Answer:
The container should be made of lead and the liquid should be water.
Explanation:
Since the volume of the container of liquid after expansion is V = V₀(1 + βΔθ) where V = initial volume, β = coefficient of volume expansion, Δθ = temperature change.
So, the volume change V₂ - V₁ where V₁ = volume of liquid and V₂ = volume of container
For steel, V₂ = V₀(1 + β₂Δθ) and V₁ = V₀(1 + β₁Δθ)
So, ΔV = V₀(1 + β₂Δθ) - V₀(1 + β₁Δθ) = V₀[1 + β₂Δθ - 1 - β₁Δθ] = V₀[β₂Δθ - β₁Δθ]
Since we want a minimum value for ΔV and V₀ and Δθ are the same, we need ΔV/V₀Δθ = β₂ - β₁ to be a minimum
where β₂ = coefficient of volume expansion of liquid and β₁ = coefficient of volume expansion of container.
So, trying each combination, with β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 171 × 10⁻⁶ (C°)⁻¹
With β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 120 × 10⁻⁶ (C°)⁻¹
With β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 36 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 1120 × 10⁻⁶ (C°)⁻¹ - 36 × 10⁻⁶ (C°)⁻¹ = 1084 × 10⁻⁶ (C°)⁻¹
With β₂ = 1120 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
β₂ - β₁ = 207 × 10⁻⁶ (C°)⁻¹ - 87 × 10⁻⁶ (C°)⁻¹ = 1033 × 10⁻⁶ (C°)⁻¹
The combination that gives the lowest value for β₂ - β₁ is β₂ = 207 × 10⁻⁶ (C°)⁻¹] and β₁ = 87 × 10⁻⁶ (C°)⁻¹
Since β₁ = 87 × 10⁻⁶ (C°)⁻¹ = coefficient of expansion for lead β₂ = 207 × 10⁻⁶ (C°)⁻¹] = coefficient of expansion for water, the container should be made of lead and the liquid should be water.
Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move of rock and soil, which Hans knows from previous experience has an average density of . Hans has available a dump truck with a capacity of and a maximum safe load of .Calculate the number of trips the dump truck will have to make to haul the customer's load away.
Complete Question:
Hans, a scheduler at Mangel-Wurzel Transport, gets a call from a regular customer needing to move 19.8m³ of rock and soil, which Hans knows from previous experience has an average density of 650/kgm³. Hans has available a dump truck with a capacity of 4m³ and a maximum safe load of 3700kg. Required: calculate the number of trips the dump truck will have to make to haul the customer's load away.
Answer:
Mangel-Wurzel Transport
The number of trips that the dump truck will have to make to haul the customer's load away is:
= 5 trips.
Explanation:
a) Data and Calculations:
Volume of customer's load (rock and soil) = 19.8m³
Density of load = 650 kg/m³
Mass of load = Volume of load * Density of load
= 19.8m³ × 650 kg/m³
= 12,870 kg
The maximum safe load (mass) of the dump truck = 3,700 kg
Volume of the dump truck = 4m³
Assuming the truck is to carry 4m³ of the load.
The mass of load that the 4m³ capacity truck can carry = 4m³ × 650kg/m³
= 2,600kg
Quick Check:
Mass = 2,600kg < 3,700 kg, satisfying required conditions.
The number of trips that the truck would make to haul the customer's load away is, therefore, calculated as follows:
Number of trips = N
N = total volume of load/ volume per trip
N = 19.8/4
N = 4.95
N = 5 trips approx.
Using 1.5 V batteries, a switch, and three lamps, devise a circuit to apply 4.5 V across eitherone lamp, two lamps in series, or three lamps in series with a single-control switch. Draw theschematic.
Answer: the attached picture is the answer.
Explanation:
Assuming:
the switch position connect to 1, hence 4.5V exist at across lamp1
the switch position connects to 2 hence 4.5 V exist across lamp 1 and lamp 2
the switch position connects to 3, hence, 4.5 V exist across lamp 1, lamp 2 and lamp 3.
The temperature gradient in a spherical (or cylindrical) wall at steady state will always decrease (in magnitude) with increasing distance from the center (line), i.e. radial distance.
A. True
B. False
Answer:
True
Explanation:
Yes it is true that the Temperature gradient would also decrease with magnitude just as the distances rise from the centre line.
