Answer:
note:
solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word due to different version of MS Office please find the attachment
The amount of work produced by 1 kg of steam and 1 kg of R–134a would be the same in a closed system at the given conditions.
Explanation:In a closed system, the amount of work produced depends on the change in internal energy of the system. The change in internal energy is given by the formula ΔU = Q - W, where Q is the heat added to the system and W is the work done by the system. Since both 1 kg of steam and 1 kg of R–134a are at the same temperature and pressure, their internal energy changes would be the same for the same amount of heat added. Therefore, the amount of work produced by both substances would also be the same in a closed system.
Learn more about work produced in a closed system here:https://brainly.com/question/34701684
#SPJ3
A successful quality strategy features which of the following elements? engaging employees in the necessary activities to implement quality an organizational culture that fosters quality an understanding of the principles of quality engaging employees in the necessary activities to implement quality and an understanding of the principles of quality engaging employees in the necessary activities to implement quality, an organizational culture that fosters quality, and an understanding of the principles of quality
Answer:
Quality strategy is an integral part of organization's strategy that ensures quality. It involves market and productivity strategies that have a very high significance. A successful quality strategy must endeavour to have features such as; engaging employees in the necessary activities to implement quality, an organizational culture that fosters quality and an understanding of the principles of quality.
All these will enable the organization to stand out and beat competition.
A successful quality strategy features includes engaging the employees:
in the necessary activities to implement qualityin an organizational culture that fosters qualityin understanding the principles of quality.A quality strategy serve as an integral part of organization's strategy that ensures quality.
The quality strategy involves market and productivity strategies that have a very high significance.
Hence, the successful quality strategy features includes engaging the employees in necessary activities to implement quality, organizational culture that fosters quality and understanding the principles of quality.
Therefore, all the option is correct.
Read more about quality strategy
brainly.com/question/24462624
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.
1- If the elongation of wire BC is 0.3 mm after the force P is applied, determine the magnitude of P. The wire is made of steel with an elastic modulus of 200 GPa and has a diameter of 2 mm.
Answer:
P = 188.496 N.
Explanation:
The axial elongation of the wire is modelled by this equation:
[tex]\delta = \frac{P \cdot L}{(\frac{\pi}{4}\cdot D^{2})\cdot E_{steel}}[/tex]
The force can be calculated by isolating it within the expression:
[tex]P=\frac{\delta \cdot (\frac{\pi}{4}\cdot D^{2})\cdot E_{steel}}{L}[/tex]
Let assume that [tex]L = 1 m[/tex]. Then:
[tex]P =\frac{(0.3 mm)\cdot (\frac{1 m}{1000 mm} )\cdot (\frac{\pi}{4}\cdot [(2 mm)\cdot(\frac{1 m}{1000 mm} )]^{2} )\cdot (200 \times 10^{9} Pa)}{1 m} \\P \approx 188.496 N[/tex]
A house has a black tar, flat, horizontal roof. The lower surface of the roof is well insulated, while the upper surface is exposed to ambient air at 300K through a convective coefficient of 10 W/m2-K. Calculate the roof equilibrium temperature for a) a clear sunny day with an incident solar radiaton flux of 500 W/m2 and the ambient sky at an effective temperature of 50K and b) a clear night with an ambient sky temperature of 50K.
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
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:
Consider the Poisson trip generation model in Example 8.4. Suppose that a household has five members with an annual income of $150,000 and lies in a neighborhood with a retail employment of 320. What is the expected number of peak-hour shopping trips, and what is the probability that the household will make more than one peak-hour shopping trip.
Process: (1) Calculate the Poisson parameter (2) Determine the probability of making zero trips and one trip; (3) subtract the values from #2 from 1 to determine probability
Answer:
The solution is given in the attachments.
Using your time efficiency function from HW1, measure the execution times of both insertion and merge sorting algorithms using the attached data files (one with 1,000 integers and the other with 1,000,000 integers ).
