Answer:
The mass flow rate is 0.27 kg/s
The exit velocity is 76.1 m/s
The exit pressure is 695 KPa
Explanation:
Assuming the flow to be steady state and the behavior of air as an ideal gas.
The mass flow rate of the air is given as:
Mass Flow Rate = ρ x A1 x V1
where,
ρ = density of air
A1 = inlet area = 3.8 cm² = 3.8 x 10^-4 m²
V1 = inlet velocity = 100 m/s
For density using general gas equation:
PV = nRT
PV = (m/M)RT
PM/RT = ρ
ρ = (680000 N/m²)(0.02897 kg/mol)/(8.314 J/mol.k)(60 + 273)k
ρ = 7.11 kg/m³
Therefore,
Mass Flow Rate = (7.11 kg/m³)(3.8 x 10^-4 m²)(100 m/s)
Mass Flow Rate = 0.27 kg/s = 270 g/s
Now, for steady flow, the mass flow rate remains constant throughout the flow. Hence, flow rate at inlet will be equal to the flow rate at outlet:
Mass Flow Rate = ρ x A2 x V2
where,
ρ = density of air = 7.11 kg/m³ (Assuming in-compressible flow)
A2 = exit area = 5 cm² = 5 x 10^-4 m²
V2 = exit velocity = ?
Therefore:
0.27 kg/s = (7.11 kg/m³)(5 x 10^-4 m²) V2
V2 = 76.1 m/s
Now, for exit pressure, we use Bernoulli's equation between inlet and exit, using subscript 1 for inlet and 2 for exit:
P1 + (1/2) ρ V1² + ρ g h1 = P2 + (1/2) ρ V2² + ρ g h2
Since, both inlet and exit are at same temperature.
Therefore, h1 = h2, and those terms will cancel out.
P1 + (1/2) ρ V1² = P2 + (1/2) ρ V2²
P2 = P1 + (1/2) ρ V1² - (1/2) ρ V2²
P2 = P1 + (1/2) ρ (V1² - V2²)
P2 = 680000 Pa + (0.5)(7.11 kg/m³)[(100m/s)² - (76.1 m/s)²]
P2 = 680000 Pa + 14962.25 Pa
P2 = 694962.25 Pa = 695 KPa
A liquid phase reaction, A → B, is to be carried out in an isothermal, well mixed batch reactor with a volume of 1L. Initially there are 6 moles of A. The rate of destruction of A is given by –rA =k1CA/ (1+k2CA), where k1=4, k2 =5. The unit of time in the rate constants is hours. Calculate the time, in hours, that the reaction must proceed in the reactor in order to result in 3 moles of A remaining in the reactor.
Answer:
the time, in hours = 4.07hrs
Explanation:
The detailed step by step derivation and appropriate integration is as shown in the attached files.
A lagoon is designed to accommodate an input flow of 0.10 m^3/s of nonconservative pollutant with concentration 30 mg/L and decay coefficient of 0.2/day. The effluent from the lagoon must have pollutant concentration of less than 10 mg/L. How large is the lagoon (assume complete mixing)?
Answer:
Volume of the lagoon required for the decay process must be larger than 86580 m³ = 8.658 × 10⁷ L
Explanation:
The lagoon can be modelled as a Mixed flow reactor.
From the value of the decay constant (0.2/day), one can deduce that the decay reaction of the pollutant is a first order reaction.
The performance equation of a Mixed flow reactor is given from the material and component balance thus:
(V/F₀) = (C₀ - C)/((C₀)(-r)) (From the Chemical Reaction Engineering textbook, authored by Prof. Octave Levenspiel)
V = volume of the reactor (The lagoon) = ?
C₀ = Initial concentration of the reactant (the pollutant concentration) = 30 mg/L = 0.03 mg/m³
F₀ = Initial flow rate of reactant in mg/s = 0.10 m³/s × C₀ = 0.1 m³/s × 0.03 mg/m³ = 0.003 mg/s
C = concentration of reactant at any time; effluent concentration < 10mg/L, this means the maximum concentration of pollutant allowed in the effluent is 10 mg/L
For the sake of easy calculation, C = the maximum value = 10 mg/L = 0.01 mg/m³
(-r) = kC (Since we know this decay process is a first order reaction)
This makes the performance equation to be:
(kVC₀/F₀) = (C₀ - C)/C
V = F₀(C₀ - C)/(kC₀C)
k = 0.2/day = 0.2/(24 × 3600s) = 2.31 × 10⁻⁶/s
V = 0.003(0.03 - 0.01)/(2.31 × 10⁻⁶ × 0.03 × 0.01)
V = 86580 m³
Since this calculation is made for the maximum concentration of 10mg/L of pollutant in the effluent, the volume obtained is the minimum volume of reactor (lagoon) to ensure a maximum volume of 10 mg/L of pollutant is contained in the effluent.