We have this cylinder equation as
[T1-T2 / ln(r1-r2)]2πKL
The radial distance is r2-r1
The gradient of temperature is T1-T2
From the equation,
The temperature gradient has a direct and proportional relationship to radial distance
T1-T2 ∝ ln(r2-r1)
1/T1-T2 = k(r2-r1)
This inverse relationship above confirms that the statement is true
Determine the pressure difference in N/m2,between two points 800m apart in horizontal pipe-line,150 mm diameter, discharging water at the rate of 12.5litres per second. Take the frictional coefficient ,f, as being 0.008
Answer: [tex]10.631\times 10^3\ N/m^2[/tex]
Explanation:
Given
Discharge is [tex]Q=12.5\ L[/tex]
Diameter of pipe [tex]d=150\ mm[/tex]
Distance between two ends of pipe [tex]L=800\ m[/tex]
friction factor [tex]f=0.008[/tex]
Average velocity is given by
[tex]\Rightarrow v_{avg}=\dfrac{12.5\times 10^{-3}}{\frac{\pi }{4}(0.15)^2}\\\\\Rightarrow v_{avg}=\dfrac{15.9134\times 10^{-3}}{2.25\times 10^{-2}}\\\\\Rightarrow v_{avg}=7.07\times 10^{-1}\\\Rightarrow v_{avg}=0.707\ m/s[/tex]
Pressure difference is given by
[tex]\Rightarrow \Delta P=f\ \dfrac{L}{d}\dfrac{\rho v_{avg}^2}{2}\\\\\Rightarrow \Delta P=0.008\times \dfrac{800}{0.15}\times \dfrac{997\times (0.707)^2}{2}\\\\\Rightarrow \Delta P=10,631.45\ N/m^2\\\Rightarrow \Delta P=10.631\ kPa[/tex]
A 20-mm-diameter steel bar is to be used as a torsion spring. If the torsional stress in the bar is not to exceed 110 MPa when one end is twisted through an angle of 15°, what must be the length of the bar?
Answer:
1.887 m
Explanation:
(15 *pi)/180
= 0.2618 rad
Polar moment
= Pi*d⁴/32
= (22/7*20⁴)/32
= 15707.96
Torque on shaft
= ((22/7)*20³*110)/16
= 172857.14
= 172.8nm
Shear modulus
G = 79.3
L = Gjθ/T
= 79.3x10⁹x(1.571*10^-8)x0.2618/172.8
= 1.887 m
The length of the bar is therefore 1.887 meters
Reinforced concrete is a raw material that has always been available, but it was unappreciated by architects until the nineteenth century.
a. True
b. False
Answer: False
Explanation:
Reinforced concrete is simply a combination of the traditional cement concrete with the steel bars which are the reinforcements.
Reinforced concrete is utilized for construction purpose mostly on a large scale. The reinforced concrete was invented by French gardener Joseph in 1849 therefore, it has always been available and appreciated by architects before the 19th century.
For a steel alloy it has been determined that a carburizing heat treatment of 3-h duration will raise the carbon concentration to 0.38 wt% at a point 2.6 mm from the surface. Estimate the time (in h) necessary to achieve the same concentration at a 6.1 mm position for an identical steel and at the same carburizing temperature.
Answer:
The right answer is "16.5 hrs".
Explanation:
Given values are:
[tex]x_1=2.6 \ mm[/tex]
[tex]t_1=3 \ hrs[/tex]
[tex]x_2=6.1 \ mm[/tex]
As we know,
⇒ [tex]\frac{x^2}{Dt}=constant[/tex]
or,
⇒ [tex]\frac{x_1^2}{t_1} =\frac{x_2^2}{t_2}[/tex]
⇒ [tex]t_2=(\frac{x_2}{x_1})^2\times t_1[/tex]
By putting the values, we get
[tex]=(\frac{6.1}{2.6} )^2\times 3[/tex]
[tex]=5.5\times 3[/tex]
[tex]=16.5 \ hrs[/tex]
(50 POINTS) How many people use pipes in the world? How do you know this?
Answer:
7.9 billion people
Explanation:
Calculate density, specific weight and weight of one litter of petrol having specific gravity 0.7
Explanation:
mass=19kg
density=800kg/m³
volume=?
as we know that
density=mass/volume
density×volume=mass
volume=mass/density
putting the values
volume=19kg/800kg/m³
so volume=0.02375≈0.02m³
what are qualifications that are required to be an architect
Answer:
Bachelor's degree
In order to become a licensed architect in the US and the District of Columbia, applicants are required to complete a professional degree in architecture, gain on-the-job experience through a paid internship.
