Answer: Tl = - 13.3°C
the lowest outdoor temperature is - 13.3°C
Explanation:
Given that;
Temperature of Th = 21°C = 21 + 273 = 294 K
the rate at which heat lost is Qh = 5400 kJ/h°C
the power input to heat pump Wnet = 6 kw
The COP of a reversible heat pump depends on the temperature limits in the cycle only, and is determined by;
COPhp = Th/(Th - Tl)
COPhp = Qh/Wnet
Qh/Wnet = Th/(Th -Tl)
the amount of heat loss is expressed as
Qh = 5400/3600(294 - Tl)
the temperature of sink
( 5400/3600(294 - Tl)) / 6 = 294 / ( 294 - Tl)
now solving the equation
Tl = 259.7 - 273
Tl = - 13.3°C
so the lowest outdoor temperature is - 13.3°C
A power screw is 30 mm in diameter and has a thread pitch of 5 mm. Find the thread depth, the thread width, the mean and root diameters, and the lead, provided that square threads are used. Assume single threads.
Answer:
thread depth = 2.5 mm
thread width = 2.5 mm
mean diameter = 27.5 mm
root diameter = 25 mm
lead of screw = 5 mm
Explanation:
given data
power screw diameter D = 30 mm
thread pitch P = 5 mm
solution
First, we get here thread depth fr square thread
thread depth = [tex]\frac{P}{2}[/tex] ......................1
thread depth = [tex]\frac{5}{2}[/tex]
thread depth = 2.5 mm
and
thread width for square thread
thread width = [tex]\frac{P}{2}[/tex] ......................2
thread width = [tex]\frac{5}{2}[/tex]
thread width = 2.5 mm
and
mean diameter is
mean diameter = D - [tex]\frac{P}{2}[/tex] ................3
mean diameter = 30 - [tex]\frac{5}{2}[/tex]
mean diameter = 27.5 mm
and
root diameter is
root diameter = D - P ....................4
root diameter = 30 - 5
root diameter = 25 mm
and
lead of screw for single thread so n = 1
so lead of screw = 1 × 5
lead of screw = 5 mm
Consider atmospheric air at 25 C and a velocity of 25 m/s flowing over both surfaces of a 1-m-long flat plate that is maintained at 125 C. Determine the rate of heat transfer per unit width from the plate for values of the critical Reynolds number corresponding to 105 , 5 105 , and 106 .
Answer:
Explanation:
Temperature of atmospheric air To = 25°C = 298 K
Free stream velocity of air Vo = 25 m/s
Length and width of plate = 1m
Temperature of plate Tp = 125°C = 398 K
We know for air, Prandtl number Pr = 1
And for air, thermal conductivity K = 24.1×10?³ W/mK
Here, charectorestic dimension D = 1m
Given value of Reynolds number Re = 105
For laminar boundary layer flow over flat plate
= 3.402
Therefore, hx = 0.08199 W/m²K
So, heat transfer rate q = hx×A×(Tp – To)
= 0.08199×1×(398 – 298)
A cylinder is to be cast out of aluminum. The diameter of the disk is 500 mm and its thickness is 20 mm. The mold constant 2.0 sec/mm2 in Chvorinov's formula to calculate the solidification time.
Required:
a. Calculate the minimum time (minutes) for the casting to solidify.
b. Discuss if the result in part (a) is the same when casting grey cast iron.
