Answer:
Vb.Net
msgbox ("Press "q" twice to quit", msgboxstyle.information)
if char.q = keypress and keypress.count = 2 then
End
End if
Explanation:
Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 300°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 3560 kJ and kinetic and potential energy effects are negligible.
Determine the final volume, in m3, and the work for the process, in kJ.
To determine the final volume and work done during the heating of water in a piston-cylinder assembly at constant pressure, tabulated data like steam tables are required since water is not an ideal gas. The work done is calculated using the formula W = PΔV.
Explanation:The student has been asked to find the final volume and the work done during a constant pressure process in which 5 kg of water is heated in a piston-cylinder assembly from an initial state of 5 bar and 300°C. To solve for the final volume and work done, one would typically use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat input minus the work output. However, since water at this state is not an ideal gas, tabulated data from steam tables or software would be used to determine the specific volume at the final state and then multiplied by the mass to find the total volume. The work done in a constant pressure process is equal to the pressure times the change in volume (W = PΔV). Without the final specific volume from the tables, we cannot compute the final volume or work directly.
In studying of traffic flow at a highway toll booth over the course of 60 minutes, it is determined that the arrival and departure rates are deterministic, but not uniform. The arrival rate is found to vary according to the function A(t) = 1.8 + 0.25t - 0.0030t^2. The departure rate function is D(t) = 1.4 + 0.11t. In both of these functions, t is in minutes after the beginning of the observation and A(t) and D(t) are in vehicles per minute. At what time does the maximum queue length occur?
a. 2.7 min
b. 9.4 min
c. 49.4 min
d. 60.0 min
The time that the maximum queue length occurs would be c. 49.4 min
To find the time at which the maximum queue length occurs, we need to determine when the arrival rate equals the departure rate.
The queue length increases when the arrival rate exceeds the departure rate and decreases when the departure rate exceeds the arrival rate. The maximum queue length occurs when the arrival rate equals the departure rate.
Substituting the given functions into the equation, we get:
[tex]\[ 1.8 + 0.25t - 0.0030t^2 = 1.4 + 0.11t \][/tex]
Rearranging the terms, we get a quadratic equation:
[tex]\[ -0.0030t^2 + 0.25t - 0.11t + 1.8 - 1.4 = 0 \][/tex]
[tex]\[ -0.0030t^2 + 0.14t + 0.4 = 0 \][/tex]
Now, we can solve this quadratic equation to find the value(s) of t at which the maximum queue length occurs. We can use the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
Where:
a = -0.0030
b = 0.14
c = 0.4
Calculate the values of t using this formula. Then, we'll choose the appropriate value based on the physical meaning of the problem.
Using the quadratic formula:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
[tex]\[ t = \frac{-0.14 \pm \sqrt{(0.14)^2 - 4(-0.0030)(0.4)}}{2(-0.0030)} \][/tex]
[tex]\[ t = \frac{-0.14 \pm \sqrt{0.0196 + 0.0048}}{-0.0060} \][/tex]
[tex]\[ t = \frac{-0.14 \pm 0.156}{-0.0060} \][/tex]
Now, we have two possible values for t :
1. [tex]\( t_1 = \frac{-0.14 + 0.156}{-0.0060} \)[/tex]
2. [tex]\( t_2 = \frac{-0.14 - 0.156}{-0.0060} \)[/tex]
Calculate these values:
1. [tex]\( t_1 = \frac{0.016}{-0.0060} = -2.67 \)[/tex]
2. [tex]\( t_2 = \frac{-0.296}{-0.0060} = 49.4[/tex]
Since t represents time, it cannot be negative. The maximum queue length occurs at approximately t = 49.4 minutes after the beginning of the observation.
In each case indicate whether the quantity in question increased, decreased or stayed the same when the string length is increased. Assume that the tension is unchanged. The function generator is kept at the same frequency, and the string is in resonance in all cases. Part A Number of antinodes ___ Number of antinodes ___ increased. decreased. stayed the same. Request Answer Part B Wavelength ___ Wavelength ___ increased. decreased. stayed the same. Request Answer Part C Fundamental frequency ___ Fundamental frequency ___ increased. decreased. stayed the same.
Answer:
Answer: No of anti-nodes increases
Answer: wavelength remains same.
Answer: fundamental frequency decreases
Explanation:
a)
The number of nodes (n) would have (n-1) anti-nodes.
The relation of Length of string with n is given below:
[tex]L = \frac{n*lambda}{2}[/tex]
Hence, n and L are directly proportional so as string length increases number of nodes and anti-nodes also increases.
Answer: No of anti-nodes increases
b)
Wavelength is dependent on the frequency:
[tex]lambda = \frac{v}{f}[/tex]
The speed v of the string remains same through-out and frequency of generator is unchanged!
Hence according to above relationship lambda is unchanged.
Answer: wavelength remains same.
C
Fundamental frequency equates to 1st harmonic that 2 nodes and 1 anti-node. Wavelength is = 2*L
Hence, if L increases wavelength increases
Using relation in part b
As wavelength increases fundamental frequency decreases
Answer: fundamental frequency decreases
A room is cooled by circulating chilled water through a heat exchanger located in the room. The air is circulated through the heat exchanger by a 0.25-hp (shaft output) fan. Typical efficiency of small electric motors driving 0.25-hp equipment is 60 percent. Determine the rate of heat supply by the fan–motor assembly to the room.
To determine the rate of heat supply by the fan-motor assembly, the electrical input power is calculated based on the 0.25-hp shaft output and 60% efficiency of the motor. The resulting heat supply to the room is the same as the electrical power input, which is 310.83 watts.