Calculate the scale and speed of the pattern in order to gain useful results for a turbine operate at 150 rev/min at height difference of 22 m and a predictable flow rate of 85 m per second. A scale pattern is made and examined with a volume flow rate of 0.1 m per second and a height difference of 5 m , the power value equal to 4.5 kW when checked at the speed evaluated . Predict the power and efficiency of the full size turbine .
Answer:
first mark me as a brainleast
1. A manufacturing cell with two workers is responsible for producing a small frying pan with a required takt time of 496 seconds. The material passes through two processes: a deep drawing process and a trimming process. The average cycle time for the deep drawing process is 450 seconds and average cycle time for trimming is 430 seconds. (2 pts.)
a. Does the work cell have adequate capacity to meet demand? Explain.
b. What is the required daily production capacity of the work cell (in number of frying pans per day)? Assume 480 minutes/workday of available time.
2. What is the total daily idle time for both workers in Problem 1? Report your answer in (a) seconds of idle time and (b) as a percentage of total working time for the cell. (2 pts.)
Answer:
Explanation:
[tex]496=\frac{480\times 60}{demand}[/tex]
demand per day = 58 pans
Due to availability of two workers we can have parallel we can have deep drawing and trimming operations simultaneously.
Hence the cycle time would be the greater time of the two operations.
cycle time = 450 seconds
[tex]\text{capacity of work cell}=\frac{\text{available working time}}{\text{cycle time}}[/tex]
[tex]\text{capacity of work cell}=\frac{480\times 60}{450}[/tex]
[tex]\text{capacity of work cell}=64 ~pans[/tex] (which is greater than the demand of 58 pans)
Therefore the work cell has sufficient capacity and time (496 sec.>cycle time 450 sec) to meet the demand.
b)
Required daily production is 58 pans
what is the best glide speed for your training airplane
1.5 nautical miles per 1,000 feet
A venturimeter of 400 mm × 200 mm is provided in a vertical pipeline carrying oil of specific gravity 0.82, flow being upward. The difference in elevation of the throat section and entrance section of the venturimeter is 300 mm. The differential U-tube mercury manometer shows a gauge deflection of 300 mm. Calculate: (i) The discharge of oil, and (ii) The pressure difference between the entrance section and the throat section.Take the coefficient of meter as 0.98 and specific gravity of mercury as 13.6
Answer:
the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Explanation:
Given:
Diameter of the pipe = 100mm = 0.1m
Contraction ratio = 0.5
thus, diameter at the throat of venturimeter = 0.5×0.1m = 0.05m
The formula for discharge through a venturimeter is given as:
Where,
is the coefficient of discharge = 0.97 (given)
A₁ = Area of the pipe
A₁ =
A₂ = Area at the throat
A₂ =
g = acceleration due to gravity = 9.8m/s²
Now,
The gauge pressure at throat = Absolute pressure - The atmospheric pressure
⇒The gauge pressure at throat = 2 - 10.3 = -8.3 m (Atmosphric pressure = 10.3 m of water)
Thus, the pressure difference at the throat and the pipe = 3- (-8.3) = 11.3m
Substituting the values in the discharge formula we get
or
or
Q = 29.28 ×10⁻³ m³/s
Hence, the rate of flow = 29.28 ×10⁻³ m³/s or 0.029 m³/s
Hope This Helps :D
Problema:
Una nevera de vinos, con un peso bruto de 50 kg., que tiene las siguientes dimensiones: .60 m Largo x .49 m ancho x .50 m altura. Para ser transportadas en un contenedor de 40 pies D.V. responder las siguientes preguntas:
• 1.Cuántas neveras de vinos de acuerdo al volumen caben en un contenedor de 40 pies?
• De acuerdo dimensiones internas (largo, ancho y alto), ¿Cuántas caben en un contenedor de 40 pies?
• De acuerdo al peso que soporta el contenedor. ¿Cuántas neveras de vinos es posible transportar?
Answer:
I can't understand this language .
Given that the system function of a third order Butterworth type analog low-pass filter with a 3 dB cut-off frequency of 2 radian/second is:
2s HS = S2 + 0.2 s +1
Answer the following questions:
1. Use the bilinear transformation to obtain H(z). Use T=2 second.
2. Give H(w) for your filter.
3. Use MATLAB to give the magnitude spectrum.
4. Comment on the quality of the design.
5. With the aid of simple sketch graphs explain how frequency warping affects the frequency response of the digital filter.
