Answer:
(a) The volume flow rate of the refrigerant at the inlet is 0.3078 m3/s
(b) The mass flow rate of the refrigerant is 2.695 kg/s
(c) The velocity and volume flow rate at the exit is 6.017 m/s
Explanation:
According to the given data we have the following:
diameter of the pipe=d=28 cm=0.28 m
inlet pressure P1=200 kPa
inlet temperature T1=20°C
inlet velocity V1=5.5 m/s
Exit pressure P2=180 kPa
Exit Temperature T2=40°C
a. To calculate the volume flow rate of the refrigerant at the inlet we would have to use the following formula:
V1=AV1
=π/4(0.28∧2)5
V1=0.3078 m3/s
b. To calculate the mass flow rate of the refrigerant we would have to use the following formula:
m=p1 V1
m=V1/v1
=0.3078/0.11418
=2.695 kg/s
c. To calculate the velocity and volume flow rate at the exit we would have to use the following formula:
m=m1=m2
V1/v1=V2/v2
V2=(v2/v1)v1
=(0.13741/0.11418)5
=6.017 m/s
What's the "most common" concern with using variable frequency drives (VFDs)? 1) carrier frequency 2) harmonic distortion 3) hertz modulation
The common" concern with using variable frequency drives (VFDs) is C. hertz modulation.
What is variable frequency drive?It should be noted that a variable frequency drive simply means a type of motor drive that us used in mechanical drive system.
In this case, common" concern with using variable frequency drives (VFDs) is hertz modulation
Learn more about frequency on:
brainly.com/question/6985885
#SPJ9
4. In its natural state, a soil weighs 2800 lb/cy, while in the loose and compacted states, it weighs 2500 lb/cy and 3300 lb/cy, respectively. a. Find the load and shrinkage factors for this soil. b. How many trucks loads with a capacity of 5 lcy/truck would be required to haul 750,000 ccy of this soil to a project
Answer:
a. load factor = 0.893
shrinkage factor = 0.848
b. Number of Trucks loads = 113,585 Trucks loads
Explanation:
Here, we start by identifying the factors as given in the question.
γn = 2800 lb/cy
γloose = 2500 lb/cy
and γcompacted = 3300 lb/cy
a. Mathematically,
Load factor = γloose/γn = 2500/2800 = 0.893
Shrinkage factor = γn/γcompacted = 2800/3300 = 0.848
b. To find the number of trucks loads with a capacity of 5 lcy/truck, we use the mathematical formula as follows;
ρlcy = 5
Load factor × Shrinkage factor = ρloose/γn × γn/γcompacted = ρlcy/ρccy
0.893 × 0.848 = 5/ρccy
ρccy =5/(0.893 × 0.848) = 6.603
The number of truck loads = 750,000/6.603 = 113,584.7 which is approximately 113,585 trucks loads
(a) Consider a message signal containing frequency components at 100, 200, and 400 Hz. This signal is applied to a SSB modulator together with a carrier at 100 kHz, with only the upper sideband retained. In the coherent detector used to recover the local oscillator supplies a sinusoidal wave of frequency 100.02 kHz. Determine the frequency components of the detector output. (b) Repeat your analysis, assuming that only the lower sideband is transmitted.
Answer:
Explanation:
The frequency components in the message signal are
f1 = 100Hz, f2 = 200Hz and f3 = 400Hz
When amplitude modulated with a carrier signal of frequency fc = 100kHz
Generates the following frequency components
Lower side band
[tex]100k - 100 = 99.9kHz\\\\100k - 200 = 99.8kHz\\\\100k - 400 = 99.6kHz\\\\[/tex]
Carrier frequency 100kHz
Upper side band
[tex]100k + 100 = 100.1kHz\\\\100k + 200 = 100.2kHz\\\\100k + 400 = 100.4kHz[/tex]
After passing through the SSB filter that filters the lower side band, the transmitted frequency component will be
[tex]100k, 100.1k, 100.2k\ \texttt {and}\ 100.4kHz[/tex]
At the receive these are mixed (superheterodyned) with local ocillator frequency whichh is 100.02KHz, the output frequencies will be
[tex]100.02 - 100.1k = 0.08k = 80Hz\\\\100.02 - 100.2k = 0.18k = 180Hz\\\\100.02 - 100.4 = 0.38k = 380Hz[/tex]
After passing through the SSB filter that filters the higher side band, the transmitted frequency component will be
[tex]100k, 99.9k, 99.8k\ \ and \ \99.6kHz[/tex]
At the receive these are mixed (superheterodyned) with local oscillator frequency which is 100.02KHz, and then fed to the detector whose output frequencies will be
[tex]100.02 - 99.9k = 0.12k = 120Hz\\\\100.02 - 99.8k = 0.22k = 220Hz\\\\100.02 - 99.6k = 0.42k = 420Hz[/tex]
A) The frequency Components of the Detector Output are;
80 Hz, 120 Hz and 380 Hz
B) The frequency Components if only the lower sideband is transmitted are; 120 Hz, 220 Hz and 420 Hz
Message SignalsA) We are given the frequency components in the message signal as;
f1 = 100Hzf2 = 200Hzf3 = 400HzWe are told that the carrier signal has a frequency; fc = 100kHz
Thus, the frequency components generated are;
Lower side band:
100 kHz - 100 Hz = 99.9 kHz100 kHz - 200 Hz = 99.8 kHz100 kHz - 400 Hz = 99.6 kHzUpper side band:
100 kHz + 100 Hz = 100.1 kHz100 kHz + 200 Hz = 100.2 kHz100 kHz + 400 Hz = 100.4 kHzWe are told that the local oscillator now supplies a sinusoidal wave of frequency 100.02 kHz.
