Answer: No, not in favor of the idea.
This is because, an increase in the refrigerator's work output is equal to work input. In real live application, the work input of the refrigerator is always greater than the additional work output. This will result to reducing the efficiency of the power plant.
Refrigerating cooling water before the condenser in a power plant is impractical due to high energy costs, added complexity, and inefficiencies, outweighing the potential efficiency gains.
While the idea of refrigerating the cooling water before it enters the condenser in a power plant to increase efficiency might seem theoretically sound, in practice it is not advisable due to several reasons:
1.Energy Cost of Refrigeration: Refrigeration itself is an energy-intensive process. The energy required to cool the water to a significantly lower temperature would likely be greater than the energy savings achieved by the increased efficiency of the heat engine. This means that the net energy gain would be negative.
2.Second Law of Thermodynamics: According to the second law of thermodynamics, any energy conversion process, including refrigeration, involves irreversibilities and losses. The additional step of refrigeration would introduce more inefficiencies and would not be 100% efficient. Thus, the overall efficiency of the system would decrease.
3.Complexity and Cost: Adding a refrigeration system to cool the condenser water would significantly increase the complexity and cost of the power plant. This includes the initial capital cost, operational costs, and maintenance costs. The benefits gained from a marginal increase in thermal efficiency would likely not justify these additional expenses.
4. Practical Alternatives: There are more practical and cost-effective ways to improve the efficiency of a power plant. These include optimizing the thermodynamic cycle, improving the insulation, using better quality fuels, and upgrading to more efficient machinery and technology. Improving the heat exchange process itself, rather than cooling the water, is generally more practical.
5.Environmental Considerations: The energy used for refrigeration would typically come from the power plant itself or the grid, potentially leading to increased fuel consumption and higher greenhouse gas emissions. This is counterproductive to the goal of improving efficiency and sustainability.
In summary, while lowering the temperature at which heat is rejected can theoretically improve the thermal efficiency of a heat engine, refrigerating the cooling water is not a practical or economical solution due to the high energy cost, increased complexity, and overall inefficiency of the refrigeration process itself. Instead, other methods to enhance efficiency should be explored.
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]
How many of the following statements are correct regarding the buckling of slender members?
(i) Buckling occurs in axially loaded members in tension;
(ii) Buckling is caused by the lateral deflection of the members;
(iii) Buckling is an instability phenomenon.
Answer:
False, True , True.
Explanation:
Buckling is defined as large lateral deflection of member of under compression with slight increase in load beyond a critical load.
1) Buckling occurs in axially loaded member in tension. is false as buckling occurs in tension.
2) Buckling is caused by lateral deflection of the member and not the axial deflection. hence true.
3) Buckling is an instability phenomenon that can lead to failure. Hence True.
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 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]
A joining process in which a filler metal is melted and distributed by capillary action between faying surfaces, the base metals does not melt, and in which the filler metal melts at a temperature less than 450ᵉC is called?a.An arc welding processb. A soldering processc. An adhesive joining processd. A brazing processe. None of the above
Answer:
A soldering process
Explanation:
Given that ,The filler metal's melting point temperature is less than 450 ° C.
Usually, the brazing material has the liquid temperature of the melting point, the full melting point of the filler material approaching 450 degrees centigrade, while the filler material is less than 450 degrees centigrade in the case of soldering.
Therefore the answer is "A soldering process".
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 film of paint 3 mm thick is applied to a surface inclined above the horizontal at an angle \theta. The paint has rheological properties of a Bingham plastic with yield stress 15 Pa and \mu[infinity]= 60 cP. With a specific gravity of 1.3, at what angle \ theta will the paint start to flow down the surface?
Answer:
[tex]\theta = 23.083 degree[/tex]
Explanation:
Given data:
yield stress [tex]\tau_y = 15 Pa[/tex]
thickness t = 3 mm
[tex]\mu = 60 cP = 60\times 10^{-2} P[/tex]
G= 1.3
[tex]\rho_{point} = G \times \rho_{water}[/tex]
[tex]=1.3 \times 1000 = 1300 kg/m^3[/tex]
[tex]\tau = \tau_y + mu \frac{du}{dy}[/tex]
for point flow [tex]\frac{du}{dy} = 0[/tex]
[tex]\tau = \tau_y = 15 N/m^2[/tex]
[tex]\tau = \frac{force}{area}[/tex]
[tex]= \frac{weight of point . sin\theta}{area} = \frac{\rho_{point} . volume. sin\theta}{area}[/tex][/tex]
[tex]15 = \frac{1300 \times 9.8 \times A\times t \times sin\theta}{A}[/tex]
solving for theta value
[tex]sin\theta = 0.392[/tex]
[tex]\theta =sin^{-1} 0.392[/tex]
[tex]\theta = 23.083 degree[/tex]
22.5-4. Minimum Liquid Flow in a Packed Tower. The gas stream from a chemical reactor contains 25 mol % ammonia and the rest are inert gases. The total flow is 181.4 kg mol/h to an absorption tower at 303 K and 1.013 × 105 Pa pressure, where water containing 0.005 mol frac ammonia is the scrubbing liquid. The outlet gas concentration is to be 2.0 mol % ammonia. What is the minimum flow L min ′ Using 1.5 times the minimum, plot the equilibrium and operating lines. Ans. L min ′ = 262.6kg mol/h
Answer:
check attachments for answers
Explanation:
. 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
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:
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
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.