Explanation:The question asks to determine the rate of heat supply by a fan-motor assembly used to circulate chilled water through a heat exchanger for cooling a room. Given that the fan has a shaft output of 0.25 horsepower (hp) and that small electric motors driving such equipment typically have an efficiency of 60 percent, we can calculate the electrical power input needed to run the fan.
The electrical power input (Pinput) can be calculated as:
Pinput = Poutput / Efficiency
Where Poutput is the shaft output power (0.25 hp) and 'Efficiency' is the efficiency of the electric motor (60%, or 0.60 in decimal form).
Pinput = (0.25 hp) / 0.60
To convert horsepower to watts, we use the conversion factor 1 hp = 746 watts.
Pinput = (0.25 hp × 746 watts/hp) / 0.60
Pinput = 310.83 watts (rounded to two decimal places)
The rate of heat supply to the room will be equal to the electrical power input to the fan, which is 310.83 watts. This accounts for both the useful work done and all inefficiencies in the system that convert electrical energy into heat.
A coal-fired power plant equipped with a SO2 scrubber is required to achieve an overall SO2 removal efficiency of 85%. The existing scrubber is 95% efficient. Rather than treating the entire gas stream to 95% removal, the plant proposes to treat part of the flue gas to 95% removal, and to bypass the remainder around the scrubber. Calculate the fraction of the flue gas stream that can be bypassed around the scrubber (i.e., Qbypass/Q) and still satisfy the regulatory requirement.
Answer:
bypassed fraction B will be B= 0.105 (10.5%)
Explanation:
doing a mass balance of SO₂ at the exit
total mass outflow of SO₂ = remaining SO₂ from the scrubber outflow + bypass stream of SO₂
F*(1-er) = Fs*(1-es) + Fb
where
er= required efficiency
es= scrubber efficiency
Fs and Fb = total mass inflow of SO₂ to the scrubber and to the bypass respectively
F= total mass inflow of SO₂
and from a mass balance at the inlet
F= Fs+ Fb
therefore the bypassed fraction B=Fb/F is
F*(1-er) = Fs*(1-es) + Fb
1-er= (1-B)*(1-es) +B
1-er = 1-es - (1-es)*B + B
(es-er) = es*B
B= (es-er)/es = 1- er/es
replacing values
B= 1- er/es=1-0.85/0.95 = 2/19 = 0.105 (10.5%)
why is the thermal conductivity of super insolation order of magnitude lower than the thermal conductivity of ordinary insulation?
Answer:
Super insulation are obtained by using layers of highly reflective sheets separated by glass fibers in an vacuumed space. Radiation heat transfer between any of the surfaces is inversely proportional to the number of sheets used and thus heat lost by radiation will be very low by using these highly reflective sheets which will an effective way of heat transfer.
Explanation:
assume a strain gage is bonded to the cylinder wall surface in the direction of the axial strain. The strain gage has nominal resistance R0 and a Gage Factor GF . It is connected in a Wheatstone bridge configuration where all resistors have the same nominal resistance; the bridge has an input voltage Vin. (The strain gage is bonded and the Wheatstone bridge balanced with the vessel already pressurized.) Develop an expression for the voltage change ?V across the bridge if the cylinder pressure changes by ?P.
Explanation:
Note: For equations refer the attached document!
The net upward pressure force per unit height p*D must be balanced by the downward tensile force per unit height 2T, a force that can also be expressed as a stress, σhoop, times area 2t. Equating and solving for σh gives:
Eq 1
Similarly, the axial stress σaxial can be calculated by dividing the total force on the end of the can, pA=pπ(D/2)2 by the cross sectional area of the wall, πDt, giving:
Eq 2
For a flat sheet in biaxial tension, the strain in a given direction such as the ‘hoop’ tangential direction is given by the following constitutive relation - with Young’s modulus E and Poisson’s ratio ν:
Eq 3
Finally, solving for unknown pressure as a function of hoop strain:
Eq 4
Resistance of a conductor of length L, cross-sectional area A, and resistivity ρ is
Eq 5
Consequently, a small differential change in ΔR/R can be expressed as
Eq 6
Where ΔL/L is longitudinal strain ε, and ΔA/A is –2νε where ν is the Poisson’s ratio of the resistive material. Substitution and factoring out ε from the right hand side leaves
Eq 7
Where Δρ/ρε can be considered nearly constant, and thus the parenthetical term effectively becomes a single constant, the gage factor, GF
Eq 8
For Wheat stone bridge:
Eq 9
Given that R1=R3=R4=Ro, and R2 (the strain gage) = Ro + ΔR, substituting into equation above:
Eq9
Substituting e with respective stress-strain relation
Eq 10
Write IEEE floating point representation of the following decimal number. Show your work.
1.25
Answer:
00111111101000000000000000000000
Explanation:
View Image
0 01111111 01000000000000000000000
The first bit is the sign bit. It's 0 for positive numbers and 1 for negative numbers.
The next 8-bits are for the exponents.
The first 0-126₁₀ (0-2⁷⁻¹) are for the negative exponent 2⁻¹-2⁻¹²⁶.
And the last 127-256₁₀ (2⁷-2⁸) are for the positive exponents 2⁰-2¹²⁶.
You have 1.25₁₀ which is 1.010₂ in binary. But IEEE wants it in scientific notation form. So its actually 1.010₂*2⁰
The exponent bit value is 127+0=127 which is 01111111 in binary.
