Answer:
option B is correct. Fracture will definitely not occur
Explanation:
The formula for fracture toughness is given by;
K_ic = σY√πa
Where,
σ is the applied stress
Y is the dimensionless parameter
a is the crack length.
Let's make σ the subject
So,
σ = [K_ic/Y√πa]
Plugging in the relevant values;
σ = [50/(1.1√π*(0.5 x 10^(-3))]
σ = 1147 MPa
Thus, the material can withstand a stress of 1147 MPa
So, if tensile stress of 1000 MPa is applied, fracture will not occur because the material can withstand a higher stress of 1147 MPa before it fractures. So option B is correct.
List the RTL (Register Transfer Language) sequence of micro-operations needed to execute the instruction
STORE X
from the MARIE instruction set architecture. Then write the corresponding signal sequence to perform these micro-operations and to reset the clock cycle counter.
You may refer to the provided "MARIE Architecture and Instruction Set" file in the Front Matter folder.
Answer:
So these are the RTL representation:
MAR<----X
MBR<-----AC
M[MAR]<------MBR
Control signal sequence are:
P3T0:MAR<----X
P2P3 P4T1:MBR<-----AC
P0P1 P3T2:M[MAR]<------MBR
Explanation:
STORE X instruction is used for storing the value of AC to the memory address pointed by X. This operation can be done by using the Register Transfers at System Level and this can be represented by using a notation called Register Transfer Language RTL. Let us see what are the register transfer operations happening at the system level.
1. First of all the address X has to be tranfered on to the Memory Address Register MAR.
MAR<----X
2. Next we have to tranfer the contents of AC into the Memory Buffer Register MBR
MBR<-----AC
3. Store the MBR into memory where MAR points to.
M[MAR]<------MBR
So these are the RTL representation:
MAR<----X
MBR<-----AC
M[MAR]<------MBR
Control signal sequence are:
P3T0:MAR<----X
P2P3 P4T1:MBR<-----AC
P0P1 P3T2:M[MAR]<------MBR
Consider a refrigeration truck traveling at 55 mph at a location where the airtemperature is 80F. The refrigerated compartment of the truck can be considered to the a 9-ft-wide,8-ft-high, and 20-ft-long rectangular box. The refrigeration system of the truck removes heat at arate of 600 Btu/min. The outer surface of the truck is coated with a low-emissivity material, andthus radiation heat transfer is very small. Determine the averagetemperature of the outer surface ofthe refrigeration compartment of the truck. Assume the air flow over the entire outer surface to beturbulent, and the heat transfer coefficient at the front and rear surfaces to be equal to that on sidesurfaces.
The question is an engineering problem that involves calculating the average temperature of a truck's refrigeration compartment's outer surface. A precise answer would require additional thermal information and properties, which are not provided in the question.
Explanation:The question is related to the field of thermodynamics and heat transfer, specific to engineering. It requires determining the average temperature of the outer surface of the refrigeration compartment of a truck. However, to provide a precise answer, additional data would be required, including the thermal properties of the truck's material, the heat transfer coefficient, and the difference between the interior and exterior temperatures. Typically, such a problem would involve calculations using the concepts of convection and possibly conduction, considering the truck as a heat exchanger with heat being removed by the refrigeration system and added from the external environment. Without specific values for heat transfer coefficients and material properties, a calculation cannot be accurately made.
You are not changing the carrier signal, only the message. In the scope the center Page 5 of 6 frequency should remain at 14kHz, the frequency span at 20kHz and the dB/div at 15dB. Here are the AM signal in time domain and its frequency spectrum. Explain what you see. (Hint: Remember the Fourier analysis of the square wave performed in Lab #3.)
Answer:
We will see a waveform displayed on the screen and it will be PWM(pulse width modulation) of sinusoidal wave,this wave should have a frequency of 14KHz and it will form a vibration spectrum.
The intercept of the CML is the origin while the intercept of the SML is RF CML consists of efficient portfolios, while the SML is concerned with all portfolios or securities CML could be downward sloping while that is impossible for the SML CML and the SML are essentially the same except in terms of the securities represented
Answer:
Here, The CML is used for efficient portfolios whereas the SML applies to all portfolios or securities
Considering the seperation theorem, The separation theorem states that the investment decision is sepaarte from the financing decision.
This implies that The SML can be used to analyze the relationship between risk and requried return for all assets.
Under the separation theorem investors should Hold the same portfolio of risky assets and the same expected return but at different levels of risk
Explanation:
See answer
A square hole is to be cut using ECM through a plate of low alloy steel that is 12 mm thick. The hole is 35 mm on each side, but the electrode used to cut the hole is slightly less than 35 mm on its sides to allow for overcut, and its shape includes a hole in its center to permit the flow of electrolyte and to reduce the area of the cut. This will produce a center core that can be removed after the tool breaks through. This tool design results in a frontal area of 245 mm2 . The applied current = 1200 amps.
Using an efficiency of 95%, how long will it take to cut the hole?
It will take approximately 0.0238 s to cut the hole.
To calculate the time required to cut the hole using Electrochemical Machining (ECM), we can use the following formula:
[tex]\text { Time }=\frac{\text { Volume of Material to be Removed }}{\text { Material Removal Rate (MRR) }}[/tex]
First, let's calculate the volume of material to be removed:
Volume of Material=Frontal Area × Thickness of Plate
Given:
[tex]Frontal Area =245 \mathrm{~mm}^2\\Thickness of Plate $=12 \mathrm{~mm}$[/tex]
Volume of Material = [tex]245 \mathrm{~mm}^2 \times 12 \mathrm{~mm}=2940 \mathrm{~mm}^3[/tex]
Next, we need to calculate the Material Removal Rate (MRR). MRR is usually given in units of volume removed per unit time per unit current density. Let's assume it is given as [tex]X \mathrm{~mm}^3 / \mathrm{min} / \mathrm{A} / \mathrm{mm}^2 .[/tex] Given the current density [tex]\text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}[/tex], we can calculate the MRR as:
MRR=X×Current Density
Given:
Applied Current = 1200 A
Efficiency = 95%
Frontal Area = 245 [tex]mm^2[/tex]
[tex]\begin{aligned}& \text { Current Density }=\frac{\text { Applied Current }}{\text { Frontal Area }}=\frac{1200 \mathrm{~A}}{245 \mathrm{~mm}^2} \\& \text { Current Density } \approx 4.897 \mathrm{~A} / \mathrm{mm}^2\end{aligned}[/tex]
Let's assume X=0.2[tex]mm^3[/tex] /s/A/[tex]mm^2[/tex](hypothetical value).
