To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.
Density can be defined as
[tex]\rho = \frac{m}{V}[/tex]
Where
m = Mass
V = Volume
For state one we know that
[tex]\rho_1 = \frac{m_1}{V}[/tex]
[tex]m_1 = \rho_1 V[/tex]
[tex]m_1 = 1.18*1[/tex]
[tex]m_1 = 1.18Kg[/tex]
For state two we have to
[tex]\rho_2 = \frac{m_2}{V}[/tex]
[tex]m_2 = \rho_2 V[/tex]
[tex]m_1 = 7.2*1[/tex]
[tex]m_1 = 7.2Kg[/tex]
Therefore the total change of mass would be
[tex]\Delta m = m_2-m_1[/tex]
[tex]\Delta m = 7.2-1.18[/tex]
[tex]\Delta m = 6.02Kg[/tex]
Therefore the mass of air that has entered to the tank is 6.02Kg
A direct shear test was performed on a specimen of dry sand. The specimen was 50 mm by 50 mm square and 25 mm thick (height). A normal stress of 192 kPa was applied to the specimen and the shear stress at failure () was 120 kPa.
a. What is the effective friction angle of the sand?
b. For a normal stress of 200 kPa, what shear force (F) will be required to cause failure of the specimen? (Report answer in N).
To solve this problem it is necessary to apply the concepts related to the normal effort and the shear effort due to failure. The shear stress can be defined based on normal stress and effective friction angle. Mathematically it can be defined as
[tex]\tau_f = \sigma_n tan\phi[/tex]
Where
[tex]\tau_f =[/tex] Shear stress at failure
[tex]\sigma_n =[/tex] Normal stress
[tex]\phi =[/tex] Effective friction angle
PART A) Our values are given as ,
[tex]\sigma_n = 192kPa[/tex]
[tex]\tau_f = 120kPa[/tex]
Replacing at the previous equation we have,
[tex]\tau_f = \sigma_n tan\phi[/tex]
[tex]120 = 192tan\phi[/tex]
[tex]\phi = tan^{-1}(\frac{120}{192})[/tex]
[tex]\phi = 32.27°[/tex]
Therefore the effective friction angle of the sand is 32.27°
B) Using the maximum possible effort and the angle previously given we can calculate the maximum shear force, from which from its definition it is possible to find the force.
[tex]\tau_f = \sigma_n tan\phi[/tex]
[tex]\tau_f = 200tan(32\°)[/tex]
[tex]\tau_f = 124.97kPa[/tex]
With the value of the shear stress from its basic definition we can find the force. By definition the shear stress is given by
[tex]\tau_f = \frac{F}{A}[/tex]
Re-arrante to find the Force
[tex]F = A\tau_f[/tex]
The shear stress is a function of the Force and the Area, therefore, the area would be the square of the sides (50mm)
Replacing in the equation we have to,
[tex]F = 124.97*10^{-3}*(50*50)[/tex]
[tex]F = 312.425N[/tex]
Therefore the shear force required to cause failure of the specimen is 312.425N
An alloy is evaluated for potential creep deformation in a short-term laboratory experiment. The creep rate is found to be 1% per hour at 800°C and 0.055% per hour at 700°C.
(a) Calculate the activation energy for creep in this temperature range.
(b) Estimate the creep rate to be expected at the service temperature of 500°C.
Answer:
a) Q = 251.758 kJ/mol
b) creep rate is [tex]= 1.751 \times 10^{-5} \% per hr[/tex]
Explanation:
we know Arrhenius expression is given as
[tex]\dot \epsilon =Ce^{\frac{-Q}{RT}[/tex]
where
Q is activation energy
C is pre- exponential constant
At 700 degree C creep rate is[tex] \dot \epsilon = 5.5\times 10^{-2} [/tex]% per hr
At 800 degree C creep rate is[tex] \dot \epsilon = 1 [/tex]% per hr
activation energy for creep is [tex]\frac{\epsilon_{800}}{\epsilon_{700}}[/tex] = [tex]= \frac{C\times e^{\frac{-Q}{R(800+273)}}}{C\times e^{\frac{-Q}{R(700+273)}}}[/tex]
[tex]\frac{1\%}{5.5 \times 10^{-2}\%} = e^{[\frac{-Q}{R(800+273)}] -[\frac{-Q}{R(800+273)}]}[/tex]
[tex]\frac{0.01}{5.5\times 10^{-4}} = ln [e^{\frac{Q}{8.314}[\frac{1}{1073} - \frac{1}{973}]}][/tex]
solving for Q we get
Q = 251.758 kJ/mol
b) creep rate at 500 degree C
we know
[tex]C = \epsilon e^{\frac{Q}{RT}}[/tex]
[tex]=- 1\% e{\frac{251758}{8.314(500+273}} = 1.804 \times 10^{12} \% per hr[/tex]
[tex]\epsilon_{500} = C e^{\frac{Q}{RT}}[/tex]
[tex]= 1.804 \times 10^{12} e{\frac{251758}{8.314(500+273}}[/tex]
[tex]= 1.751 \times 10^{-5} \% per hr[/tex]
Assign deliveryCost with the cost (in dollars) to deliver a piece of baggage weighing baggageWeight. The baggage delivery service charges twenty dollars for the first 50 pounds and one dolllar for each additional pound. The baggage delivery service calculates delivery charge by rounding to the next pound. Assume baggageWeight is always greater than 50 pounds.