Thus, the output frequencies are;
100.02 kHz - 100.1 kHz = 80 Hz
100.02 kHz - 100.2 kHz = 180 Hz
100.02 kHz - 100.4 kHz = 380 Hz
B) Repeating the analysis assuming only the lower sideband is repeated gives us the frequencies as;
100.02 kHz - 99.9 kHz = 120 Hz
100.02 kHz - 99.8 kHz = 220 Hz
100.02 kHz - 99.6 kHz = 420 Hz
Read more about Message Signals at; https://brainly.com/question/25904079
An eddy current separator is to separate aluminum product from an input streamshredded MSW. The feed rate to the separator is 2,500 kg/hr. The feed is known to contain174 kg of aluminum and 2,326 kg of reject. After operating for 1 hour, a total of 256 kg ofmaterials is collected in the product stream. On close inspection, it is found that 140 kg ofproduct is aluminum. Estimate the % recovery of aluminum product and the % purity of thealuminum produc
Answer:
the % recovery of aluminum product is 80.5%
the % purity of the aluminum product is 54.7%
Explanation:
feed rate to separator = 2500 kg/hr
in one hour, there will be 2500 kg/hr x 1 hr = 2500 kg of material is fed into the machine
of this 2500 kg, the feed is known to contain 174 kg of aluminium and 2326 kg of rejects.
After the separation, 256 kg is collected in the product stream.
of this 256 kg, 140 kg is aluminium.
% recovery of aluminium will be = mass of aluminium in material collected in the product stream ÷ mass of aluminium contained in the feed material
% recovery of aluminium = 140kg/174kg x 100% = 80.5%
% purity of the aluminium product = mass of aluminium in final product ÷ total mass of product collected in product stream
% purity of the aluminium product = 140kg/256kg
x 100% = 54.7%
Describe with an example how corroded structures can lead to environment pollution?
list everything wrong with 2020
Answer:
George Floyd (BLACK LIFES MATTER)
C O V I D - 19
Quarantine
no sports
wearing a mask
and a whole lot of other stuff
Explanation:
A hot air balloon is used as an air-vehicle to carry passengers. It is assumed that this balloon is sealed and has a spherical shape. Initially, the balloon is filled up with air at the pressure and temperature of 100 kPa and 27°C respectively and the initial diameter (D) of the balloon is 10 m. Then the balloon is heated up to the point that the volume is 1.2 times greater than the original volume (V2 =1.2V1 ). Due to elastic material used in this balloon, the inside pressure ( P ) is proportional to balloonâs diameter, i.e. P = ð¼D, where ð¼ is a constant.
Required:
a. Show that the process is polytropic (i.e. PV" = Constant) and find the exponent n and the constant.
b. Find the temperature at the end of the process by assuming air to be ideal gas.
c. Find the total amount of work that is done by the balloon's boundaries and the fraction of this work that is done on the surrounding atmospheric air at the pressure of 100 kPa.
Answer:
a. [tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex] which is constant therefore, n = constant
b. The temperature at the end of the process is 109.6°C
c. The work done by the balloon boundaries = 10.81 MJ
The work done on the surrounding atmospheric air = 10.6 MJ
Explanation:
p₁ = 100 kPa
T₁ = 27°C
D₁ = 10 m
v₂ = 1.2 × v₁
p ∝ α·D
α = Constant
[tex]v_1 = \dfrac{4}{3} \times \pi \times r^3[/tex]
[tex]\therefore v_1 = \dfrac{4}{3} \times \pi \times \left (\dfrac{10}{2} \right )^3 = 523.6 \ m^3[/tex]
v₂ = 1.2 × v₁ = 1.2 × 523.6 = 628.32 m³
Therefore, D₂ = 10.63 m
We check the following relation for a polytropic process;
[tex]\dfrac{p_{1}}{p_{2}} = \left (\dfrac{V_{2}}{V_{1}} \right )^{n} = \left (\dfrac{T_{1}}{T_{2}} \right )^{\dfrac{n}{n-1}}[/tex]
We have;
[tex]\dfrac{\alpha \times D_{1}}{\alpha \times D_{2}} = \left (\dfrac{ \dfrac{4}{3} \times \pi \times \left (\dfrac{D_2}{2} \right )^3}{\dfrac{4}{3} \times \pi \times \left (\dfrac{D_1}{2} \right )^3} \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]\dfrac{D_{1}}{ D_{2}} = \left (\dfrac{ \left{D_2} }{ {D_1}} \right )^{3\times n} = \left (\dfrac{ \left{D_1} }{ {D_2}} \right )^{-3\times n}[/tex]
[tex]\dfrac{ D_{1}}{ D_{2}} = \left ( 1.2 \right )^{n} = \left (\dfrac{ \left{D_2} ^3}{ {D_1}^3} \right )^{n}[/tex]
[tex]log \left (\dfrac{D_{1}}{ D_{2}}\right ) = -3\times n \times log\left (\dfrac{ \left{D_1} }{ {D_2}} \right )[/tex]
n = -1/3
Therefore, the relation, pVⁿ = Constant
b. The temperature T₂ is found as follows;
[tex]\left (\dfrac{628.32 }{523.6} \right )^{-\dfrac{1}{3} } = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{-\dfrac{1}{3}}{-\dfrac{1}{3}-1}} = \left (\dfrac{300.15}{T_{2}} \right )^{\dfrac{1}{4}}[/tex]
T₂ = 300.15/0.784 = 382.75 K = 109.6°C
c. [tex]W_{pdv} = \dfrac{p_1 \times v_1 -p_2 \times v_2 }{n-1}[/tex]
[tex]p_2 = \dfrac{p_{1}}{ \left (\dfrac{V_{2}}{V_{1}} \right )^{n} } = \dfrac{100\times 10^3}{ \left (1.2) \right ^{-\dfrac{1}{3} } }[/tex]
p₂ = 100000/0.941 = 106.265 kPa
[tex]W_{pdv} = \dfrac{100 \times 10^3 \times 523.6 -106.265 \times 10^3 \times 628.32 }{-\dfrac{1}{3} -1} = 10806697.1433 \ J[/tex]
The work done by the balloon boundaries = 10.81 MJ
Work done against atmospheric pressure, Pₐ, is given by the relation;
Pₐ × (V₂ - V₁) = 1.01×10⁵×(628.32 - 523.6) = 10576695.3 J
The work done on the surrounding atmospheric air = 10.6 MJ
Sensors are used to monitor the pressure and the temperature of a chemical solution stored in a vat. The circuitry for each sensor produces a HIGH voltage when a specified maximum value is exceeded. An alarm requiring a LOW voltage input must be activated when either the pressure or the temperature is excessive. Design a circuit for this application
The temperature of a flowing gas is to be measured with a thermocouple junction and wire stretched between two legs of a sting, a wind tunnel test fixture. The junction is formed by butt-welding two wires of different material. For wires of diameter D = 125 m and a convection coefficient of h = 700 W/m^2 K, determine the minimum separation distance between the two legs of the sting, L=L1+L2, to ensure that the sting temperature does not influence the junction temperature and, in turn, invalidate the gas temperature measurement. Consider two different types of thermocouple junctions consisting of (i) copper and constantan wires and (ii) chromel and aluminel wires. Evaluate the thermal conductivity of copper and constantan at T300 K. Use kCh =19 W/mK and kA = l29 W/mK for the thermal conductivities of the chromel and alumel wires, respectively.