[tex]\mathrm{MRR}=0.2 \times 4.897 \approx 0.979 \mathrm{~mm}^3 / \mathrm{s}[/tex]
To convert mm³/s to g/s, we need to know the density of the material. Let's assume the density of low alloy steel is approximately [tex]7.85 \mathrm{~g} / \mathrm{cm}^3[/tex], which is [tex]7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3 .[/tex]
Now, we can calculate the MRR in grams per second:
[tex]MRR in $\mathrm{g} / \mathrm{s}=\mathrm{MRR} \times$ Density of material\\MRR in $\mathrm{g} / \mathrm{s}=0.979 \mathrm{~mm}^3 / \mathrm{s} \times 7.85 \times 10^{-6} \mathrm{~g} / \mathrm{mm}^3$\\MRR in $\mathrm{g} / \mathrm{s} \approx 0.00000768515 \mathrm{~g} / \mathrm{s}$[/tex]
Now, we can use these values to calculate the time required:
[tex]\text { Time }=\frac{2940 }{0.00000768515} \\& \text { Time } \approx 0.022594341[/tex]
However, we need to consider the efficiency. Since the efficiency is 95%, the actual time required will be:
[tex]\begin{aligned}& \text { Actual Time }=\frac{\text { Time }}{\text { Efficiency }} \\& \text { Actual Time }=\frac{0.022594341}{0.95}\end{aligned}[/tex]
≈ 0.0238 s
The piping system that connects one reservoir to a second reservoir consists of 150-ft of 3-in. cast iron pipe that has four flanged elbows, a well-rounded entrance, and sharp-edged exit, and a fully open gate valve. For 75 gal/min of water at 50 °F, determine the elevation difference between the two reservoirs (in ft).
The elevation difference between the two reservoirs is 10.28 ft
Given information:
- Length of the 3-in. cast iron pipe = 150 ft
- Number of flanged elbows = 4
- Entrance condition: well-rounded
- Exit condition: sharp-edged
- Fully open gate valve
- Flow rate = 75 gal/min
- Water temperature = 50 °F
Step 1: Convert the flow rate from gal/min to ft³/s.
Flow rate in ft³/s = (75 gal/min) × (1 ft³/7.48 gal) × (1 min/60 s) = 0.167 ft³/s
Step 2: Calculate the velocity of flow.
Velocity of flow, V = Flow rate / Cross-sectional area of the pipe
Cross-sectional area of the 3-in. cast iron pipe = (π/4) × (3/12)² ft² = 0.0491 ft²
Velocity of flow, V = 0.167 ft³/s / 0.0491 ft² = 3.4 ft/s
Step 3: Calculate the Reynolds number to determine the flow regime.
Reynolds number, Re = (ρVD) / μ
Where,
ρ = density of water at 50 °F = 1.94 slugs/ft³ (from tables)
V = velocity of flow = 3.4 ft/s
D = diameter of the pipe = 3/12 ft
μ = dynamic viscosity of water at 50 °F = 2.73 × 10⁻⁵ lb.s/ft² (from tables)
Re = (1.94 slugs/ft³) × (3.4 ft/s) × (3/12 ft) / (2.73 × 10⁻⁵ lb.s/ft²) = 1.9 × 10⁵
Since Re > 4000, the flow is turbulent.
Step 4: Calculate the head loss due to friction in the straight pipe.
Head loss due to friction, hf = f × (L/D) × (V²/2g)
Where,
f = friction factor (depends on Reynolds number and pipe roughness)
L = length of the pipe = 150 ft
D = diameter of the pipe = 3/12 ft
V = velocity of flow = 3.4 ft/s
g = acceleration due to gravity = 32.2 ft/s²
From the Moody diagram or friction factor tables for cast iron pipe, assuming a relative roughness of 0.00085, and Re = 1.9 × 10⁵, the friction factor, f = 0.021.
Substituting the values, hf = 0.021 × (150 × 12/3) × [(3.4)²/(2 × 32.2)] = 8.4 ft
Step 5: Calculate the head loss due to fittings (elbows, entrance, exit, and valve).
Head loss due to fittings, hf_fittings = Σ(K × V²/2g)
Where K is the loss coefficient for each fitting.
For 4 flanged elbows, K = 0.3 (for each elbow)
For a well-rounded entrance, K = 0.03
For sharp-edged exit, K = 1.0
For a fully open gate valve, K = 0.2
Substituting the values, hf_fittings = [4 × 0.3 + 0.03 + 1.0 + 0.2] × [(3.4)²/(2 × 32.2)] = 1.7 ft
Step 6: Calculate the velocity head.
Velocity head, hv = V²/2g = (3.4)²/(2 × 32.2) = 0.18 ft
Step 7: Calculate the total head loss.
Total head loss, H = hf + hf_fittings + hv
H = 8.4 ft + 1.7 ft + 0.18 ft = 10.28 ft
Therefore, the elevation difference between the two reservoirs is 10.28 ft for the given piping system and a flow rate of 75 gal/min of water at 50 °F.
One method of reducing an aircarft’s landing distance is through the use of thrust reversers. Consider the turbofan engine in Fig. P2.5 with thrust reverser of the bypass airstream. It is given that 1500 lbm / s of air at 60 8 F and 14.7 psia enters the engine at a velocity of 450 ft / s and that 1250 lbm / s of bypass air leaves the engine at 60 deg to the horizontal, velocity of 890ft / s, and pressure of 14.7 psia. The remaining 250 lbm / s leaves the engine core at a velocity of 1200 fps and a pressure of 14.7 psia.
Determine the force on the strut [Fx]. Assume the outside of the engine sees a pressure of 14.7 psia.
Answer:
The force on the strut is 909000 lbm ft/s²
Explanation:
please look at the solution in the attached Word file
A 2-lane highway is to be constructed across a 6-ft diameter metal culvert which is oriented perpendicular to the highway center line. The culvert is located at station 33+00, and the invert (bottom) of the culvert has an elevation of 150 ft. The elevation of station 21+00 on the tangent alignment is 140 ft. The alignment gradient is +4.0% from station 21+00 to station 29+00, and the gradient from station 29+00 to 37+00 is -3%.
a. Find the longest full station vertical curve that will give approximately 3 ft (3 plusminus 0.5 ft) of cover over the top of the culvert.
b. Compute the station and elevation of the PVC and PVT and the elevation of each full station on the vertical curve.