To determine the delivery cost for baggage weighing more than 50 pounds, charge twenty dollars for the first 50 pounds and one dollar for each additional pound, rounding the weight to the nearest whole number.
Explanation:To calculate the cost to deliver a piece of baggage weighing more than 50 pounds, we must understand the pricing structure of the baggage delivery service. The service charges twenty dollars for the first 50 pounds and one dollar for each additional pound. Since the weight is rounded to the next pound, if the baggage weight is greater than 50 pounds but not a whole number, we round up to the nearest whole number before calculating the additional cost.
For example, if the baggage weighs 53.2 pounds, we round the weight to 54 pounds. The first 50 pounds cost twenty dollars, so we only need to calculate the cost for the additional weight over 50 pounds, which in this case is 4 pounds (54 - 50 = 4). Thus, the additional cost is 4 dollars (4 additional pounds × $1 per additional pound = $4). The total delivery cost would be $24 ($20 for the first 50 pounds + $4 for the additional 4 pounds).
Moist air enters a duct at 10 oC, 70% relative humidity, and a volumetric flow rate of 150 m3/min. The mixture is heated as it flows in the duct and exits at 40oC. No moisture is added or removed, and the mixture pressure remains approximately constant at 1 bar. For steady-state operation, determine (a) the rate of heat transfer, in kJ/min, and (b) the relative humidity at the exit. Change in kinetic and potential energy can be ignored.
Write a function call with arguments tensPlace, onesPlace, and userInt. Be sure to pass the first two arguments as pointers. Sample output for the given program: tensPlace = 4, onesPlace = 1
The question pertains to the use of functions and pointers in programming. The task involves writing a function call that manipulates the tens and ones digit of an integer by passing the first two arguments as pointers. An example solution involves creating a function named ExtractDigits to perform this task and using pointers to directly modify specified variables.
Explanation:The question refers to the concept of functions and pointers in programming, which is an advanced topic typically covered at the college level in computer science or engineering courses. The task is to write a function call that uses pointers for the first two arguments, tensPlace and onesPlace, and a regular argument for userInt. This function could be crafted to manipulate or assess the tens and ones place of a given integer, userInt.
Example Solution:Suppose we have a function named ExtractDigits designed to extract the tens and ones place of an integer. This function might look like:
void ExtractDigits(int* tensPlace, int* onesPlace, int userInt){
*tensPlace = (userInt / 10) % 10;
*onesPlace = userInt % 10;
}
To call this function with the variables tensPlace, onesPlace, and a specific integer (for example, 41), you can use the following code:
int main() {
int tens, ones;
ExtractDigits(&tens, &ones, 41);
printf("tensPlace = %d, onesPlace = %d", tens, ones);
return 0;
}
This function call passes the addresses of tens and ones to the function so that their values can be modified directly, which is a fundamental use of pointers in C and C++ programming.
In a p + − njunction, the n side has a donor concentration of 1 × 1016 cm−3 . If ni = 1 × 1010 cm−3 , the relative dielectric constant Pr = 12, Dn = 50 cm2/s, Dp = 20 cm2/s, and the electron and hole minority carriers have lifetimes of τ = 100 ns and 50 ns respectively, and a forward bias of 0.6 V, calculate the hole diffusion current density 2µm away from the depletion edge on the n-side. If we double the p + doping, what effect will it have on the hole diffusion current?
Answer:
0.3
Explanation:
Please see attachment.
Unidirectional and continuous carbon fibers with a diameter of 6.5 micron are embedded within an epoxy that offers a yield point of 73 MPa. If the bond strength across the fiber-epoxy interface is 32 MPa, compute the minimum fiber length to guarantee that the fibers are conveying an optimum fraction of force that is applied to the composite. Answer Format: X.XX Unit: mm
Answer:
L=0.0074 mm
Explanation:
Given that
d= 6.5 micron meter
d= 6.5 x 10⁻³ mm
Yield strength ,Sy = 73 MPa
Bond strength ,σ = 32 MPa
The fiber length given as
[tex]L=\dfrac{S_y \times d}{2 \times \sigma }[/tex]
By putting the values
[tex]L=\dfrac{S_y \times d}{2 \times \sigma }[/tex]
[tex]L=\dfrac{73 \times 6.5\times 10^{-3}}{2\times 32 }\ mm[/tex]
L=0.0074 mm
Therefore the minimum fiber length is 0.0074 mm.