Answer:
a) the minimum time (minutes) for the aluminium casting to solidify is 2.86 min
b) the minimum time (minutes) for the grey iron casting to solidify is 2.13 min. Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)
Explanation:
Given that; diameter of Disk = 500 mm, thickness t = 20, mold constant Cm = 2.0 sec/mm²
first we find the volume and Area;
Volume V = πD²t / 4
Volume V = π × (500)² × 20 / 4 = 3,926,991 mm³
Area A = 2πD²/ 4 + πDt
Area A = {[π × (500)²] / 2} +{ π × (500) × (20)}
Area A = 392,699.08 + 31,415.93
Area A = 424,115 mm²
a)
Chvorinov’s rule
T(aluminium) = Cm (V/A) ²
T(aluminium) = 2.0 × (3,926,991 / 424,115) ²
T(aluminium) = 171.5 s = 2.86 min
∴ the minimum time (minutes) for the casting to solidify is 2.86 min
b)
For cast iron
Cm (mold constant = 1.488 sec/mm²)
Chvorinov’s rule
T(iron) = Cm (V/A) ²
T(iron) = 1.488 × (3,926,991 / 424,115) ²
T(iron) = 127.5720s = 2.13 min
Therefore solidification of grey iron cast will take less time (2.13 min) compared to the solidification of the aluminium cast (2.86 min)
A four-cylinder four-stroke engine is modelled using the air standard Otto cycle (two engine revolutions per cycle). Given the conditions at state 1, total volume (V1) of each cylinder, compression ratio (r), rate of heat addition (Q), and engine speed in RPM, determine the efficiency and other values listed below. The gas constant for air is R =0.287 kJ/kg-K.
T1 = 300 K
P1 = 100 kPa
V1 = 500 cm^3
r = 10
Q = 60 kW
Speed = 5600 RPM
Required:
a. Determine the total mass (kg) of air in the engine.
b. Determine the specific internal energy (kJ/kg) at state 1.
c. Determine the specific volume (m^3/kg) at state 1.
d. Determine the relative specific volume at state 1.
Answer:
a) Mt = 0.0023229
b) = U1 = 214.07
c) = V₁ = 0.861 m³/kg
d) = Vr1 = 621.2
Explanation:
Given that
R = 0.287 KJ/kg.K, T1 = 300 K , P1 = 100 kPa , V1 = 500 cm³, r = 10 , Q = 60 kW , Speed N = 5600 RPM, Number of cylinders K = 4
specific heat at constant volume Cv = 0.7174 kJ/kg.K
Specific heat at constant pressure is 1.0045 Kj/kg.K
a) To determine the total mass (kg) of air in the engine.
we say
P1V1 = mRT1
we the figures substitute
(100 x 10³) ( 500 x 10⁻⁶) = m ( 0.287 x 10³) ( 300 )
50 = m x 86100
m = 0.00005 / 86100 = 0.0005807 ( mass of one cylinder)
Total mass of 4 cylinder
Mt = m x k
Mt = 0.0005807 x 4
Mt = 0.0023229
b) To determine the specific internal energy (kJ/kg) at state 1
i.e at T1 = 300
we obtain the value of specific internal energy U1 at 300 K ( state 1) from the table ideal gas properties of air.
U1 = 214.07
c) To determine the specific volume (m³/kg) at state 1.
we say
V₁ = V1/m
V₁ = (500 x 10⁻⁶) / 0.0005807
V₁ = 0.861 m³/kg
d) To determine the relative specific volume at state 1.
To obtain the value of relative specific volume at 300 K ( i.e state 1) from the table ideal gas properties of air.
At T1 = 300 k
Vr1 = 621.2
Strain gage is a device that senses the strain of the structure. The property of the strain gage that is used to correlate with the strain to be measured is
Answer:
resistance
Explanation:
A strain gauge changes resistance with applied strain.
An AC generator supplies an rms voltage of 120 V at 50.0 Hz. It is connected in series with a 0.650 H inductor, a 4.80 μF capacitor and a 301 Ω resistor.
(a) What is the impedance of the circuit?
(b) What is the rms current through the resistor?
(c) What is the average power dissipated in the circuit?
(d) What is the peak current through the resistor?
(e) What is the peak voltage across the inductor?
(f) What is the peak voltage across the capacitor?
The generator frequency is now changed so that the circuit is in resonance. What is that new (resonance) frequency?