Answer:
Dude just like forget the highway and drive on the interstate bruh
Explanation:
A turbine operates at steady state, and experiences a heat loss. 1.1 kg/s of water flows through the system. The inlet is maintained at 100 bar, 520 Celsius. The outlet is maintained at 10 bar, 280 Celsius. A rate of heat loss of 60 kW is measured. Determine the rate of work output from the turbine, in kW.
Answer:
[tex]\dot W_{out} = 399.47\,kW[/tex]
Explanation:
The turbine is modelled after the First Law of Thermodynamics:
[tex]-\dot Q_{out} -\dot W_{out} + \dot m\cdot (h_{in}-h_{out}) = 0[/tex]
The work done by the turbine is:
[tex]\dot W_{out} = \dot m \cdot (h_{in}-h_{out})-\dot Q_{out}[/tex]
The properties of the water are obtained from property tables:
Inlet (Superheated Steam)
[tex]P = 10\,MPa[/tex]
[tex]T = 520\,^{\textdegree}C[/tex]
[tex]h = 3425.9\,\frac{kJ}{kg}[/tex]
Outlet (Superheated Steam)
[tex]P = 1\,MPa[/tex]
[tex]T = 280\,^{\textdegree}C[/tex]
[tex]h = 3008.2\,\frac{kJ}{kg}[/tex]
The work output is:
[tex]\dot W_{out} = \left(1.1\,\frac{kg}{s}\right)\cdot \left(3425.9\,\frac{kJ}{kg} -3008.2\,\frac{kJ}{kg}\right) - 60\,kW[/tex]
[tex]\dot W_{out} = 399.47\,kW[/tex]
A solenoid with a length of 20 cm and a radius of 5 cm consists of 400 turns and carries a current of 12 A. If z = 0 represents the midpoint of the solenoid, generate a plot for |H(z)| along the axis of the solenoid for a range −20 cm ≤ z ≤ 20 cm. Please use MATLAB to generate the solution numerically and indicate how you did it. Include a copy of the computer code that you create to solve the problem
Answer:
Copy MATLAB code to plot the magnitude of magnetic field strength with respect to z on the axis of solenoid:
z=-20:0.01:20;
H=120.*(((20-(2.*z))./sqrt((20-(2.*z)).^2+100))+((20+(2.*z))./sqrt((20+(2.*z)).^2+100)));
plot(z,H)
title('plot of |H| vs z along the axis of solenoid')
ylabel('Magnitude of magnetic field 'H")
xlabel('position on axis of solenoid 'z")
Explanation:
full explanation is attached as picture and the resultant plot also.
A pendulum is made of two rods that are firmly welded together to make a cross. The rod AB is 2 m long with a mass of 20 kg. The rod CD is 1 m long with a mass of 10 kg. Recall that for a rod the moment of inertia about its center of mass is given by IG = 1/12 m L2 cross2 What is the moment of inertia about A for member AB?
Answer:
IA = 80/3 kgm^2
Explanation:
Given:-
- The mass of rod AB, m1 = 20 kg
- The length of rod AB, L1 = 2m
- The mass of rod CD, m2 = 10 kg
- The length of rod CD, L2 = 1m
Find:-
What is the moment of inertia about A for member AB?
Solution:-
- The moment of inertia About point "O" the center of rod AB is given as:
IG = 1/12*m1*L^2
- To shift the axis of moment of inertia for any object at a distance "d" from the center of mass of that particular object we apply the parallel axis theorem. The new moment of inertia about any arbitrary point, which in our case A end of rod AB is:
IA = IG + m1*d^2
- Where the distance "d" from center of rod AB to its ends is 1/2*L1 = 1 m.
So the moment of inertia for rod AB at point A would be:
IA = 1/12*m1*L^2 + m1*0.5*L1^2
IA = 1/3 * m1*L1^2
IA = 1/3*20*2^2
IA = 80/3 kgm^2
Evaporation in Double-Effect Reverse-Feed Evaporators. A feed containing 2 wt % dissolved organic solids in water is fed to a double-effect evaporator with reverse feed. The feed enters at 100°F and is concentrated to 25% solids. The boiling-point rise can be considered negligible as well as the heat of solution. Each evaporator has a 1000-ft2 surface area and the heat-transfer 700 btu/h -ft2-°F. The feed enters evaporator number 2 and steam at 100 psia is fed to number 1. The pressure iin the vapor space of evaporator number 2 is 0.98 psia. Assume that the heat capacity of all liquid solutions is that of liquid water. Calculate the feed rate F and the product rate Li of a solution containing 25% solids. (Hint: Assume a feed rate of, say, F 1000 lbm/h. Calculate the area. Then calculate the actual coefficients are U1 500 and U2- feed rate by multiplying 1000 by 1000/calculated area.) Ans. F- 133 800 lbm/h (60 691 kg/h), L,- 10700lb»/h (4853 kg/h)
Answer:
472,826 lb/hr
Explanation:
As per the given data:
Solids in feed= 2%
Solids in concentrate= 25%
HTA1 = HTA2 = 1000 ft^2
U1 = 500 Btu/ h ft^2 F & U2 = 700 Btu/ h ft^2 F
Overall material balance: Feed= Distillate + concentrate ----> Eq-1
Component balance: Feed * 0.02 = Distillate * 0 + concentrate * 0.25
Feed = 12.5 * concentrate ---> Eq-2
Boiling point rise = negligible, so solution & solvent vapor temperature will be same.
Assumed that the 1st effect is operating under atmospheric pressure (Boiling point - 212F).
As per the data:
Latent heat 212F = 300 Btu/lb
Latent heat 100F = 320 Btu/lb
As per material balance:
Vapor flowrate * latent heat = Overall HT coefficient * HTA * DT
1st effect: M-1= (500 * 1000 * (326-212)) / 300 = 190,000 lb/hr
2nd effect: M-2= (700*1000 * (212-100)) / 320 = 245,000 lb/hr
Distillate = M-1 + M-2 = 190,000+245,000 = 435,000 lb/hr
Substituting the above in Eq-1
Feed = 435,000 + concentrate
Substitute Eq-2 in the above
12.5 * concentrate = 435,000 + concentrate
Concentrate, L1 = 435,000/11.5 = 37826 lb /hr
Feed, F = 435,000 + 37826 = 472,826 lb/hr
1st effect operating pressure is not given, That may be the reason we are not getting the given answer. But procedure is right.
