Answer: (a). W = 4KJ and Q = 4.25KJ
(b). W = -2KJ and ΔPE = 2KJ
Explanation:
(a).
i. We are asked to calculate the work done during the expansion process considering gas as system.
from W = [tex]\int\limits^a_b {p} \, dV[/tex] where a = V₂ and b = V₁
so W = P(V₂-V₁)
W = (2 × 10²) (0.12 - 0.10)
W = 4 KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
since motion of the system is constrained, there is no change in both the potential and kinetic energy i.e. ΔPE = ΔKE = 0
∴ Q - W = ΔU
Q = ΔU + W
Q = 0.25 + 4
Q = 4.25 KJ
(b).
i. to calculate the work done during the expansion process considering piston as system;
W = [tex]\int\limits^a_b {(Patm - Pgas)} \, dV[/tex]where a and b represent V₂ and V₁ respectively.
W = (Patm - Pgas)(V₂ - V₁)
W = (1-2) ×10² × (0.12-0.1)
W = -2KJ
ii. We apply the energy balance to gas as system
given Q - W = ΔE
Where ΔE = ΔU + ΔKE + ΔPE
Q = 0 since the piston and cylinder walls are perfectly insulated.
for piston, we neglect the change in internal energy and kinetic energy
ΔU = ΔKE = 0
from Q - W = ΔU + ΔKE + ΔPE
0 - (-2) = 0 + 0 + ΔPE
∴ ΔPE = 2KJ
What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C
Answer:
331809.5gallon/hr or 92.16gallon/s
Explanation:
What is the peak runoff discharge for a 1.3 in/hr storm event from a 9.4-acre concrete-paved parking lot (C
convert 9.4 acre to inches we have=5.896*10^7
How to calculate Peak runoff discharge
1. take the dimension of the roof
2. multiply the dimension by the n umber of inches of rainfall
3. Divide by 231 to get gallon equivalence (because 1 gallon = 231 cubic inches)
5.896*10^7*1.3
7.66*10^7 cubic inches/hr
1 gallon=231 cubic inches
7.66*10^7 cubic inches=331809.5gallon/hr or 92.16gallon/s
this is gotten by converting 1 hr to seconds
331809.5gallon/hr /3600s=92.16gallon/s
Water is the working fluid in an ideal Rankine cycle. The condenser pressure is 8 kPa, and saturated vapor enters the turbine at (a) 18 MPa and (b) 4 MPa. The net power output of the cycle is 100 MW.
Determine for each case a) the mass flow rate of steam, in kg/h, b) the heat transfer rates for the working fluid passing through the boiler and condenser, each in kW, and c) the thermal efficiency.
Explanation:
The obtained data from water properties tables are:
Point 1 (condenser exit) @ 8 KPa, saturated fluid
[tex]h_{f} = 173.358 \\h_{fg} = 2402.522[/tex]
Point 2 (Pump exit) @ 18 MPa, saturated fluid & @ 4 MPa, saturated fluid
[tex]h_{2a} = 489.752\\h_{2b} = 313.2[/tex]
Point 3 (Boiler exit) @ 18 MPa, saturated steam & @ 4 MPa, saturated steam
[tex]h_{3a} = 2701.26 \\s_{3a} = 7.1656\\h_{3b} = 2634.14\\s_{3b} = 7.6876[/tex]
Point 4 (Turbine exit) @ 8 KPa, mixed fluid
[tex]x_{a} = 0.8608\\h_{4a} = 2241.448938\\x_{b} = 0.9291\\h_{4b} = 2405.54119[/tex]
Calculate mass flow rates
Part a) @ 18 MPa
mass flow
[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3a} - h_{4a}) - (h_{2a} - h_{f})}\\\\= \frac{100*10^ 3}{(2701.26 - 2241.448938 ) - (489.752 - 173.358)}\\\\= 697.2671076 \frac{kg}{s} = 2510161.587 \frac{kg}{hr}[/tex]
Heat transfer rate through boiler
[tex]Q_{in} = mass flow * (h_{3a} - h_{2a})\\Q_{in} = (697.2671076)*(2701.26-489.752)\\\\Q_{in} = 1542011.787 W[/tex]
Heat transfer rate through condenser
[tex]Q_{out} = mass flow * (h_{4a} - h_{f})\\Q_{out} = (697.2671076)*(2241.448938-173.358)\\\\Q_{out} = 1442011.787 W[/tex]
Thermal Efficiency
[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.06485[/tex]
Part b) @ 4 MPa
mass flow
[tex]\frac{100*10^6 }{w_{T} - w_{P}} = \frac{100*10^3 }{(h_{3b} - h_{4b}) - (h_{2b} - h_{f})}\\\\= \frac{100*10^ 3}{(2634.14 - 2405.54119 ) - (313.12 - 173.358)}\\\\= 1125 \frac{kg}{s} = 4052374.235 \frac{kg}{hr}[/tex]
Heat transfer rate through boiler
[tex]Q_{in} = mass flow * (h_{3b} - h_{2b})\\Q_{in} = (1125.65951)*(2634.14-313.12)\\\\Q_{in} = 2612678.236 W[/tex]
Heat transfer rate through condenser
[tex]Q_{out} = mass flow * (h_{4b} - h_{f})\\Q_{out} = (1125)*(2405.54119-173.358)\\\\Q_{out} = 2511206.089 W[/tex]
Thermal Efficiency
[tex]n = \frac{W_{net} }{Q_{in} } = \frac{100*10^3}{1542011.787} \\\\n = 0.038275[/tex]
For (a) 18 MPa:
- The mass flow rate of steam is approximately 238,050 kg/h.
- The heat transfer rates are approximately 674,970 kW (boiler) and 481,360 kW (condenser).