Suppose the Bookstore is processing an input file containing the titles of books in order to remove duplicates from their list. Write a program that reads all of the titles from an input file called bookTitles.txt and writes them to an output file called noDuplicates.txt. When complete, the output files should contain all unique titles found in the input file.
Answer:
books = []
fp = open("bookTitles.txt")
for line in fp.readlines():
title = line.strip()
if title not in books:
books.append(title)
fp.close()
fout = open("noDuplicates.txt", "w")
for title in books:
print(tile, file=fout)
fout.close()
except FileNotFoundError:
print("Unable to open bookTitles.txt")
Two pipes of identical diameter and material are connected in parallel. The length of pipe A is five times the length of pipe B. Assuming the flow is fully turbulent in both pipes and thus the friction factor is independent of the Reynolds number and disregarding minor losses, determine the ratio of the flow rates in the two pipes.
Answer:
[tex]\dfrac{Q_B}{Q_A}=\sqrt{5}[/tex]
Explanation:
Lets take
Length of pipe B = L
Length of pipe A = 5 L
Discharge in pipe A = Q₁
Discharge in pipe B = Q₂
We know that head loss in the pipe given as
[tex]h_f=\dfrac{FLQ^2}{12.1d^5}[/tex]
F=Friction factor, Q=Discharge,L=length
d=Diameter of pipe
here all only Q and L is varying and all other quantity is constant
So we can say that
LQ²= Constant
L₁Q₁²=L₂Q²₂
By putting the values
5LQ₁²=LQ²₂
[tex]\dfrac{Q_2}{Q_1}=\sqrt{5}[/tex]
Therefore
[tex]\dfrac{Q_B}{Q_A}=\sqrt{5}[/tex]
A program is seeded with 30 faults. During testing, 21 faults are detected, 15 of which are seeded faults and 6 of which are indigenous faults. What is the Mill's estimate of the number of indigenous faults remaining undetected in the program?
Answer:
Estimated number of indigenous faults remaining undetected is 6
Explanation:
The maximum likelihood estimate of indigenous faults is given by,
[tex]N_F=n_F\times \frac{N_S}{n_S}[/tex] here,
[tex]n_F[/tex] = the number of unseeded faults = 6
[tex]N_S[/tex] = number of seeded faults = 30
[tex]n_s[/tex] = number of seeded faults found = 15
So NF will be calculated as,
[tex]N_F=6\times \frac{30}{15}=12[/tex]
And the estimate of faults remaining is [tex]N_F-n_F[/tex] = 12 - 6 = 6
. Using the Newton Raphson method, determine the uniform flow depth in a trapezoidal channel with a bottom width of 3.0 m and side slope of 1:2 (v:h) with a discharge of 10 (m3/s). The slope is 0.0004 and Manning's roughness is 0.015.
Answer:
y ≈ 2.5
Explanation:
Given data:
bottom width is 3 m
side slope is 1:2
discharge is 10 m^3/s
slope is 0.004
manning roughness coefficient is 0.015
manning equation is written as
[tex]v =1/n R^{2/3} s^{1/2}[/tex]
where R is hydraulic radius
S = bed slope
[tex]Q = Av =A 1/n R^{2/3} s^{1/2}[/tex]
[tex]A = 1/2 \times (B+B+4y) \times y =(B+2y) y[/tex]
[tex]R =\frac{A}{P}[/tex]
P is perimeter [tex]= (B+2\sqrt{5} y)[/tex]
[tex]R =\frac{(3+2y) y}{(3+2\sqrt{5} y)}[/tex]
[tex]Q = (2+2y) y) \times 1/0.015 [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} 0.004^{1/2}[/tex]
solving for y[tex]100 =(2+2y) y) \times (1/0.015) [\frac{(3+2y) y}{(3+2\sqrt{5} y)}]^{2/3} \times 0.004^{1/2}[/tex]
solving for y value by using iteration method ,we get
y ≈ 2.5
// This program is supposed to display every fifth year // starting with 2017; that is, 2017, 2022, 2027, 2032, // and so on, for 30 years. start Declarations num year num START_YEAR = 2017 num FACTOR = 5 num END_YEAR = 30 year = START_YEAR while year <= END_YEAR output year endif stop 1. What problems do you see in this logic? 2. Show corrected pseudocode to fix the problems. 3. Is there another way the code can be corrected and still have the same outcome? If so, please describe. 4. Implement your pseudocode in either Raptor or VBA. (submit this file) 5. Does it work? You must test your code before submitting and indicate if the program functions according to the assignment description. Explain.