[ Find the figure in the attachment ]
Determine the enthalpy of foA novel gaseous hydrocarbon fuel CxHy is proposed for use in spark-ignition engines. An analysis of a sample of this fuel revealed that its molecular weight is 140 and its molar H/C ratio is 2.0. In order to evaluate some of its properties, the fuel was burned with stoichiometric standard air in a constant pressure, steady-flow reactor. The fuel and air entered the reactor at 25°C, and the products of complete combustion were cooled to 25°C. At the exit condition, water in the products was a liquid. It was measuredrmation of the fuel at 25°C
Answer:
a. The fuel formula is C10H20 (dec-1-ene)
b. The heat given-off to the cooling system of the reactor represents the Higher Heating Value (HHV) of the gaseous hydrocarbon fuel
c. LHV = 44.36 KJ/kg
d. ΔH (formation) = -6669.6 KJ/mol
Explanation:
Molecular weight = 140, Molar H/C ratio = 2.0
a) To determine the fuel formula, we represent the information we have as equations
Molecular weight of C = 12, Molecular weight of H = 1
CxHy = 140g
12 · x + 1 · y = 140 ·········Eqn 1
y = 2x ·········Eqn 2
Substitute y = 2x into Eqn 1, we have:
12 (x) + 1 (2x) = 140
12x + 2x = 140
14x = 140
x = 10
Substitute x into Eqns 2, we have:
y = 2 x 10 = 20
∴ the fuel formula is C10H20 (dec-1-ene)
b) The heat given-off to the cooling system of the reactor represents the Higher Heating Value (HHV) of the gaseous hydrocarbon fuel . HHV refers to the amount of heat given-off in the combustion of a specified amount of fuel which was initially at 25°C and cooled back to a temperature of 25°C.
c) We will use LHV formula given as:
LHV = HHV – [2.441 (0.09H + M) ÷ 1000]
where:
LHV = lower heating value of fuel in MJ/kg
HHV = higher heating value of fuel in MJ/kg
M = Weight % of moisture in fuel
H = Weight % of hydrogen in fuel
HHV = 47.5 MJ/kg, M = 0, H = mass of hydrogen ÷ total mass of fuel
⇒ H = 20 ÷ 140 = 0.143 * 100 = 14.3 kg per 100 kg, Latent heat of steam (at 25°C) = 2441 KJ/kg
LHV = 47.5 - [2441 * (0.09 * 14.3 + 0) ÷ 1000]
LHV = 47.5 - 3.14 = 44.36
LHV = 44.36 KJ/kg
d) To get the enthalpy of the combustion of the fuel, we have:
fuel + air - - - - > carbon dioxide + water
C10H20 + 15 O2 - - - - > 10 CO2 + 10 H20
Enthalpy (formation) = Enthalpy (product) - Enthalpy (reactant)
ΔH (CO2) = -393.5 kJ/mol, ΔH (H2O) = -285.8 kJ/mol, ΔH (C10h20) = - 123.4KJ/mol
ΔH (formation) = 10 * (-393.5) + 10 * (-285.8) - (-123.4)
ΔH (formation) = -3935 - 2858 + 123.4
ΔH (formation) = -6669.6 KJ/mol
A pipe 300 m long has a slope of 1 in 100 and tapers from 1.2 m diameter at the high end to 0.6 m diameter at the low end. Quantity of water flowing is 5400 liters per minute. If the pressure at the high end is 68.67 kPa, find the pressure at the low end. Neglect head loss.
Answer:
P = 98052.64 Pa or 98.05 kPa
Explanation:
Using Bernoulli's equation;
P+rho*v^2/2+ rho*g*z = constant
at high end;
dia= 1.2 m
flow, Q = 5400 L/min = 0.09 m^3/s
therefore, velocity at the high end, v1 = Q/A =0.09/(pi()*(1.2/2)^2) = 0.08 m/s
pressure, P = 68.67 kPa
Solving for elevation, z
assume lower end is reference line. then higher end will be 'x' m high wrt to lower end.
x= 300*sin(tan^-1(1/100)) = 3 m
that means higher end will be 3 m above with respect to lower end
Similarly for lower end;
dia= 0.6 m
flow, Q = 5400 L/min = 0.09 m^3/s
therefore, velocity at the high end, v2 = Q/A =0.09/(pi()*(0.6/2)^2) = 0.318 m/s
assume pressure, P
z=0
put all values in the formula, we get;
68.67*10^3+1000*0.08^2/2+ 1000*9.81*3 =P+1000*0.318^2/2+ 0
solving this, we get;
P = 98052.64 Pa or 98.05 kPa
According to Bernoulli's principle, a lower pressure than expected at the
lower end because the velocity of the fluid is higher.
The pressure at the low end is approximately 97.964 kPa.Reasons:The given parameter are;
Length of the pipe, L = 300 m
Slope of the pipe = 1 in 100
Diameter at the high end, d₁ = 1.2 m
Diameter at the low end, d₂ = 0.6 m
The volume flowrate, Q = 5,400 L/min
Pressure at the high end, P = 68.67 kPa
Required:Pressure at the low end
Solution:The elevation of the pipe, z₁ = [tex]300 \, m \times \frac{1}{100} = 3 \, m[/tex]
z₂ = 0
The continuity equation is given as follows;
Q = A₁·v₁ = A₂·v₂
[tex]\sqrt[n]{x} \displaystyle Q = 5,400 \, L/min = 5,400 \, \frac{L}{min} \times \frac{1 \, m^3}{1,000 \, L} \times \frac{1 \, min}{60 \, seconds} = \mathbf{ 0.09 \, m^3/s}[/tex]
Therefore;
[tex]\displaystyle 0.09 = \frac{\pi}{4} \times 1.2^2 \times v_1 = \mathbf{\frac{\pi}{4} \times 0.6^2 \times v_2}[/tex]
[tex]\displaystyle v_1 = \frac{0.09}{\frac{\pi}{4} \times 1.2^2 } \approx 0.0796[/tex]
[tex]\displaystyle v_2 =\frac{0.09}{\frac{\pi}{4} \times 0.6^2 } \approx \mathbf{0.318}[/tex]
The Bernoulli's equation is given as follows;[tex]\displaystyle \frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 = \mathbf{ \frac{p_2}{\rho \cdot g} + \frac{v_2^2}{2 \cdpt g} + z_2}[/tex]
Therefore;
[tex]\displaystyle p_2 = \left(\frac{p_1}{\rho \cdot g} + \frac{v_1^2}{2 \cdot g} + z_1 - \left( \frac{v_2^2}{2 \cdot g} + z_2 \right)\right) \times \rho \cdot g = p_1 + \frac{\rho}{2} \cdot \left(v_1^2} -v_2^2 \right)+ \rho \cdot g \left(z_1 - z_2\right)[/tex]
The density of water, ρ = 997 kg/m³
Which gives;
[tex]\displaystyle p_2 = 68670 + \frac{997 }{2} \times \left(0.0796^2- 0.318^2 \right) + 997 \times 9.81 \times \left( 3 - 0 \right) \approx \mathbf{97,964.46}[/tex]The pressure at the low end, p₂ ≈ 97,964.46 Pa ≈ 97.964 kPaLearn more about Bernoulli's principle here:
https://brainly.com/question/6207420
6msection of 150lossless line is driven by a source with vg(t) = 5 cos(8π × 107t − 30◦ ) (V) and Zg = 150 . If the line, which has a relative permittivity r = 2.25, is terminated in a load ZL = (150 − j50) , determine: (a) λ on the line. ∗ (b) The reflection coefficient at the load. (c) The input impedance. (d) The input voltage Vi. (e) The time-domain input voltage vi(t).
