Answer:
The answer is below
Explanation:
1. Proportional Control is a form of control engineering in which an output is directly proportional to the error signal.
Characteristics of proportional control are:
* It is utilized when the deviation between the input and output is small
* It is also utilized when the deviation is not sudden.
* It reduces steady-state error
* It speeds up the response of the overdamped system
2. Proportional plus Integral Control is a form of control engineering in which a collective proportion and integral control of the output is equivalent to the combined proportion and integral of the error signal.
Characteristics of proportional plus integral control are:
* it can revert the controlled variable to the original set point
* It decreases steady-state error
* It quickens up the reaction of the overdamped system
3. Proportional plus integral plus derivative control is mostly applicable in operating the process elements such as temperature, pressure, speed, etc. It is recommended for industrial use.
Characteristics of Proportional plus integral plus derivative control are:
* It enhances the temporary reaction of the system.
* It also lessens steady-state error
* It accelerates the response of the overdamped system
Estimate (a) the maximum, and (b) the minimum thermal conductivity values (in W/m-K) for a cermet that contains 76 vol% carbide particles in a metal matrix. Assume thermal conductivities of 30 and 67 W/m-K for the carbide and metal, respectively.
Answer:
The answer is below
Explanation:
Given that:
Volume of carbide ([tex]V_{C}[/tex]) = 76% = 0.76, Volume of Nickel ([tex]V_{M}[/tex]) = 100% - 76% = 24% = 0.24, thermal conductivities of carbide ([tex]E_{C}[/tex]) = 30 W/m-K and thermal conductivities of meta. ([tex]E_M[/tex]) = 67 W/m-K
a) The maximum thermal conductivity is given by:
Max = [tex]E_CV_c+E_mV_m=(0.76*30) + (0.24*67) = 38.88\ W/mK[/tex]
b) The minimum thermal conductivity is given by:
Min = [tex]\frac{E_ME_C}{E_MV_C+E_CV_M}=\frac{30*67}{(0.76*67)+(0.24*30)}=34.58\ W/mK[/tex]
a cantilever beam 1.5m long has a square box cross section with the outer width and height being 100mm and a wall thickness of 8mm. the beam carries a uniform load of 6.5kN/m along the entire length and, in the same direction, a concentrated force of 4kN at the free end. (a) determine the max bending stress (b) determine the max transervse shear stress (c) determine the max shear stress in the beam
Answer:
a) 159.07 MPa
b) 10.45 MPa
c) 79.535 MPa
Explanation:
Given data :
length of cantilever beam = 1.5m
outer width and height = 100 mm
wall thickness = 8mm
uniform load carried by beam along entire length= 6.5 kN/m
concentrated force at free end = 4kN
first we determine these values :
Mmax = ( 6.5 *(1.5) * (1.5/2) + 4 * 1.5 ) = 13312.5 N.m
Vmax = ( 6.5 * (1.5) + 4 ) = 13750 N
A) determine max bending stress
б = [tex]\frac{MC}{I}[/tex] = [tex]\frac{13312.5 ( 0.112)}{1/12(0.1^4-0.084^4)}[/tex] = 159.07 MPa
B) Determine max transverse shear stress
attached below
ζ = 10.45 MPa
C) Determine max shear stress in the beam
This occurs at the top of the beam or at the centroidal axis
hence max stress in the beam = 159.07 / 2 = 79.535 MPa
attached below is the remaining solution
A person walks into a refrigerated warehouse with head uncovered. Model the head as a 25- cm diameter sphere at 35°C with a surface emissivity of 0.95. Heat is lost from the head to the surrounding air at 25°C by convection with a convection coefficient of ???????????????? ???????? ????????????????∙???????? , and by radiation to the surrounding black walls at 15°C. Determine the total rate of heat loss. StefanBoltzmann Constant, ???????? = ????????. ???????????????? × ????????????????−???????? ???????? ????????????????∙???????????????? . (10 points)
Answer:
Hello some parts of your question is missing below is the missing part
Convection coefficient = 11 w/m^2. °c
answer : 44.83 watts
Explanation:
Given data :
surface emissivity ( ε )= 0.95
head ( sphere) diameter( D ) = 0.25 m
Temperature of sphere( T ) = 35° C
Temperature of surrounding ( T∞ ) = 25°C
Temperature of surrounding surface ( Ts ) = 15°C
б = ( 5.67 * 10^-8 )
Determine the total rate of heat loss
First we calculate the surface area of the sphere
As = [tex]\pi D^{2}[/tex]
= [tex]\pi * 0.25^2[/tex] = 0.2 m^2
next we calculate heat loss due to radiation
Qrad = ε * б * As( [tex]T^{4} - T^{4} _{s}[/tex] ) ---- ( 1 )
where ;
ε = 0.95
б = ( 5.67 * 10^-8 )
As = 0.2 m^2
T = 35 + 273 = 308 k
Ts = 15 + 273 = 288 k
input values into equation 1
Qrad = 0.95 * ( 5.67 * 10^-8 ) * 0.2 ( (308)^4 - ( 288)^4 )
= 22.83 watts
Qrad ( heat loss due to radiation ) = 22.83 watts
calculate the heat loss due to convection
Qconv = h* As ( ΔT )
= 11*0.2 ( 35 -25 ) = 22 watts
Hence total rate of heat loss
= 22 + 22.83
= 44.83 watts
Help please and please
Determine the size of memory needed for CD recording of a piece of music, which lasts for 26 minutes, is done with a 20-bit Analog-to-Digital Converter (ADC) in stereo (2 channels), at the rate of 44.1 kSa/s, with the compression factor 6 (allow 10% error margin).