The lower the concentration required for the pollutant in the effluent, the larger the volume of reactor (lagoon) required for this decay reaction. (Provided all the other parameters stay the same)
Hope this helps!
The volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".
lagoon relation:The volume for the lagoon relation:
[tex]\to Q_1C_1 = Q_2C_2+KC_2V\\\\[/tex]
[tex]\to Q_1=0.10\\\\\to C_1=30\\\\\to Q_2=0.2\\\\\to C_2=10\\\\\to K=0.10[/tex]
Putting the value into the formula and calculating the volume:
[tex]\to (0.10 \times 30)=(0.2\times 10)+(0.10\times 10\times V \times (\frac{1}{24 \ hrs}) \times (\frac{1 \ hr}{3,600 \ sec})) \\\\\to (3)=(2)+(1\times V \times \frac{1}{86400}) \\\\\to 3-2=(1\times V \times \frac{1}{86400}) \\\\\to 1=(1\times V \times \frac{1}{86400}) \\\\\to 1 \times 86400 = V\\\\\to V= 8.64 \times 10^4 \ m^3\\\\[/tex]
Therefore, the calculated volume is "[tex]8.64 \times 10^4 \ m^3[/tex]".
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c.Task 3: an application program where the user enters the price of an item and the program computes shipping costs. If the item price is $100 or more, then shipping is free otherwise it is 2% of the price. The program should output the shipping cost and the total price.
Answer:
//Program is written in C++ language
// Comments explains difficult lines
#include<iostream>
using namespace std;
int main (){
//Variable declaration
double price, shipping;
cout<<"Item Price: ";
cin>>price;
//Test price for shipping fee
shipping = 0; //Initialise shipping fee to 0 (free)
if(price <100)
{
shipping = 0.2 * price;
}
//The above if statement tests if item price is less than 100.
//If yes, then the shipping fee is calculated as 20% of the item price
//Else, (if item price is at least 100), shipping fee is 0, which means free
if(shipping != 0)
{
cout<<"Shipping fee is "<<shipping;
}
else
{
cout<<"Item will be shipped for free";
}
return 0;
}
Consider a large plane wall of thickness L = 0.3 m, thermal conductivity k = 2.5 W/m · °C, and surface area A =12 m2. The left side of the wall at x = 0 is subjected to a net heat flux of q0 = 700 W/m2 while the temperature at that surface is measured to be T1 =80°C. Assuming constant thermal conductivity and no heat generation in the wall, a. Express the differential equation and the boundary conditions for steady onedimensional heat conduction through the wall. b. Obtain a relation for the variation of temperature in the wall by solving the differential equation. c. Evaluate the temperature of the right surface of the wall at x = L.
Answer:
a) -k* dT / dx = q_o
b) T(x) = -280*x + 80
c) T(L) = -4 C
Explanation:
Given:
- large plane wall of thickness L = 0.3 m
- thermal conductivity k = 2.5 W/m · °C
- surface area A =12 m2.
- left side of the wall at net heat flux q_o = 700 W/m2 @ x = 0
- temperature at that surface is measured to be T1 =80°C.
Find:
- Express the differential equation and the boundary conditions for steady one dimensional heat conduction through the wall.
- Obtain a relation for the variation of temperature in the wall by solving the differential equation
- Evaluate the temperature of the right surface of the wall at x = L.
Solution:
- The mathematical formulation of Rate of change of temperature is as follows:
d^2T / dx^2 = 0
- Using energy balance:
E_out = E_in
-k* dT / dx = q_o
- Integrate the ODE with respect to x:
T(x) = - (q_o / k)*x + C
- Use the boundary conditions, T(0) = T_1 = 80C
80 = - (q_o / k)*0 + C
C = 80 C
-Hence the Temperature distribution in the wall along the thickness is:
T(x) = - (q_o / k)*x + 80
T(x) = -(700/2.5)*x + 80
T(x) = -280*x + 80
- Use the above relation and compute T(L):
T(L) = -280*0.3 + 80
T(L) = -84 + 80 = -4 C
Differential equation: [tex]\(\frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0\).[/tex]Temperature variation: [tex]\(T(x) = T_1\).[/tex]Temperature at x = L is [tex]\(T(L) = T_1\)[/tex].
a. The differential equation for steady one-dimensional heat conduction through the wall is given by Fourier's law:
[tex]\[ \frac{{d}}{{dx}} \left( k \frac{{dT}}{{dx}} \right) = 0 \][/tex]
This equation states that the rate of change of heat flux with respect to distance ( x ) is constant and equal to zero in steady-state conditions.