- The thermal efficiency is approximately 14.81%.
For (b) 4 MPa:
- The mass flow rate of steam is approximately 187,320 kg/h.
- The heat transfer rates are approximately 480,040 kW (boiler) and 380,450 kW (condenser).
- The thermal efficiency is approximately 20.83%.
Explanation and Calculation
To analyze the ideal Rankine cycle, we need to follow these steps for each case:
Case (a): 18 MPa
1. Determine Specific Enthalpies
- Condenser exit: Saturated liquid at 8 kPa.
From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]
-Pump exit: Compressed liquid at 18 MPa.
Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]
[tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]
[tex]\[ h_2 \approx 191.81 + 0.001 \times (18000 - 8) \approx 209.81 \, \text{kJ/kg} \][/tex]
- Boiler exit: Saturated vapor at 18 MPa.
From steam tables: [tex]\( h_3 \approx 2821.6 \, \text{kJ/kg} \)[/tex]
-Turbine exit: Expanding to 8 kPa.
From steam tables: [tex]\( h_4 \approx 2201.4 \, \text{kJ/kg} \)[/tex]
2. Mass Flow Rate of Steam
Using the power output:
[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} (2821.6 - 2201.4 - (209.81 - 191.81)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} \times 420.2 \][/tex]
[tex]\[ \dot{m} \approx 238050 \, \text{kg/h} \][/tex]
3. Heat Transfer Rates
- Boiler:
[tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]
[tex]\[ \dot{Q}_{\text{in}} = 238050 \times (2821.6 - 209.81) \approx 674.97 \times 10^3 \, \text{kW} \][/tex]
- Condenser:
[tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]
[tex]\[ \dot{Q}_{\text{out}} = 238050 \times (2201.4 - 191.81) \approx 481.36 \times 10^3 \, \text{kW} \][/tex]
4. Thermal Efficiency
[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]
[tex]\[ \eta = \frac{100 \times 10^3}{674.97 \times 10^3} \approx 14.81\% \][/tex]
Case (b): 4 MPa
1. Determine Specific Enthalpies
- Condenser exit: Saturated liquid at 8 kPa.
From steam tables: [tex]\( h_1 \approx 191.81 \, \text{kJ/kg} \)[/tex]
- Pump exit: Compressed liquid at 4 MPa.
Using [tex]\( h_2 = h_1 + v_1 \Delta P \)[/tex]
[tex]\[ v_1 \approx 0.001 \, \text{m}^3/\text{kg} \][/tex]
[tex]\[ h_2 \approx 191.81 + 0.001 \times (4000 - 8) \approx 195.81 \, \text{kJ/kg} \][/tex]
- Boiler exit: Saturated vapor at 4 MPa.
From steam tables: [tex]\( h_3 \approx 2749.7 \, \text{kJ/kg} \)[/tex]
- Turbine exit: Expanding to 8 kPa.
From steam tables: [tex]\( h_4 \approx 2222.0 \, \text{kJ/kg} \)[/tex]
2. Mass Flow Rate of Steam
Using the power output:
[tex]\[ \dot{W}_{\text{net}} = \dot{m} (h_3 - h_4 - (h_2 - h_1)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} (2749.7 - 2222.0 - (195.81 - 191.81)) \][/tex]
[tex]\[ 100 \times 10^6 = \dot{m} \times 531.9 \][/tex]
[tex]\[ \dot{m} \approx 187320 \, \text{kg/h} \][/tex]
3. Heat Transfer Rates
- Boiler:
[tex]\[ \dot{Q}_{\text{in}} = \dot{m} (h_3 - h_2) \][/tex]
[tex]\[ \dot{Q}_{\text{in}} = 187320 \times (2749.7 - 195.81) \approx 480.04 \times 10^3 \, \text{kW} \][/tex]
- Condenser:
[tex]\[ \dot{Q}_{\text{out}} = \dot{m} (h_4 - h_1) \][/tex]
[tex]\[ \dot{Q}_{\text{out}} = 187320 \times (2222.0 - 191.81) \approx 380.45 \times 10^3 \, \text{kW} \][/tex]
4. Thermal Efficiency
[tex]\[ \eta = \frac{\dot{W}_{\text{net}}}{\dot{Q}_{\text{in}}} \][/tex]
[tex]\[ \eta = \frac{100 \times 10^3}{480.04 \times 10^3} \approx 20.83\% \][/tex]
A six- lane freeway ( three lanes in each direction) in a scenic area has a measured free- flow speed of 55 mi/ h. The peak- hour factor is 0.80, and there are 8% large trucks and buses and 6% recreational vehicles in the traffic stream. One upgrade is 5% and 0.5 mi long. An analyst has determined that the freeway is operating at capacity on this upgrade during the peak hour. If the peak- hour traffic volume is 3900 vehicles, what value for the driver population factor was used?
Answer:
0.867
Explanation:
The driver population factor ([tex]f_{p}[/tex])can be estimated using the equation below:
[tex]f_{p} = \frac{V}{PHF*N*f_{HV}*v_{p}}[/tex]
The value of the heavy vehicle factor ([tex]f_{HV}[/tex]) is determined below:
The values of the [tex]E_{T}[/tex] = 2 and [tex]E_{R}[/tex] = 3 are gotten from the tables for the RVs, trucks and buses upgrades for passenger-car equivalents. Therefore:
[tex]f_{HV}[/tex] = 1/[1+0.08(2-1)+0.06(3-1)] = 1/[1+0.08+0.12] = 1/1.2 = 0.833
Furthermore, the vp is taken as 2250 pc/(h*In) from the table of LOS criteria for lane freeway using the 15 minutes flow rate. Therefore:
[tex]f_{p}[/tex] = 3900/[0.8*3*0.833*2250] = 3900/4498.2 = 0.867
the net work output and the thermal efficiency for the Carnot and the simple ideal Rankine cycles with steam as the working fluid are to be calculated and compared. Steam enters the turbine in both cases at 5 MPa as a saturated vapor and the condenser pressure is 50 kPa. in the Rankine cycle, the condenser exit state is saturated liquid and in the Carnot cycle, the boiler inlet state is saturated liquid. Draw the T-s diagrams for both cycles.