A window-mounted air conditioner supplies 29 m3/min of air at 15°C, 1 bar to a room. Air returns from the room to the evaporator of the unit at 22°C. The air conditioner operates at steady state on a vapor-compression refrigeration cycle with Refrigerant 22 entering the compressor at 4 bar, 10°C. Saturated liquid refrigerant at 14 bar leaves the condenser. The compressor has an isentropic efficiency of 70%, and refrigerant exits the compressor at 14 bar.
Determine the compressor power, in KW, the refrigeration capacity, in tons, and the coefficient of performance.
Answer:
Please see explanation below .
Explanation:
Since you have not provided the db schema, I am providing the queries by guessing the table and column names. Feel free to reach out in case of any concerns.
1. SELECT ISBN, TITLE, PUBLICATION_YEAR FROM BOOKS WHERE AUTHOR_ID = (SELECT AUTHOR_ID FROM AUTHOR WHERE AUTHOR_NAME = 'C.J. Cherryh');
2. SELECT COUNT(*) FROM BOOKS;
3. SELECT COUNT(*) FROM ORDERS WHERE CUSTOMER_ID = 11 AND order_filled = 'No';
The article "Some Parameters of the Population Biology of Spotted Flounder (Ciutharus linguatula Linnaeus, 1758) in Edremit Bay (North Aegean Sea)" (D. Turker, B. Bayhan, etal., Turkish Journal of Veterinary and Animal Science, 2005:1013-1018) reports that a sample of 482 female spotted flounder had an average weight of 20.95 g with a standard deviation of 14.5 g, and a sample of 614 male spotted flounder had an average weight of 22.79 g with a standard deviation of 15.6 g. Can you conclude that the mearn weight of male spotted flounder is greater than that of females?
Based on the higher male average weight and large sample sizes, we expect a relatively large positive t-statistic and a statistically significant p-value (low p-value).
Formulate the hypothesis:
Null hypothesis (H0): μm = μf (The mean weight of male spotted flounder (μm) is equal to the mean weight of female spotted flounder (μf))
Alternative hypothesis (H1): μm > μf (The mean weight of male spotted flounder is greater than the mean weight of female spotted flounder) (One-tailed test)
Calculate the pooled variance (assuming unequal variances):
Sp^2 = [(nf - 1) * σf^2 + (nm - 1) * σm^2] / (nf + nm - 2)
nf = Female sample size (482)
nm = Male sample size (614)
σf = Female standard deviation (14.5 g)
σm = Male standard deviation (15.6 g)
3. Calculate the t-statistic:
t = (xm - xf) / sqrt(Sp^2 * (1/nf + 1/nm))
xm = Male average weight (22.79 g)
xf = Female average weight (20.95 g)
Sp^2 (pooled variance) - calculated in step 2
4. Determine the p-value:
You'll need statistical software to find the p-value associated with the calculated t-statistic and the degrees of freedom (df = nf + nm - 2).
Statistical Measures:
t-statistic: This measures the difference between the means relative to the standard error of the sampling distribution. A positive value suggests the male mean is higher.
p-value: This represents the probability of observing a t-statistic as extreme or more extreme than the one calculated, assuming the null hypothesis is true.
A statistically significant p-value (typically less than 0.05) indicates we can reject the null hypothesis and conclude a difference between the means.
Based on the higher male average weight and large sample sizes, we expect a relatively large positive t-statistic and a statistically significant p-value (low p-value).
This would allow us to reject the null hypothesis (equal means) and conclude that male spotted flounder, on average, have a greater weight than females based on this data sample.
Given a dictionary d and a list lst, remove all elements from the dictionary whose key is an element of lst. For example, given the dictionary {1:2, 3:4, 5:6, 7:8} and the list [1, 7], the resulting dictionary would be {3:4, 5:6}. Assume every element of the list is a key in the dictionary.