Answer:
a. 5m
b. r = 0.16 e^-80.5◦
c. Zpn = (115.7 + j27.4) ohms
d. Vi = 2.2e^-j22.56◦ volts
e. Vi(t) = 2.2 cos (8π × 107t − 22.56◦ ) Volts
Explanation:
In this question, we are tasked with calculating a series of terms.
Please check attachment for complete solution and step by step explanation
Q2: Determine the number of 8-inch-high by 8-inch-wide by 16-inch-long concrete blocks required to complete the wall in Figures below. The overhead doors are 10 feet wide by 12 feet high. If lintel blocks are required wherever the #4 horizontal bars are located and above the doors, how many plain blocks and how many lintel blocks are needed for the wall?
The image of the elevation and wall section is missing, so i have attached them.
Answer:
Number of concrete blocks = 1020 blocks
Number of lintel blocks = 240
Number of plain blocks = 780
Explanation:
First of all, we'll find the net area of the wall as follows:
The height of the wall is 17' - 4", so we need to convert it to ft, thus, h = 17' + (4/12)' = 17.33'
So, Gross wall area = 80′ × 17.333' = 1,387 ft²
From the image, there are 4 doors, thus, Area of doors = 4 × 10′ × 12′ = 480 ft²
Thus, Net area = Gross wall area - area of doors
Net area = 1,387 – 480 = 907 ft²
We are told that the block is 8 inches length by 16inches width. Thus, converting to ft; (8/12)ft by (16/12)ft.
So area of one block = (8/12) x (16/12) = 128/144 ft² = 0.88889 ft²
So, number of blocks per ft² = 1/0.88889 = 1.125 blocks
So, there are 1.125 blocks per ft²
Thus, Number of concrete blocks = 907 ft² × 1.125 blocks per ft² = 1,020 blocks
From the image attached, there are five rows of lintel blocks, located at 4′, 8′, 12′, 16′, and 17′-4″. The bottom three rows pass through the doors and are only a total of (80′ – (4 × 10′) ) = 40′ long .
The top two rows are 80′ long. Thus, there is 40 feet (4 × 10′) of lintel block above the doors.
Lintel blocks = (3 × 40′) + (2 × 80′) + 40′ = 320 ft
Number of Lintel blocks = 320′ × 12 in per ft / 16″ = 240 blocks
Number of Plain blocks = 1,020 blocks – 240 blocks = 780 blocks
Without specific wall dimensions provided, it is not possible to calculate the exact number of plain and lintel concrete blocks required for the construction of the wall. However, the general procedure involves calculating the wall's total area, subtracting the area of doors or windows, and then dividing by the size of one block, while also considering the requirements for lintel blocks above openings and where supports are needed.
Explanation:Determining the Quantity of Concrete Blocks for a Wall
To determine the number of 8-inch-high by 8-inch-wide by 16-inch-long concrete blocks required to complete a wall, one must calculate the total volume of the wall and then divide by the volume of a single block. Unfortunately, the dimensions of the wall are not provided in the question, so we cannot calculate the exact number of blocks needed. As for the lintel blocks, which are used above the doors and wherever the #4 horizontal bars are located, the number would depend on the linear footage that needs to be covered by the lintels and the length of each lintel block.
Without specific wall dimensions, we must refer to the typical process where one would calculate the wall's total surface area, subtract the area occupied by doors or windows, and divide by the area covered by one block. Then, identify the areas requiring lintel blocks and count the number of lintels needed based on their standard lengths.
The plain blocks and lintel blocks calculation requires detailed measurements of the wall and openings. Since the details provided are related to the overhead doors with known dimensions, one would need to consider the door's dimensions when calculating the plane wall areas and the lintel requirements above them.
Consider a very long, slender rod. One end of the rod is attached to a base surface maintained at Tb, while the surface of the rod is exposed to an air temperature of 400°C. Thermocouples imbedded in the rod at locations 25 mm and 120 mm from the base surface register temperatures of 335°C and 375°C, respectively.
(a) Calculate the rod base temperature (°C).
(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent.
Answer:
(a) Calculate the rod base temperature (°C). = 299.86°C
(b) Determine the rod length (mm) for the case where the ratio of the heat transfer from a finite length fin to the heat transfer from a very long fin under the same conditions is 99 percent. = 0.4325m
Explanation:
see attached file below
Ronny has a hydraulic jack. The input force is 250 N, while the output force is 7,500 N. If the area of the pipe below the input is 0.2 m2, what is the area of the pipe below the load?
Answer:
6 m²
Explanation:
application of fluid pressure according to Pascal's principle for the two pistons is given as:
[tex]P_1=P_2[/tex]
Where P₁ is the pressure at the input and P₂ is the pressure at the output.
But P₁ = F₁ / A₁ and P₂ = F₂ / A₂
Where F₁ and F₂ are the forces applied at the input and output respectively and A₁ and A₂ are the area of the input pipe and output pipe respectively
Since, [tex]P_1=P_2[/tex]
[tex]\frac{F_1}{A_1} =\frac{F_2}{A_2}\\[/tex]
But A₁ = 0.2 m², F₁ = 250 N, F₂ = 7500 N. Substituting values to get:
[tex]\frac{F_1}{A_1} =\frac{F_2}{A_2}\\\frac{250}{0.2}=\frac{7500}{A_2}\\ A_2=\frac{7500*0.2}{250} = 6m^2[/tex]
Therefore, the area of the pipe below the load is 6 m²
The attached program (studentsGpa.cpp) uses dynamic allocation to create an array of strings. It asks the user to enter a number and based on the entered number it allocates the array size. Then based on that number it asks the user that many times to enter student’s names. What you need to do:Add another array of doubles to store the gpa of each student as you enter themYou need to display both the student’s name and the gpa.NOTE: must use pointer notation not array subscript. Any submission that uses array subscript won’t be graded
Question: The program was not attached to your question. Find attached of the program and the answer.