Answer:
a) Rankine
Net work output = 719.1 KJ/kg
Thermal Eff = 0.294
a) Carnot
Net work output = 563.2 KJ/kg
Thermal Eff = 0.294
For T-s diagrams see attachments
Explanation:
Part a Rankine Cycle
The obtained data from water property tables:
[tex]P_{L,sat liquid} = 50 KPa \\v_{1} = 0.00103m^3/kg\\\\ h_{1} = 340.54KJ/kg\\\\P_{H} = 5000KPa\\h_{2} = h_{1} + v_{1} *(P_{H} - P_{L} )\\h_{2} =350.54 + (0.00103)*(5000 - 50)\\\\h_{2} = 345.64KJ/kg\\\\P_{H,satsteam} = 5000KPa\\s_{3} = 5.9737KJ/kgK\\\\h_{3} = 2794.2KJ/kg\\\\s_{3} = s_{4} = 5.9739KJ/kgK\\P_{L}= 50KPa\\\\h_{4}= 2070KJ/kg\\\\[/tex]
Heat transferred from boiler
[tex]q_{b} = h_{3}-h_{2}\\q_{b}=2794.2-345.64\\\\q_{b} =2448.56KJ/kg\\\\[/tex]
Heat transferred from condenser
[tex]q_{c} = h_{4}-h_{1}\\q_{b}=2070-340.54\\\\q_{b} =1729.46KJ/kg\\\\[/tex]
Thermal Efficiency
[tex]u_{R} = 1- \frac{q_{c}}{q_{b}}\\\\u_{R} = 1 - \frac{1729.46}{2448.56}\\\\u_{R} =0.294[/tex]
Net work output
[tex]w_{R} = q_{b}-q_{c}\\w_{R} = 2448.56-1729.46\\\\w_{R}=719.1KJ/kg[/tex]
Part b Carnot Cycle
The obtained data from water property tables:
[tex]P_{H,sat-steam} = 5000KPa\\T_{3} = 263.94 C\\s_{3} = 5.9737KJ/kgK\\\\h_{3} = 2794.2KJ/kg\\\\T_{2,sat-liquid} = T_{3} = 263.94C\\s_{2} = 2.920KJ/kgK\\\\h_{2} = 1150KJ/kg\\\\P_{L} = 50KPa\\s_{1}=s_{2} = 2.920KJ/kgK\\\\h_{1} = 989KJ/kg\\\\s_{3} = s_{4} = 5.9737KJ/kgK\\P_{L} = 50KPa\\\\h_{4} = 2070KJ/kg[/tex]
Heat transferred from boiler
[tex]q_{b} = h_{3}-h_{2}\\q_{b}=2794.2-1150\\\\q_{b} =1644.2KJ/kg\\\\[/tex]
Heat transferred from condenser
[tex]q_{c} = h_{4}-h_{1}\\q_{b}=2070-989\\\\q_{b} =1081KJ/kg\\\\[/tex]
Thermal Efficiency
[tex]u_{C} = 1- \frac{q_{c}}{q_{b}}\\\\u_{C} = 1 - \frac{1081}{1644.2}\\\\u_{C} =0.343[/tex]
Net work output
[tex]w_{C} = q_{b}-q_{c}\\w_{C} = 1644.2-1081\\\\w_{C}=563.2KJ/kg[/tex]
Given a Pane object appPane and a TextField object nameField, write a statement that adds nameField to the pane.
Answer:
appPane.getChildren().add(nameField);
Explanation:
The question is related to JavaFx which is a set of packages used for making interactive graphical user interface that contains graphical components for better user experience.
Pane is a JavaFx component which is used for adjusting the position and size of a graphical component for a given scene.
We can create a pane object by appPane = new Pane(); command.
TextField is a JavaFx object that is used for displaying a line of text. We have various functions available to set properties of TextField object.
We can create a TextField object by nameField = new TextField(); command.
A pane can have multiple set of graphical components for various parts of an application. These are called children.
Hence we can write the following statement to add a TextField object nameField to a Pane object appPane.
appPane.getChildren().add(nameField);
You are provided with a fuel cell that is designed to operate at j = 3 A∕cm2 and P = 1.5 W∕cm2. How much fuel cell active area (in cm2) is required to deliver 2 kW of electrical power? (This is approximately enough to provide power to the average American home.)
(a) 296.3 cm2
(b) 1333.3 cm2
(c) 444.4cm2
(d) 666.6 cm2
To provide 2 kW of electrical power, the fuel cell requires an active area of approximately 1333.3 cm2, calculated by dividing the total power required by the fuel cell's power density.
Explanation:The question involves calculating the area of a fuel cell required to deliver a specific power output, given its operational parameters. This problem is grounded in the principles of electrochemical engineering and requires an understanding of how power density influences the design and performance of fuel cells. Given that the fuel cell operates at a current density (j) of 3 A/cm2 and a power density (P) of 1.5 W/cm2, we are tasked with determining the active area needed to supply 2 kW of electrical power.
The formula to calculate the area is straightforward:
Start by identifying the total power output required, which is 2 kW or 2000 W.Use the provided power density of the fuel cell, 1.5 W/cm2.Calculate the area by dividing the total power by the power density: Area = 2000 W / 1.5 W/cm2 = 1333.3 cm2.Therefore, to deliver 2 kW of electrical power, a fuel cell with an active area of approximately 1333.3 cm2 is required.