This function remove_elements takes a dictionary d and a list lst as input and removes all key-value pairs from d where the key is an element of lst is written below
Writing the program in Python
The program can be written by iterating through the list and removing the corresponding key-value pairs from the dictionary.
The python function that removes all required elements from the dictionary is as follows
def remove_elements(d, lst):
for key in lst:
if key in d:
del d[key]
return d
# Example usage
d = {1: 2, 3: 4, 5: 6, 7: 8}
lst = [1, 7]
result = remove_elements(d, lst)
print(result)
Question
Given a dictionary d and a list lst, remove all elements from the dictionary whose key is an element of lst.
For example, given the dictionary {1:2, 3:4, 5:6, 7:8} and the list [1, 7], the resulting dictionary would be {3:4, 5:6}.
Assume every element of the list is a key in the dictionary.
Write the program in Python
You are using a Geiger counter to measure the activity of a radioactive substance over the course of several minutes. If the reading of 400. counts has diminished to 100. counts after 66.7 minutes , what is the half-life of this substance? Express your answer with the appropriate units.
Answer: 33.35 minutes
Explanation:
A(t) = A(o) *(.5)^[t/(t1/2)]....equ1
Where
A(t) = geiger count after time t = 100
A(o) = initial geiger count = 400
(t1/2) = the half life of decay
t = time between geiger count = 66.7 minutes
Sub into equ 1
100=400(.5)^[66.7/(t1/2)
Equ becomes
.25= (.5)^[66.7/(t1/2)]
Take log of both sides
Log 0.25 = [66.7/(t1/2)] * log 0.5
66.7/(t1/2) = 2
(t1/2) = (66.7/2 ) = 33.35 minutes
A pressure gage and a manometer are connected to a compressed air tank to measure its pressure. If the reading on the pressure gage is 11 kPa, determine the distance h between the two fluid levels of the water filled manometer. Assume the density of water is 1000 kg/m3 and the atmospheric pressure is 101 kPa.Quiz 3
Answer:
h=1.122652m
Explanation:
Assuming density of air = 1.2kg/m³
the differential pressure is given by:
[tex]h^{i} =h(\frac{density of manometer}{density of flowing air}-1)\\h^{i} =h(\frac{1000}{1.2}-1)\\ h^{i}=832.33h...(1)\\\\but\\ h^{i} =\frac{change in pressure}{air density*g} \\\\h^{i} =\frac{11*10^3}{1.2*9.81}\\\\h^{i}= 934.42...(2)\\\\equating, \\\\934.42=832.33h\\\\h=1.122652m[/tex]
The distance h between the two fluid levels of the water-filled manometer is approximately 1.121 meters.
To determine the distance h between the two fluid levels of the water-filled manometer, we need to use the pressure readings from both the pressure gauge and the manometer. Here's how we can approach the problem:
Given:
- Reading on the pressure gauge: [tex]\( P_{gauge} = 11 \, \text{kPa} \)[/tex]- Atmospheric pressure: [tex]\( P_{atm} = 101 \, \text{kPa} \)[/tex]
- Density of water: [tex]\( \rho = 1000 \, \text{kg/m}^3 \)[/tex]
- Acceleration due to gravity: [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
Step-by-Step Solution:
1. Convert the gauge pressure to absolute pressure:
[tex]\[ P_{absolute} = P_{gauge} + P_{atm} \][/tex]
[tex]\[ P_{absolute} = 11 \, \text{kPa} + 101 \, \text{kPa} = 112 \, \text{kPa} \][/tex]
2. Determine the pressure difference between the air tank and atmospheric pressure:
The pressure difference [tex](\( \Delta P \))[/tex] that the manometer measures is equal to the gauge pressure:
[tex]\[ \Delta P = P_{absolute} - P_{atm} = 112 \, \text{kPa} - 101 \, \text{kPa} = 11 \, \text{kPa} \][/tex]
3. Use the pressure difference to find the height ( h ):
The pressure difference is related to the height ( h ) of the water column by the hydrostatic equation:
[tex]\[ \Delta P = \rho g h \][/tex]
Solving for ( h ):
[tex]\[ h = \frac{\Delta P}{\rho g} \][/tex]
Convert [tex]\(\Delta P\)[/tex] to Pascals [tex](since \(1 \, \text{kPa} = 1000 \, \text{Pa}\))[/tex]:
[tex]\[ \Delta P = 11 \, \text{kPa} = 11000 \, \text{Pa} \][/tex]
Now, calculate ( h ):
[tex]\[ h = \frac{11000 \, \text{Pa}}{1000 \, \text{kg/m}^3 \times 9.81 \, \text{m/s}^2} \][/tex]
[tex]\[ h \approx \frac{11000}{9810} \, \text{m} \][/tex]
[tex]\[ h \approx 1.121 \, \text{m} \][/tex]
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
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).