Answer:
minimum separation distance between the two legs of the sting L = L 1 + L 2 therefore L = 9.48 + 4.68 = 14.16 mL = 1.14 mExplanation:
D ( diameter ) = 125 m
convection coefficient of h = 700 W/m^2
Calculate THE CROSS SECTIONAL AREA
Ac = [tex]\frac{\pi }{4} * D^2[/tex] = [tex]\frac{\pi }{4} * ( 125 )^2[/tex] = 0.79 * 15625 = 12343.75 m^2
perimeter
p = [tex]\pi * D[/tex] = 3.14 * 125 = 392.5 m
at 300k temperature the thermal conductivity of copper and constantan from the thermodynamic property table are :
Kcu = 401 w/m.k
Kconstantan = 23 W/m.k
To calculate the length of copper wire of the thermocouple junction
L 1 = 4.6 ([tex]\frac{Kcv Ac}{h P}[/tex]) ^ 1/2 = 4.6 [tex](\frac{401 *12343.75 }{700 *392.5})^\frac{1}{2}[/tex]
L 1 = 4.6 ( 4949843.75 / 274750 )^1/2
L 1 = 9.48 m
calculate length of constantan wire
L 2 = 4.6 [tex](\frac{kcons*Ac}{hp} )^\frac{1}{2}[/tex]
= 4.6 ( (23 * 12343.75) / ( 700 * 392.5) ) ^1/2
L 2 = 4.6 ( 283906.25 / 274750 ) ^ 1/2
L 2 = 4.68 m
I) therefore the minimum separation distance between the two legs of the sting L = L 1 + L 2
L = 9.48 + 4.68 = 14.16 m
ii) Evaluating the thermal conductivity of copper and constantan
Kc ( thermal conductivity of chromel) = 19 w/m.k
Ka ( thermal conductivity of alumel ) = 29 W/m.k
distance between the legs L = L 1 + L 2
THEREFORE
L = 4.6 ( (Kcn * Ac ) / ( hp ) )^1/2 + 4.6 ( (Kac * Ac)/(hp) )^1/2
L = 4.6 [tex](\frac{Ac}{hp} )^\frac{1}{2} [ (Kcn)^\frac{1}{2} + (Kal)^\frac{1}{2} ][/tex]
L = 4.6 ( 12343.75 /( 700 * 392.5) )^1/2 * [ 19^1/2 + 29^1/2 ]
L = 4.6 ( 12343.75 / 274750 ) ^1/2 * 5.39
L = 1.14 m
The uniform sign has a weight of 1500 lb and is supported by the pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in. If the face of the sign is subjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at points C and D. Show the results on a differential volume element located at each of these points. Neglect the thickness of the sign, and assume that it is supported along the outside edge of the pipe.The uniform sign has a weight of 1500 lb and is supported bythe pipe AB, which has an inner radius of 2.75 in. and anouter radius of 3.00 in.. If the face of the sign issubjected to a uniform wind pressure of p = 150lb/ft2, determine the state of stress at pointsC and D. Show the results on a differentialvolume element located at each of these points. Neglect the thickness of the sign, and assume that it issupported along the outside edge of the pipe.
Answer:
See explanation
Explanation:
See the document for the complete FBD and the introductory part of the solution.
Static Balance ( Sum of Forces = 0 ) in all three directions
∑[tex]F_G_X = W - G_x = 0[/tex]
[tex]G_X = W = 1500 lb[/tex]
∑[tex]F_G_Y = P - G_Y = 0[/tex]
[tex]G_Y = P = -10,800 lb[/tex]
∑[tex]F_G_Z = - G_Z = 0[/tex]
Where, ( [tex]G_X, G_Y, G_Z[/tex] ) are internal forces at section ( G ) along the defined coordinate axes.
Static Balance ( Sum of Moments about G = 0 ) in all three directions
[tex]M_G = r_O_G x F_O[/tex]
Where,
r_OG: The vector from point O to point G
F_OG: The force vector at point O
- The vector ( r_OG ) and ( F_OG ) can be written as follows:
[tex]r_O_G = [ -( 3 + \frac{H}{2} ) i + (\frac{r_o}{12})j - ( \frac{r_o}{12} + \frac{L}{2})k ] \\\\r_O_G = [ -( 6 ) i + (0.25)j - (6)k ] \\[/tex]
[tex]F_O_G = [ ( W ) i + ( P ) k ]\\\\F_O_G = [ (1500) i - ( 10,800 ) k ] lb[/tex]
- Then perform the cross product of the two vectors ( r_OG ) and ( F_OG ):
[tex]( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = \left[\begin{array}{ccc}i&j&k\\-6&0.25&-6\\1500&-10,800&0\end{array}\right] \\\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = -( 6*10,800 ) i - ( 6*1500 ) j + [ ( 10,800*6) - ( 0.25*1500) ] k\\\\( M_G_X )i + (M_G_Y)j+ (M_G_Z)k = - (64,800)i - (9,000)j + (64,425)k[/tex]
- The internal torque ( T ) and shear force ( V ) that act on slice ( G ) are due to pressure force ( P ) as follows:
[tex]T = P*[\frac{L}{2}] = (10,800)*(6) = 64,800 lb.ft[/tex]
[tex]V = P = -10,800 lb[/tex]
- For the state of stress at point "C" we need to determine the the normal stress along x direction ( σ_x ) and planar stress ( τ_xy ) as follows:-
σ_x = [tex]-\frac{G_x}{A} - \frac{M_G_Y. z*}{I_Y_Y} + \frac{M_G_Z. y*}{I_Z_Z}[/tex]
Where,
A: The area of pipe cross section
[tex]A = \pi * [ ( \frac{r_o}{12})^2 - ( \frac{r_i}{12})^2 ] = \pi * [ ( \frac{3}{12})^2 - ( \frac{2.75}{12})^2 ] = 0.03136 ft^2[/tex]
z*: The distance of point "C" along z-direction from central axis ( x )
[tex]z*= [\frac{r_i}{12} ] = [\frac{2.75}{12} ] = 0.22916 ft[/tex]
I_YY: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
y*: The distance of point "C" along y-direction from central axis ( x )
[tex]y* = 0[/tex]
- The normal stress ( σ_x ) becomes:
σ_x = [tex][-\frac{1500}{0.03136} - \frac{-9,000*0.22916}{0.00090} + \frac{64,425*0}{0.00090} ] * (\frac{1}{12})^2 = 15.5 ksi[/tex]
- The planar stress is ( τ_xy ) is a contribution of torsion ( T ) and shear force ( V ):
τ_xy = [tex]- \frac{T.c}{J} + \frac{V.Q}{I.t}[/tex]
Where,
c: The radial distance from central axis ( x ) and point "C".