Answer:
Explanation:
1. The problems in the above pseudocode include (but not limited to) the following
1. The end year is not properly represented.
2. Though, the factor of 5 is declared, it's not implemented as increment in the pseudocode
3. Incorrect use of control structures. (While ...... And .......Endif)
2.
Start
Start_Year = 2017
Kount = 1
While Kount <= 30
Display Start_Year
Start_Year = Start_Year + 5
Kount = Kount + 1
End While
Stop
3. Yes, by doing the following
* Apply increment of 5 to start year within the whole loop, where necessary
* Use matching control structures
While statement ends with End While (not end if, as it is in the question)
4. Using VBA
Dim Start_Year: Start_Year = 2017 ' Declare and initialise year to 2017
Dim Counter: Counter = 1 ' Declare and Initialise Counter to 1
While Count <= 30 ' Test Value of Counter; to check if the desired number of year has gotten to 30. While the condition remains value, the following code will be executed.
msgbox Start_Year ' Display the value of start year
Start_Year = Start_Year + 5 'Increase value of start year by a factor of 5
Counter = Counter + 1 'Increment Counter
Wend
5. Yes
A thick-walled tube of stainless steel having a k = 21.63 W/m∙K with dimensions of 0.0254 m ID and 0.0508 m OD is covered with 0.0254 m thick layer of an insulation (k = 0.2423 W/m∙K). The inside-wall temperature of the pipe is 811 K and the outside surface of the insulation is 310.8 K. For a 0.305 m length of pipe, calculate the heat loss and the temperature at the interface between the metal and the insulation.
Answer:
Q=339.5W
T2=805.3K
Explanation:
Hi!
To solve this problem follow the steps below, the procedure is attached in an image
1. Draw the complete outline of the problem.
2.to find the heat Raise the heat transfer equation for cylinders from the inside of the metal tube, to the outside of the insulation.
3. Once the heat is found, Pose the heat transfer equation for cylinders from the inner part of the metal tube to the outside of the metal tube and solve to find the temperature
Compute the longitudinal modulus of elasticity (in GPa) for a continuous and aligned hybrid composite consisting of aramid and glass fibers in volume fractions of 0.22 and 0.30, respectively, within a polyester resin matrix, given the following data:
Modulus of Elasticity:
(GPa)
Glass fibers 72.5
Aramid fibers 131
Polyester 2.5
An uncharged capacitor and a resistor are connected in series to a source of voltage. If the voltage = 7.41 Volts, C = 11.5 µFarad, and R = 89.4 Ohms, find (a) the time constant of the circuit, (b) the maximum charge on the capacitor, and (c) the charge on the capacitor at a time equal to one time constant after the battery is connected
Answer:
a) RC = 1.03 mseg.
b) Qmax = CV = 85.2 μC
c) Q = 53.9 μC
Explanation:
a) In a RC circuit, during the transient period, the capacitor charges exponentially (starting from 0 due to the voltage in the capacitor can´t change instantaneously) with time, being the exponent -t/RC.
This product RC, which defines the rate at which the capacitor charges, is called the time constant of the circuit.
In this case , it can be calculated as follows:
ζ = R C = 89.4 Ω . 11.5 μF = 1.03 mseg.
b) As the charge begins to build up the capacitor plates, a voltage establishes between plates, that opposes to the battery voltage. When this voltage is equal to the battery one, the capacitor reaches to the maximum charge, which is, by definition, as follows:
Q = C V = 11.5 μF . 7.41 V = 85.2 μC
c) During the charging process, the charge increases following this equation:
Q = CV (1 - e⁻t/RC)
When t = RC, the expression for Q is as follows:
Q = CV ( 1- e⁻¹) = 0.63 x CV = 53.9 μC
A 240 V, 60 Hz squirrel-cage induction motor has a full-load slip of 0.02 and a full-load speed of 1764 rpm. The winding resistance of the rotor is 0.6 Ω and a winding reactance of 5 Ω at 60 Hz. Determine the rotor current at full-load. a. Ir =-3.53 A b. Ir= 7.89 A c. Ir= 237 A d. Ir= 1.18 A
Answer:
full load current = 7.891151 A
so correct option is b. Ir= 7.89
Explanation:
given data
Energy E = V = 240 V
frequency = 60 Hz
full-load slip = 0.02
full-load speed = 1764 rpm
winding resistance = 0.6 Ω
winding reactance = 5 Ω
to find out
rotor current at full-load
solution
we will apply here full load current formula that is express as
full load current = [tex]\frac{S*E}{\sqrt{R^2+ (S*X)^2}}[/tex] ...................1
here S is full-load slip and E is energy given and R is winding resistance and X is winding reactance
put here value we get
full load current = [tex]\frac{0.02*240}{\sqrt{0.6^2+ (0.02*5)^2}}[/tex]
full load current = 7.891151 A
so correct option is b. Ir= 7.89
Water flows through a horizontal 60 mm diameter galvanized iron pipe at a rate of 0.02 m3/s. If the pressure drop is 135 kPa per 10 m of pipe, do you think this pipe is
a) a new pipe,
b) an old pipe with somewhat increased roughness due to aging or
c) a very old pipe that is partially clogged by deposits. Justify your answer.