Answer:
Explanation:
f = 50.0 Hz, L = 0.650 H, π = 3.14
C = 4.80 μF, R = 301 Ω resistor. V = 120volts
XL = wL = 2πfL
= 2×3.14×50* 0.650
= 204.1 Ohm
Xc= 1/wC
Xc = 1/2πfC
Xc = 1/2×3.14×50×4.80μF
= 1/0.0015072
= 663.48Ohms
1. Total impedance, Z = sqrt (R^2 + (Xc-XL)^2)= √ 301^2+ (663.48Ohms - 204.1 Ohm)^2
√ 90601 + (459.38)^2
√ 90601+211029.98
√ 301630.9844
= 549.209
Z = 549.21Ohms
2. I=V/Z = 120/ 549.21Ohms =0.218Ampere
3. P=V×I = 120* 0.218 = 26.16Watt
Note that
I rms = Vrms/Xc
= 120/663.48Ohms
= 0.18086A
4. I(max) = I(rms) × √2
= 0.18086A × 1.4142
= 0.2557
= 0.256A
5. V=I(max) * XL
= 0.256A ×204.1
=52.2496
= 52.250volts
6. V=I(max) × Xc
= 0.256A × 663.48Ohms
= 169.85volts
7. Xc=XL
1/2πfC = 2πfL
1/2πfC = 2πf× 0.650
1/2×3.14×f×4.80μF = 2×3.14×f×0.650
1/6.28×f×4.8×10^-6 = 4.082f
1/0.000030144× f = 4.082×f
1 = 0.000030144×f×4.082×f
1 = 0.000123f^2
f^2 = 1/0.000123048
f^2 = 8126.922
f =√8126.922
f = 90.14 Hz
You have accumulated several parking tickets while at school, but you are graduating later in the year and plan to return to your home in another jurisdiction. A friend tells you that the authorities in your home jurisdiction will never find out about the tickets when you re-register your car and apply for a new license. What should you do?
Answer:
pay off the parking tickets
Explanation:
In the scenario being described, the best thing to do would be to pay off the parking tickets. The parking tickets stay under your name, and if they are not paid in time can cause problems down the road. For starters, if they are not paid in time the amount will increase largely which will be harder to pay. If that increased amount is also not paid, then the government will suspend your licence indefinitely which can later lead to higher insurance rates.
Water at 20oC, with a free-stream velocity of 1.5 m/s, flows over a circular pipe with diameter of 2.0 cm and surface temperature of 80oC. Calculate the average heat transfer coefficient and the heat transfer rate per meter length of pipe.\
Answer:
Average heat transfer coefficient = 31 kw/m^2 k
Heat transfer rate per meter length of pipe = 116.808 KW
Explanation:
water temperature = 20⁰c,
free-stream velocity = 1.5 m/s
circular pipe diameter = 2.0 cm = 0.02 m
surface temperature = 80⁰c
A) calculate average heat transfer coefficient
we apply the formula below :
m = αAv
A (area) = [tex]\pi /4 (d)^2[/tex]
m = 10^3 * [tex]\pi / 4 ( 0.02)^2[/tex] * 1.5
= 10^3 * 0.7857( 0.0004) * 1.5
= 0.4714 kg/s
Average heat transfer coefficient
h = [tex]\frac{m(cp)}{A}[/tex] , A = [tex]\pi DL[/tex]
L = 1 m , m = 0.4714 kgs , cp = 4.18
back to equation
h = [tex]\frac{0.4714*4.18}{\pi * 0.02 }[/tex] = 1.970 / 0.0628 = 31.369 ≈ 31 kw/m^2 k
B) Heat transfer rate per meter length of pipe
Q = ha( ΔT ), a = [tex]\pi DL[/tex]
= 31 * 0.0628 * ( 80 - 20 )
= 31 * 0.0628 * 60 = 116.808 KW
Consider an ideal gas undergoing a constant pressure process from state 1 to state
2 in a closed system. The specific heat capacities for this material depend on temperature in
the following way, cv = aT^b , cp = cT^d , where the constants a, b, c and d are known. Calculate
the specific entropy change, (s2 − s1), from state 1 to state 2.
