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]
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
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
. Write a MATLAB program that calculates the arithmetic mean, the geometric mean, and the root-mean-square average for a given set of values. Your program must use 3 functions to calculate the 3 required means. The output should be formatted as follows:
Your name Statistical Package arithmetic mean = x.xxxxx geometric mean = x.xxxxx RMS average = x.xxxxx
Test your program on the following values: 1.1, 3.3, 3.00, 2.22, 2.00, 2.72, 4.00, 4.62 and 5.37. Your main program calls 3 functions.
The data should be read by the main program from a text file, stored in an array and then passed to the functions.
The results and any other output should be printed by the main program. Notice how the output results are aligned. Also notice that the results are printed accurate to 5 decimal places.
DO NOT USE MATLAB BUILT-IN FUNCTIONS.
Answer:
Mean.m
fid = fopen('input.txt');
Array = fscanf(fid, '%g');
Amean = AM(Array);
Gmean = GM(Array);
Rmean = RMS(Array);
display('Amlan Das Statistical Package')
display(['arithmetic mean = ', sprintf('%.5f',Amean)]);
disp(['geometric mean = ', sprintf('%.5f',Gmean)]);
disp(['RMS average = ', sprintf('%.5f',Rmean)]);
AM.m
function [ AMean ] = AM( Array )
n = length(Array);
AMean = 0;
for i = 1:n
AMean = AMean + Array(i);
end
AMean = AMean/n;
end
GM.m
function [ GMean ] = GM(Array)
n = length(Array);
GMean = 1;
for i = 1:n
GMean = GMean*Array(i);
end
GMean = GMean^(1/n);
end
RMS.m
function [ RMean ] = RMS( Array)
n = length(Array);
RMean = 0;
for i = 1:n
RMean = RMean + Array(i)^2;
end
RMean = (RMean/n)^(0.5);
end
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]
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.
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
3. (20 points) Suppose we wish to search a linked list of length n, where each element contains a key k along with a hash value h(k). Each key is a long character string. How might we take advantage of the hash values when searching the list for an element with a given key?
Answer:
Alternatively we produce a complex (hash) value for key which mean that "to obtain a numerical value for every single string" that we are looking for. Then compare that values along the range of list, that turns out be numerical values so that comparison becomes faster.
Explanation:
Every individual key is a big character so to compare every keys, it is required to conduct a quite time consuming string reference procedure at every node. Alternatively we produce a complex (hash) value for key which mean that "to obtain a numerical value for every single string" that we are looking for. Then compare that values along the range of list, that turns out be numerical values so that comparison becomes faster.
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.
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]
A small circular plate has a diameter of 2 cm and can be approximated as a blackbody. To determine the radiation from the plate, a radiometer is placed normal to the direction of viewing from the plate at a distance of 50 cm. If the radiometer measured an irradiation of 129 W/m2 from the plate, determine the temperature of the plate. Take the Stefan-Boltzmann constant as σ = 5.67 x 10-8 W/m2·K4.The temperature of the plate is______________.
Answer: Temperature T = 218.399K.
Explanation:
Q= 129W/m2
σ = 5.67 EXP -8W/m2K4
Q= σ x t^4
t = (Q/σ)^0.25
t= 218.399k
A rapid sand filter has a loading rate of 8.00 m/h, surface dimensions of 10 m ´ 8 m, an effective filtration rate of 7.70 m/h. A complete filter cycle duration is 52 h and the filter is rinsed for 20 minutes at the start of each cycle. a. What flow rate (m3 /s) does the filter handle during production? b. What volume of water is used for backwashing plus rinsing the filter in each filter cycle?
Answer:
Explanation:
given data
loading rate = 8.00 m/h
filtration rate = 7.70 m/h
dimensions = 10 m × 8 m
filter cycle duration = 52 h
time = 20 min
to find out
flow rate and volume of water is used for back washing plus rinsing the filter
solution
we consider here production efficiency is 96%
so here flow rate will be
flow rate = area × rate of filtration
flow rate = 10 × 8 × 7.7
flow rate = 616 m³/h
and
we know back washing generally 3 to 5 % of total volume of water per cycle so
volume of water is = 616 × 52
volume of water is 32032 m³
and
volume of water of back washing is = 4% of 32032
volume of water of back washing is 1281.2 m³
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).
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.
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.
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");
}
}
}
A pipe leads from a storage tank on the roof of a building to the ground floor. The absolute pressure of the water in the storage tank where it connects to the pipe is 3.0×105Pa, the pipe has a radius of 1.0 cm where it connects to the storage tank, and the speed of flow in this pipe is 1.6 m/s. The pipe on the ground floor has a radius of 0.50 cm and is 9.0 m below the storage tank. Find the speed of flow in the pipe on the ground floor.