Answer:
pipe is old one with increased roughness
Explanation:
discharge is given as
[tex]V =\frac{Q}{A} = \frac{ 0.02}{\pi \4 \times (60\times 10^{-3})^2}[/tex]
V = 7.07 m/s
from bernou;ii's theorem we have
[tex]\frac{p_1}{\gamma} +\frac{V_1^2}{2g} + z_1 = \frac{p_2}{\gamma} +\frac{V_2^2}{2g} + z_2 + h_l[/tex]
as we know pipe is horizontal and with constant velocity so we have
[tex]\frac{P_1}{\gamma } + \frac{P_2 {\gamma } + \frac{flv^2}{2gD}[/tex]
[tex]P_1 -P_2 = \frac{flv^2}{2gD} \times \gamma[/tex]
[tex]135 \times 10^3 = \frac{f \times 10\times 7.07^2}{2\times 9.81 \times 60 \times 10^{-5}} \times 1000 \times 9.81[/tex]
solving for friction factor f
f = 0.0324
fro galvanized iron pipe we have [tex]\epsilon = 0.15 mm[/tex]
[tex]\frac{\epsilon}{d} = \frac{0.15}{60} = 0.0025[/tex]
reynold number is
[tex]Re =\frac{Vd}{\nu} = \frac{7.07 \times 60\times 10^{-3}}{1.12\times 10^{-6}}[/tex]
Re = 378750
from moody chart
[tex]For Re = 378750 and \frac{\epsilon}{d} = 0.0025[/tex]
[tex]f_{new} = 0.025[/tex]
therefore new friction factor is less than old friction factoer hence pipe is not new one
now for Re = 378750 and f = 0.0324
from moody chart
we have [tex]\frac{\epsilon}{d} =0.006[/tex]
[tex]\epsilon = 0.006 \times 60[/tex]
[tex]\epsilon = 0.36 mm[/tex]
thus pipe is old one with increased roughness
We wish to find roots of the following equation: cos(x) = x 3
(a) Perform two iterations of Bisection method by hand to find the roots of the nonlinear equation using the initial interval of 0.5 to 1.
(b) Perform two iterations of Regula Falsi method by hand to find the roots of the nonlinear equation using the initial interval of 0.5 to 1.
(c) Perform two iterations of Newton-Raphson method by hand to find the roots of the nonlinear equation using the initial guess of 0.75.
An air conditioner removes heat steadily from a house at a rate of 750 kJ/min while drawing electric power at a rate of 6 kW. Determine (a) the COP of this air conditioner and (b) the rate of heat transfer to the outside air.
Answer:
a. 2.08, b. 1110 kJ/min
Explanation:
The power consumption and the cooling rate of an air conditioner are given. The COP or Coefficient of Performance and the rate of heat rejection are to be determined. Assume that the air conditioner operates steadily.
a. The coefficient of performance of the air conditioner (refrigerator) is determined from its definition, which is
COP(r) = Q(L)/W(net in), where Q(L) is the rate of heat removed and W(net in) is the work done to remove said heat
COP(r) = (750 kJ/min/6 kW) x (1 kW/60kJ/min) = 2.08
The COP of this air conditioner is 2.08.
b. The rate of heat discharged to the outside air is determined from the energy balance.
Q(H) = Q(L) + W(net in)
Q(H) = 750 kJ/min + 6 x 60 kJ/min = 1110 kJ/min
The rate of heat transfer to the outside air is 1110 kJ for every minute.
a) The COP of the air conditioner is 2.083.
b) Rate of heat transfer to the outside air is 18.5 kW.