Answer:
[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]
Explanation:
Hello,
In this case by combining the first and second law of thermodynamics for this ideal gas, we can obtain the following expression for the differential of the specific entropy at constant pressure:
[tex]ds=c_p\frac{dT}{T}-Rg\ \frac{dP}{P}[/tex]
Whereas Rg is the specific ideal gas constant for the studied gas; thus, integrating:
[tex]\int\limits^{s_2}_{s_1} {} \, ds=c\int\limits^{T_2}_{T_1} {T^{d-1}dT} \,-Rg\ \int\limits^{P_2}_{P_1} {\frac{dP}{P}} \,[/tex]
We obtain the expression to compute the specific entropy change:
[tex]s_2-s_1=c\frac{T^d}{d}-Rg\ ln(\frac{P_2}{P_1})[/tex]
Best regards.
In general, MOSFET'S:___________.
A) are mostly used in switching circuits
B) can be fabricated in much higher densities than BJT'S
C) produce simpler circuits than BJTS
D) all of the above
Answer:
A. Are mostly used in switching circuitsExplanation:
MOSFET: The acronym for "metal oxide semiconductor field-effect transistor" are mostly used in switching circuits.
There are two classes of MOSFET
1. Depletion mode
2. Enhancement mode
Generally a MOSFET is a kind of transistor, it is actually a field effect transistor with tree terminals gate, source and drain terminals, also the MOSFET can be used as an amplifier for the amplification of electronic signals in the electronic circuit/devices
After clamping a buret to a ring stand, you notice that the set-up is tippy and unstable. What should you do to stabilize the set-up
Answer:
Move the buret clamp to a ring stand with a larger base.
Explanation:
A right stand is used for titration experiments in the laboratory. It holds the burette firmly during experiments so that accurate readings can be taken.
The right stand is made up of support base, vertical stainless steel, clamp with adjustable screw that holds on to the vertical rod.
The clamp is used to hold the burette in place.
If after clamping a buret to a ring stand, you notice that the set-up is tippy and unstable, the best action will be to move the buret clamp to a ring stand with a larger base.
The larger base provides a better center of gravity and stabilises the setup
A two-lane, one-way ramp from an urban expressway with a design speed of 30 mi/h connects with a local road at a T-intersection. The turning roadway has a vertical curb on both sides. Determine the width of the turning roadway if the predominant turning vehicles are single unit trucks with some semi-trailers. Use 0.08 for super-elevation if applicable.
Answer:
30 feet
Explanation:
Given data :
design speed = 30 miles/h
super elevation = 0.08
determine the width of the turning roadway
calculate the value of R = V^2 / 15( e + p)
e = 0.08 , p = 0.2 , v = 30
R = (30)^2 / 15 ( 0.08 + 0.2 )
= 900 / 15 ( 0.28 )
≈ 215 ft
pavement width from the calculation above = 28 ft
width of the turning roadway = pavement width + 2 = 30 feet ( because there are two vertical widths joining up the main road at the T junction )
Summary of Possible Weather and Associated Aviation Impacts for Geographic/Topographic Categories Common in the Western United States.
Geographic/Topographic Descriptive Summary of Potential Aviation Impacts
Category of a Possible Weather That Could Impact Based on Weather
of Airport Location Aviation Operations
Along the US West coast,
with steep mountains to the east
(An example of this category is
Santa Barbara Airport, located
on the Southern California Coast,
at an elevation of 10 feet).
Within a valley in elevated terrain
surrounded by high mountains
(An example of this category is
Friedman Memorial Airport, located
in Central Idaho, at an elevation of 5300 feet).
In elevated terrain on the leeside of
high mountains
(An example of this category is Northern Colorado
Regional Airport, located in northern Colorado,
at an elevation of 5000 feet, on the leeside
of the Rocky mountains).
Answer: answer provided in the explanation section.
Explanation:
Weather phenomenons that would impart Aviation Operations in Santa Barbara -
1. Although winters are cold, wet, and partly cloudy here. It is in general favorable for flying. But sometimes strong winds damage this pleasant weather.