Because assembly language is so close in nature to machine language, it is referred to as a ____________.
low-level language
Answer:
symbolic machine code.
Explanation:
The instructions in the language are closely linked to the machine's architecture.
An automotive Battery is rated at120 A-h. This means that under certain test conditions, it canoutput 1 A at 12 V for 120 hours.
(a) How much total energy is storedin battery?
(b) If the headlights are left on overnight (8h),how much energy will still be stored in the battery in the morning(Assume a 150 w total power rating for both headlightstogether)
Answer:
Part A:
In W-h:
Energy Stored=1440 W-h
In Joules:
[tex]Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ[/tex]
Part B:
In W-h:
Energy left=240 W-h
In Joules:
Energy left= 8.64*10^5 J
Explanation:
Part A:
We are given rating 120A-h and voltage 12 V
Energy Stored= Rating*Voltage (Gives us units W-h)
Energy Stored=120A-h*12V
Energy Stored=1440 W-h
Converting it into joules (watt=joules/sec)
Energy Stored=[tex]1440 Joules * \frac{3600sec}{h}h[/tex]
Energy Stored=5184000 Joules
[tex]Energy Stored=5.184*10^6Joules\\Energy Stored=5.184 MJ[/tex]
Part B:
Energy used by lights for 8h=150*8
Energy used by lights for 8h=1200W-h
Energy left= Energy Stored(Calculated above)- Energy used by lights for 8h
Energy left=1440-1200
Energy left=240 W-h
Energy left=[tex]240 Joules * \frac{3600sec}{h}h[/tex]
Energy left=864000 Joules
Energy left= 8.64*10^5 J
The total energy stored in the battery is 1440 Wh. If the headlights are left on overnight (8h), the energy remaining in the battery in the morning is 1320 Wh.
Explanation:(a) The total energy stored in the battery can be calculated by multiplying the battery's capacity (120 A-h) by the voltage (12 V). So, the total energy stored in the battery is 1440 Wh.
(b) To calculate the energy consumed by the headlights in 8 hours, we need to find the power (P) using the formula P = V × I, where V is the voltage (12 V) and I is the current (the total power rating divided by the voltage). Then, we can find the energy consumed by multiplying the power by the time (8 hours). We can subtract this energy consumption from the total energy stored in the battery to find the energy remaining in the morning. So, the energy remaining in the battery is 1440 Wh - 12 kWh = 1320 Wh.
Learn more about Energy stored in an automotive battery here:https://brainly.com/question/34222184
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The displacement of an oscillating mass is 0.004*sin(7t) meters. What is the peak amplitude of its velocity in meters/second?
Answer:
The peak amplitude velocity is found to be 0.028 m/s.
Explanation:
Given that the displacement of an oscillating mass is:
Displacement = x = 0.004 Sin(7t)
Now, to find out the velocity of this particle, me have to take derivative of 'x' with respect to 't'.
Velocity = V = dx/dt = (7)(0.004)Cos (7t)
V = 0.028 Cos (7t)
Now, for the maximum displacement, the value of the Cos function must be maximum. And we know that the maximum value of Cos function is 1. Thus, to get maximum displacement, we set the value of Cos (7t) to be equal to 1.
Vmax = 0.028(1)
Vmax = 0.028 m/s (Peak Amplitude of Velocity)
Assess the capabilities of a hydroelectric power plant from the following field data: Estimated water flow rate, 40 m3/s River inlet at 1 atm, 10 oC Discharge at 1 atm, 10.2 oC, 200 m below the intake?
Answer:
This hydroelectric power plant has the capability of producing around 78.4 MW of electric energy.
Explanation:
First, we calculate the pressure the water exerts or gains when it travels 200 m below the intake. W have the data:
Density of water = ρ = 1000 kg/m^3
Acceleration due to gravity = g = 9.8 m/s²
Height lost = h = 200 m
Volume flow rate = 40 m^3/s
The pressure difference, is now given as:
ΔP = ρgh
ΔP = (1000 kg/m^3)(9.8 m/s²)(200 m)
ΔP = 1960000 Pa = 1.96 MPa
Now, we calculate the power capacity of this flow:
Power = ΔP(Volume flow rate)
Power = (1960000 N/m²)(40 m^3/s)
Power = 78400000 Watt
Power = 78.4 MW
Thus, this plant can produce at max 78.4 MW. The actual power produced will be slightly less than this due of the losses and efficiency of turbine system used.
Previously , you wrote a program named Hurricane that classified hurricanes into five categories using the Saffir-Simpson Hurricane Scale. Now, create a modified version named Hurricane Modularized that passes a user’s input wind speed to a method that returns the hurricane category.
Answer:
Answer of the Java class explained below with appropriate comments
Explanation:
using System;
class MainClass {
public static void Main (string[] args) {
//entering wind speed of the hurricane
Console.WriteLine ("Enter wind speed in mph: ");
int windSpeed = Convert.ToInt32(Console.ReadLine());
//checking for the category with the credited wind speed
int category = GetCategory(windSpeed);
Console.WriteLine ("Category: " + category);
}
public static int GetCategory(int wind) {
//function for classifying the wind speeds
if(wind >= 157) {
return 5;
}
if(wind >= 130) {
return 4;
}
if(wind >= 111) {
return 3;
}
if(wind >= 96) {
return 2 ;
}
return 1;
}
}
The motion of a particle is defined by the relation x = t3 – 6t2 + 9t + 3, where x and t are expressed in feet and seconds, respectively. Determine When the velocity is zero The position, acceleration and total distance traveled when t= 5 sec
The displacement of a particle when its velocity is zero can be found by differentiating the displacement function to get the velocity function, then determining when this velocity is zero. By substituting the corresponding time back into the displacement equation, we calculate the displacement at that time. Additionally, for acceleration, we can integrate to find velocity and displacement with given initial conditions.