6. Comment on the need for prewarping, i.e. give conditions when prewarping is needed.
answer
d just too the test
Steam enters an adiabatic turbine at 6 MPa, 600°C, and 80 m/s and leaves at 50 kPa, 100°C, and 140 m/s. If the power output of the turbine is 5 MW, determine (a) the reversible power output and (b) the second-law efficiency of the turbine. Assume the surroundings to be at 25°C.
Answer:
(a) the reversible power output of turbine is 5810 kw
(b) The second-law efficiency of he turbine = 86.05%
Explanation:
In state 1: the steam has a pressure of 6 MPa and 600°C. Obtain the enthalpy and entropy at this state.
h1 = 3658 kJ/kg s1=7.167 kJ/kgK
In state 2: the steam has a pressure of 50 kPa and 100°C. Obtain the enthalpy and entropy at this state
h2 = 2682kl/kg S2= 7.694 kJ/kg
Assuming that the energy balance equation given
Wout=m [h1-h2+(v1²-v2²) /2]
Let
W =5 MW
V1= 80 m/s V2= 140 m/s
h1 = 3658kJ/kg h2 = 2682 kJ/kg
∴5 MW x1000 kW/ 1 MW =m [(3658-2682)+ ((80m/s)²-(140m/s)²)/2](1N /1kg m/ s²) *(1KJ/1000 Nm)
m = 5.158kg/s
Consider the energy balance equation given
Wrev,out =Wout-mT0(s1-s2)
Substitute Wout =5 MW m = 5.158kg/s 7
s1= 7.167 kJ/kg-K s2= 7.694kJ/kg-K and 25°C .
Wrev,out=(5 MW x 1000 kW /1 MW) -5.158x(273+25) Kx(7.167-7.694)
= 5810 kW
(a) Therefore, the reversible power output of turbine is 5810 kw.
The given values of quantities were substituted and the reversible power output are calculated.
(b) Calculating the second law efficiency of the turbine:
η=Wout/W rev,out
Let Wout = 5 MW and Wrev,out = 5810 kW
η=(5 MW x 1000 kW)/(1 MW *5810)
η= 86.05%
Identify the first step in preparing a spectrophotometer for use.
A. Make sure all samples and the blank are ready for measurement.
B. Prepare a calibration curve.
C. Measure the absorbance of the blank.
D. Turn on the light source and the spectrophotometer.
Answer:
D. Turn on the light source and the spectrophotometer.
Explanation:
A spectrophotometer is a machine used to measure the presence of any light-absorbing particle in a solution as well as its concentration. To prepare a spectrophotometer for use, the first step is to turn on the spectrophotometer and allow it to warm up for at least 15 minutes. After this is done, the next step will be to ensure that the samples and blank are ready. Next, an appropriate wavelength is set for the solute being determined. Finally, the absorbance is measured of both the blank and samples.
A circular rod with a gage length of 3.1 m and a diameter of 3 cm is subjected to an axial load of 68 kN . If the modulus of elasticity is 200 GPa , what is the change in length
Answer:
1.49 mm
Explanation:
The modulus of elasticity, Y = stress/strain = σ/ε
σ = F/A where F = load = 68 kN = 68 × 10³ N and A = cross-sectional area of rod = πd²/4 where d = diameter of rod = 3 cm = 3 × 10⁻² m.
ε = ΔL/L where ΔL = change in length of the circular rod and L = length of circular rod = 3.1 ,
So, Y = σ/ε
Y = F/A ÷ ΔL/L
Y = FL/AΔL
making the change in length ΔL subject of the formula, we have
ΔL = FL/AY
substituting the value of A into the equation, we have
So, ΔL = FL/(πd²/4)Y
ΔL = 4FL/πd²Y
Since Y = 200 GPa = 200 × 10⁹ Pa
Substituting the values of the variables into the equation, we have
ΔL = 4FL/πd²Y
ΔL = 4 × 68 × 10³ N × ×3.1 m/[π(3 × 10⁻²m)² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[9π × 10⁻⁴m² × 200 × 10⁹ Pa]
ΔL = 843.2 × 10³ Nm/[1800π × 10⁵ N]
ΔL = 843.2 × 10³ Nm/5654.87 × 10⁵ N
ΔL = 0.149 × 10⁻² m
ΔL = 1.49 × 10⁻³ m
ΔL = 1.49 mm
The change in length of the circular rod is 1.49 mm
) Please label the following statements as either True (T) or False (F). (a) In general, the greater the % of cold work, the smaller the recrystallization grain size. (b) The higher the annealing temperature, the smaller the recrystallization grain size. (c) The greater the % of cold work, the lower the recrystallization temperature.
