Answer:
7.15
Explanation:
Firstly, the COP of such heat pump must be measured that is,
[tex]COP_{HP}=\frac{T_H}{T_H-T_L}[/tex]
Therefore, the temperature relationship, [tex]T_H=1.15\;T_L[/tex]
Then, we should apply the values in the COP.
[tex]=\frac{1.15\;T_L}{1.15-1}[/tex]
[tex]=7.67[/tex]
The number of heat rejected by the heat pump must then be calculated.
[tex]Q_H=COP_{HP}\times W_{nst}[/tex]
[tex]=7.67\times5=38.35[/tex]
We must then calculate the refrigerant mass flow rate.
[tex]m=0.264\;kg/s[/tex]
[tex]q_H=\frac{Q_H}{m}[/tex]
[tex]=\frac{38.35}{0.264}=145.27[/tex]
The [tex]h_g[/tex] value is 145.27 and therefore the hot reservoir temperature is 64° C.
The pressure at 64 ° C is thus 1849.36 kPa by interpolation.
And, the lowest reservoir temperature must be calculated.
[tex]T_L=\frac{T_H}{1.15}[/tex]
[tex]=\frac{64+273}{1.15}=293.04[/tex]
[tex]=19.89\°C[/tex]
the lowest reservoir temperature = 258.703 kpa
So, the pressure ratio should be = 7.15
Two Electric field vectors E1 and E2 are perpendicular to each other; obtain its base
vectors.
Answer:
<E1, E2>.
Explanation:
So, in the question above we are given that the Two Electric field vectors E1 and E2 are perpendicular to each other. Thus, we are going to have the i and the j components for the two Electric Field that is E1 and E2 respectively. That is to say the addition we give us a resultant E which is an arbitrary vector;
E = |E| cos θi + |E| sin θj. -------------------(1).
Therefore, if we make use of the components division rule we will have something like what we have below;
x = |E2|/ |E| cos θ and y = |E1|/|E| sin θ
Therefore, we will now have;
E = x |E2| i + y |E1| j.
The base vectors is then Given as <E1, E2>.
Sludge wasting rate (Qw) from the solids residence time (Thetac = mcrt) calculation. Given the following information from the previous problem. The total design flow is 15,000 m3/day. Theoretical hydraulic detention time (Theta) = 8 hours. The NPDES limit is 25 mg/L BOD/30 mg/L TSS.
Assume that the waste strength is 170 mg/L BOD after primary clarification.
XA=MLSS = 2200 mg/L,
Xw = Xu = XR = 6,600 mg/L,
qc = 8 days.
Make sure you account for the solids in the discharge.
What volume of sludge (Qw=m3/day) is wasted each day from the secondary clarifiers?
Answer:
The volume of sludge wasted each day from the secondary classifiers is Qw = 208.33 m^3 / day
Explanation:
Check the file attached for a complete solution.
The volume of the aeration tank was first calculated, V = 5000 m^3 / day.
The value of V was consequently substituted into the formula for the wasted sludge flow. The value of the wasted sludge flow was calculated to be Qw = 208.33 m^3 / day.
Determine the drag on a small circular disk of 0.02-ft diameter moving 0.01 ft/s through oil with a specific gravity of 0.89 and a viscosity 10000 times that of water. The disk is oriented normal to the upstream velocity. By what percent is the drag reduced if the disk is oriented parallel to the flow?