Explanation:Finding the Displacement When Velocity is ZeroThe question involves determining the displacement of a particle when its velocity is zero. Since velocity is the derivative of displacement with respect to time, we first need to find the velocity function by differentiating the displacement function x = t3
- 6t2 + 9t + 3. Once we have the velocity function, we can locate the time t at which the velocity is zero. The displacement at this time will be our required value. Similarly, if given an acceleration function, a(t) = t - 1, and initial conditions for velocity and position, we integrate the acceleration to find the velocity, and again to find the displacement.
To address the specific question of the student regarding the particle's position, acceleration, and total distance traveled at t = 5 seconds, we would substitute t into our established functions for position and acceleration accordingly.
If the initial acceleration of a particle is to be calculated from the provided differential equation dv(t)/dt = 6.0 - 3v(t), with the initial condition of the particle being at rest, we would substitute t = 0 into this equation to find the initial acceleration.
A branched molecular structure is stronger in the solid-state and more viscous in the molten state than a linear structure for the same polymer.
True or False?
Answer:
True
Explanation:
The speed of an aircraft is given to be 260 m/s in air. If the speed of sound at that location is 330 m/s, the flight of the aircraft is _____.
Answer:
[tex]Ma=\frac{260}{330} \\Ma=0.7878[/tex]
So, Ma < 1 Flow is Subsonic
Explanation:
Mach Number:
Mach Number is the ratio of speed of the object to the speed of the sound. It is used to categorize the speed of the object on the basis of mach number as sonic, supersonic and hyper sonic. (It is a unit less quantity)
Mach < 1 Subsonic
Mach > 1 Supersonic
Ma= Speed of the object/Speed of the sound
[tex]Ma=\frac{260}{330} \\Ma=0.7878[/tex]
So, Ma < 1 Flow is Subsonic
Real resistors can only be manufactured to a specific tolerance, so that in effect the value of the resistance is uncertain. For example, a 1Ω resistor specified as 5% tolerance could in practice be found to have a value anywhere in the range of 0.95 to 1.05Ω. Calculate the potential voltage range across a 2.2 kΩ 10% tolerance resistor if the current flowing through the element is 4 sin 44t mA.
Answer:
The potential voltage range across a 2.2 kΩ 10% tolerance resistor when current of 4 sin 44t mA is flowing through the element is between a range of 7.92sin44t and 9.68sin44t volts.
Explanation:
Given that there is 10% tolerance for the 2.2 kΩ resistor, this implies that the resistance would range between 2,200 — 10% of 2,200 and 2,200 + 10% of 2,200, which is:
(i) 2,200 — 10% of 2,200 = 2,200 — 220 = 1,980 Ω, and
(ii) 2,200 + 10% of 2,200 = 2,200 + 220 = 2,420 Ω
Therefore, we will calculate the potential voltage for 1,980 Ω and 2,420 Ω if the current flowing through the element is 4sin44t mA:
(a) The potential voltage for a resistance of 1,980 Ω: we will use the formula: potential voltage v = i × R
Where i = 4sin44t mA = 0.004sin44t A, and R = 1,980 Ω
The potential voltage = v = 1,980 × 0.004sin44t = 7.92sin44t (in volts)
(b) The potential voltage for a resistance of 2,420 Ω: we will use the formula: potential voltage v = i × R
Where i = 4sin44t mA = 0.004sin44t A, and R = 2,420 Ω
The potential voltage = v = 2,420 × 0.004sin44t = 9.68sin44t (in volts)
The pressure in a water line is 1500 kPa. What is the line pressure in (a) lb/ft2units and (b) lbf/in2(psi) units?
Answer:
Part A:
[tex]1500\ KPa= 31328.145 \frac{lb}{ft^2}[/tex]
Part B:
[tex]1500 KPa=217.55656 \frac{lb}{in^2}[/tex](Psi)
Explanation:
Part A:
Line Pressure is 1500 KPa
We need a conversion factor which converts KPa to lb/ft^2.
[tex]20.88543 \frac{lb}{ft^2}= 1\ KPa[/tex]
In order to convert 1500 KPa to lb/ft^2, we proceed as:
[tex]1\ KPa=20.88543 \frac{lb}{ft^2} \\1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa}\\1500\ KPa= 31328.145 \frac{lb}{ft^2}[/tex]
1500 KPa is 31328.145 lb/ft^2
Part B:
We will use the same procedure we did in Part A:
1 ft= 12 in
[tex](1\ ft)^2=(12\ in)^2\\1 ft^2=144 in^2[/tex]
Converting [tex]1500 KPa\ into\ \frac{lb}{in^2}[/tex]
[tex]1500\ KPa= 1500 KPa*20.88543 \frac{lb}{ft^2.KPa} * \frac{ft^2}{144\ in^2}[/tex]
[tex]1500 KPa=217.55656 \frac{lb}{in^2}[/tex](Psi)
1500 KPa is 217.55656 lb/in^2 (psi)
The concentration of carbon monoxide (co) in an exhaust gas is 1x10^4 ppmv. what is the concentration in mg/m^3 at 25 and 1 atm
Answer:
1.1451 x 104 (11451.13)mg/m3
Explanation:
1 ppmv is defined as one volume of a contaminant or solid(CO)(mL) in 1 x 106 volume of solvent/water.