Answer: the size of memory needed for the CD recording is 28.7 MB
Explanation:
so in the case of stereo, the bitrate is;
⇒ 26 × 60 × 44.1 × 10³ × 2
= 137592 × 10³
for 10 bit
⇒ 137592 × 10³ × 10
= 1375920 × 10³ bits
now divide by 8 (convert to bytes)
⇒ (1375920 × 10³) / 8
= 171,990,000 BYTE
divide by 1000 (convert to kilobytes)
= 171,990,000 / 1000
= 171,990 KILOBYTES
now Given that, the compression ratio is 6
so
171,990 / 6
= 28665 KB
we know that. 1 MB = 1000 KB
x MB = 28665 KB
x MB = 28665 / 1000
⇒ 28.665 MB ≈ 28.7 MB
Therefore the size of memory needed for the CD recording is 28.7 MB
Write 83,120 in expanded form using powers of 10.
Answer:
8*10000+3*1000+1*00+2*10+2
Explanation:
7. The process of separating a milk’s solids from its liquids is called
A. Homogenization
B. Curdling
C. Creaming
D. Baking
Answer: B is correct
Explanation:
the process in which dairies make cheese by separating a milk's solids from its liquid is called. curdling.
In a compression test, a steel test specimen (modulus of elasticity 30 106 lb/in2 ) has a starting height 2.0 in and diameter 1.5 in. The metal yields (0.2% offset) at a load 140,000 lb. At a load of 260,000 lb, the height has been reduced to 1.6 in. Determine (a) yield strength and (b) fl ow curve parameters (strength coeffi cient and strain-hardening exponent). Assume that the cross-sectional area increases uniformly during the test.
Answer:
A) σ_y = 79096 lb/in² = 79.1 ksi
B) strain-hardening exponent = 0.102
(strength coefficient = 137838.78 lb/in²
Explanation:
A) Formula for volume is;
V = πd²h/4
We are given;
height 2.0 in and diameter 1.5 in
Thus;
V = (π × 1.5² × 2)/4
V = 3.53 in³
Area is;
A = πd²/4
A = (π × 1.5²)/4
A = 1.77 in²
Yield strength is gotten from the formula;
σ_y = Force/Area
We are given load = 140,000 lb
Thus;
σ_y = 140000/1.77
σ_y = 79096 lb/in²
B) We are given
modulus of elasticity: E = 30 × 10^(6) lb/in²
Formula for strain is;
ε = σ_y/E
ε = 79096/(30 × 10^(6))
ε = 0.00264
The metal yields (0.2% offset), thus;
strain offsets = 0.00264 + 0.002
strain offsets: ε1 = 0.00464
Thus;
(h_i - h_o)/h_o = 0.00464
(h_i/h_o) - 1 = 0.00464
(h_i/h_o) = 1.00464
h_i = h_o(1.00464)
h_o = 2 in
Thus; h_i = 2(1.00464) = 2.00928 in
Area = Volume/height = 3.53/2.00928 = 1.757 in²
True stress is;
σ = force/area = 140000/1.757
σ1 = 79681.27 lb/in²
At a load of 260,000 lb, the height has been reduced to 1.6 in. Thus;
Area = 3.53/1.6 = 2.206 in²
True stress is;
σ2 = 260000/2.206
σ2 = 117860.38 lb/in²
True strain;
ε2 = In(2/1.6)
ε2 = 0.223
From flow curve;
σ = kεⁿ
Thus;
σ1 = k(ε1)ⁿ
79681.27 = k(0.00464ⁿ) - - - (eq 1)
Also for σ2 = k(ε2)ⁿ;
117860.38 = k(0.223ⁿ) - - - - - (eq 2)
From eq 1,
k = 79681.27/0.00464ⁿ
Putting this for k in eq2 to get;
117860.38 = (0.223ⁿ) × 79681.27/0.00464ⁿ
117860.38/79681.27 = 0.223ⁿ/0.00464ⁿ
Solving for n, we have ≈ 0.102
Thus,K is;
k = 79681.27/0.00464^(0.102)
k = 137838.78 lb/in²
How many flip-flop values are complemented in an 8-bit binary ripple counter to reach the next count value after: 0110111 and 01010110?
Answer:
- Four (4) flip-flop values will complemented
- one (1) flip-flop value will complemented
Explanation:
To find how many flip flop number of bits complemented, we just need to figure out what the next count in the sequence is and find how many bits have changed.
taking a look at the a) 00110111
we need to just 1 to the value,
so
00110111 + 0000001 = 00111000
So here, only the first four bits are complemented.
Therefore Four (4) flip-flop values will complemented
Next
b) 01010110
we also add 1 to the value
01010110 + 00000001 = 01010111
only the first bit is complemented.
Therefore one (1) flip-flop value will complemented
is a street the same as a avenue
they're essentially the same thing so i'd say yes
Five identical keys are suspended from a balance, which is held horizontally as shown. The two keys on the left are attached to the balance 6 cm from the pivot and the three keys on the right are attached 5 cm from the pivot. What will happen when the person lets go of the balance beam?
Answer:
movement in clockwise direction.
Explanation:
The following parameters or information are given from the question above, they are:
[1]. There are two identical keys, [2]. two out of the five keys are attached to 6cm from the pivot, [3]. three keys out of the five keys on the right are attached 5 cm.
Therefore, considering the moment of force, the two keys on the left = 2 × 6 = 12.
Also, considering the moment of force, the 3 keys on the right = 3 × 5 = 15.
Therefore, we have more weight on the right keys. So, in order to balance the force there must be movement in clockwise direction.