Answer:
33.3%
Explanation:
Given that:
specific gravity (SG) = 0.89
Diameter (D) = 0.01 ft/s
Density of oil [tex]\rho= SG\rho _{h20} = 0.89 * 1.94=1.7266\frac{sl}{ft^3}[/tex]
Since the viscosity 10000 times that of water, The reynold number [tex]R_E=\frac{\rho VD}{\mu} =\frac{1.7266*0.01*0.01}{0.234}=7.38*10^{-4}[/tex]
Since RE < 1, the drag coefficient for normal flow is given as: [tex]C_{D1}=\frac{24.4}{R_E}= \frac{20.4}{7.38*10^{-4}}=2.76*10^4[/tex]
the drag coefficient for parallel flow is given as: [tex]C_{D2}=\frac{13.6}{R_E}= \frac{13.6}{7.38*10^{-4}}=1.84*10^4[/tex]
Percent reduced = [tex]\frac{D_1-D_2}{D_2} *100=\frac{2.76-1.84}{3.3}=33.3[/tex] = 33.3%
A steady green traffic light means
Answer:
Its C. you may proceed, but only if the path is clear
Explanation:
I just gave Quiz and its correct
Create an abstract class DiscountPolicy. It should have a single abstract method computeDiscount that will return the discount for the purchase of a given number of a single item. The method has two parameters, count and itemCost. 2. Derive a class BulkDiscount from DiscountPolicy, as described in the previous exercise. It should have a constructor that has two parameters, minimum and percent. It should define the method computeDiscount so that if the quantity purchased of an item is more than minimum, the discount is percent percent. 3. Derive a class BuyNItemsGetOneFree from DiscountPolicy, as described in Exercise 1. The class should have a constructor that has a single parameter n. In addition, the class should define the method computeDiscount so that every nth item is free. For example, the following table gives the discount for the purchase of various counts of an item that costs $10, when n is 3: count 1 2 3 4 5 6 7 Discount 0 0 10 10 10 20 20
4. Derive a class CombinedDiscount from DiscountPolicy, as described in Exercise 1. It should have a constructor that has two parameters of type DiscountPolicy. It should define the method computeDiscount to return the maximum value returned by computeDiscount for each of its two private discount policies. The two discount policies are described in Exercises 2 and 3. 5. Define DiscountPolicy as an interface instead of the abstract class described in Exercise 1.
Answer:
Java Code was used to define classes in the abstract discount policy,The bulk discount, The buy items get one free and the combined discount
Explanation:
Solution
Code:
Main.java
public class Main {
public static void main(String[] args) {
BulkDiscount bd=new BulkDiscount(10,5);
BuyNItemsGetOneFree bnd=new BuyNItemsGetOneFree(5);
CombinedDiscount cd=new CombinedDiscount(bd,bnd);
System.out.println("Bulk Discount :"+bd.computeDiscount(20, 20));
System.out.println("Nth item discount :"+bnd.computeDiscount(20, 20));
System.out.println("Combined discount :"+cd.computeDiscount(20, 20));
}
}
discountPolicy.java
public abstract class DiscountPolicy
{
public abstract double computeDiscount(int count, double itemCost);
}
BulkDiscount.java
public class BulkDiscount extends DiscountPolicy
{
private double percent;
private double minimum;
public BulkDiscount(int minimum, double percent)
{
this.minimum = minimum;
this.percent = percent;
}
at Override
public double computeDiscount(int count, double itemCost)
{
if (count >= minimum)
{
return (percent/100)*(count*itemCost); //discount is total price * percentage discount
}
return 0;
}
}
BuyNItemsGetOneFree.java
public class BuyNItemsGetOneFree extends DiscountPolicy
{
private int itemNumberForFree;
public BuyNItemsGetOneFree(int n)
{
itemNumberForFree = n;
}
at Override
public double computeDiscount(int count, double itemCost)
{
if(count > itemNumberForFree)
return (count/itemNumberForFree)*itemCost;
else
return 0;
}
}
CombinedDiscount.java
public class CombinedDiscount extends DiscountPolicy
{
private DiscountPolicy first, second;
public CombinedDiscount(DiscountPolicy firstDiscount, DiscountPolicy secondDiscount)
{
first = firstDiscount;
second = secondDiscount;
}
at Override
public double computeDiscount(int count, double itemCost)
{
double firstDiscount=first.computeDiscount(count, itemCost);
double secondDiscount=second.computeDiscount(count, itemCost);
if(firstDiscount>secondDiscount){
return firstDiscount;
}else{
return secondDiscount;
}
}
}
A refrigerator uses refrigerant-134a as the working fluid and operates on the ideal vapor-compression refrigeration cycle except for the compression process. The refrigerant enters the evaporator at 120 kPa with a quality of 34 percent and leaves the compressor at 70°C. If the compressor consumes 450 W of power, determine (a) the mass flow rate of the refrigerant, (b) the condenser pressure, and (c) the COP of the refrigerator
Answer:
(a) 0.0064 kg/s
(b) 800 KPa
(c) 2.03
Explanation:
The ideal vapor compression cycle consists of following processes:
Process 1-2 Isentropic compression in a compressor
Process 2-3 Constant-pressure heat rejection in a condenser
Process 3-4 Throttling in an expansion device
Process 4-1 Constant-pressure heat absorption in an evaporator
For state 4 (while entering compressor):
x₄ = 34% = 0.34
P₄ = 120 KPa
from saturated table:
h₄ = hf + x hfg = 22.4 KJ/kg + (0.34)(214.52 KJ/kg)
h₄ = 95.34 KJ/kg
For State 1 (Entering Compressor):
h₁ = hg at 120 KPa
h₁ = 236.99 KJ/kg
s₁ = sg at 120 KPa = 0.94789 KJ/kg.k
For State 3 (Entering Expansion Valve)
Since 3 - 4 is an isenthalpic process.