The last 23-bits are for the mantissa, which is the fraction part of your number. 0.25₁₀ in binary is 010₂... so your mantissa will be:
010...00000000000000000000
A simple ideal Rankine cycle with water as the working fluid operates between the pressure limits of 4 MPa in the boiler and 20 kPa in the condenser and a turbine inlet temperature of 700°C. The boiler is sized to provide a steam flow of 50 kg/s. Determine the power produced by the turbine and consumed by the pump.
Answer:
a) 69,630KW
b) 203 KW
Explanation:
The data obtained from Tables A-4, A-5 and A-6 is as follows:
[tex]h_{1} = h_{f,@20KPa} = 251.42 KJ/kg\\v_{1} = v_{f,@20KPa} = 0.001017 KJ/kgK\\\\w_{p,in} = v_{1} * (P_{2} - P_{1})\\w_{p,in} = (0.001017)*(4000-20)\\\\w_{p,in} = 4.05 KJ/kg\\\\h_{2} = h_{1} - w_{p,in} \\h_{2} = 251.42 + 4.05\\\\h_{2} = 255.47KJ/kg\\\\P_{3} = 4000KPa\\T_{3} = 700 C\\s_{3} = 7.6214 KJ/kgK\\\\h_{3} = 3906.3 KJ/kg\\\\P_{4} = 20 KPa\\s_{3} = s_{4} = 7.6214KJ/kgK\\s_{f} = 0.8320 KJ/kgK\\s_{fg} = 7.0752 KJ/kgK\\\\[/tex]
[tex]x_{4} = \frac{s_{4} - s_{f} }{s_{fg} } \\\\x_{4} = \frac{7.6214-0.8320}{7.0752} = 0.9596\\\\h_{f} = 251.42KJ/kg \\h_{fg} = 2357.5KJ/kg \\\\h_{4} = h_{f} + x_{4}*h_{fg} = 251.42 + 0.9596*2357.5 = 2513.7KJ/kg\\\\[/tex]
The power produced and consumed by turbine and pump respectively are:
[tex]W_{T,out} = flow(m) *(h_{3} - h_{4}) \\W_{T,out} = 50 *(3906.3-2513.7)\\\\W_{T,out} = 69,630 KW\\\\W_{p,in} = flow(m) *w_{p,in} = 50*4.05 = 203 KW[/tex]
The idling engines of a landing turbojet produce forward thrust when operating in a normal manner, but they can produce reverse thrust if the jet is properly deflected. Suppose that while the aircraft rolls down the runway at 150 km/h the idling engine consumes air at 50 kg/s and produces an exhaust velocity of 150 m/s.
a. What is the forward thrust of this engine?
b. What are the magnitude and direction (i.e., forward or reverse) if the exhaust is deflected 90 degree without affecting the mass flow?
c. What are the magnitude and direction of the thrust (forward or reverse) after the plane has come to a stop, with 90 degree exhaust deflection and an airflow of 40 kg/s?
Answer:
T = 5416.67 N
T = -2083.5 N
T = 0
Explanation:
Forward thrust has positive values and reverse thrust has negative values.
part a
Flight speed u = ( 150 km / h ) / 3.6 = 41.67 km / s
The thrust force represents the horizontal or x-component of momentum equation:
[tex]T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (150 - 41.67)\\\\T = 5416.67 N[/tex]
Answer: The thrust force T = 5416.67 N
part b
Now the exhaust velocity is now vertical due to reverse thrust application, then it has a zero horizontal component, thus thrust equation is:
[tex]T = flow(m_{exhaust})*(u_{exhaust} - u_{flight} )\\T = (50 kg/s ) * (0 - 41.67)\\\\T = -2083.5 N[/tex]
Answer: The thrust force T = -2083.5 N reverse direction
part c
Now the exhaust velocity and flight velocity is zero, then it has a zero horizontal component, thus thrust is also zero as there is no difference in two velocities in x direction.
Answer: T = 0 N
A signal whose timing is completely influenced by the traffic volumes, when detected, on all of the approaches operates in the following mode: a. Pretimed b. Semi-actuated c. Fully-actuated
Answer: c. Fully actuated
Explanation: fully actuated signals are completely influenced by volumes if traffic and employs sensors at all approaches for its detection.
Primary U.S. interstate highways are numbered 1-99. Odd numbers (like the 5 or 95) go north/south, and evens (like the 10 or 90) go east/west. Auxiliary highways are numbered 100-999, and service the primary highway indicated by the rightmost two digits. Thus, the 405 services the 5, and the 290 services the 90 Given a highway number, indicate whether it is a primary or auxiliary highway. If auxiliary, indicate what primary highway it serves. Also indicate if the (primary) highway runs north/south or east/west. Ex: If the input is: 90 the output is: The 90 is primary, going east/west. Ex: If the input is: 290 the output is: The 290 is auxiliary, serving the 90, going east/west Ex: If the input is G here to search The 290 is auxiliary, serving the 90, going eas Ex: If the input is: 0 or any number not between 1 and 999, the output is: 0 is not a valid interstate highway number. See Wikipedia for more info on highway numbering LAB 4.16.1: LAB: Interstate highway numbers ACTIVITY LabProgr- 1 import java.util.Scanner; 2 3 public class LabProgram public static void main(Stringl] args) t Scanner scnr new Scanner(System.in); int highwayNumber; int primaryNumber; 4 5 6 7 highwayNumber scnr.nextInt(); 1e Type your code
Answer:
The Java code is given below with appropriate variable names and tags for better understanding
Explanation:
import java.util.Scanner;
public class LabProgram{
public static void main(String[] args) {
Scanner scnr = new Scanner(System.in);
int highwayNumber;
int primaryNumber;
highwayNumber = scnr.nextInt();
if(highwayNumber<1 || highwayNumber>999)
System.out.println(highwayNumber+" is not a valid interstate highway number.");
else{
if(highwayNumber>=1 && highwayNumber<=99){
System.out.print("The "+highwayNumber+" is primary, going ");
if(highwayNumber%2==1)
System.out.println("north/south.");
else
System.out.println("east/west.");
}
else{
primaryNumber = highwayNumber%100;
System.out.print("The "+highwayNumber+" is auxillary, serving the "+primaryNumber+", going ");
if(primaryNumber%2==1)
System.out.println("north/south.");
else
System.out.println("east/west.");
}
}
}
}
The program is an illustration of conditional statements.