Answer:
See the explanation for the answer.
Explanation:
#include <iostream>
using namespace std;
int main()
{
cout << "How many students will you enter? ";
int n;
cin>>n;
string *name = new string[n];
double *gpa = new double[n];
for(int i=0;i<n;i++){
cout<<"Enter student"<<(i+1)<<"'s name: ";
cin>>name[i];
cout<<"Enter student"<<(i+1)<<"'s gpa: ";
cin>>gpa[i];
}
cout<<"The list students"<<endl;
cout<<"Name GPA"<<endl;
cout<<"----------------------"<<endl;
for(int i=0;i<n;i++){
cout<<name[i]<<" "<<gpa[i]<<endl;
}
return 0;
}
OUTPUT : See the attached file.
The diameter of an extruder barrel = 85 mm and its length = 2.00 m. The screw rotates at 55 rev/min, its channel depth = 8.0 mm, and its flight angle = 18°. Head pressure at the die end of the barrel = 10.0(10^6) Pa. Viscosity of the polymer melt = 100 Pa- s
(a) Find the volume flow rate of plastic at the die end of the barrel.
________ x 10^-6 m^3/s
Answer:
Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s
Explanation:
given data
diameter = 85 mm
length = 2 m
depth = 9mm
N = 60 rev/min
pressure p = 11 × [tex]10^6[/tex] Pa
viscosity n = 100 Pas
angle = 18°
so Qd will be
Qd = 0.5 × π² ×D²×dc × sinA × cosA ..............1
put here value and we get
Qd = 0.5 × π² × ( 85 [tex]\times 10^{-3}[/tex] )²× 9 [tex]\times 10^{-3}[/tex] × sin18 × cos18
Qd = 94.305 × [tex]10^{-6}[/tex] m³/s
and
Qb = p × π × D × dc³ × sin²A ÷ 12 × n × L ............2
Qb = 11 × [tex]10^{6}[/tex] × π × 85 [tex]\times 10^{-3}[/tex] × ( 9 [tex]\times 10^{-3}[/tex] )³ × sin²18 ÷ 12 × 100 × 2
Qb = 85.2 × [tex]10^{-6}[/tex] m³/s
so here
volume flow rate Qx = Qd - Qb ..............3
Qx = 94.305 × [tex]10^{-6}[/tex] - 85.2 × [tex]10^{-6}[/tex]
Qx = 9.10[tex]9.10^5 \times 10^{-6}[/tex] m³/s
An offset h must be introduced into a metal tube of 0.75-in. outer diameter and 0.08-in. wall thickness. Knowing that the maximum stress after the offset is introduced must not exceed 4 times the stress in the tube when it is straight, determine the largest offset that can be used
Complete Question
The diagram for this question is shown on the first uploaded image
Answer:
The largest offset that can be used is [tex]h = 0.455 \ in[/tex]
Explanation:
From the question we are told that
The diameter of the metal tube is [tex]d_m = 0.75 \ in[/tex]
The thickness of the wall is [tex]D = 0.08 \ in[/tex]
Generally the inner diameter is mathematically evaluated as
[tex]d_i = d_m -2D[/tex]
[tex]= 0.75 - 2(0.08)[/tex]
[tex]= 0.59 \ in[/tex]
Generally the tube's cross-sectional area can be evaluated as
[tex]a = \frac{\pi}{4} (d_m^2 - d_i^2)[/tex]
[tex]= \frac{\pi}{4} (0.75^2 - 0.59^2)[/tex]
[tex]= 0.1684 \ in^2[/tex]
Generally the maximum stress of the metal is mathematically evaluated as
[tex]\sigma = \frac{P}{A}[/tex]
[tex]\sigma = \frac{P}{ 0.1684}[/tex]
The diagram showing when the stress is been applied is shown on the second uploaded image
Since the internal forces in the cross section are the same with the force P and the bending couple M then
[tex]M = P * h[/tex]
Where h is the offset
The maximum stress becomes
[tex]\sigma_n = \frac{P}{A} + \frac{M r_m }{I}[/tex]
Where [tex]r_m[/tex] is the radius of the outer diameter which is evaluated as
[tex]r_m = \frac{0.75}{2}[/tex]
[tex]r_m = 0.375 \ in[/tex]
and I is the moment of inertia which is evaluated as
[tex]I = \frac{\pi}{64} (d_m^4 - d_i^4 )[/tex]
[tex]= \frac{\pi}{64}(0.75^4 - 0.59^4)[/tex]
[tex]= 0.009583 \ in^4[/tex]
So the maximum stress becomes
[tex]\sigma' = \frac{P}{0.1684} + \frac{Phr}{0.009583}[/tex]
Now the question made us to understand that the maximum stress when the offset was introduced must not exceed the 4 times the original stress
So
[tex]\sigma ' = 4 \sigma[/tex]
=> [tex]\frac{P}{0.1684} + \frac{Phr_m }{0.009583} = 4 [\frac{P}{0.1684} ][/tex]
The P would cancel out
[tex]\frac{1}{0.1684} + \frac{h(0.375)}{0.009583} = \frac{4}{0.1684}[/tex]
[tex]5.94 + 39.13h = 23.753[/tex]
[tex]39.13h = 17. 813[/tex]
[tex]h = 0.455 \ in[/tex]
How does the resistance in the circuit impact the height and width of the resonance curve? (If the resistance were to increase would the height change? Would the width? If so, how?)
Answer:
The reactances vary with frequency, with large XL at high frequencies and large Xc at low frequencies, as we have seen in three previous examples. At some intermediate frequency fo, the reactances will be the same and will cancel, giving Z = R; this is a minimum value for impedance and a maximum value for Irms results. We can get an expression for fo by taking
XL=Xc
Substituting the definitions of XL and XC,
2[tex]\pi[/tex]foL=1/2[tex]\pi[/tex]foC
Solving this expression for fo yields
fo=1/2[tex]\pi[/tex][tex]\sqrt{LC}[/tex]
where fo is the resonant frequency of an RLC series circuit. This is also the natural frequency at which the circuit would oscillate if it were not driven by the voltage source. In fo, the effects of the inductor and capacitor are canceled, so that Z = R and Irms is a maximum.