2. The Sundowner winds cause rapid warming and a decrease in relative humidity. The wind speed is very high surrounding this area for this type of wind.
3. Cloud is an important factor that affects aviation operations. Starting from April, here the sky is clouded up to November. The sky is overcast (80 to 100 percent cloud cover) or mostly cloudy (60 to 80 percent) 44% on a yearly basis. Thus extra cloud cover can trouble aviation operations.
4. The average hourly wind speed can also be a factor. This also experiences seasonal variations, these variations are studied carefully in the aviation industry. The windier part of the year starts in January and ends in June. In April, the wind speed can reach 9.5 miles per hour.
This and more are some factors to look into when considering wheather conditions that would affect aviation operations.
I hope this was a bit helpful. cheers
In a typical transmission line, the current I is very small and the voltage V is very large. A unit length of line has resistance R. For a power line that supplies power to 10,000 households, we can conclude that:________
Answer:
IV > [tex]I^{2} R[/tex]
Explanation:
The current in the power line = I
The voltage in the power line = V
The resistance of the power line = R
Power supplied from the power house = P
power delivered to the households = [tex]p[/tex]
We know that the power supplied to a power line system is proportional to
P = IV ....1
we also know that according to Ohm's law, the relationship between the voltage, resistance, and current through an electrical system is given as
V = IR ....2
substituting equation 2 into equation 1, the power delivered to the households is proportional to the square of the current.
[tex]p[/tex] = [tex]I^{2} R[/tex] ....3
The problem is that when power is delivered across a transmission line, some of the power is loss due to Joules heating effect of the power lines. This energy and power loss is proportional to [tex]I^{2}[/tex] therefore, the electrical power delivered to the households will be less than the electrical power supplied from the power station. This means that
P > [tex]p[/tex]
equating these two powers from equations 1 and equation 3, we have
IV > [tex]I^{2} R[/tex]
Q1) Determine the force in each member of the
truss and state if the members are in tension or
compression.
Set P1 = 10 kN, P2=15 KN
Answer:
CD = DE = DF = 0BC = CE = 15 N tensionFA = 15 N compressionCF = 15√2 N compressionBF = 25 N tensionBG = 55/2 N tensionAB = (25√5)/2 N compressionExplanation:
The only vertical force that can be applied at joint D is that of link CD. Since joint D is stationary, there must be no vertical force. Hence the force in link CD must be zero, as must the force in link DE.
At joint E, the only horizontal force is that applied by link EF, so it, too, must be zero.
Then link CE has 15 N tension.
The downward force in CE must be balanced by an upward force in CF. Of that force, only 1/√2 of it will be vertical, so the force in CF is a compression of 15√2 N.
In order for the horizontal forces at C to be balanced the 15 N horizontal compression in CF must be balanced by a 15 N tension in BC.
At joint F, the 15 N horizontal compression in CF must be balanced by a 15 N compression in FA. CF contributes a downward force of 15 N at joint F. Together with the external load of 10 N, the total downward force at F is 25 N. Then the tension in BF must be 25 N to balance that.
At joint B, the 25 N downward vertical force in BF must be balanced by the vertical component of the compressive force in AB. That component is 2/√5 of the total force in AB, which must be a compression of 25√5/2 N.
The horizontal forces at joint B include the 15 N tension in BC and the 25/2 N compression in AB. These are balanced by a (25/2+15) N = 55/2 N tension in BG.
In summary, the link forces are ...
(25√5)/2 N compression in AB15 N tension in BC25 N tension in BF0 N in CD, DE, and EF15 N tension in CE15√2 compression in CF15 N compression in FA_____
Note that the forces at the pins of G and A are in accordance with those that give a net torque about those point of 0, serving as a check on the above calculations.