Step 1
Given Data:
- Heat removal rate from the house, [tex]\( \dot{Q}_{in} = 750 \, \text{kJ/min} \)[/tex]
- Electric power input, [tex]\( \dot{W}_{in} = 6 \, \text{kW} \)[/tex]
Converting Units:
[tex]\[ \dot{Q}_{in} = 750 \, \text{kJ/min} = \frac{750 \, \text{kJ}}{60 \, \text{s}} = 12.5 \, \text{kW} \][/tex]
Calculations:
(a) Coefficient of Performance (COP):
The COP of an air conditioner is defined as the ratio of the heat removed from the house to the work input (electric power):
[tex]\[ \text{COP} = \frac{\dot{Q}_{in}}{\dot{W}_{in}} \][/tex]
Step 2
Substituting the given values:
[tex]\[ \text{COP} = \frac{12.5 \, \text{kW}}{6 \, \text{kW}} = 2.083 \][/tex]
(b) Rate of Heat Transfer to the Outside Air:
The rate of heat transfer to the outside air, [tex]\( \dot{Q}_{out} \)[/tex], can be determined using the first law of thermodynamics for a steady-state system:
[tex]\[ \dot{Q}_{out} = \dot{Q}_{in} + \dot{W}_{in} \][/tex]
Substituting the given values:
[tex]\[ \dot{Q}_{out} = 12.5 \, \text{kW} + 6 \, \text{kW} = 18.5 \, \text{kW} \][/tex]
Write a simple phonebook program that reads in a series of name-number pairs from the user (that is, name and number on one line separated by whitespace) and stores them in a Map from Strings to Integers. Then ask the user for a name and return the matching number, or tell the user that the name wasn’t found.
Answer:
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class PhoneBook {
public static void main(String[] args) {
Scanner in = new Scanner(System.in);
Map<String, String> map = new HashMap<>();
String name, number, choice;
do {
System.out.print("Enter name: ");
name = in.next();
System.out.print("Enter number: ");
number = in.next();
map.put(name, number);
System.out.print("Do you want to try again(y or n): ");
choice = in.next();
} while (!choice.equalsIgnoreCase("n"));
System.out.print("Enter name to search for: ");
name = in.next();
if (map.containsKey(name)) {
System.out.println(map.get(name));
} else {
System.out.println(name + " is not in the phone book");
}
}
}
Water flows in a constant diameter pipe with the following conditions measured:
At section (a) pa = 31.1 psi and za = 56.7 ft; at section (b) pb = 27.3 psi and zb = 68.8 ft.
(a) Determine the head loss from section (a) to section (b).
(b) Is the flow from (a) to (b) or from (b) to (a)?
Answer:
a) [tex]h_L=-3.331ft[/tex]
b) The flow would be going from section (b) to section (a)
Explanation:
1) Notation
[tex]p_a =31.1psi=4478.4\frac{lb}{ft^2}[/tex]
[tex]p_b =27.3psi=3931.2\frac{lb}{ft^2}[/tex]
For above conversions we use the conversion factor [tex]1psi=144\frac{lb}{ft^2}[/tex]
[tex]z_a =56.7ft[/tex]
[tex]z_a =68.8ft[/tex]
[tex]h_L =?[/tex] head loss from section
2) Formulas and definitions
For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.
The formula is given by:
[tex]\frac{p_a}{\gamma}+\frac{V_a^2}{2g}+z_a =\frac{p_b}{\gamma}+\frac{V_b^2}{2g}+z_b +h_L[/tex]
Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:
[tex]\frac{p_a}{\gamma}+z_a =\frac{p_b}{\gamma}+z_b +h_L[/tex]
3)Part a
And on this case we have all the values in order to replace and solve for [tex]h_L[/tex]
[tex]\frac{4478.4\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+56.7ft=\frac{3931.2\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+68.8ft +h_L[/tex]
[tex]h_L=(71.769+56.7-63-68.8)ft=-3.331ft[/tex]
4)Part b
Analyzing the value obtained for [tex]\h_L[/tex] is a negative value, so on this case this means that the flow would be going from section (b) to section (a).
Air is to be heated steadily by an 8-kW electric resistance heater as it flows through an insulated duct. If the air enters at 55°C at a rate of 2 kg/s, determine the exit temperature of air. Solve using appropriate software.
To solve this problem it is necessary to apply the concepts related to the heat exchange of a body.
By definition heat exchange in terms of mass flow can be expressed as
[tex]W = \dot{m}c_p \Delta T[/tex]
Where
[tex]C_p =[/tex] Specific heat
[tex]\dot{m}[/tex]= Mass flow rate
[tex]\Delta T[/tex] = Change in Temperature
Our values are given as
[tex]C_p = 1.005kJ/kgK \rightarrow[/tex] Specific heat of air
[tex]T_1 = 50\°C[/tex]
[tex]\dot{m} = 2kg/s[/tex]
[tex]W = 8kW[/tex]
From our equation we have that
[tex]W = \dot{m}c_p \Delta T[/tex]
[tex]W = \dot{m}c_p (T_2-T_1)[/tex]
Rearrange to find [tex]T_2[/tex]
[tex]T_2 = \frac{W}{\dot{m}c_p}+T_1[/tex]
Replacing
[tex]T_2 = \frac{8}{2*1.005}+(50+273)[/tex]
[tex]T_2 = 326.98K \approx 53.98\°C[/tex]
Therefore the exit temperature of air is 53.98°C
The wall shear stress in a fully developed flow portion of a 12-in.-diameter pipe is 1.85 lb/ft^2.