Explanation:
Resonance in AC circuits is analogous to mechanical resonance, where resonance is defined as a forced oscillation, in this case, forced by the voltage source, at the natural frequency of the system. The receiver on a radio is an RLC circuit that oscillates best at its {f} 0. A variable capacitor is often used to adjust fo to receive a desired frequency and reject others is a graph of current versus frequency, illustrating a resonant peak at Irms at fo. The two arcs are for two dissimilar circuits, which vary only in the amount of resistance in them. The peak is lower and wider for the highest resistance circuit. Thus, the circuit of higher resistance does not resonate as strongly and would not be as selective in a radio receiver, for example.
A current versus frequency graph for two RLC series circuits that differ only in the amount of resistance. Both have resonance at fo, but for the highest resistance it is lower and wider. The conductive AC voltage source has a fixed amplitude Vo.
Under certain conditions, wind blowing past a rectangular speed limit Sign can cause the sing to oscillate with a frequency omega. Assume that omega is a function of the Sign width, b, Sign height, h, wind velocity, V, air density p, and an elastic constant, K, for the supporting pole. Hint: The constant, k, has dimensions of [force x length]. (i) How many (non-dimensional) Pi-groups are there? (ii) Find these non-dimensional groups. (iii) Can one of the pi-groups be considered a Reynolds number?
Answer:
Explanation:
Given that,
Omega is a function of the following
ω = f(b, h, v, ρ, k)
Where, all unit have a dimension of
ω = T^-1
b = L
h = L
V = LT^-1
ρ = FL^-4T²
k = FL
Then,
From the pie theorem
The required pi term is 6—3 = 3 terms,
So, we use V, p and b as a repeating term.
For first pi
π1 = ω•b^a•v^b•ρ^c.
Since
ω= T^-1, b = L, v = LT^-1 and
ρ= FL^-4T²
Since π is dimensionless then,
π = F^0•L^0•T^0
(T^-1)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0
Rearranging
F^0•L^0•T^0 = F^c•T^(2c-b-1)• L^(a+b-4c)
Comparing coefficient
c = 0
2c - b - 1 = 0
b = 2c - 1 = 0 - 1 = -1
a + b - 4c = 0
a = 4c - b = 0 - -1 = 0+1
a = 1
Then, a = 1, b = -1 and c = 0
So, π1 = ω•b^a•v^b•ρ^c.
π1 = ω•b^1•v^-1•ρ^0
π1 = ω•b / v
Check dimensions
ωb/v = (T^-1)L / LT^-1 = L^0•T^0 = 1
Then, π1 is dimensionless
For second pi
π2 = h•b^a•v^b•ρ^c.
Since
h = L, b = L, v = LT^-1 and ρ= FL^-4T²
Since π is dimensionless then,
π = F^0•L^0•T^0
(L)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0
Rearranging
F^0•L^0•T^0 = F^c•T^(2c-b)• L^(1+a+b-4c)
Comparing coefficient
c = 0
2c - b = 0
b = 2c = 0
1 + a + b - 4c = 0
a = 4c - b - 1 = 0 -0 - 1 = -1
a = -1
Then, a = -1, b = 0 and c = 0
So, π2 = h•b^a•v^b•ρ^c.
π2 = h•b^-1•v^0•ρ^0
π2 = h / b
Check dimensions
h / b = L / L = 1
Then, π2 is dimensionless
For third pi
π3 = k•b^a•v^b•ρ^c.
Since
k= FL, b = L, v = LT^-1 and ρ=FL^-4T²
Since π is dimensionless then,
π = F^0•L^0•T^0
(FL)•(L^a)•(LT^-1)^b•(FL^-4T²)^c = F^0•L^0•T^0
Rearranging
F^0•L^0•T^0 = F^(c+1)•T^(2c-b)• L^(1+a+b-4c)
Comparing coefficient
c + 1= 0
Then, c = -1
2c - b = 0
b = 2c = -2
1 + a + b - 4c = 0
a = 4c - b - 1 = -4 +2 - 1 = -3
a = -3
Then, a = -3, b = -2 and c = -1
So, π3 = k•b^a•v^b•ρ^c.
π3 = k•b^-3•v^-2•ρ^-1
Therefore,
π3 = k / b³•v²•ρ
Let check for dimension
π3 = FL / (L³• L²T^-2 • FL^-4T²)
π3 = FL / (L^(3+2-4) • T^(-2+2) •F)
π3 = FL / (L• T^(0) •F)
π3 = FL / LF = 1
π3 is also dimensions less
So.
I. There are three none dimensional pi
II. The none dimensional group are
π1 = ω•b / v
π2 = h / b
π3 = k / b³•v²•ρ
III. Reynolds Number. The Reynolds number is the ratio of inertial forces to viscous forces and it is dimensionless
So, the π3 can be considered as a Reynolds number
Ray L. Zapp is thinking about testing strategies for his new HashTable class, which uses Rainforest's cloud storage service to make hash tables persistent. For the next 4 questions, write the letter of the one BEST testing technique (from the list below) to achieve the listed testing goal. (Some testing techniques may be used more than once or not at all.) A. Black-box testing B. Fuzz testing C. Stubbing& mocking D. Mutation testing E. White-box/glass-box testing 4 13.(1 pts) Making sure the hash table is robust when presented with nonsensical input 15. (1 pts) Making sure his service behaves correctly when Rainforest's cloud storage service is down 16. (1 pts) Improving the thoroughness of coverage of his test suite by uncovering test cases he missed 一( 1 pts) Making sure hash collisions are handled properly
Answer:
As Ray L zapp is thinking about new strategies to test his new hash Table,hence therefore the best testing technique is Stubbing& mocking .
For the following transfer function, derive expressions for the real and imaginary part for s = jω in terms of the frequency variable ω. Then write a MATLAB script to plot the imaginary part versus the real part (and its reflection about the real axis) for a frequency ω range of 10−2 to 102 radians per second. Verify that your plots match the output of the nyquist function in MATLAB. • G(s) = 1 /(s+0.5)(s+1)(s+2) Suppose G represents an open-loop plant transfer function. Use your plot to determine the Gain Margin for the closed-loop system, i.e., determine how much the loop gain could be increased before the closed-loop becomes unstable
Answer:
See all solutions attached as picture.