A charge is distributed uniformly along a long straight wire. The electric field 2 cm from the wire is 36 N/C. The electric field 4 cm from the wire is:
Answer:
New electric field = 18 N/C
Explanation:
Given:
Length (E1) = 2 cm
New length (E2) = 4 cm
Electric field = 36 N/C
Find:
New electric field
Computation:
New electric field = 36 [2 / 4]
New electric field = 36 [1/2]
New electric field = 18 N/C
Input resistance of a FET is very high due to A) forward-biased junctions have high impedance B) gate-source junction is reverse-biased C) drain-source junction is reverse-biased D) none of the above
Answer:
B) gate-source junction is reverse-biased
Explanation:
FET is described as an electric field that controls the specific current and is being applied to a "third electrode" which is generally known as "gate". However, only the electric field is responsible for controlling the "current flow" in a specific channel and then the particular device is being "voltage operated" that consists of high "input impedance".
In FET, the different "charge carriers" tend to enter a particular channel via "source" and exits through "drain".
The fins attached to a heat exchanger-surface are determined to have an effectiveness of 0.9. Do you think the rate of heat transfer from the surface has increased or decreased as a result of the addition of these fins?
Answer:
The rate of heat transfer has increased.
Explanation:
Heat transfer rate is the rate at which heat energy is dissipated to the ambient from a hot body. The rate of heat transfer is proportional to the available surface area for heat exchange. This means that the greater the exposed surface area for heat exchange, the greater the rate at which heat is lost to the ambient. In introducing the fins to the heat exchange system (fins have a large surface area to volume ratio for maximum exposure to the ambient), one maximizes the available surface area for heat exchange between the material and the ambient, increasing the rate of heat transfer.
/ Air enters a 20-cm-diameter 12-m-long underwater duct at 50°C and 1 atm at a
mean velocity of 7 m/s, and is cooled by the water outside. If the average heat
transfer coefficient is 85 W/m2
°C and the tube temperature is nearly equal to the
water temperature of 5°C, determine the exit temperature of air and the rate of heat
transfer.
Answer:
A) EXIT TEMPERATURE = 14⁰C
b) rate of heat transfer of air = - 13475.78 = - 13.5 kw
Explanation:
Given data :
diameter of duct = 20-cm = 0.2 m
length of duct = 12-m
temperature of air at inlet= 50⁰c
pressure = 1 atm
mean velocity = 7 m/s
average heat transfer coefficient = 85 w/m^2⁰c
water temperature = 5⁰c
surface temperature ( Ts) = 5⁰c
properties of air at 50⁰c and at 1 atm
= 1.092 kg/m^3
Cp = 1007 j/kg⁰c
k = 0.02735 W/m⁰c
Pr = 0.7228
v = 1.798 * 10^-5 m^2/s
determine the exit temperature of air and the rate of heat transfer
attached below is the detailed solution
Calculate the mass flow rate
= p*Ac*Vmean
= 1.092 * 0.0314 * 7 = 0.24 kg/s
Assume that the heat is transferred from the cold reservoir to the hot reservoir contrary to the Clausis statement of the second law. Prove that this violates the increase of entropy principle—as it should according to Clausius.
Answer: hello attached below is the diagram which is part of your question
Total entropy change = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k it violates Clausius increase of entropy which is Sgen > 0
Explanation:
Clausius statement states that it is impossible to transfer heat energy from a cooler body to a hotter body in a cycle or region without any other external factors affecting it .
applying the increase in entropy principle to prove this
temp of cold reservoir (t hot)= 600 k
temp of hot reservoir(t cold) = 1220 k
energy (q) = 100 kj
total entropy change = entropy change in cold reservoir + entropy change in hot reservoir = -0.166 + 0.083 = -0.0837 kj/k
entropy change in cold reservoir = Q/t cold = 100 / 600 = -0.166 kj/k
entropy change in hot reservoir = Q / t hot = 100 / 1220 = 0.083 kj/k
hence it violates Clausius inequality of increase of entropy principle which is states that generated entropy has to be > 0
Q1: You have to select an idea developing an application like web/mobile or industrial, it should be based on innovative idea, not just a simple CRUD application. After selecting the idea do the following: 1) How your project will be helpful and what problem this project addresses. (10-Marks) 2) Write down the requirements. (10Marks) 3) List the functional and non-functional requirements of your project. (10marks) 4) Which process model you will follow for this project and why? (10marks) 5) Draw the Level 0, and level 1 DFD of your application. (20marks)
Answer:
Creating an app is both an expression of our self and a reflection of what we see is missing in the world. We find ourselves digging deep into who we are, what we would enjoy working on, and what needs still need to be fulfilled. Generating an app idea for the first time can be extremely daunting. Especially with an endless amount of possibilities such as building a church app.