[tex]c = \frac{r_i}{12} = \frac{2.75}{12} = 0.22916 ft[/tex]
J: The polar moment of inertia of the annular cross section of pipe:
[tex]J = \frac{\pi }{2}* [ ( \frac{r_o}{12})^4 - ( \frac{r_i}{12})^4 ] = \frac{\pi }{2}* [ ( \frac{3}{12})^4 - ( \frac{2.75}{12})^4 ] = 0.00180 ft^4[/tex]
Q: The first moment of area for point "C" = semi-circle
[tex]Q = Y_c*A_c = \frac{4*( r_m)}{3\pi } * \frac{\pi*( r_m)^2 }{2} = \frac{2. ( r_m)^3}{3} \\\\Q = \frac{2. [ ( \frac{r_o}{12})^3 - ( \frac{r_i}{12})^3] }{3} = \frac{2. [ ( \frac{3}{12})^3 - ( \frac{2.75}{12})^3] }{3} = 0.00239ft^3[/tex]
I: The second area moment of pipe along and about "y" axis:
[tex]I_Y_Y = \frac{\pi }{4} * [ (\frac{r_o}{12})^4 - (\frac{r_i}{12})^4 ]=\frac{\pi }{4} * [ (\frac{3}{12})^4 - (\frac{2.75}{12})^4 ] \\\\I_Y_Y = 0.00090 ft^4[/tex]
t: The effective thickness of thin walled pipe:
[tex]t = 2* [ \frac{r_o}{12} - \frac{r_i}{12} ] = 2* [ \frac{3}{12} - \frac{2.75}{12} ] = 0.04166 ft[/tex]
- The planar stress is ( τ_xy ) becomes:
τ_xy = [tex][ - \frac{-64,800*0.22916}{0.0018} + \frac{-10,800*0.00239}{0.0009*0.04166} ] * [ \frac{1}{12}]^2 = 52.4 ksi[/tex]
- The principal stresses at point "C" can be determined from the following formula:-
σ_x = 15.55 ksi, σ_y = 0 ksi , τ_xy = 52.4 ksi
σ_1 =[tex]\frac{sigma_x+sigma_y}{2} + \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_2 = [tex]\frac{sigma_x+sigma_y}{2} - \sqrt{(\frac{sigma_x+sigma_y}{2})^2 + (tow_x_y)^2 }[/tex]
σ_1 = [tex]\frac{15.55+0}{2} + \sqrt{(\frac{15.55+0}{2})^2 + (52.4)^2 } = 60.75 ksi[/tex]
σ_2 =[tex]-\sqrt{\left(\frac{15.55+0}{2}\right)^2\:+\:\left(52.4\right)^2\:}+\frac{15.55+0}{2} = -45.20 ksi[/tex]
- The angle of maximum plane stress ( θ ):
θ = [tex]0.5*arctan ( \frac{tow_x_y}{\frac{sigma_x-sigma_y}{2} } )= 0.5*arctan*( \frac{52.4}{7.8} ) = 40.8 deg[/tex]
Note: The plane stresses at point D are evaluated using the following procedure given above. Due to 5,000 character limit at Brainly, i'm unable to post here.
While having a discussion about O-rings at the bottom of filters, Technician A says that the Automotive Filter Manufacturers Council recommends that the filter O-ring be lubricated with oil after installing the filter. Technician B says that the filter O-ring should be lubricated before installation. Who is correct
Answer:
Technician B is correct
Explanation:
O- rings are used with oil transmission filters to avoid transmission failures but some people use lip seals as well. either of them is inserted onto the outer part of the transmission system i.e it is inserted/found in-between Transmission filters and the transmission systems and it main purpose is to avoid leaks and transmission failure in the short and long term.
0-rings should be lubricated before installation this is because the o-rings are usually super tight when installing and would require lubrication to ease the installation process else the rubber might get ruptured and this would lead to instant transmission failure.
Which statements describe how the Fed responds to high inflation? Check all that apply.
It charges banks more interest.
It pays banks less interest.
It sells more securities.
It decreases the money supply.
It increases the money supply.
Answer:
Answer:
• it charges banks more interest
• it sells more securities
• it decreases the money supply
Explanation:
hope this help edge 21
Caulking is recommended around the edges of partitions between apartments to... Group of answer choices reduce the need for trim. reduce sound transmission. reduce heat loss. increase the fire rating of the partition
Answer:
Reduce sound transmission.
Explanation:
A caulking is a flexible material used to seal joints, cracks or gaps formed between building materials and pipes against leakage.
Caulking is recommended around the edges of partitions between apartments to reduce sound transmission.
Hence, in the event that an individual notices that air or sound is gaining entrance into their apartment, a caulking can be used to mitigate this noise or unwanted sound.
The caulking when applied to the gap or edges of partitions between apartments would create a tight seal and block the flow or entry of air, thereby reducing sound transmission.
A phone charger requires 0.5 A at 5V. It is connected to a transformer with 100 % of efficiency whose primary contains 2200 turns and is connected to 220-V household outlet.