Conditional statements are statements whose execution is dependent on its truth value.
The missing code segment in Java where comments are used to explain each line is as follows:
import java.util.Scanner;
public class Main {
public static void main(String [] args){
Scanner scnr = new Scanner(System.in);
int highwayNumber; int primaryNumber;
highwayNumber = scnr.nextInt();
//This checks if the highwayNumber is not between 1 and 999 (inclusive)
if (highwayNumber <1 || highwayNumber > 999){
//If yes, the highwayNumber is invalid
System.out.print(highwayNumber+" is not a valid interstate highwayNumber number.");
}
//If otherwise
else{
//This checks if highwayNumber is less than 100
if (highwayNumber< 100){
if (highwayNumber%2 == 0){
//Even highwayNumber are primary going east/west
System.out.print("I-"+highwayNumber+" is primary, going east/west.");
}
else{
//Odd highwayNumber are primary going north/south
System.out.print("I-"+highwayNumber+" is primary, going north/south.");
}
}
//Otherwise
else{
if ((highwayNumber%100) % 2 == 0){
//Even highwayNumber are auxiliary going east/west
System.out.print("I-"+highwayNumber+" is auxiliary, going east/west.");
}
else{
//Even highwayNumber are auxiliary going north/south
System.out.print("I-"+highwayNumber+" is auxiliary, going north/south.");
}
}
}
}
}
Read more about similar programs at:
https://brainly.com/question/22078310
You are given a semiconductor resistor made from silicon with an impurity concentration of resistivity 1.00×10−3Ωm1.00×10−3Ωm. The resistor has a height of HH =0.5 mmmm, a length of LL = 2 mmmm, and a width of WW = 1.25 mmmm. The resistor can absorb (dissipate) up to PP = 7.81WW. What is the resistance of the resistor (RR), the maximum voltage (VV), and the maximum current (II)?
Answer:
The resistance (R) of the resistor is 2.4 ohm
The maximum voltage (V) is 4.33V
The maximum current (I) is 1.80A
Explanation:
Resistance (R) = resistivity×length/area
Resistivity = 0.003 ohm meter, length = 2mm = 2/1000 = 0.002m, width = 1.25mm = 1.25/1000 = 0.00125m, height = 0.5mm = 0.0005m, area = width × height = 0.00125m × 0.0005m = 6.25×10^-7m^2
R = 0.003×0.002/6.25×10^-7 = 3.2 ohm
Power (P) = V^2/R
V^2 = P × R = 7.81 × 3.2= 24.992
V = √24.992 = 4.99V
P = IV
I = P/V = 7.81/4.99 = 1.57A
The viscosity of a fluid is to be measured by an viscometer constructed of two 5 ft long concentric cylinders. The inner diameter of the outer cylinder is 6 in, and the gap between the two cylinders is 0.035 in. The outer cylinder is rotated at 250 rpm, and the torque is measured to be 1.2lbf *ft.
A) Determine the viscosity of the fluid.
Answer:
[tex] \mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}[/tex]
Explanation:
For this case we need to remember first thet the torque T is defined as:
[tex] T = FR[/tex]
Where T represent the torque, F the force acting in the inner cylinder and R the radius for the inner cylinder.
For the inner cylinder the force acting can be expressed as:
[tex] F = \mu A \frac{v}{l}[/tex]
Where [tex] \mu[/tex] represent the viscosity of the fluid, A the area of the inner cylinder, v represent the tangential velocity and l the thickness of fluid between the two cylinders.
And the tangential velocity for this case can be esxpressed as [tex] v = wR[/tex]
The info given is:
[tex] l = 0.035 in *\frac{1ft}{12in}=0.00292 ft[/tex]
[tex] R= \frac{D}{2} =\frac{6 in}{2}= 3 in*\frac{1ft}{12 in}=0.25 ft[/tex]
[tex] L = 5 ft[/tex] from the info given
N= 250 rpm represent the reveolutions per minute
[tex] T = 1.2 lbf ft[/tex] represent the torque given
We can find the surface area for the cylinder with this formula:
[tex] A= 2\pi R L[/tex]
And if we replace we got:
[tex] A= 2\pi 0.25 ft *5 ft= 7.85 ft^2[/tex]
Now we can find the tangential velocity like this:
[tex] v=wR= \frac2\pi *250 rpm* \frac{1min}{60s} * 0.25 ft=6.55\frac{ft}{s}[/tex]
Now we can set up the following equation for the torque:
[tex] T = FR[/tex]
[tex] T = \mu A \frac{v}{l} R[/tex]
And we can find the value for the viscosity [tex]\mu[/tex] like this:
[tex] \mu = \frac{T}{A \frac{v}{l} R}[/tex]
And if we replace we got:
[tex] \mu= \frac{1.2 lb ft}{7.85ft^2 * \frac{6.545 ft/s}{0.00292ft} *0.25 ft}=2.728x10^{-4} \frac{lbf s}{ft^2}[/tex]
A gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m^3, undergoes a constant-pressure expansion at 2 bar to a final volume of 0.12 m^3, while being slowly heated through the base. The change in internal energy of the gas is 0.25 kJ. the piston and cylinder walls are fabricated from heat-resistant material and the piston moves smoothly in the cylinder. The local atmosphere pressure is 1 bar.
a.) For the gas as the system, evaluate the work and the heat transfer, each in
b.) For the piston as a syatem, evautate the work and change in potential energy, in kJ.