Answer:
A. This option is true
B. This option is false
C. This option is true
Explanation:
A. Generally speaking, the greater percentage of cold, the recrystallization grain size would turn out to be smaller. Therefore this true.
B. A higher annealing temperature does not result in smaller recrystallization grain size. Therefore this is false.
C. As the percentage of cold work is greater, the recrystallization temperature would tend to be lower. Therefore this is true.
Set the leak rate to zero and choose a non-zero value for the proportional feedback gain.Restart the simulation and turn on the outflow valve.What happens to the liquid level in the tank?Repeat this process with higher and lower values for the proportional feedback gain.What happens when the proportional feedback gain is increased?What happens when it is decreased?Find the proportional gain that will reach steady state the quickest without oscillationin the state of the valve and restart the simulation.What is the system time constant, as determined from the tank level versus time plot.
Answer:
Explanation:
The proportional gain K is usually a fixed property of the controller . If proportional gain is increased , The sensitivity of the controller to error is increased but the stability is impaired. The system approaches the behaviour of on off controlled system and it response become oscillatory
If a heat engine has an efficiency of 30% and its power output is 600 W, what is the rate of heat input from the combustion phase
Answer:
The heat input from the combustion phase is 2000 watts.
Explanation:
The energy efficiency of the heat engine ([tex]\eta[/tex]), no unit, is defined by this formula:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex] (1)
Where:
[tex]\dot Q[/tex] - Heat input, in watts.
[tex]\dot W[/tex] - Power output, in watts.
If we know that [tex]\dot W = 600\,W[/tex] and [tex]\eta = 0.3[/tex], then the heat input from the combustion phase is:
[tex]\eta = \frac{\dot W}{\dot Q}[/tex]
[tex]\dot Q = \frac{\dot W}{\eta}[/tex]
[tex]\dot Q = \frac{600\,W}{0.3}[/tex]
[tex]\dot Q = 2000\,W[/tex]
The heat input from the combustion phase is 2000 watts.
Match the test to the property it measures.
a. Rockwell
b. Inston
c. Charpy
d. Fatigue
e. Brinell
f. Izod
1. impact strength
2. stress vs strain
3. hardness
4. Endurance Limit
Answer:
a. Rockwell 3. hardness
b. Instron 2. stress vs strain
c. Charpy 1. impact strength
d. Fatigue 4. Endurance Limit
e. Brinell 3. hardness
f. Izod 1. impact strength
Explanation:
Izod and Charpy are the impact strength testing procedure of a material in which a heavy hammer is attached to an arm is released to impact on the test specimen. In Izod test the specimen with v-notch is held vertical with the notch facing outward while in Charpy test the specimen is supported horizontally with notch facing inward to the impacting hammer.
Instron testing system does universal testing of the material which gradually applies the load recording all the stresses and the corresponding strains until the material fails.
Fatigue is the property of a material due to which it fails under the repeated cyclic loading by the initiation and propagation of cracks. The property of a material resist failure subjected to infinite number of repeated cyclic loads below a certain stress limit.
Rockwell and Brinell are the hardness testing methods. In Rockwell test an intender ball is firstly pressed against the specimen using minor load for a certain time and then a major load is pressed against it for a certain time. After the intender is removed the depth of impression on the surface is measured while in case of Brinell hardness we apply only one load against the intender ball for a certain time and after its removal the radius of impression is measured.
An ideal neon sign transformer provides 9130 V at 51.0 mA with an input voltage of 240 V. Calculate the transformer's input power and current.
Answer:
Input power = 465.63 W
current = 1.94 A
Explanation:
we have the following data to answer this question
V = 9130
i = 0.051
the input power = VI
I = 51.0 mA = 0.051
= 9130 * 0.051
= 465.63 watts
the current = 465.63/240
= 1.94A
therefore the input power is 465.63 wwatts
while the current is 1.94A
the input power is the same thing as the output power.
In a true Brayton cycle, the pressure ratio is 9. Air input temperature to the cycle 300 K pressure is 100 kPa. The maximum temperature in the cycle is 1300 K. Compressor and turbine their yields are equal to each other. Net work obtained from the cycle is 225 kJ / kg. Accordingly, the cycle find the overall yield. The specific temperatures are variable.