(a) How many turns should there be in the secondary?
(b) What is the current in the primary?
(c) What would be the output current and output voltage values if number of secondary turns (N2) doubled of its initial value?
Answer:
Explanation:
a ) for transformer which steps down voltage , if V₁ and V₂ be voltage of primary and secondary coil and n₁ and n₂ be the no of turns of wire in them
V₁ /V₂ = n₁ / n₂
Here V₁ = 220 V , V₂ = 5V , n₁ = 2200 n₂ = ?
220 /5 = 2200 / n₂
n₂ = 2200 x 5 / 220
= 50
b )
for 100 % efficiency
input power = output power
V₁ I₁ = V₂I₂
I₁ and I₂ are current in primary and secondary coil
220 x I₁ = 5 x .5
I₁ = .01136 A .
c )
If n₂ = 100
V₁ /V₂ = n₁ / n₂
220 / V₂ = 2200 / 100
V₂ = 10 V
V₁ I₁ = V₂I₂
220 x .01136 = 10 I₂
I₂ = .25 A.
A non-inductive load takes a current of 15 A at 125 V. An inductor is then connected in series in order that the same current shall be supplied from 240 V, 50 Hz mains. Ignore the resistance of the inductor and calculate: i. the inductance of the inductor; ii. the impedance of the circuit; iii. the phase difference between the current and the applied voltage.
Answer:
(i) The inductance of the inductor is = 43.43 mH (ii) the impedance of the circuit is = 16∠58.61° Ω (iii) the phase difference for current and the voltage applied is Q = 58.61°
Explanation:
Solution
Given that:
I= 5 A
V = 125V
Resistance R= Not known yet
Thus
To find the resistance we have the following formula which is shown below:
R = V/I
=125/15
R =8.333Ω
Now,
Voltage = 240
Frequency = 50Hz
Current (I) remain at = 15A
Z= not known (impedance)
so,
To find the impedance we have the formula which is shown below:
Z = V/I =240/15
Z= 16Ω⇒ Z = R + jXL
Z = 8.333 + jXL = 16
Thus
√8.333² + XL² = 16²
8.333² + XL² = 16²
XL² = 186.561
XL = 13.658Ω
Now
We find the inductance of the Inductor and the impedance of the circuit.
(i) In solving for the inductance of the inductor, a formula is applied here, which is shown below:
L = XL/w
=13.658/ 2π * 50
=13.658/314.15 = 0.043 = 43.43 mH
Note: w= 2πf
(ii) For the impedance of the circuit we have the following:
z = 8.333 + j 13.658
z = 16∠58.61° Ω
(iii) The next step is to find the phase difference between the applied voltage and current.
Q = this is the voltage across the inductor in a series of resonant circuit.
Q can also be called the applied voltage
Thus,
Q is described as an Impedance angle
Therefore, Q = 58.81°
A particle oscillates between the points x=40 mm and x=160 mm with
an acceleration a =
k(100 - x), where a and x are expressed in mm/s2 and
respectively, and k is a constant. The velocity of the particle is 18 mm/s when x = 100 mm
and is zero at both x = 40 mm and x = 160 mm. Determine (a) the value of k,
(b) the velocity when x = 120 mm.
Answer:
(a) k = 0.09 s⁻¹
(b) The velocity= ± 16.97 mm/s
Explanation:
(a) Given that the acceleration = a = k(100 - x)
Therefore;
[tex]a = \dfrac{dv}{dt} = \dfrac{dv}{dx} \times \dfrac{dx}{dt} = \dfrac{dv}{dx} \times v = k(100 - x)[/tex]
When x = 40 mm, v = 0 mm/s hence;
[tex]\int\limits^v_0 {v } \, dv = \int\limits^x_{40} {k(100 - x)} \, dx[/tex]
[tex]\dfrac{1}{2} v^2 = k \cdot \left [100\cdot x-\frac{1}{2}\cdot x^{2} \right ]_{x}^{40}[/tex]
[tex]\dfrac{1}{2} v^2 = -\dfrac{ k\cdot \left (x^{2}-200\cdot x+6400 \right ) }{2}[/tex]
At x = 100 mm, v = 18 mm/s hence we have;
[tex]\dfrac{1}{2} 18^2 = -\dfrac{ k\cdot \left (100^{2}-200\times 100+6400 \right ) }{2} = 1800\cdot k[/tex]
[tex]\dfrac{1}{2} 18^2 =162 = 1800\cdot k[/tex]
k = 162/1800 = 9/100 = 0.09 s⁻¹
(b) When x = 120 mm, we have
[tex]\dfrac{1}{2} v^2 = -\dfrac{ 0.09\times \left (120^{2}-200\times 120+6400 \right ) }{2} = 144[/tex]
Therefore;
v² = 2 × 144 = 288
The velocity, v = √288 = ±12·√2 = ± 16.97 mm/s.
In a hydroelectric power plant, water enters the turbine nozzles at 800 kPa absolute with a low velocity. If the nozzle outlets are exposed to atmospheric pressure of 100 kPa, determine the maximum velocity (m/s) to which water can be accelerated by the nozzles before striking the turbine blades.
Answer:
The answer is VN =37.416 m/s
Explanation:
Recall that:
Pressure (atmospheric) = 100 kPa
So. we solve for the maximum velocity (m/s) to which water can be accelerated by the nozzles
Now,
Pabs =Patm + Pgauge = 800 KN/m²
Thus
PT/9.81 + VT²/2g =PN/9.81 + VN²/2g
Here
Acceleration due to gravity = 9.81 m/s
800/9.81 + 0
= 100/9.81 + VN²/19.62
Here,
9.81 * 2= 19.62
Thus,
VN²/19.62 = 700/9.81
So,
VN² =1400
VN =37.416 m/s
Note: (800 - 100) = 700
Answer:
[tex]V2 = 37.417ms^{-1}[/tex]
Explanation:
Given the following data;
Water enters the turbine nozzles (inlet) = 800kPa = 800000pa.
Nozzle outlets = 100kPa = 100000pa.
Density of water = 1000kg/m³.