Write a program that uses the function isPalindrome given below. Test your program on the following strings: madam, abba, 22, 67876, 444244, trymeuemyrt. Modify the function isPalindrome given so that when determining whether a string is a palindrome, cases are ignored, that is, uppercase and lowercase letters are considered the same. bool isPalindrome(string str)
{
int length = str.length();
for (int i = 0; i < length / 2; i++)
{
if (str[i] != str[length - 1 - i] )
return false;
}
return true;
}
The revised isPalindrome function now considers uppercase and lowercase letters as the same, correctly identifying palindromes regardless of case. It first converts the input string to all lowercase and then checks for palindrome properties.
Explanation:Understanding Palindromes in Programming
A palindrome is a sequence of characters that reads the same forwards and backwards. The provided isPalindrome function is designed to check if a given string is a palindrome. However, the function needs to be modified to disregard the case of the letters, allowing 'Madam' and 'madam' to both be recognized as palindromes.
To achieve this, we can modify the function by converting the input string to all lowercase or uppercase before the comparison. Here's the revised version of the function:
bool isPalindrome(string str) {Now, when this modified function is tested with the strings 'madam', 'abba', '22', '67876', '444244', and 'trymeuemyrt', it will correctly identify the palindromes regardless of case.
Palindromes, isPalindrome function modification for case-insensitivity, Examples of palindromes.
Palindromes are sequences of letters or words that read the same forwards and backwards. They can be found in various contexts, including DNA sequences. The objective of the given program is to determine if a given string is a palindrome, which remains the same when read backward.
The modified function isPalindrome checks if a string is a palindrome while ignoring case differences, treating uppercase and lowercase letters as the same. This modification allows the function to correctly identify palindromes regardless of letter casing.
Examples such as 'madam', 'abba', 'trymeuemyrt' correspond to palindromes while considering the modified function's case-insensitive behavior, showcasing the effectiveness of the updated algorithm
// 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 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
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.
A 20 mm 3 20 mm silicon chip is mounted such thatthe edges are flush in a substrate. The substrate provides anunheated starting length of 20 mm that acts as turbulator. Airflow at 25°C (1 atm) with a velocity of 25 m/s is used to coolthe upper surface of the chip. If the maximum surface temperature of the chip cannot exceed 75°C, determine the maximumallowable power dissipation on the chip surface.
Answer:
q= 1.77 W
Explanation:
Given data, Dimensions = 20mm x 20mm, Unheated starting length (ε) = 20 mm, Air flow temperature = 25 C, Velocity of flow = 25m/s, Surface Temperature = 75 C
To find maximum allowed power dissipated use the formula Q = hA(ΔT), where Q = Maximum allowed power dissipated from surface , h = heat transfer coefficiant, A = Area of Surface, ΔT = Temperature difference between surface and surrounding
we need to find h, which is given by (Nu x K)/x, where Nu is the Nusselt's Number, k is the thermal conductivity at film temperature and x is the length of the substrate.
For Nu use the Churchill and Ozoe relation used for parallel flow over the flat plate , Nu = (Nux (ε=0))/[1-(ε/x)^3/4]^1/3
Nux (ε=0) = 0.453 x Re^0.5 x Pr^1/3, where Re is the Reynolds number calculated by [Rex = Vx/v, where V is the velocity and v is the kinematic velocity, x being the length of the substrate] and Pr being the Prandtl Number
The constants, namel Pr, k and v are temperature dependent, so we need to find the film temperature
Tfilm = (Tsubstrate + Tmax)/2 = (25 + 75)/2 = 50 C
At 50 C, Pr = 0.7228, k = 0.02735w/m.k , v = 1.798 x 10^-5 m2/s
First find Rex and keep using the value in the subsequent formulas until we reach Q
Rex = Vx/v = 25 x (0.02 + 0.02)/ 1.798 x 10^-5 = 55617.35 < 5 x 10^5, thus flow is laminar, (x = L + ε)
Nux = (Nux (ε=0))/[1-(ε/x)^3/4]^1/3 = 0.453 x Re^0.5 x Pr^1/3/[1-(ε/x)^3/4]^1/3
= 0.453 x 55617.35 ^0.5 x 0.7228^1/3/ [1 - (0.02/0.04)^3/4]^1/3 = 129.54
h = (Nu x K)/x = (129.54 x 0.02735)/0.04 = 88.75W/m2.k
Q = 88.75 x (0.02)^2 x (75-25) = 1.77 W
A large-particle composite consisting of tungsten particles within a copper matrix is to be prepared. If the volume fractions of tungsten and copper are 0.90 and 0.10, respectively, estimate the upper limit for the specific stiffness of this composite given the data that follow. Specific Gravity Modulus of Elasticity (GPa)
Copper 8.9 110
Tungsten 19.3 407
The upper limit for the specific stiffness of this composite is approximately 20.208 GPa.