The uncertainty has always spawned a certain fear inside creators. The fear of creating something no one will enjoy. Spending hundreds of dollars and hours building something which might not bring back any real tangible results. The fear of losing our investment to a poor concept is daunting but not random. But simple app ideas are actually pretty easy to come by.
Great app idea generation is not a gift given to a selected few, instead, it is a process by which any of us are able to carefully explore step by step methods to find our own solution to any problem. Whether you are a seasoned creator or a novice, we have provided a few recommendations to challenge and aid you as you create your next masterpiece.
if I am right then make me brainliest
Define centrifugal pump. Give the construction and working of centrifugal pump.
Water discharging into a 10-m-wide rectangular horizontal channel from a sluice gate is observed to have undergone a hydraulic jump. The flow depth and velocity before the jump are 0.8m and 7m/s, respectively. Determine (a) the flow depth and the Froude number after the jump (b) the head loss (c) the dissipation ratio.
Answer:
A) Flow depth = 2.46 m, Froude number after jump = 0.464
B) head loss = 0.572 m
C) dissipation ratio = 0.173
Explanation:
Given data :
Velocity before jump ( v1 ) = 7 m/s
flow depth before jump ( y1 ) = 0.8 m
g = 9.81 m/s
Esi = 3.3 m ( calculated )
attached below is a detailed solution of the problem
what scale model proves the initial concept?
Answer: A prototype
Explanation:
The scale model that proves the initial concept is called a domain model.
What is a scale model?A copy or depiction of something where all parts have the same dimensions as the original. A scale model is an image or copy of an object that is either larger or smaller than the object being represented's actual size.
A domain model is a type of conceptual model that is used to depict the structural elements and conceptual constraints within a domain of interest.
A domain model will include all of the entities, their attributes, and relationships, as well as the constraints that govern the conceptual integrity of the structural model elements that comprise that problem domain.
Therefore, a domain model is the scale model that proves the initial concept.
To learn more about the scale model, refer to the below link:
https://brainly.com/question/14341149
#SPJ2
As the asteroid falls closer to the Earth's surface its _______ energy decreases and its _______ energy increases.
Answer:
As the asteroid falls closer to the Earth's surface its Gravitational Potential energy decreases and its Kinetic energy increases.
An air-conditioner which uses R-134a operates on the ideal vapor compression refrigeration cycle with a given compressor efficiency.
--Given Values--
Evaporator Temperature: T1 (C) = 9
Condenser Temperature: T3 (C) = 39
Mass flow rate of refrigerant: mdot (kg/s) = 0.027
Compressor Efficiency: nc (%) = 90
a) Determine the specific enthalpy (kJ/kg) at the compressor inlet.
Your Answer =
b) Determine the specific entropy (kJ/kg-K) at the compressor inlet
Your Answer =
c) Determine the specific enthalpy (kJ/kg) at the compressor exit
Your Answer =
d) Determine the specific enthalpy (kJ/kg) at the condenser exit.
Your Answer =
e) Determine the specific enthalpy (kJ/kg) at the evaporator inlet.
Your Answer =
f) Determine the coefficient of performance for the system.
Your Answer =
g) Determine the cooling capacity (kW) of the system.
Your Answer =
h) Determine the power input (kW)to the compressor.
Your Answer =
Answer:
A) 251.8 kj/kg
B) 0.9150 kj/kg-k
C) 155.4 kj/kg
F) 1.50
G) 3.95 kw
H) 2.6 kw
Explanation:
Given conditions :
air conditioner : R -134a
compressor efficiency (nc) = 90%.