Answer: (a). W = 4KJ and Q = 4.25KJ
(b). W = -2KJ and ΔPE = 2KJ
Explanation:
(a).
i. We are asked to calculate the work done during the expansion process considering gas as system.
from W = [tex]\int\limits^a_b {p} \, dV[/tex] where a = V₂ and b = V₁
so W = P(V₂-V₁)
W = (2 × 10²) (0.12 - 0.10)
W = 4 KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
since motion of the system is constrained, there is no change in both the potential and kinetic energy i.e. ΔPE = ΔKE = 0
∴ Q - W = ΔU
Q = ΔU + W
Q = 0.25 + 4
Q = 4.25 KJ
(b).
i. to calculate the work done during the expansion process considering piston as system;
W = [tex]\int\limits^a_b {(Patm - Pgas)} \, dV[/tex]where a and b represent V₂ and V₁ respectively.
W = (Patm - Pgas)(V₂ - V₁)
W = (1-2) ×10² × (0.12-0.1)
W = -2KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
Q = 0 since the piston and cylinder walls are perfectly insulated.
for piston, we neglect the change in internal energy and kinetic energy
ΔU = ΔKE = 0
from Q - W = ΔU + ΔKE + ΔPE
0 - (-2) = 0 + 0 + ΔPE
∴ ΔPE = 2KJ
When generating a hierarchical cluster, a tree structure called a _______.
Answer:
Dendrogram
Explanation:
Dendrogram is referred to as the tree structure that represents the hierarchy between the object in a cluster. it is also referred to as an output that is drawn from clustering.
The dendrogram is interpreted by observing the object situated at the higher side in a scatter plot. By joining the object at the same level it represents the order of cluster. another purpose of modifying form of Dendrogram is to help in calculating the distance between the objects in clusters.
A pipe in a district heating network is transporting over-pressurized hot water (10 atm) at a mass flow of 0.5 kg/s. The pipe is 5 m long, has an inner radius of 50 cm and pipe wall thickness of 50 mm. The pipe has a thermal conductivity of 20 W/m-K, and the inner pipe surface is at a uniform temperature of 110 ºC. The convection heat transfer coefficient of the air surrounding the pipe is 100W/m2 -K. The temperature of the water at inlet of pipe is 130 ºC and the constant pressure specific heat of hot water is 4000 J/kg-ºC. If the temperature of the air surrounding the pipe is 20 ºC, determine the exit temperature of the water at the end of the pipe.
The calculation of the exit temperature of water in the heated pipe involves using the energy balance equation, considering the heat lost through the pipe walls by convection, and then finding the change in thermal energy of the water via the heat transfer equation to solve for the exit temperature.
Explanation:Exit Temperature of Water in a Heated Pipe
To determine the exit temperature of water at the end of an over-pressurized heated pipe, we must consider the energy balance for the water flowing through the pipe. Based on the first law of thermodynamics, the change in thermal energy of the water will be equal to the heat lost through the pipe walls by convection:
Q = mc_p
(Exit Temperature - Inlet Temperature)
In this case, Q will be negative, since the water is losing heat to the surrounding air. The heat transfer from the pipe to the air is given by:
Q = hA(T_surface - T_air)
The area A for heat transfer is the external surface area of the pipe (
2
classes Math.PI
* radius * length of the pipe). Since we have the heat transfer coefficient h, the surface temperature of the pipe T_surface, and the air temperature T_air, we can calculate Q. Then we can use the mass flow rate m and the constant pressure specific heat c_p to find the exit temperature of the water.
To solve, we first calculate Q, then rearrange the first equation to solve for the Exit Temperature.
Water flowing through both a small pipe and a large pipe can fill a water tank in 4 hours. Water flowing through the small pipe alone can fill the tank in 15 more hours. How many hours would it take to fill the tank using only the small pipe?
Answer with explanation:
As is the question the answer would be 19 hours, and the key to solving it is in the phrase in 15 more hours, basically what they are saying is that the small pipe takes 15 hours more than both the big and the small to fill the tank. Since both pipes working together can fill the tank in 4 hours we need to add 4 and 15 to solve the problem.
If the question is how many hours would it take to fill the tank using only the big pipe? Then we could solve t for the following equation:
[tex]\frac{1}{4+15} + \frac{1}{t} = \frac{1}{4}[/tex]
Getting as a result: 5.06
Note that the equation is the result of taking the rate of the small pipe (what we solved before), plus the unknown rate of the big one equals the rate of both.
The lattice constant of a simple cubic primitive cell is 5.28 Å. Determine the distancebetween the nearest parallel ( a ) (100), ( b ) (110), and ( c ) (111) planes.
Answer:
a)5.28 Å , b)3.73 Å , c)3.048 Å
Explanation:
the atoms are situated only at the corners of cube.Each and every atom in simple cubic primitive at the corner is shared with 8 adjacent unit cells.