Answer:
i did not known answer but anobody help you
Explain by Research how a basic generator works ? using diagram
Atmospheric pressure is 101 kPa. Pressure inside a tire is measured using a typical tire pressure gage to be 900 kPa. Find gage pressure and absolute pressure in the tire. ___________________________________________________________________
Answer:
The gage and absolute pressures are 900 and 1001 kilopascals, respectively.
Explanation:
The gage pressure ([tex]P_{g}[/tex]), in kilopascals, is the difference between absolute ([tex]P_{abs}[/tex]) and atmospheric pressures ([tex]P_{atm}[/tex]), measured in kilopascals. If we know that [tex]P_{g} = 900\,kPa[/tex] and [tex]P_{atm} = 101\,kPa[/tex], then the gage and absolute pressures are, respectively:
[tex]P_{g} = 900\,kPa[/tex]
[tex]P_{abs} = P_{atm} + P_{g}[/tex]
[tex]P_{abs} = 101\,kPa + 900\,kPa[/tex]
[tex]P_{abs} = 1001\,kPa[/tex]
The gage and absolute pressures are 900 and 1001 kilopascals, respectively.
g The inside surface of a 17 mm inner diameter tube with a 2.4 mm thick wall indicates a temperature of 46 deg C. The outside temperature is 43 deg C. The tube is 5 m long. If the tube material has a conductivity of 0.15 W/m/K, estimate the heat transfer rate through the tube wall assuming SS 1D conduction. Indicate the direction of heat transfer with a or - sign ( meaning outward and vice versa). Express your answer using two significant digits in W.
Answer:
-50 W
Explanation:
The heat transfer rate Q = kA(T₂ - T₁)/d where k = thermal conductivity of material = 0.15 W/m-K, A = surface area of tube = πdL where d = diameter of tube = 17 mm = 0.017 m and L = length of tube = 5 m, T₁ = inside temperature = 46 °C, T₂ = outside temperature = 43 °C and d = thickness of tube = 2.4 mm = 0.0024 m
Since Q = kA(T₂ - T₁)/d ,
Q = kπdL(T₂ - T₁)/d
substituting the values of the variables into the equation, we have
Q = 0.15 W/m-K × π × 0.017 m × 5 m(43 °C - 46 °C )/0.0024 m
Q = 0.01275π Wm/K(-3 K )/0.0024 m
Q = -0.03825π Wm/0.0024 m
Q = -0.1202 Wm/0.0024 m
Q = -50.07 W
Q = -50 W
So, the heat transfer rate is -50 W meaning heat transfer out of the tube.
The following is a correlation for the average Nusselt number for natural convection over spherical surface. As can be seen in the above, the Nusselt number approaches 2 as Rayleigh number approaches zero. Prove that this situation corresponds to conduction heat transfer and in conduction heat transfer over sphere, the Nusselt number becomes 2. Hint: First step: Write an expression for heat transfer between two spherical shells that share the same center. Second step: Assume the outer spherical shell is infinitely large.
Answer:
Explanation:
[tex]r_2=[/tex]∞
[tex]q=4\pi kT_1(T_2-T_1)\\[/tex]
[tex]q=2\pi kD.[/tex]ΔT--------(1)
[tex]q=hA[/tex] ΔT[tex]=4\pi r_1^2(T_2_s-T_1_s)\\[/tex]
[tex]N_u=\frac{hD}{k} = 2+\frac{0.589 R_a^\frac{1}{4} }{[1+(\frac{0.046}{p_r}\frac{9}{16} )^\frac{4}{9} } ------(3)[/tex]
By equation (1) and (2)
[tex]2\pi kD.[/tex]ΔT=h.4[tex]\pi r_1^2[/tex]ΔT
[tex]2kD=hD^2\\\frac{hD}{k} =2\\N_u=\frac{hD}{k}=2\\[/tex]-------(4)
From equation (3) and (4)
So for sphere [tex]R_a[/tex]→0
The heat transfer surface area of a fin is equal to the sum of all surfaces of the fin exposed to the surrounding medium, including the surface area of the fin tip. Under what conditions can we neglect heat transfer from the fin tip?
Answer:
The explanation according to the given query is summarized in the explanation segment below.
Explanation:
If somehow the fin has become too lengthy, this same fin tip temperature approaches the temperature gradient and maybe we'll ignore heat transmission out from end tips.Additionally, effective heat transmission as well from the tip could be ignored unless the end tip surface is relatively tiny throughout comparison to its overall surface.