Determine the pressure gradient ∂p/∂x, wherexis the flow direction when (a) the pipe is horizontal, (b) the pipe is vertical with the flow up, and (c) the pipe is vertical with the flowdown.
Answer:
a) [tex]-7.4\frac{lb}{ft^3}[/tex]
b) [tex]-69.8\frac{lb}{ft^3}[/tex]
c) [tex]55 \frac{lb}{ft^3}[/tex]
Explanation:
1) Notation
[tex]\tau=1.85\frac{lb}{ft^2}[/tex] represent the shear stress defined as "the external force acting on an object or surface parallel to the slope or plane in which it lies"
R represent the radial distance
L the longitude
[tex]\theta=0\degree[/tex] since at the begin we have a horizontal pipe, but for parts b and c the angle would change.
D represent the diameter for the pipe
[tex]\gamma=62.4\frac{lb}{ft^3}[/tex] is the specific weight for the water
2) Part a
For this case we can use the shear stress and the radial distance to find the pressure difference per unit of lenght, with the following formula
[tex]\frac{2\tau}{r}=\frac{\Delta p -\gamma Lsin\theta}{L}[/tex]
[tex]\frac{2\tau}{r}=\frac{\Delta p}{L}-\gamma sin\theta[/tex]
If we convert the difference's into differentials we have this:
[tex]-\frac{dp}{dx}=\frac{2\tau}{r}+\gamma sin\theta[/tex]
We can replace [tex]r=\frac{D}{2}[/tex] and we have this:
[tex]\frac{dp}{dx}=-[\frac{4\tau}{D}+\gamma sin\theta][/tex]
Replacing the values given we have:
[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin0]=-7.4\frac{lb}{ft^3}[/tex]
3) Part b
When the pipe is on vertical upward position the new angle would be [tex]\theta=\pi/2[/tex], and replacing into the formula we got this:
[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin90]=-69.8\frac{lb}{ft^3}[/tex]
4) Part c
When the pipe is on vertical downward position the new angle would be [tex]\theta=-\pi/2[/tex], and replacing into the formula we got this:
[tex]\frac{dp}{dx}=-[\frac{4x1.85\frac{lb}{ft^2}}{1ft}+62.4\frac{lb}{ft^3} sin(-90)]=55 \frac{lb}{ft^3}[/tex]
Water at 120oC boils inside a channel with a flat surface measuring 45 cm x 45 cm. Air at 62 m/s and 20oC flows over the channel parallel to the surface. Determine the heat transfer rate to the air. Neglect wall thermal resistance.
Answer:
Q = 2.532 x 10³ W
Explanation:
The properties of air at the film temperature of
T(f) = (T(s) + T()∞)/2
T(f) = (120 + 20)/2 = 70 °C
From the table of properties of air at T = 20 °C we know that ρ = 1.028 kg/m³ , k = 0.02881W/mK, Pr = 0.7177 , v = 1.995 x 10⁻⁵ m²/s
Calculate the Reynold’s Number
Re(l) = V x L/ν, where V is the speed of air flowing, L is the length of the surface in meters and ν is the kinematic viscosity
Re(l) = 62 x 0.45/(1.995 x 10⁻⁵) = 1.398 x 10⁶
Re(l) is greater than the critical Reynold Number. Thus, we have combined laminar and turbulent flow and the average Nusselt number for the entire plate is determined by
Nu = (0.037 x (Re(l)^0.8) - 871) x Pr^1/3, where Pr is the Prandtl’s Number
Nu = (0.037 x (1.398 x 10⁶)^0.8) – 871) x (0.7177)^1/3
Nu = 1952.85
We also know that
Nu = h x L/k, where h is the heat transfer coefficient, L is the length of the surface and k is the thermal conductivity
Rearranging to solve for h
h = (Nu x k)/l
h = 1952.85 x 0.02881/0.45 = 125.03 W/m²C
The heat transfer rate Q, may be determined by
Q = h x A x (T(s) - T(∞))
Q = 125.03 x 0.45 x 0.45 x (120 - 20)
Q = 2.532 x 10³ W
The heat transfer rate is 2.532 x 10³ W to the air.
Consider a pump operating adiabatically at steady state. Liquid water enters at 20◦C, 100 kPa with a mass flow rate of 53 kg/min. The pressure at the pump exit is 5 MPa. The pump isentropic efficiency is 75%. Assumer negligible changes in kinetic and potential energy and the water behaves as an incompressible substance. Determine the power required by the pump, in kW.