1ppmv = 1mL/m3
Concentration in mg/m3 = volume in ppm x molecular weight x pressure(kPa)/( gas constant x temperature(K)
Molecular weight of CO = 12 + 16
= 28g/mol
Temperature = 273.15 + 25
= 298.15K
Pressure = 1 x 101.325kPa
= 101.325kPa
Ppmv = 1 x 10-4ppmv
Gas constant, R = 8.3144 L.kPa/mol.K
Concentration in mg/m3 = (1 x 104 * 28 * 101.325)/(8.3144 * 298)
= 1.1451 x 104mg/m3
= 11451.13 mg/m3
Consider as a system the gas in a vertical cylinder; the cylinder is fitted with a piston onwhich a number of small weights are placed. The initial pressure is 200 kPa, and the initial volume ofthe gas is 0.04 m^3. Let a bunsen burner be placed under the cylinder, and let thevolume of the gas increase 0.1m^3 while the pressure remainsconstant.
A. Calculate the work done by the system during this process.
Answer:
Work done = 12,000J = 12KJ
Explanation:
This process is an Isobaric one i.e a process at constant pressure
P = 200KPa = 200 x 1000 Pa = 200,000PaVi = 0.04m3Vf = 0.1m3From the formula for Workdone = Integral (PdV)
Work done = P ( Vf - Vi)
Plugging the values in the equation,
Work done = 200,000 ( 0.1 - 0.04)
Work done = 12,000J = 12KJ
Why is the contractor normally required to submit a bid bond when making a proposal to an owner on a competitively bid contract?
Answer:
It serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.
Explanation:
A bid bond is a type of construction bond that protects the obligee in a construction bidding process.
A bid bond typically involves three parties:
The obligee; the owner or developer of the construction project under bid. The principal; the bidder or proposed contractor.
The surety; the agency that issues the bid bond to the principal example insurance company or bank.
A bid bond generally serves as a guarantee that the contractor who wins the bid will honor the terms of the bid after the contract is signed.
Contractors are required to submit a bid bond to ensure they are financially able and willing to complete the project they bid on, preserving the integrity of the bidding process by preventing cost overruns, favoritism, and corruption.
Explanation:A contractor is normally required to submit a bid bond when making a proposal to an owner on a competitively bid contract to ensure that the contractor is serious and financially able to carry out the project. The bid bond is a form of guarantee that the contractor will enter into the contract at the bid price if awarded the project and provides a financial penalty if the contractor fails to do so. Municipalities and other entities require this bond to prevent situations where a contractor might win a bid and then refuse to complete the work or ask for additional funds to finish the project (“cost overruns”). This is significant in a competitive bidding environment where fairness and the avoidance of favoritism or corruption are essential, as it stops anyone from being favored over others and ensures the project will be completed at the bid price.
Tenders, which are bid documents that outline the project plan and cost, play a crucial role in the construction industry. A well-prepared tender demonstrates that the projected cost reflects a balance between safety, environmental friendliness, and profitability. In contrast, a sealed-bid auction often used in these processes, requires strategic bidding and is susceptible to collusion or bribery to uncover competitor's bid details. Hence, a bid bond safeguards the process by providing a financial deterrent against such unethical practices.
Write a program with total change amount as an integer input, and output the change using the fewest coins, one coin type per line. The coin types are Dollars, Quarters, Dimes, Nickels, and Pennies. Use singular and plural coin names as appropriate, like 1 Penny vs. 2 Pennies.
A program with total change amount as an integer input, and output the change using the fewest coins, one coin type per line is in the explanation part below.
Python program that outputs the change using the fewest coins after receiving the entire amount of change as an integer input:
def calculate_change(change_amount):
coin_types = {
"Dollar": 100,
"Quarter": 25,
"Dime": 10,
"Nickel": 5,
"Penny": 1
}
print("Change breakdown:")
for coin_name, coin_value in coin_types.items():
if change_amount >= coin_value:
num_coins = change_amount // coin_value
change_amount -= num_coins * coin_value
coin_name_plural = coin_name + "s" if num_coins > 1 else coin_name
print(f"{num_coins} {coin_name_plural}")
# Input the total change amount
total_change = int(input("Enter total change amount: "))
calculate_change(total_change)
Thus, this is the Python program asked.
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For H2O, determine the specified property at the indicated state.
(a) T = 140°C, v = 0.5 m3/kg. Find p, in bar.
(b) p = 30 MPa, T = 80°C. Find v, in m3/kg.
(c) p = 10 MPa, T = 600°C. Find v, in m3/kg.
(d) T = 80°C, x = 0.4. Find p, in bar, and v, in m3/kg.
For H2O, determine the specific volume at the indicated state, in m3/kg.
(a) T = 440°C, p = 20 MPa.
(b) T = 160°C, p = 20 MPa.
(c) T = 40°C, p = 2 MPa.
The properties of water and steam used to do work can be determined from the steam table, given input of the temperature, pressure, specific volume, as well as other thermodynamic variables
The responses to the required values of pressure and specific volume are as follows;
Part I
(a) Given T = 140°C, v = 0.5 m³/kg, to find p
From the steam tables, we have, at 140°, [tex]v_f[/tex] = 0.00107976, [tex]v_g[/tex] = 0.508519, [tex]p_s[/tex] = 3.61501
Given that the [tex]v_f[/tex] < v < [tex]v_g[/tex], the fluid has two phases and the experienced pressure [tex]p_s[/tex] = 3.61501 bar
(b) Given that at 30 MPa, and 80°C the steam is in the superheated heated region
From the single phase region of the steam tables, we have;
v = 1.01553 × 10⁻³ m³/kg
(c) Given that at p = 10 MPa = 100 bar, and T = 600°C, we have from the single phase region of the steam table
v = 3.83775 × 10⁻² m³/kg
(d) At T = 80°C, and x = 0.4, we have;
v = 0.00102904 + 0.4 × (3.40527 - 0.00102904) ≈ 1.363
v = 1.363 m³/kg
Part II
(a) At T = 440 °C, p = 20 MPa. from the single phase region of the steam table, we have;
v = 1·22459 × 10⁻² m³/kg
(b) At T = 160°C, p = 20 MPa = 200 bar, the steam is in the superheated region, and we have;
v = 1.08865 × 10⁻³ m³/kg
(c) At T = 40°C and p = 2 MPa = 20 bar, we have
v = 0.00100700 m³/kg = 1.007 × 10⁻³ m³/kg
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Write a program that adds the numbers 1 through 5 into the numbers ArrayList and then prints out the first element in the list.