Explanation:
It is well explanatory
The rotor is driven at ????m = 120π rad/sec. If the stator winding carries a current of 5 A (rms) at 60 Hz, determine the instantaneous voltage and rms voltage induced in the rotor coil. Determine the frequency of the rotor induced voltage. (b) Suppose the stator and rotor coils are connected in series and a current of 5
Answer:
(a) Frequency induced voltage of motor is 120 Hz
(b) The speed at which the machine will produce an average torque is 240π rad/s. Maximum Torque is 1.0 Nm
Explanation:
• Build upon the results of problem 3-85 to determine the minimum factor of safety for fatigue based on infinite life, using the modified : Goodman criterion. The shaft CD rotates at a constant speed, has a constant diameter of 1.13 in, and is made from cold-drawn AISI 1018 steel. From problem 3-85, the critical stress element in shaft CD experiences a completely reversed bending stress due to the rotation, as well as steady torsional and axial stresses. Thus, a,bend= 12 kpsi, Om, bend= 0 kpsi, Oa,axial= 0 kpsi, om, axial= -0.9 kpsi, Ta = 0 kpsi, and Im = 10 kpsi. The minimum factor of safety for fatigue is
Answer:
minimum factor of safety for fatigue is = 1.5432
Explanation:
given data
AISI 1018 steel cold drawn as table
ultimate strength Sut = 63.800 kpsi
yield strength Syt = 53.700 kpsi
modulus of elasticity E = 29.700 kpsi
we get here
[tex]\sigma a[/tex] = [tex]\sqrt{(\sigma a \times kb)^2+3\times (za\times kt)^2}[/tex] ...........1
here kb and kt = 1 combined bending and torsion fatigue factor
put here value and we get
[tex]\sigma a[/tex] = [tex]\sqrt{(12 \times 1)^2+3\times (0\times 1)^2}[/tex]
[tex]\sigma a[/tex] = 12 kpsi
and
[tex]\sigma m[/tex] = [tex]\sqrt{(\sigma m \times kb)^2+3\times (zm\times kt)^2}[/tex] ...........2
put here value and we get
[tex]\sigma m[/tex] = [tex]\sqrt{(-0.9 \times 1)^2+3\times (10\times 1)^2}[/tex]
[tex]\sigma m[/tex] = 17.34 kpsi
now we apply here goodman line equation here that is
[tex]\frac{\sigma m}{Sut} + \frac{\sigma a}{Se} = \frac{1}{FOS}[/tex] ...................3
here Se = 0.5 × Sut
Se = 0.5 × 63.800 = 31.9 kspi
put value in equation 3 we get
[tex]\frac{17.34}{63.800} + \frac{12}{31.9} = \frac{1}{FOS}[/tex]
solve it we get
FOS = 1.5432
Two monitoring wells were constructed in an unconfined aquifer. The wells are separated by a distance of 250 ft. The water surface elevations in the up-gradient and down-gradient wells were 101.00 ft and 100.85 ft, respectively. The aquifer hydraulic conductivity is 5 ft/day. The fluid velocity (ft/day) in the aquifer is most nearly:
Answer:
0.003
Explanation:
? → ? = −??ℎ??→ ?
= ??ℎ??? = ?ℎ2− ℎ1∆?→ ? = 5????.101 ?? − 100.85 ??250 ??= 0.003???
Air is cooled and dehumidified as it flows over the coils of refrigeration system at 100 kPa from 30 ºC and relative humidityof 84%to 15 ºC and relative humidity of 100%. The mass flow rate of dry air is 0.4 kg/s.
Using the formulaand not the Psychrometric chart, determine:
a) the mass flow rate of water,
b)the heat removal from the air.
Answer:
(a) The mass glow rate of water is 0.0032kg/s
(b) The heat removal from air is 14.17KJ/sec
Explanation:
In this question, we are asked to use the formula to calculate the mass flow rate and the heat removal from water based on the data in the question.
To answer this question, we take values of absolute humidity and enthalpy values from table from corresponding temperature
Please check attachment for complete solution and step by step explanation
Consider a rectangular fin that is used to cool a motorcycle engine. The fin is 0.15m long and at a temperature of 250C, while the motorcycle is moving at 80 km/h in air at 27 C. The air is in parallel flow over both surfaces of the fin, and turbulent flow conditions may be assumed to exist throughout. What is the rate of heat removal per unit width of the fin?
Answer:
q' = 5826 W/m
Explanation:
Given:-
- The length of the rectangular fin, L = 0.15 m
- The surface temperature of fin, Ts = 250°C
- The free stream velocity of air, U = 80 km/h
- The temperature of air, Ta = 27°C
- Parallel flow over both surface of the fin, assuming turbulent conditions through out.
Find:-
What is the rate of heat removal per unit width of the fin?
Solution:-
- Assume steady state conditions, Negligible radiation and flow conditions to be turbulent.
- From Table A-4, evaluate air properties (T = 412 K, P = 1 atm ):
Dynamic viscosity , v = 27.85 * 10^-6 m^2/s
Thermal conductivity, k = 0.0346 W / m.K
Prandlt number Pr = 0.69
- Compute the Nusselt Number (Nu) for the - turbulent conditions - the appropriate relation is as follows:
[tex]Nu = 0.037*Re_L^\frac{4}{5} * Pr^\frac{1}{3}[/tex]
Where, Re_L: The average Reynolds number for the entire length of fin:
[tex]Re_L = \frac{U*L}{v} \\\\Re_L = \frac{80*\frac{1000}{3600} * 0.15}{27.85*10^-^6} \\\\Re_L = 119688.80909[/tex]
Therefore,
[tex]Nu = 0.037*(119688.80909)^\frac{4}{5} * 0.69^\frac{1}{3}\\\\Nu = 378[/tex]
- The convection coefficient (h) can now be determined from:
[tex]h = \frac{k*Nu}{L} \\\\h = \frac{0.0346*378}{0.15} \\\\h = 87 \frac{W}{m^2K}[/tex]
- The rate of heat loss q' per unit width can be determined from convection heat transfer relation, Remember to multiply by (x2) because the flow of air persists on both side of the fin:
[tex]q' = 2*[h*L*(T_s - T_a)]\\\\q' = 2*[87*0.15*(250 - 27)]\\\\q' = 5826\frac{W}{m}[/tex]
- The rate of heat loss per unit width from the rectangular fin is q' = 5826 W/m
- The heat loss per unit width (q') due to radiation:
[tex]q' = 2*a*T_s^4*L[/tex]
Where, a: Stefan boltzman constant = 5.67*10^-8
[tex]q' = 2*5.67*10^-^8*(523)^4*0.15\\\\q' = 1273 \frac{W}{m}[/tex]
- We see that radiation loss is not negligible, it account for 20% of the heat loss due to convection. Since the emissivity (e) of the fin has not been given. So, in the context of the given data this value is omitted from calculations.