Therefore, a particular unit cell consist only 1/8th part of an atom.
The lattice constant of a simple cubic primitive cell is 5.28 Å
We know formula of distance,
d = [tex]\frac{a}{\sqrt{h^{2}+k^{2}+l^{2}}}[/tex]
a)(100)
a=5.28 Å
Distance = [tex]\frac{5.28 Å}{\sqrt{1^{2}+0^{2}+0^{2}}}[/tex]=5.28 Å
b)(110)
Distance = [tex]\frac{5.28}{\sqrt{1^{2}+1^{2}+0^{2}}}[/tex] = 3.73 Å
c)(111)
Distance= [tex]\frac{5.28}{\sqrt{1^{2}+1^{2}+1^{2}}}[/tex]= 3.048 Å
Assume a strain gage is bonded to the cylinder wall surface in the direction of the hoop strain. The strain gage has nominal resistance R0 and a Gage Factor GF. It is connected in a Wheatstone bridge configuration where all resistors have the same nominal resistance; the bridge has an input voltage, Vin. (The strain gage is bonded and the Wheatstone bridge balanced with the vessel already pressurized.)Calculate the voltage change ∆V across the Wheatstone bridge when the cylinder is pressurized to ∆P = 2.5 atm. Assume the vessel is made of 3004 aluminum with height h = 21 cm, diameter d = 9 cm, and thickness t = 65 µm. The Gage Factor is GF = 2 and the Wheatstone bridge has Vin = 6 V. The strain gage has nominal resistance R0 = 120 Ω.
Answer:
5.994 V
Explanation:
The pressure as a function of hoop strain is given:
[tex]P = \frac{4*E*t}{D}*\frac{e_{h} }{2-v}[/tex]
[tex]e_{h} = \frac{D*P*(2-v)}{4*E*t} .... Eq1[/tex]
For wheat-stone bridge with equal nominal resistance of resistors:
[tex]V_{out} = \frac{GF*e*V_{in} }{4} .... Eq2[/tex]
Hence, input Eq1 into Eq2
[tex]V_{out} = \frac{GF*e*V_{in}*D*P*(2-v) }{16*E*t} .....Eq3\\[/tex]
Given data:
P = 253313 Pa
D = d + 2t = 0.09013 m
t = 65 um
GF = 2
E = 75 GPa
v = 0.33
Use the data above and compute Vout using Eq3
[tex]V_{out} = \frac{2*6*0.09013*253313*(2-0.33) }{16*75*10^9*65*10^-6} \\\\V_{out} = 0.006285 V\\\\change in V = 6 - 0.006285 = 5.994 V[/tex]
If E = 94.2 mJ of energy is transferred when Q = 1.66 C of charge flows through a circuit element, what is the voltage across the circuit element?
Answer:
V = 56.8 mV
Explanation:
When a current I flows across a circuit element, if we assume that the dimensions of the circuit are much less than the wavelength of the power source creating this current, there exists a fixed relationship between the power dissipated in the circuit element, the current I and the voltage V across it, as follows:
P = V*I
By definition, power is the rate of change of energy, and current, the rate of change of the charge Q, so we can replace P and I, as follows:
E/t = V*q/t ⇒ E = V*Q
Solving for V:
V = E/Q = 94.2 mJ /1.66 C = 56.8 mV
If the electric field just outside a thin conducting sheet is equal to 1.5 N/C, determine the surface charge density on the conductor.
Answer:
The surface charge density on the conductor is found to be 26.55 x 1-6-12 C/m²
Explanation:
The electric field intensity due to a thin conducting sheet is given by the following formula:
Electric Field Intensity = (Surface Charge Density)/2(Permittivity of free space)
From this formula:
Surface Charge Density = 2(Electric Field Intensity)(Permittivity of free space)
We have the following data:
Electric Field Intensity = 1.5 N/C
Permittivity of free space = 8.85 x 10^-12 C²/N.m²
Therefore,
Surface Charge Density = 2(1.5 N/C)(8.85 x 10^-12 C²/Nm²)
Surface Charge Density = 26.55 x 10^-12 C/m²
Hence, the surface charge density on the conducting thin sheet will be 26.55 x 10^ -12 C/m².
The surface charge density on the conductor is; σ = 13.275 × 10⁻¹² C/m²
What is the surface charge density?The formula for surface charge density on a conductor in an electric field just outside the surface of conductor is;
σ = E * ϵ₀
where;
E is electric field = 1.5 N/C
ϵ₀ is permittivity of space = 8.85 × 10⁻¹² C²/N.m²
Thus;
σ = 1.5 * 8.85 × 10⁻¹²
σ = 13.275 × 10⁻¹² C/m²
Read more about Surface Charge Density at; https://brainly.com/question/14306160
Niobium has a BCC crystal structure, an atomic radius of 0.143 nm and an atomic weight of 92.91 g/mol. Calculate the theoretical density for Nb.
To calculate the theoretical density of Niobium with a BCC structure, we use its atomic radius and atomic weight, converting these into the cube's edge length using the BCC relation, then apply the density formula.
Explanation:To calculate the theoretical density of Niobium (Nb), which has a body-centered cubic (BCC) crystal structure, we first use the known values: atomic radius = 0.143 nm (or 0.143 × 10-9 m) and atomic weight = 92.91 g/mol. The formula for the density (ρ) in a BCC structure is ρ = (2 × M) / (a3 × NA), where M is the atomic mass, a is the edge length of the cube, and NA is Avogadro's number (6.022 × 1023 atoms/mol).