We would apply, the Bernoulli equation between the inlet and outlet;
[tex]\frac{P_{1} }{d}+\frac{V1^{2} }{2} +gz_{1} = \frac{P_{2} }{d}+\frac{V2^{2} }{2} +gz_{2}[/tex]
Where, V1 is approximately equal to zero(0).
Z[tex]z_{1} = z_{2}[/tex]
Therefore, to find the maximum velocity, V2;
[tex]V2 = \sqrt{2(\frac{P_{1} }{d}-\frac{P_{2} }{d}) }[/tex]
[tex]V2 = \sqrt{2(\frac{800000}{1000}-\frac{100000}{1000}) }[/tex]
[tex]V2 = \sqrt{2(800-100)}[/tex]
[tex]V2 = \sqrt{2(700)}[/tex]
[tex]V2 = \sqrt{1400}[/tex]
[tex]V2 = 37.417ms^{-1}[/tex]
Hence, the maximum velocity, V2 is 37.417m/s
Talc and graphite are two of the lowest minerals on the hardness scale. They are also described by terms like greasy or soapy. Both have a crystal structure characterized by sheet-structures at the atomic level, yet they don't behave like micas. What accounts for their unusual physical properties
Answer:
The reason for their unusual properties of the greasy feel and low hardness is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets.
Explanation:
Talc is a monoclinic mineral with a sheet structure similar to the micas and also has perfect cleavage that follows planes between the weakly bonded sheets.
Now, these sheets are held together only by van der Waals bonds and this allows them to slip past each other easily. Thus, this unique characteristic is responsible for talc's extreme softness, its greasy, soapy feel, and its value as a high-temperature lubricant.
While for graphite, it's carbon atoms are linked in a hexagonal network which forms sheets that are one atom thick. It's sheets are poorly connected and easily cleave or slide over one another when subjected to a small amount of force. Thus, gives graphite its very low hardness, its perfect cleavage, and its slippery feel.
So, we can conclude that the reason for their unusual properties is that the chemical bonds between the sheets is so weak that very low stresses can allow slip between the sheets; hence, the greasy feel and low hardness.
For the pipe-fl ow-reducing section of Fig. P3.54, D 1 5 8 cm, D 2 5 5 cm, and p 2 5 1 atm. All fl uids are at 20 8 C. If V 1 5 5 m/s and the manometer reading is h 5 58 cm, estimate the total force resisted by the fl ange bolts.
Answer:
The total force resisted by the flange bolts is 163.98 N
Explanation:
Solution
The first step is to find the pipe cross section at the inlet section
Now,
A₁ = π /4 D₁²
D₁ = diameter of the pipe at the inlet section
Now we insert 8 cm for D₁ which gives us A₁ = π /4 D (8)²
=50.265 cm² * ( 1 m²/100² cm²)
= 5.0265 * 10^⁻³ m²
Secondly, we find cross section area of the pipe at the inlet section
A₂ = π /4 D₂²
D₂ = diameter of the pipe at the inlet section
Now we insert 5 cm for D₁ which gives us A₁ = π /4 D (5)²
= 19.63 cm² * ( 1 m²/100² cm²)
= 1.963 * 10^⁻³ m²
Now,
we write down the conversation mass relation which is stated as follows:
Q₁ = Q₂
Where Q₁ and Q₂ are both the flow rate at the exist and inlet.
We now insert A₁V₁ for Q₁ and A₂V₂ for Q₂
So,
V₁ and V₂ are defined as the velocities at the inlet and exit
We now insert 5.0265 * 10^⁻³ m² for A₁ 5 m/s for V₁ and 1.963 * 10^⁻³ m² for A₂
= 5.0265 * 5 = 1.963 * V₂
V₂ = 12.8 m/s
Note: Kindly find an attached copy of the part of the solution to the given question below
Design a decimal arithmetic unit with two selection variables, V1, and Vo, and two BCD digits, A and B. The unit should have four arithmetic operations which depend on the values of the selection variables as shown below. V1=0011, V0=0101 and output functions are as follows;
1- A+9's complement of B
2- A+B
3- A+10's complement of B
4- A+1 (add 1 to A)
(You can see question number 3 in the attached file)
You are tasked with designing a thin-walled vessel to contain a pressurized gas. You are given the parameters that the inner diameter of the tank will be 60 inches and the tank wall thickness will be 5/8" (0.625 inches). The allowable circumferential (hoop) stress and longitudinal stresses cannot exceed 30 ksi.
(1) What is the maximum pressure that can be applied within the tank before failure? = psi(2) If you had the opportunity to construct a spherical tank having an inside diameter of 60 inches and a wall thickness of 5/8" (instead of the thin-walled cylindrical tank as described above), what is the maximum pressure that can be applied to the spherical tank? = psi
Answer:
Explanation:
For cylinder
Diameter d = 60 inches
thickness t = 0.625 inches
circumferential (hoop) stress = 30 ksi
[tex]hoop \ \ stress =\sigma_1=\frac{P_1d}{2t}\\\\\sigma_1=30ksi\\\\30000=\frac{P_1\times 60}{2\times0.625}\\\\P_1=624psi[/tex]
[tex]longitudinal \ \ stress =\sigma_2=\frac{P_2d}{2t}\\\\\sigma_2=30ksi\\\\30000=\frac{P_2\times 60}{4\times0.625}\\\\30000=\frac{P_2\times 60}{2.5}\\\\75000=P_2\times60\\\\P_2=\frac{75000}{60} \\\\P_1=1250psi[/tex]
Therefore maximum pressure without failure is P₁ = 625 psi
ii) For Sphere
[tex]\sigma_1=\sigma_2=\frac{Pd}{4t} \\\\P=\frac{30000\times 4 \times 0.625}{60} \\\\=\frac{75000}{60}\\\\=1250\ \ psi[/tex]
g A rectangular bar of length L has a slot in the central half of its length. The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. The overall length of the bar is L = 570 mm, and the elastic modulus of the material is 77 GPa. If the average normal stress in the central portion of the bar is 200 MPa, calculate the overall elongation δ of the bar.