Answer:
6.4 m/s
Explanation:
From the equation of continuity
A1V1=A2V2 where A1 and V1 are area and velocity of inlet respectively while A2 and V2 are the area and velocity of outlet respectively
[tex]A1=\pi (r1)^{2}[/tex]
[tex]A2=\pi (r2)^{2}[/tex]
where r1 and r2 are radius of inlet and outlet respectively
v1 is given as 1.6 m/s
Therefore
[tex]\pi (0.01)^{2}\times 1.6 = \pi (0.005)^{2}v2[/tex]
[tex]V2=\frac {\pi (0.01)^{2}\times 1.6}{\pi (0.005)^{2}}=6.4 m/s[/tex]
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
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.
Problem 5) Water is pumped through a 60 m long, 0.3 m diameter pipe from a lower reservoir to a higher reservoir whose surface is 10 m above the lower one. The sum of the minor loss coefficient for the system is kL = 14.5. When the pump adds 40 kW to the water, the flow rate is 0.2 m3/s. Determine the pipe roughness.
Answer:
\epsilon = 0.028*0.3 = 0.0084
Explanation:
\frac{P_1}{\rho} + \frac{v_1^2}{2g} +z_1 +h_p - h_l =\frac{P_2}{\rho} + \frac{v_2^2}{2g} +z_2
where P_1 = P_2 = 0
V1 AND V2 =0
Z1 =0
h_P = \frac{w_p}{\rho Q}
=\frac{40}{9.8*10^3*0.2} = 20.4 m
20.4 - (f [\frac{l}{d}] +kl) \frac{v_1^2}{2g} = 10
we know thaTV =\frac{Q}{A}
V = \frac{0.2}{\pi \frac{0.3^2}{4}} =2.82 m/sec
20.4 - (f \frac{60}{0.3} +14.5) \frac{2.82^2}{2*9.81} = 10
f = 0.0560
Re =\frac{\rho v D}{\mu}
Re =\frac{10^2*2.82*0.3}{1.12*10^{-3}} =7.53*10^5
fro Re = 7.53*10^5 and f = 0.0560
\frac{\epsilon}{D] = 0.028
\epsilon = 0.028*0.3 = 0.0084
A 15 ft high vertical wall retains an overconsolidated soil where OCR-1.5, c'-: O, ф , --33°, and 1 1 5.0 lb/ft3.
Determine the magnitude and location of the thrust on the wall, assuming that the soil is at rest.
Answer:
magnitude of thrust uis 11061.65 lb/ft
location is 5 ft from bottom
Explanation:
Given data:
Height of vertical wall is 15 ft
OCR is 1.5
[tex]\phi = 33^o[/tex]
saturated uit weight[tex] \gamma_{sat} = 115.0 lb/ft^3[/tex]
coeeficent of earth pressure [tex]K_o[/tex]
[tex]K_o = 1 -sin \phi[/tex]
= 1 - sin 33 = 0.455
for over consolidate
[tex]K_{con} = K_o \times OCR[/tex]
[tex] = 0.455 \times 1.5 = 0.683[/tex]
Pressure at bottom of wall is
[tex]P =K_{con} \times (\gamma_{sat} - \gamma_{w}) + \gamma_w \times H[/tex]
[tex]= 0.683 \times (115 - 62.4) \times 15 + 62.4 \times 15[/tex]
P = 1474.88 lb/ft^3
Magnitude pf thrust is
[tex]F= \frac{1}{2} PH[/tex]
[tex]=\frac{1}{2} 1474.88\times 15 = 11061.65 lb/ft[/tex]
the location must H/3 from bottom so
[tex]x = \frac{15}{3} = 5 ft[/tex]
Ammonia enters an adiabatic compressor operating at steady state as saturated vapor at 300 kPa and exits at 1400 kPa, 140◦C. Kinetic and potential energy effects are negligible. Determine:
a. power input required [kJ/kg]
b. isentropic compressor efficiency
c. rate of entropy production per unit mass [kJ/kg K] in the compressor
Answer:
a. 149.74 KJ/KG
b. 97.9%
c. 0.81 kJ/kg K
Explanation:
Which of these strategies can maximize insulation levels?
Use of vacuum-insulation panels
A. Adding rigid insulation outside of a stud wall to reduce thermal bridging
B. Increasing insulation thickness
C. Using spray-foam instead of batt insulation
D. Any of the above
Answer:
D. Any of the above
Explanation:
All of the following strategies maximize insulation levels:
Use of vacuum-insulation panels Adding rigid insulation outside of a stud wall to reduce thermal bridging Increasing insulation thickness Using spray-foam instead of batt insulationAn 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]
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
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
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.
// 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
For the unity negative feedback system G(s) = K(s+6)/ (s + 1)(s + 2)(s + 5) It's known that the system is operating with a dominant-pole damping ratio of 0.5. Design a PD controller so that the settling time is reduced by a factor of 3, compared with the uncompensated unity negative feedback system. Compare the percentage overshoot and resonant frequencies of the uncompensated and compensated systems.
Answer:The awnser is 5
Explanation:Just divide all of it
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
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.