Answer:
The data file with 1000 integers
for merge sort the time efficiency is (1000×㏒1000) = 1000 × 3 = 3000
for insertion sort the time efficiency is (1000 × 2) = 2000
The data file with 1,000,000 integers
for merge sort the time efficiency is (1,000,000×㏒1,000,000) = 1,000,000 × 6 = 6,000,000
for insertion sort the time efficiency is (1,000,000 × 2) = 2,000,000
Explanation:
The execution time or temporal complexity refers to how the execution time of an algorithm increases as the size of input increases
For merge sort, the time efficiency is given by the formula
O(n㏒n)For insertion sort the time efficiency is given by the formula
O(n×2)Where n refers to the size of input
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]
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.
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 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
declare integer product declare integer number product = 0 do while product < 100 display ""Type your number"" input number product = number * 10 loop display product End While
Full Question
1. Correct the following code and
2. Convert the do while loop the following code to a while loop
declare integer product
declare integer number
product = 0
do while product < 100
display ""Type your number""
input number
product = number * 10
loop
display product
End While
Answer:
1. Code Correction
The errors in the code segment are:
a. The use of do while on line 4
You either use do or while product < 100
b. The use of double "" as open and end quotes for the string literal on line 5
c. The use of "loop" statement on line 7
The correction of the code segment is as follows:
declare integer product
declare integer number
product = 0
while product < 100
display "Type your number"
input number
product = number * 10
display product
End While
2. The same code segment using a do-while statement
declare integer product
declare integer number
product = 0
Do
display "Type your number"
input number
product = number * 10
display product
while product < 100
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.
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
Gas at a temperature of 250°C with a convective coefficient 75 W/m2·K flows through a packed bed of aluminum (2024) spheres that is used as a thermal energy storage system. If the initial temperature of the spheres is 25°C and the diameter of each sphere is 75-mm, find the time, in seconds, required to for a sphere to reach 85% of the maximum possible thermal energy as well as the surface temperature and the centerline temperature at this time, both in °C. Evaluate the properties of the aluminum (2024) at 500 K.
Answer:
See explanation
Explanation:
Airflow t
T¥, h D
Ts
w
Heater
h
(q•, k ) L
(a) Under conditions for which a uniform surface temperature Ts is maintained around the circum- ference of the heater and the temperature Too and convection coefficient h of the airflow are known, obtain an expression for the rate of heat transfer per unit length to the air. Evaluate the
heat rate for Ts = 300°C, D = 20 mm, an alu- minum sleeve (ks = 240 W/m · K), w = 40 mm, N = 16, t = 4 mm, L = 20 mm, Too = 50°C, and h = 500 W/m2 · K.
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
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]
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.
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
An asphalt mixture is placed and compacted using normal rolling procedures. Two tests are taken from the compacted mixture and checked for density. Core 1 had a density of 150 pcf and an air void content of 6.5 %. Core 2 had a density of 151 pcf. What is the air void content of core 2.
Answer:
Air void content of core 2 = 5.87%
Explanation:
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.
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
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.
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 enrichment plant has a throughput of 32,000 kg U/day and produces 26,000 kg U as tails. What is the enrichment of the product if the feed is natural uranium and the tails are 0.25%
Answer:
1.10 %
Explanation:
The enrichment of the product can be calculated using the following equation:
[tex] \frac{W}{P} = \frac{x_{p} - x_{f}}{x_{f} - x_{w}} [/tex] (1)
where W: is waste or tails = 26000, P: is the product = 32000, xp: is the wt. fraction of ²³⁵U in product, xf: is the wt. fraction of ²³⁵U in feed = 0.72% and xw: is the wt. fraction of ²³⁵U in waste or tails = 0.25%.
By solving equation (1) for xp, we can find the enrichment of the product:
[tex] x_{p} = \frac {W}{P} \cdot (x_{f} - x_{w}) + x_{f} [/tex]
[tex] x_{p} = \frac {26000}{32000} \cdot (0.72 - 0.25) + 0.72 = 1.10% [/tex]
Therefore, the enrichment of the product is 1.10 %.
I hope it helps you!
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;
}
**************************************************
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]
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
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.
Learn more about Manometer Oil Levels here:https://brainly.com/question/33422583
#SPJ3
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