Answer:
the overall elongation δ of the bar is 1.2337 mm
Explanation:
From the information given :
According to the principle of superposition being applied to the axial load P of the system; we have:
[tex]\delta = \delta_{AB} +\delta_{BC} + \delta_{CD}[/tex]
where;
δ = overall elongation
[tex]\delta _{AB}[/tex] = elongation of bar AB
[tex]\delta _{BC}[/tex] = elongation of bar BC
[tex]\delta _{CD} =[/tex] elongation of bar CD]
If we replace; [tex]\dfrac{PL}{AE}[/tex] for δ and bt for area;
we have:
[tex]\delta = \dfrac{P_{AB}L_{AB}}{(b_{AB}t)E} +\dfrac{P_{BC}L_{BC}}{(b_{BC}t)E}+\dfrac{P_{CD}L_{CD}}{(b_{CD}t)E}[/tex]
where ;
P = load
L = length of the bar
A = area of the cross-section
E = young modulus of elasticity
Let once again replace:
P for [tex]P_{AB}, P_{BC} , P_{CD}[/tex] (since load in all member of AB, BC and CD will remain the same )
[tex]\dfrac{L}{4}[/tex] for [tex]L_{AB}[/tex],
[tex]\dfrac{L}{2}[/tex] for [tex]L_{BC}[/tex] and
[tex]\dfrac{L}{4}[/tex] for [tex]L_{CD}[/tex]
[tex]2\dfrac{b}{3}[/tex] for [tex]b_{BC}[/tex]
b for [tex]b_{CD}[/tex]
[tex]\delta = \dfrac{P (\dfrac{L}{4})}{btE}+ \dfrac{P (\dfrac{L}{2})}{2 \dfrac{b}{3}tE}+\dfrac{P (\dfrac{L}{4})}{btE}[/tex]
[tex]\delta = \dfrac{PL}{btE}[\dfrac{1}{4}+ \dfrac{1}{2}*\dfrac{3}{2}+ \dfrac{1}{4}][/tex]
[tex]\delta = \dfrac{5}{4}\dfrac{PL}{btE} --- \ (1)[/tex]
The stress in the central portion can be calculated as:
[tex]\sigma = \dfrac{P}{A}[/tex]
[tex]\sigma = \dfrac{P}{\dfrac{2}{3}bt}[/tex]
[tex]\sigma = \dfrac{3P}{2bt}[/tex]
So; Now:
[tex]\delta = \dfrac{5}{4}* \dfrac{2 * \sigma}{3}*\dfrac{L}{E}[/tex]
[tex]\delta= \dfrac{5}{4}* \dfrac{2 * 200}{3}*\dfrac{570}{77*10^3 \ MPa}[/tex]
δ = 1.2337 mm
Therefore, the overall elongation δ of the bar is 1.2337 mm
An amplifier which needs a high input resistance and a high output resistance is : Select one: a. A voltage amplifier b. None of these c. A transresistance amplifier d. A current amplifier e. A transconductance amplifier Clear my choice
Answer:
None of these
Explanation:
There are different types of amplifiers, and each has different characteristics.
Voltage amplifier needs high input and low output resistance.Current amplifier needs Low Input and High Output resistance.Trans-conductance amplifier Low Input and High Output resistance.Trans-Resistance amplifier requires High Input and Low output resistance.Therefore, the correct answer is "None of these "
The drum has a mass of 50 kg and a radius of gyration about the pin at O of 0.23 o k m = . If the 15kg block is moving downward at 3 / m s , and a force of P N =100 is applied to the brake arm, determine how far the block descends from the instant the brake is applied until it stops. Neglect the thickness of the handle. The coefficient of kinetic friction at the brake pad is 0.5 k = .
Note: The diagram referred to in this question is attached as a file below.
Answer:
The block descended a distance of 9.75m from the instant the brake is applied until it stops.
Explanation:
For clarity and easiness of expression, the calculations and the Free Body Diagram are contained in the attached file. Check the attached file below.
The block descended a distance of 9.75 m
Two blocks of rubber (B) with a modulus of rigidity G = 14 MPa are bonded to rigid supports and to a rigid metal plate A. Knowing that c = 80 mm and P = 46 kN, determine the smallest allowable dimensions a and b of the blocks if the shearing stress in the rubber is not to exceed 1.4 MPa and the deflection of the plate is to be at least 7 mm.
Answer:
a = 0.07m or 70mm
b = 0.205m or 205mm
Explanation:
Given the following data;
Modulus of rigidity, G = 14MPa=14000000Pa.
c = 80mm = 0.08m.
P = 46kN=46000N.
Shearing stress (r) in the rubber shouldn't exceed 1.4MPa=1400000Pa.
Deflection (d) of the plate is to be at least 7mm = 0.007m.
From shearing strain;
[
[tex]Modulus Of Elasticity, E = \frac{d}{a} =\frac{r}{G}[/tex]
Making a the subject formula;
[tex]a = \frac{Gd}{r}[/tex]
Substituting into the above formula;
[tex]a = \frac{14000000*0.007}{1400000}[/tex]
[tex]a = \frac{98000}{1400000}[/tex]
[tex]a = 0.07m or 70mm[/tex]
a = 0.07m or 70mm.
Also, shearing stress;
[tex]r = \frac{P}{2bc}[/tex]
Making b the subject formula;
[tex]b = \frac{P}{2cr}[/tex]
Substituting into the above equation;
[tex]b = \frac{46000}{2*0.08*1400000}[/tex]
[tex]b = \frac{46000}{224000}[/tex]
[tex]b = 0.205m or 205mm[/tex]
b = 0.205m or 205mm
The basic behind equal driving is to
Follow traffic signs , Keep distance between cars , Be patient in traffic.