Therefore,
h₃ = h₄ = 95.34 KJ/kg
Since this state lies at liquid side of saturation line, therefore, h₃ must be hf. Hence from saturation table we find the pressure by interpolation.
P₃ = 800 KPa
For State 2 (Leaving Compressor)
Since, process 2-3 is at constant pressure. Therefore,
P₂ = P₃ = 800 KPa
T₂ = 70°C (given)
Saturation temperature at 800 KPa is 31.31°C, which is less than T₂. Thus, this is super heated state. From super heated property table:
h₂ = 306.9 KJ/kg
(a)
Compressor Power = m(h₂ - h₁)
where,
m = mass flow rate of refrigerant.
m = Compressor Power/(h₂ - h₁)
m = (0.450 KJ/s)/(306.9 KJ/kg - 236.99 KJ/kg)
m = 0.0064 kg/s
(b)
Condenser Pressure = P₂ = P₃ = 800 KPa
(c)
The COP of ideal vapor compression cycle is given as:
COP = (h₁ - h₄)/(h₂ - h₁)
COP = (236.99 - 95.34)/(306.9 - 236.99)
COP = 2.03
The Ph diagram is attached
How does a car batteray NOT die?
Answer:
bye hooking plugs up to it to amp it up
is sampled at a rate of to produce the sampled vector and then quantized. Assume, as usual, the minimum voltage of the dynamic range is represented by all zeros and the maximum value with all ones. The numbers should increase in binary order from bottom to top. Find the bit combination used to store each sample when rounded to the nearest integer between and (clipping may occur). Note: A partially-correct answer will not be recognized. You must answer all three correctly on the same
Answer:
d[0] = 11111111
d[1] = 11011101
d[2] = 1111011
Explanation:
Assume that the number of bits is 8. The voltage range input is -8 to 7 volts. The range is thus 15V, and the resolution is 15/2^8 = 0.0586 volts. We will first add +8 to the input to convert it to a 0-15v signal. Then find the equivalent bit representation. For 7.8 volts, the binary signal will be all 1's, since the max input voltage for the ADC is 7 volts. For 4.95, we have 4.95+8 = 12.95 volts. Thus, N = 12.95/0.0586 = 221. The binary representation is 11011101. For -0.8, we have -0.8 + 8 = 7.2. Thus, N = 7.2/0.0586 = 123. The binary representation is 1111011.
Thus,
d[0] = 11111111
d[1] = 11011101
d[2] = 1111011
For what type of metal is high speed steel drill best suited?
Answer:
high speed steel I believe
An air conditioning unit is used to provide cooling during summer for a house. If the air conditioner provides 450 kW cooling by using 150 kW electrical power, determine the coefficient of performance (COP) of the air conditioner. The outside temperature and inside temperature are 40 and 20°C, respectively. Using the inequality of Clausius determine if the cycle is possible. Determine the COP of an air conditioner working based on the Carnot cycle between the same temperature difference. Compare the COPs of the Carnot and actual air conditioners and comment on them based on your answer for the previous part (the inequality of
Answer:
See explaination
Explanation:
Please kindly check attachment for the step by step solution of the given problem.
what is the Economic
g In the above water treatment facility, chemical concentration (mg/gal) within the tank can be considered uniform. The initial chemical concentration inside the tank was 0 mg/gal, the concentration of effluent coming in is 10 mg/gal. The volume of the tank is 10,000 gallons. The fluid coming in rate is equal to fluid going out is equal to 50 gal/min. Establish a dynamic model of how the concentration of the chemical inside the tank increases over time.