Since it's a BCC structure, the atomic radius relates to the cube's edge length (a) as a = 4r / √3. Substituting the given atomic radius, we find a = 4 * 0.143 × 10-9 m / √3. Then, to find the density, we substitute M (92.91 g/mol), a, and NA into the density formula. This calculation will give us the theoretical density of Niobium in g/cm3.
At Westtown High School, the mean score on the French final examination was with a standard deviation of , while the mean score on the Spanish final examination was with a standard deviation of . To earn a language honor at graduation, students must score in the th percentile on all their language final exams. Brynne scored on both the French exam and the Spanish exam. Is Brynne qualified for honors?
Answer:
The score for both exams 88 is above the 90 percentile, so then Brynne qualified for honors. See the explanation below.
Explanation:
Assuming the following question:"At Westtown High School, the mean score on the French final examination was 81 with a standard deviation of 5, while the mean score on the Spanish final examination was 72 with a standard deviation of 12.
To earn a language honor at graduation, students must score in the 90th percentile on all their language final exams. Brynne scored 88 on both the French exam and the Spanish exam. Is Brynne qualified for honors?"
Previous concepts
Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".
The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".
French case
Let X the random variable that represent the scores for the French exam of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(81,5)[/tex]
Where [tex]\mu=81[/tex] and [tex]\sigma=5[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] (a)
[tex]P(X<a)=0.90[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-81}{5}[/tex]
And if we solve for a we got
[tex]a=81 +1.28*5=87.4[/tex]
So the value for the scores that separates the bottom 90% of data from the top 10% is 87.4 (90th percentile).
And since the score of Brynne is 88 is above the 90 percentile
Spanish case
Let X the random variable that represent the scores for the Spanish exam of a population, and for this case we know the distribution for X is given by:
[tex]X \sim N(72,12)[/tex]
Where [tex]\mu=72[/tex] and [tex]\sigma=12[/tex]
For this part we want to find a value a, such that we satisfy this condition:
[tex]P(X>a)=0.1[/tex] (a)
[tex]P(X<a)=0.90[/tex] (b)
Both conditions are equivalent on this case. We can use the z score again in order to find the value a.
As we can see on the figure attached the z value that satisfy the condition with 0.90 of the area on the left and 0.1 of the area on the right it's z=1.28. On this case P(Z<1.28)=0.90 and P(z>1.28)=0.1
If we use condition (b) from previous we have this:
[tex]P(X<a)=P(\frac{X-\mu}{\sigma}<\frac{a-\mu}{\sigma})=0.90[/tex]
[tex]P(z<\frac{a-\mu}{\sigma})=0.90[/tex]
But we know which value of z satisfy the previous equation so then we can do this:
[tex]z=1.28<\frac{a-72}{12}[/tex]
And if we solve for a we got
[tex]a=72 +1.28*12=87.36[/tex]
So the value for the scores that separates the bottom 90% of data from the top 10% is 87.36 (90th percentile).
And since the score of Brynne is 88 is above the 90 percentile
Check my work Check My Work button is now disabledItem 16Item 16 3 points As a spherical ammonia vapor bubble rises in liquid ammonia, its diameter changes from 1 cm to 3 cm. Calculate the amount of work produced by this bubble, in kJ, if the surface tension of ammonia is 0.07 N/m.
Answer:
W = 1.7593 * 10 ^ (-7) KJ
Explanation:
The work done by the bubble is given:
[tex]W = sigma*\int\limits^2_1 {} \, dA \\\\W = sigma*( {A_{2} - A_{1} } ) \\\\A = pi*D^2\\\\W = sigma*pi*(D^2_{2} - D^2_{1})\\\\W = 0.07 * pi * (0.03^2 - 0.01^2)*10^(-3)\\\\W = 1.7593 *10^(-7) KJ[/tex]
Answer: W = 1.7593 * 10 ^ (-7) KJ
If a barrel of oil weighs 1.5 kN, calculate the specific weight, density, and specific gravity of the oil. The barrel weighs 110 N
Answer
given,
oil barrel weight = 1.5 k N = 1500 N
weight of the barrel = 110 N
Assuming volume of barrel = 0.159 m³
weight of oil = 1500-110
= 1390 N
[tex]specific\ weight = \dfrac{weight}{volume}[/tex]
[tex]specific\ weight = \dfrac{1390}{0.159}[/tex]
= 8742.14 N/m³
[tex]mass = \dfrac{weight}{g}[/tex]
[tex]mass = \dfrac{1390}{9.8}[/tex]
= 141.84 kg
[tex]density = \dfrac{mass}{volume}[/tex]
[tex]density = \dfrac{141.84}{0.159}[/tex]
= 892.05 kg/m³
[tex]Specific\ gravity = \dfrac{density\ of\ oil}{density\ of\ water}[/tex]
[tex]Specific\ gravity = \dfrac{892.05}{1000}[/tex]
= 0.892
Recycled materials content is environmental information that is typically: a. self-declared by the manufacturer. b. independently certified by third-party entities. c. listed in the building code. d. part of an environmental label. e. None of the above
Answer:
a. Self declared by the manufacturer
Explanation:
Recycled content refers to the portion of materials used in a product that have been diverted from the solid waste stream. If those materials are diverted during the manufacturing process, they are be referred to as pre-consumer recycled content (sometimes referred to as post-industrial). If they are diverted after consumer use, they are
post-consumer .