The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa
Complete Question:
Grain diameter 1 (mm) = 4.4E-02
Yield stress 1 (MPa) = 131
Grain diameter 2 (mm) = 7.7E-03
Yield Stress 2 (MPa) = 268
The yield strength for an alloy that has an average grain diameter, d1, is listed above as Yield Stress 1 . At a grain diameter of d2, the yield strength increases Yield Stress 2. At what grain diameter, in mm, will the yield strength be 217 MPa
Answer:
d = 1.3 * 10⁻² m
Explanation:
According to the Hall Petch equation:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
At [tex]d_{1} = 4.4 * 10^{-2} mm[/tex], [tex]\sigma_{y1} = 131 MPa = 131 N/ mm^2[/tex]
[tex]131 = \sigma_0 + k/\sqrt{4.4 * 10^{-2}} \\k = 27.45 - 0.2096 \sigma_0[/tex]
At [tex]d_{2} = 7.7 * 10^{-3} mm[/tex], [tex]\sigma_{y2} = 131 MPa = 268 N/ mm^2[/tex]
[tex]268 = \sigma_0 + (27.45 - 0.2096 \sigma_0)/\sqrt{7.7 * 10^{-3}} \\23.5036 = 27.47 - 0.1219 \sigma_0\\ \sigma_0 = 32.45 N/mm^2[/tex]
k = 27.45 - 0.2096(32.45)
k = 20.64
At [tex]\sigma_y = 217 MPa[/tex], reapplying Hall Petch law:
[tex]\sigma_y = \sigma_0 + k/\sqrt{d} \\[/tex]
[tex]217 =32.45 + 20.64/\sqrt{d} \\217 - 32.45 = 20.64/\sqrt{d}\\184.55 = 20.64/ \sqrt{d} \\\sqrt{d} = 20.64/184.55\\\sqrt{d} = 0.11184\\d = 0.013 mm[/tex]
d = 1.3 * 10⁻² m
xpress the negative value -22 as a 2's complement integer, using eight bits. Repeat it for 16 bits and 32 bits. What does this illustrate with respect to the properties of sign extension as they pertain to 2's complement representation? 8 bit The 8-bit binary representation of 22 is 00010110. So, -22 in 2’s complement form is (NOT (00010110) + 1) = (11101001 + 1) = 11101010
Answer:
Explanation:
A negative binary number is represeneted by its 2's complement value. To get 2's complement, you just need to invert the bits and add 1 to it. So the formula is:
twos_complement = ~val + 1
So you start out with 22 and you want to make it negative.
22₁₀ = 0001 0110₂
~22₁₀ = 1110 1001₂ inverting the bits
~22₁₀ + 1 = 1110 1010₂ adding 1 to it.
so -22₁₀ == ~22₁₀ + 1 == 1110 1010₂
Do the same process for 16-bits and 32-bits and you'll find that the most significant bits will be padded with 1's.
-22₁₀ = 1110 1010₂ 8-bits
-22₁₀ = 1111 1111 1110 1010₂ 16-bits
-22₁₀ = 1111 1111 1111 1111 1111 1111 1110 1010₂ 32-bits
Effluents from metal-finishing plants have the potential of discharging undesirable quantities of metals, such as cadmium, nickel, lead, manganese, and chromium, in forms that are detrimental to water and air quality. A local metal-finishing plant has identified a wastewater stream that contains 5.15 wt% chromium (Cr) and devised the following approach to lowering risk and recovering the valuable metal. The wastewater stream is fed to a treatment unit that removes 95% of the chromium in the feed and recycles it to the plant. The residual liquid stream leaving the treatment unit is sent to a waste lagoon. The treatment unit has a maximum capacity of 4500 kg wastewater/h. If wastewater leaves the finishing plant at a rate higher than the capacity of the treatment unit, the excess (anything above 4500 kg/h) bypasses the unit and combines with the residual liquid leaving the unit, and the combined stream goes to the waste lagoon.
(a) Without assuming a basis of calculation, draw and label a flowchart of the process. (b) Waste water leaves the finishing plant at a rate m_ 1 ? 6000 kg/h. Calculate the flow rate of liquid to
the waste lagoon, m_ 6?kg/h?, and the mass fraction of Cr in this liquid, x6(kg Cr/kg). (c) Calculate the flow rate of the liquid to the waste lagoon and the mass fraction of Cr in this liquid for m_1 varying from 1000 kg/h to 10,000 kg/h in 1000 kg/h increments. Generate a plot of x6 versus m_ 1 .
(Suggestion: Use a spreadsheet for these calculations.) (d) The company has hired you as a consultant to help them determine whether or not to add capacity to the treatment unit to increase the recovery of chromium. What would you need to know to make this determination? (e) What concerns might need to be addressed regarding the waste lagoon?
Answer:
Explanation:
The solution of all the four parts is provided in the attached figures
A wall in a house contains a single window. The window consists of a single pane of glass whose area is 0.11 m2 and whose thickness is 4 mm. Treat the wall as a slab of the insulating material Styrofoam whose area and thickness are 11 m2 and 0.20 m, respectively. Heat is lost via conduction through the wall and the window. The temperature difference between the inside and outside is the same for the wall and the window. Of the total heat lost by the wall and the window, what is the percentage lost by the window
Answer:
Explanation:
Given that,
The area of glass [tex]A_g[/tex] = [tex]0.11m^2[/tex]
The thickness of the glass [tex]t_g=4mm=4\times10^-^3m[/tex]
The area of the styrofoam [tex]A_s=11m^2[/tex]
The thickness of the styrofoam [tex]t_s=0.20m[/tex]
The thermal conductivity of the glass [tex]k_g=0.80J(s.m.C^o)[/tex]
The thermal conductivity of the styrofoam [tex]k_s=0.010J(s.m.C^o)[/tex]
Inside and outside temperature difference is ΔT
The heat loss due to conduction in the window is
[tex]Q_g=\frac{k_gA_g\Delta T t}{t_g} \\\\=\frac{(0.8)(0.11)(\Delta T)t}{4.0\times 10^-^3}\\\\=(22\Delta Tt)j[/tex]
The heat loss due to conduction in the wall is
[tex]Q_s=\frac{k_sA_s\Delta T t}{t_g} \\\\=\frac{(0.010)(11)(\Delta T)t}{0.20}\\\\=(0.55\Delta Tt)j[/tex]
The net heat loss of the wall and the window is
[tex]Q=Q_g+Q_s\\\\=\frac{k_gA_g\Delta T t}{t_g}+\frac{k_sA_s\Delta T t}{t_g}\\\\=(22\Delta Tt)j +(0.55\Delta Tt)j \\\\=(22.55\Delta Tt)j[/tex]
The percentage of heat lost by the window is
[tex]=\frac{Q_g}{Q}\times 100\\\\=\frac{22\Delta T t}{22.55\Delta T t}\times 100\\\\=97.6 \%[/tex]