Answer:
Answer code is given below along with comments to explain each step
Explanation:
Method 1:
// size of the array numbers to store 1 to 5 integers
int Size = 5;
// here we have initialized our array numbers, the type of array is integers and size is 5, right now it is empty.
ArrayList<Integer> numbers = new ArrayList<Integer>(Size);
// lets store values into the array numbers one by one
numbers.add(1)
numbers.add(2)
numbers.add(3)
numbers.add(4)
numbers.add(5)
// here we are finding the first element in the array numbers by get() function, the first element is at the 0th position
int firstElement = numbers.get(0);
// to print the first element in the array numbers
System.out.println(firstElement);
Method 2:
int Size = 5;
ArrayList<Integer> numbers = new ArrayList<Integer>(Size);
// here we are using a for loop instead of adding manually one by one to store the values from 1 to 5
for (int i = 0; i < Size; i++)
{
// adding the values 1 to 5 into array numbers
numbers.add(i);
}
//here we are finding the first element in the array numbers
int firstElement = numbers.get(0);
// to print the first element in the array
System.out.println(firstElement);
The loss of power a signal suffers as it travels from the transmitting computer to a receivingcomputer is: a.white noiseb.spikingc.attenuation d.intermodulation noisee.echo
Answer:
c.attenuation
Explanation:
In telecommunication, it is called attenuation of a signal, be it acoustic, electrical or optical, to the loss of power suffered by it when passing through any transmission medium.
Attenuation is usually not expressed as a difference in powers but in logarithmic units such as decibels, which are more comfortable to use when calculating.
The attenuation is expressed in decibels (db) and the power is measured with this formula:
α = 10 * log (P1 / P2)
where P is the power both initial and final.
Calculate the resulting power factor if a synchronous motor rated at 500 hp with 90% efficiency operating at rated load and at unity power factor is added to the plant instead of the capacitor.
Assume constant voltage (1 hp = 0.746 kW).
Answer:
The question is incomplete. Below is the complete question
"An industrial plant consisting primarily of induction motor loads absorbs 500 kW at 0.6 power factor lagging. (a) Compute the required kVA rating of a shunt capacitor to improve the power factor to 0.9 lagging. (b) Calculate the resulting power factor if a synchronous motor rated 500 hp with 90% efficiency operating at rated load and at unity power factor is added to the plant instead of the capacitor. Assume constant voltage. (1 hp = 0.746 kW)"
Answer:
a. 424.5KVA
b. 0.808 lagging.
Explanation:
Let's first determine the real power, reactive and the apparent power delivered.
Ql= Ptan(the real power angle)
Ql=500*tan53.13°
Ql=666.7 Kvar
The reactive angle is
Arccos(0.9)=25.84°
Now we calculate the reactive power
Qs=Ptan25. 84°
Qs=242.4KVAR
Hence the apparent power is
Qc=Ql-Qs
Qc=666.7-242.2
Qc=424.5KVAR
The required KVA rating of the shunt capacitor is 424.5KVA
b. To calculate the resulting power factor we first determine the power absorbs by the synchronous motor.
Pm=(500*0.746)/0.9
Pm=414.4KW
The total reactive power is
Ps=P+Pm
Ps=414.4+500=914.4KW
Hence we compute the source power factor as
PF=cos[arctan(Qs/Ps)]
PF=cos[arctan(666.7/914.4)
From careful calculation, we arrive at
PF=0.808.
Note this power factor will be lagging.
A 1000 KVA three phase transformer has a secondary voltage of 208/120. What is the secondary full load amperage?
Answer:
The three phase full load secondary amperage is 2775.7 A
Explanation:
Following data is given,
S = Apparent Power = 1000 kVA
No. of phases = 3
Secondary Voltage: 208 V/120 V (Here 208 V is three phase voltage and 120 V is single phase voltage)
Since,
[tex]V_{1ph} =\frac{ V_{3ph}}{\sqrt{3} }\\V_{1ph) = \frac{208}{\sqrt{3} }\\[/tex]
[tex]V_{1ph} = 120 V[/tex]
The formula for apparent power in three phase system is given as:
[tex]S = \sqrt{3} VI[/tex]
Where:
S = Apparent Power
V = Line Voltage
I = Line Current
In order to calculate the Current on Secondary Side, substituting values in above formula,
[tex]1000 kVA = \sqrt{3} * (208) * (I)\\1000 * 1000 = \sqrt{3} * (208) * (I)\\I = \frac{1000 * 1000}{\sqrt{3} * (208) }\\ I = 2775.7 A[/tex]
A 4.60 L vessel contains 24.5 g of PCl3 and 3.00 g of O2 at 15.0 ⁰C. The vessel is heated to 195 ⁰C, and the contents react to give POCl3.What is the final pressure in the vessel,assuming that the reaction goes to completion and that all reactants and products are in the gas phase?
