles, 8th Edition [VitalSource Bookshelf version]. Retrieved from vbk://9781259200144 Always check citation for accuracy before use.
Answer:
The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.
Explanation:
A thick steel slab ( 7800 kg/m3, 480 J/kg·K, 50 W/m·K) is initially at 300°C and is cooled by water jets impinging on one of its surfaces. The temperature of the water is 25°C, and the jets maintain an extremely large, approximately uniform convection coefficient at the surface. Assuming that the surface is maintained at the temperature of the water throughout the cooling, how long will it take for the temperature to reach 50°C at a distance of 30 mm from the surface?
Answer: 67.392s
Explanation: detailed calculation is shown below
Answer:
Time(t) = 2592.91 seconds
Explanation:
From the question, we have;
ρ = 7800 kg/m
c =480 J/kg⋅K
k = 50 W/m⋅K
Ti = 300°C
Ts = 25°C
x = 25 mm or in meters; = 0.025 m
From formula, we know that;
(T(x,t) - T(s))/(T(i) - T(s)) = erf(x/(2√(αt))
Where erf is error function
And α = k/(ρc)
So, solving we have;
(50 - 25)/(300 - 25) = erf(x/(2√(αt))
erf(x/(2√(αt)) = 0.0909
From the error function table which i attached, 0.0909 will give us approximately 0.0806 upon interpolation.
Thus, (x/(2√(αt)) = 0.0806
Let's make "t" the subject of the formula;
x² = 0.0806²(2²)(αt)
t = x²/0.026α
Let's find α
From earlier, we saw that;
α = k/(ρc)
Thus; α = 50/(7800 x 480) = 1.335 x 10^(-5) m²/s
Also, from the question, x = 30mm or 0.03m
So, t = 0.03²/(0.026 x 1.335 x 10^(-5)) = 2592.91 seconds