T1 = 9⁰c, T3 = 39⁰c, mass flow rate = 0.027 kg/s
A) Specific enthalpy at the compressor inlet
at T = 9⁰c the saturated vapor (x) = 1
from the R-134a property table
h1 = 251.8 kj/kg
B ) specific entropy ( kj/kg-k) at the compressor inlet
at T = 9⁰c the saturated vapor (x) = 1
s = 0.9150 kj/kg-k ( from the R-134a property table )
C) specific enthalpy at the compressor exit
at T3 = 39⁰c , s2 = s1
has = 165.12 kj/kg
h2 = 155.4 kj/kg
attached below is the remaining solution to some of the problems
"The transistor base-emitter voltage (VBE) a. increases with an increase in temperature. b. is not affected by temperature change. c. decreases with an increase in temperature. d. has no effect on collector current."
Answer:
C) Decreases with an increase in temperature
Explanation:
As the temperature of a transistor increases, the thermal runaway property of the transistor becomes more significant and the transistors, conducting more freely as a result of the rise in temperature, causes an increase in the collector current or leakage current. The transistor base-emitter voltage decreases as a result.
With increased heating due to heavy current flow, the transistor is damaged.
B1) 20 pts. The thickness of each of the two sheets to be resistance spot welded is 3.5 mm. It is desired to form a weld nugget that is 5.5 mm in diameter and 5.0 mm thick after 0.3 sec welding time. The unit melting energy for a certain sheet metal is 9.5 J/mm3 . The electrical resistance between the surfaces is 140 micro ohms, and only one third of the electrical energy generated will be used to form the weld nugget (the rest being dissipated), determine the minimum current level required.
Answer:
minimum current level required = 8975.95 amperes
Explanation:
Given data:
diameter = 5.5 mm
length = 5.0 mm
T = 0.3
unit melting energy = 9.5 j/mm^3
electrical resistance = 140 micro ohms
thickness of each of the two sheets = 3.5mm
Determine the minimum current level required
first we calculate the volume of the weld nugget
v = [tex]\frac{\pi }{4} * D^2 * l[/tex] = [tex]\frac{\pi }{4} * 5.5^2 * 5[/tex] = 118.73 mm^3
next calculate the required melting energy
= volume of weld nugget * unit melting energy
= 118.73 * 9.5 = 1127.94 joules
next find the actual required electric energy
= required melting energy / efficiency
= 1127 .94 / ( 1/3 ) = 3383.84 J
TO DETERMINE THE CURRENT LEVEL REQUIRED use the relation below
electrical energy = I^2 * R * T
3383.84 / R*T = I^2
3383.84 / (( 140 * 10^-6 ) * 0.3 ) = I^2
therefore 8975.95 = I ( current )
Write about traditional brick production in Pakistan
Answer:
Clay bricks are manufactured by mining and clay moulded blocks. There are 20,000 brick klins in Pakistan.
Explanation:
In Pakistan, the clay bricks are manufactured by mining and baking the clay moulded blocks in brick kilns. According to an estimate, the baking process emits about 1.4 pounds of carbon per brick made, but in Pakistan, because the systems are outdated, brick kilns are used, which is producing more than the average amount of the pollution.
There are about 20,000 brick kilns in Pakistan. The traditional brick production in Pakistan is consists of hand-made bricks which are first baked in Fixed Chimney Bull's Trench Kilns (FCBTK), this is the most widely used brick firing technology in South Asia.
When replacing a timing belt, many experts and vehicle manufacturers recommend that all of the following should be replaced except the
A. water pump
B. camshaft oil seal(s).
C. camshalt sprocket
D. tensioner assembly
Answer:
Correct Answer:
A. water pump
Explanation:
Timing belt in a vehicle helps to ensure that crankshaft, pistons and valves operate together in proper sequence. Timing belts are lighter, quieter and more efficient than chains that was previously used in vehicles.
Most car manufacturers recommended that, when replacing timing belt, tension assembly, water pump, camshaft oil seal should also be replaced with it at same time.