Answer:
The final pressure in the vessel is 2.270atm
Explanation:
Mass of PCl3 = 24.5g, MW of PCl3 = 137.5g/mole, number of moles of PCl3 = 24.5/137.5 = 0.178mole
Mass of O2 = 3g, MW of O2 = 32g/mole, number of moles of O2 = 3/32 = 0.094moles
Total moles of rectant (n) = 0.178mole + 0.094mole = 0.272mole
From ideal gas equation
PV = nRT
P (pressure of reactants) = nRT/V
n = 0.272mole, R = 82.057cm^3.atm/gmol.K, T = 15°C = 15+273K = 288K, V = 4.6L = 4.6×1000cm^3 = 4600cm^3
P = 0.272×82.057×288/4600 = 1.397atm
From pressure law
P1/T1 = P2/T2
P2 (final pressure in the vessel) = P1T2/T1
P1 = 1.397atm, T1 = 288K, T2 = 195°C = 195+273K = 468K
P2 = 1.397×468/288 = 2.270atm
Which of the following has nothing to do with insulating glass? Group of answer choices
a. triple glazing
b. double glazing
c. low emissivity coatings
d. low conductivity
e. gas between sheets of glass
f. tempered glass
Answer:
Tempered glass
Explanation:
Tempered or toughened glass is a type of safety glass processed by controlled
thermal or chemical treatments to increase its strength compared with normal glass. Tempering puts the outer surfaces into compression and the interior into tension . Such stresses cause the glass, when broken, to crumble into small granular chunks instead of splintering into jagged shards as plate glass (a.k.a. annealed glass) does. The granular chunks are less likely to cause injury.
Answer: tempered glass
The characteristics listed in the options are the characteristics of an insulating glass, a type of glass with double or more panes used to reduce heat loss within a building.
Explanation:
A tempered glass is also known as a toughened glass. It is a type of glass that is produced through controlled thermal or chemical treatment which helps to increase its quality in terms of toughness. A tempered glass is a safety glass and as a result of this, its surface is four times toughened than that of a normal glass.
A hydraulic cylinder has a 125-mm diameter piston with hydraulic fluid inside the cylinder and an ambient pressure of 1 bar. Assuming standard gravity, find the piston mass that will create a pressure inside of 2500 kPa.
Answer: 3002.86kg
Explanation:
Hydraulic cylinder diameter =125mm
Ambient pressure =1bar
Pressure =2500kpa
Piston Mass (MP) =?
F|(when it moves upward )=PA=F|(when it moves downward) =PoA+Mpg
Po=1 bar=100kpa
A=(π/4)D^2=(π/4)*0.125^2=0.01227m^2
Mp=(P-Po) A/g=(2500-100)*1000*0.01227/9.80665
Mp=3002.86kg.
5) A 80-kg man has a total foot imprint area of 480 cm2. Determine the pressure this man exerts on the ground if (a) he stands on both feet and (b) he stands on one foot.
Answer:
The pressure exerted by this man on ground
(a) if he stands on both feet is 8.17 KPa
(b) if he stands on one foot is 16.33 KPa
Explanation:
(a)
When the man stand on both feet, the weight of his body is uniformly distributed around the foot imprint of both feet. Thus, total area in this case will be:
Area = A = 2 x 480 cm²
A = 960 cm²
A = 0.096 m²
The force exerted by man on his area will be equal to his weight.
Force = F = Weight
F = mg
F = (80 kg)(9.8 m/s²)
F = 784 N
Now, the pressure exerted by man on ground will be:
Pressure = P = F/A
P = 784 N/0.096 m²
P = 8166.67 Pa = 8.17 KPa
(b)
When the man stand on one foot, the weight of his body is uniformly distributed around the foot imprint of that foot only. Thus, total area in this case will be:
Area = A = 480 cm²
A = 0.048 m²
The force exerted by man on his area will be equal to his weight, in this case, as well.
Force = F = Weight
F = mg
F = (80 kg)(9.8 m/s²)
F = 784 N
Now, the pressure exerted by man on ground will be:
Pressure = P = F/A
P = 784 N/0.048 m²
P = 16333.33 Pa = 16.33 KPa
A wastewater treatment plant discharges 1.0 m3/s of effluent having an ultimate BOD of 40.0 mg/ L into a stream flowingat 10.0 m3/s. Just upstream from the discharge point, the stream has an ultimate BOD of 3.0 mg/L The deoxygenationconstant kd is estimated at 0.22/day.(a) Assuming complete and instantaneous mixing. find the ultimate BOD of the mixture of waste and river just downstreamfrom the outfall.(b) Assuming a constant cross-sectional area for the stream equal to 55 m2, what ultimate BOD would you expect to find ata point 10.000 m downstream?
Answer:
a) 6.4 mg/l
b) 5.6 mg/l
Explanation:
Given data:
effluent Discharge Q_w = 1.0 m^3.s
Ultimate BOD L_w = 40 mg/l
Discharge of stream Q_r = 10 m^3.s
Stream ultimate BOD L_r = 3 mg/l
a) Ultimate BOD of mixture[tex] = \frac{Q_w l_w + Q_r L_r}{Q_w + Q_r}[/tex]
[tex] = \frac{1*40 + 10*3}{10 +1} = 6.4 mg/l[/tex]
b) utlimate BOD at 10,000 m downstream
[tex]t =\frac{distance}{speed} = \frac{10,000}{\frac{Q_r +Q+w}{55}} \times \frac{hr}{3600} \times \frac{day}{24 hr}[/tex]
putting [tex]Q_r + Q_w = 1+ 10 = 11 m^3/s[/tex]
t = 0.578 days
we know
[tex]L_t = L_o e^{-kt}[/tex]
[tex]L_t = 6.4 \times e^{-0.22 \times 0.578}[/tex]
[tex]L_t = 5.6 mg/l[/tex]