Answer:
5778.86W
Explanation:
Hi!
To solve this problem follow the steps below, the procedure is attached in an image
1. Draw the complete outline of the problem.
2. find the specific weights of the water and its density using T = 20C, using thermodynamic tables
note=Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
3. use the Bernoulli equation to find the height of the pump's power, taking into account that the potential and kinetic energy changes are insignificant
4. find the ideal pump power
5. find the real power of the pump using the efficiency equation
Compute the longitudinal modulus of elasticity (in GPa) for a continuous and aligned hybrid composite consisting of aramid and glass fibers in volume fractions of 0.22 and 0.30, respectively, within a polyester resin matrix, given the following data:
Modulus of Elasticity
(GPa)
Glass fibers
72.5
Aramid fibers
131
Polyester
2.5
=_______GPa?
Answer:
51.77 GPa
Explanation:
For elasticity for a continous and aligned hybrid composite
[tex]E_{ci}=E_{poly} (1-V_{A} -V_{G} )+E(V_{A})+E(V_{G} )\\=(2.5GPa)(1-0.22-0.3)+(131GPa)(0.22)+(72.5GPa)(0.3)\\=1.2+28.82+21.75\\=51.77GPa[/tex]
Consider laminar, fully developed flow in a channel of constant surface temperature Ts. For a given mass flow rate and channel length, determine which rectangular channel, b/a = 1.0, 1.43, or 2.0, will provide the highest heat transfer rate. Is this heat transfer rate greater than, equal to, or less than the heat transfer rate associated with a circular tube? answer: b/a=2
Answer:
Please see attachment.
Explanation:
A 800-MW steam power plant, which is cooled by a nearby river, has a thermal efficiency of 40 percent. Determine the rate of heat transfer to the river water. Will the actual heat transfer rate be higher or lower than this value? Why?(Round the final answer to the nearest whole number.)
Answer:
Rate of heat transfer to river=1200MW
So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes
Explanation:
In order to find the actual heat transfer rate is lower or higher than its value we will first find the rate of heat transfer to power plant:
[tex]Efficiency=\frac{work}{heat transfer to power plant}[/tex]
[tex]Heat transfer=\frac{work}{Efficiency\\} \\\\Heat transfer=\frac{800}{0.40}\\\\Heat transfer=2000MW[/tex]
From First law of thermodynamics:
Rate of heat transfer to river=heat transfer to power plant-work done
Rate of heat transfer to river=2000-800
Rate of heat transfer to river=1200MW
So the actual amount of heat rejected ti the river will be less as there will some heat loss to surrounding and in pipes.
As Jamar prepares for a face-to-face interview, he begins to brainstorm a list of stories he can use to highlight his skills and qualifications. What types of stories should he rehearse? Check all that apply. Stories that criticize his previous employers Stories that discuss how he handled a tough interpersonal situation Stories that describe his religious beliefs and marital status Stories that discuss how he dealt with a crisis Stories that illustrate how he went above and beyond expectations
Answer:
Stories that discuss how he dealt with a crisis
Stories that illustrate how he went above and beyond expectations
Stories that discuss how he handled a tough interpersonal situation
Explanation:
For effective story-telling during an interview, Jamar needs to highlight stories by discussing how he dealt with certain crisis which relates to the job and tell the panel why he opted for a particular solution. The solution should provide a long-term solution to the crisis. He also needs to illustrate stories on how he went above and beyond expectations to prove that he's creative. Moreover, it's important that he highlights how he handled a tough interpersonal situation to show how he deals with situations at a personal level.
What is the magnitude of the maximum stress that exists at the tip of an internal crack having a radius of curvature of 5.5 × 10-4 mm (2.165 × 10-5 in.) and a crack length of 5 × 10-2 mm (1.969 × 10-3 in.) when a tensile stress of 220 MPa (31910 psi) is applied?
Answer:
magnitude of the maximum stress is 3263 MPa
Explanation:
given data
radius of curvature = 5.5 × [tex]10^{-4}[/tex] mm
crack length = 5 × [tex]10^{-2}[/tex] mm
tensile stress = 220 MPa
to find out
magnitude of the maximum stress
solution
we know that magnitude of the maximum stress is express as
magnitude of the maximum stress = [tex]2\sigma_o ( \frac{\alpha }{2 \rho} )^{0.5}[/tex] ..........................1
here σo is tensile stress and α is crack length and ρ is radius of curvature
so put all value in equation 1 we get
magnitude of the maximum stress = [tex]2*220 ( \frac{5.5*10^{-2}}{2 *5*10^{-4}} )^{0.5}[/tex]
solve it we get
magnitude of the maximum stress = 3263 MPa
so magnitude of the maximum stress is 3263 MPa