Using phasors, the value of 37 sin 50t + 30 cos(50t – 45°) is _________ cos(50t+(_____°)). Please report your answer so the magnitude is positive and all angles are in the range of negative 180 degrees to positive 80 degrees
Answer: 62 cos(50t - 70°)
Explanation:
First we need to convert all sines into cosines because phasor forms are represented through cosine. For that we will use the fact that sin(wt + ∅) = cos(wt + ∅ - 90°)
Therefore, 37sin50t = 37cos(50t - 90°)
Now we have 37cos(50t - 90°)+ 30 cos(50t – 45°). We need to convert them into phasor form to add the terms. For that we will use the fact Acos(wt+∅)=A∠∅ which can be represented using real and imaginary parts as A [cos(∅)+jsin(∅)].
So,
37cos(50t - 90°)
= 37∠-90°
= 37[cos(-90°)+jsin(-90°)
=37[0+j(-1)]
= -j37
Similarly,
30 cos(50t – 45°)
=30∠-45
=30[cos(-45)+jsin(-45)
=30[0.707-j0.707]
=21.21 - j21.21
37cos(50t - 90)+ 30 cos(50t – 45°) = -j37+21.21 - j21.21 = 21.21 - j58.21
Now we need to convert the real and imaginary parts back to cosine form. We will first calculate the magnitude by the formula √a²+b² where a and b are the real and imaginary parts respectively.
Here a=21.21 and b=58.21
magnitude = √(21.21)²+(58.21)²=61.95≅62
For calculating phase ∅ the formula is ∅=inversetan (b/a) where a and b are the real and imaginary parts respectively.
∅=inversetan(-58.21/21.21)
= -69.9°≅-70°
So the final answer is 62cos(50t-70°)
The value of [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] is [tex]\(61.97 \cos(50t - 70^\circ)\)[/tex].
To solve [tex]\(37 \sin 50t + 30 \cos(50t - 45^\circ)\)[/tex] using phasors, we follow these steps:
Step-by-Step Calculation:
1. Express each term as a phasor:
- The term [tex]\(37 \sin 50t\)[/tex] :
- Convert to cosine form: [tex]\(37 \sin 50t = 37 \cos(50t - 90^\circ)\)[/tex]
- Phasor form: [tex]\(37 \angle -90^\circ\)[/tex]
- The term [tex]\(30 \cos(50t - 45^\circ)\)[/tex] :
- Phasor form: [tex]\(30 \angle -45^\circ\)[/tex]
2. Convert the phasors to rectangular form:
- [tex]\(37 \angle -90^\circ\)[/tex] :
[tex]\[ 37 \cos(-90^\circ) + j 37 \sin(-90^\circ) = 0 - j 37 = -j 37 \][/tex]
- [tex]\(30 \angle -45^\circ\)[/tex]:
[tex]\[ 30 \cos(-45^\circ) + j 30 \sin(-45^\circ) = 30 \left(\frac{\sqrt{2}}{2}\right) - j 30 \left(\frac{\sqrt{2}}{2}\right) = 21.2132 - j 21.2132 \][/tex]
3. Add the rectangular components:
[tex]\[ -j 37 + (21.2132 - j 21.2132) \][/tex][tex]\[ = 21.2132 - j (37 + 21.2132) \][/tex]
[tex]\[ = 21.2132 - j 58.2132 \][/tex]
4. Convert the result back to polar form:
- Magnitude:
[tex]\[ R = \sqrt{21.2132^2 + 58.2132^2} = \sqrt{449.54 + 3388.78} = \sqrt{3838.32} = 61.97 \][/tex]
- Angle:
[tex]\[ \theta = \tan^{-1}\left(\frac{-58.2132}{21.2132}\right) = \tan^{-1}(-2.743) \approx -70^\circ \][/tex]
5. Express the final result:
Using the positive magnitude, the expression in cosine form is:
[tex]\[ 37 \sin 50t + 30 \cos(50t - 45^\circ) = 61.97 \cos(50t - 70^\circ) \][/tex]
dentify a semiconducting material and provide the value of its band gap) that could be used in: (a) (1 point) red LED (b) (1 point) UV LED c) (1 point) infrared detector (d) (1 point) blue LED (e) (1 point) green LED
Answer:
(a) Aluminum Indium Gallium Phosphide (AlInGaP). Band gap = 1.81eV ≈ 2eV
(b) Gallium Nitride (GaN). Band Gap = 3.4eV
(c) Aluminium Gallium Arsenide (AlGaA). Band Gap = 1.42eV ≈ 2.16eV
(d) Zinc Selenide (ZnSe). Band Gap = 2.82eV
(e) Gallium Phosphide (GaP). Band Gap = 2.24eV
Explanation:
LED's are semi-conducting materials that convert electrical energy to light energy. The light color emitted from the LED depends on the semi-conducting material and other compositions.
The band gap of the semi-conductor determines its wavelength. High band gap semi-conductors emit lower wavelengths which means greater power(UV semi-conducting macterials fall under this category).
What is the standard half-cell potential for the oxidation of methane under acidic conditions? The reaction for methane is as follows:
CH4(g) + 2H20(l) → CO2 +8H⁺ +8e⁻
What element is oxidized and how does its oxidation state change?
Answer:
The element that is oxidized is carbon.
Its oxidation state increased. It increased from -4 to +4
Explanation:
Oxidation is a process that involves increase in oxidation number.
The oxidation number of carbon in CH4 is -4
C + (1×4) = 0
C + 4 = 0
C = 0 - 4 = -4
The oxidation number of carbon in CO2 is +4
C + (2×-2) = 0
C - 4 = 0
C = 0+4 = 4
Increase in the oxidation number of carbon from -4 to +4 means carbon is oxidized