Answer:
0.05 mg / gallon
Explanation:
mass of chemecila coming in per minute = 50*10 = 500 mg/min
at a time t min , M = mass of chemical = 500*t mg
conecntartion of chemecal = 500t/10000 = 0.05 mg / gallon
(a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2 (b) Show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs. (c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs
Answer:
Explanation:
a) Show how two 2-to-1 multiplexers (with no added gates) could be connected to form a 3-to-1 MUX. Input selection should be as follows: If AB = 00, select I0 If AB = 01, select I1 If AB = 1− (B is a don’t-care), select I2
We are to show how Two-2-to -1 multiplexers could be connected to form 3-to-1 MUX
If AB = 00 select [tex]I_o[/tex]
If AB = 01 select [tex]I_1[/tex]
If AB = 1_(B is don't care), select [tex]I_2[/tex]
However, the truth table is attached and shown in the first file below.
Also, the free- body diagram for 2- to - 1 MUX is shown in the second diagram attached below.
b) We are show how two 4-to-1 and one 2-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
The perfect illustration showing how they are connected in displayed in the third free-body diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Show how four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs.
For four 2-to-1 and one 4-to-1 multiplexers could be connected to form an 8-to-1 MUX with three control inputs, we have a perfect illustration of the diagram in the last( which is the fourth) diagram attached below.
Where ; [tex]I_o , I_1, I_2, I_3, I_4, I_5, I_6, I_7[/tex] are the inputs of the multiplexer and Z is the output
This question is a multiplexer which is a topic in digital circuit.
Multiplexer is a type of combination circuit that consist of a maximum of [tex]2^n[/tex] data inputs 'n' selection lines and single output line. One of these data inputs will be connected to the output based on the values of selection lines. Another name for multiplexers is MUX.
If we have 'n' selection lines, we will get [tex]2^n[/tex] possible combinations zero and ones. Each combination will select a maximum of only one data input.
a)
Two 2-to-1 multiplexers to form a 3-to-1 MUX.
If AB = 00, select [tex]I_o[/tex]
If AB = 01, select [tex]I_1[/tex]
If AB = 1- (B is don't care) select I
The truth table for the above scenario is in the attached document below.
Figure 1 and 2 represents the solution to this question.
b).
Two 4-to-1 multiplexers and one 2-to-1 multiplexers and one 2-to-1 multiplexers are used to form an 8-to-1 MUX.
In the attached diagram, figure 3 shows a comprehensive detail of how it is structured.
Where [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output.
c) Four 2-to-1 multiplexers and one 4-to -1 multiplexer are used to form 8-to-1 MUX.
In the attached diagram, figure 4 shows how it is structured.
We would see that [tex]I_o[/tex] to [tex]I_7[/tex] are the inputs of the multiplexer and Z is the output in the system.
Learn more about multiplexers here;
https://brainly.com/question/25953942
Air at 100°F, 1 atm, and 10% relative humidity enters an evaporative cooler operating at steady state. The volumetric flow rate of the incoming air is 1765 ft3/min. Liquid water at 68°F enters the cooler and fully evaporates. Moist air exits the cooler at 70°F, 1 atm. There is no significant heat transfer between the device and its surroundings and kinetic and potential energy effects can be neglected. Determine the mass flow rate at which liquid enters, in lb(water)/min.
Answer:
Check the explanation
Explanation:
Kindly check the attached image below to see the step by step explanation to the question above.
Find a negative feedback controller with at least two tunable gains that (1) results in zero steady state error when the input is a unit step (1/s). (and show why it works); (2) Gives a settling time of 4 seconds; (3) has 10% overshoot. Use the standard 2nd order approximation. Plot the step response of the system and compare the standard approximation with the plot.
Answer:
Gc(s) = [tex]\frac{0.1s + 0.28727}{s}[/tex]
Explanation:
comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.
attached is the detailed solution and the plot in Matlab