Specific stiffness of composite = Volume fraction of copper * Specific stiffness of copper + Volume fraction of tungsten * Specific stiffness of tungsten
Specific stiffness of composite = [tex](0.10 * (110 / 8.9)) + (0.90 * (407 / 19.3))[/tex]
Now, calculate the specific stiffness of the composite.
Specific stiffness of composite = [tex](0.10 * (110 / 8.9)) + (0.90 * (407 / 19.3))[/tex]
Specific stiffness of composite ≈[tex](0.10 * 12.36) + (0.90 * 21.08)[/tex]
Specific stiffness of composite ≈ [tex]1.236 + 18.972[/tex]
Specific stiffness of composite ≈ 20.208 GPa
So, the upper limit for the specific stiffness of this composite is approximately 20.208 GPa.
Write a program that stores lists of names (the last name first) and ages in parallel arrays and sorts the names into alphabetical order keeping the ages with the correct names. The original arrays of names and ages should remain no changes. Therefore, you need to create an array of character pointers to store the addresses of the names in the name array initially. Apply the selection sort to this array of pointers so that the corresponding names are in alphabetical order.
Answer:
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#define STRSIZ 30
#define MAXAPP 50
int alpha_first(char *list[], int min_sub, int max_sub);
void select_sort_str(char *list[], int * numbers[], int n);
int main()
{
char applicants[MAXAPP][STRSIZ];
int ages[MAXAPP];
char* alpha[MAXAPP];
int* numeric[MAXAPP];
int num_app, i, j;
char one_char;
printf("Enter number of people (0 . . %d)\n> ", MAXAPP);
scanf("%d", &num_app);
for (i = 0; i < num_app; i++)
{
scanf("%c", &one_char);
printf("\nEnter name %d (lastname, firstname): ", i + 1);
j = 0;
while ((one_char = (char)getchar()) != '\n' && j < STRSIZ)
applicants[i][j++] = one_char;
applicants[i][j] = '\0';
printf("Enter age %d: ", (i + 1));
scanf("%d", &ages[i]);
alpha[i] = (char*)malloc(strlen(applicants[i]) * sizeof(char));
strcpy(alpha[i], applicants[i]);
numeric[i] = &ages[i];
}
printf("\n\nOriginal list");
printf("\n---------------------------\n");
for (i = 0; i < num_app; ++i)
printf("%-30s%2d\n", applicants[i], ages[i]);
select_sort_str(alpha, numeric, num_app);
printf("\n\nAlphabetized list");
printf("\n---------------------------\n");
for (i = 0; i < num_app; ++i)
printf("%-30s%2d\n", alpha[i], *numeric[i]);
printf("\n\nOriginal list");
printf("\n---------------------------\n");
for (i = 0; i < num_app; ++i)
printf("%-30s%2d\n", applicants[i], ages[i]);
return 0;
}
int alpha_first(char* list[], int min_sub, int max_sub)
{
int first, i;
first = min_sub;
for (i = min_sub + 1; i <= max_sub; ++i)
if (strcmp(list[i], list[first]) < 0)
first = i;
return (first);
}
void select_sort_str(char* list[], int* numbers[], int n)
{
int fill, index_of_min;
char* temp;
int* tempNum;
for (fill = 0; fill < n - 1; ++fill)
{
index_of_min = alpha_first(list, fill, n - 1);
if (index_of_min != fill)
{
temp = (char*)malloc(strlen(list[fill]) * sizeof(char));
strcpy(temp, list[index_of_min]);
strcpy(list[index_of_min], list[fill]);
strcpy(list[fill], temp);
tempNum = numbers[index_of_min];
numbers[index_of_min] = numbers[fill];
numbers[fill] = tempNum;
}
}
}
Write a program that counts the number of bits equal to one in R1, and stores the result in R2. For example, if R1=x0A0C, then R2=x0004 after the program runs. To receive credit, you can only modify the contents of R1 and R2, and no other register, and your solution should have no more than 9 lines of code.
Answer:
MOV R1, $0A0C ;
MOV R2, #0 ;
L1 : MOV R1, R1, LSL #1 ;
ADC R2, R2, #1 ;
CMP R1, #0 ;
BNE L1 ;
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 insulation3. (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.
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