The boundary conditions are:
1. At x = 0 : [tex]\( q = q_0 \)[/tex], [tex]\( T = T_1 \)[/tex]
2. At x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex], as there is no heat flux across the right surface of the wall.
b. To solve the differential equation, integrate it twice:
[tex]\[ k \frac{{dT}}{{dx}} = C_1 \][/tex]
[tex]\[ \frac{{dT}}{{dx}} = \frac{{C_1}}{{k}} \][/tex]
[tex]\[ T = \frac{{C_1}}{{k}} x + C_2 \][/tex]
Apply the boundary conditions:
At x = 0 : [tex]\( T = T_1 \)[/tex]
[tex]\[ C_2 = T_1 \][/tex]
At x = L : [tex]\( \frac{{dT}}{{dx}} = 0 \)[/tex]
[tex]\[ \frac{{C_1}}{{k}} = 0 \][/tex]
[tex]\[ C_1 = 0 \][/tex]
Therefore, the temperature variation in the wall is given by:
[tex]\[ T(x) = T_1 \][/tex]
c. The temperature of the right surface of the wall at x = L is equal to [tex]( T(L) = T_1 \)[/tex], as there is no variation in temperature along the wall according to the solution obtained in part b.
Charge of uniform surface density (4.0 nC/m2 ) is distributed on a spherical surface (radius = 2.0 cm). What is the total electric flux through a concentric spherical surface with a radius of 4.0 cm?
To solve this problem we will use the values of the density on surface, which relates the total load and the area of the surface. From there we will get the total charge, which will allow us to find the electric flow.
Surface density is defined as,
[tex]\mu= \frac{Q_{tot}}{A_{surface}}[/tex]
Where,
[tex]A_{surface} = 4\pi r^2[/tex]
Replacing,
[tex]4*10^{-9} = \frac{Q_{tot}}{(4\pi 0.02^2)}[/tex]
[tex]Q_{tot}= 20.11*10^{-2} C[/tex]
At the same time electric flux can be defined as,
[tex]\Phi = \frac{Qtot}{\epsilon_0}[/tex]
Here,
[tex]\epsilon_0 =[/tex] Permittivity vacuum constant
Replacing,
[tex]\Phi =\frac{(20.11 * 10-12 )}{(8.85 * 10-12)}[/tex]
[tex]\Phi =2.27N \cdot m^2 \cdot C^{-1}[/tex]
Therefore the total electric flux through the concentric spherical surface is [tex]2.27Nm^2/C[/tex]
Through conc. spherical surface, the total electric flux will be:
"2.27 N.m².C⁻¹".
Electric fluxUniform surface density charge, [tex]A_{surface}[/tex] = 4.0 nC/m²
Radius, r = 2.0 cm, or
= 0.02
We know the relation,
→ Surface density, μ = [tex]\frac{Q_{tot}}{A_{surface}}[/tex]
Here, [tex]A_{surface}[/tex] = 4πr²
[tex]Q_{tot}[/tex] = 20.11 × 10⁻² C
Now,
→ Electric flux, [tex]\Phi[/tex] = [tex]\frac{Q_{tot}}{\epsilon_0}[/tex]
By substituting the values,
= [tex]\frac{20.11\times 10^{-12}}{8.85\times 10^{-12}}[/tex]
= 2.27 N.m².C⁻¹
Thus the above response is correct.
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If the wire has a diameter of 0.6 in., determine how much it stretches when a distributed load of w=100lb/ft acts on the beam. The wire remains elastic. Use Est=29.0(103)ksi.
To determine how much a wire stretches under a load, one must consider Hooke's Law and use Young's modulus, but the original length of the wire and the type of load distribution must be known.
Explanation:The student's question relates to the concept of elasticity and specifically the calculation of how much a wire stretches under a given load within the elastic limit.
Using the provided diameter of the wire, which is 0.6 inches, and the fact that an elastic modulus (Est) of 29.0(103) ksi has been given, the stretch can be calculated using Hooke's Law, which relates stress and strain through Young's modulus. However, to calculate the stretch, we would require the initial length of the wire and the format of the distributed load (whether it's uniform or varies along the length of the wire) which have not been provided in the student’s query.
Assuming uniform distributed load and an initial length L, the stretch (delta L) can be found using delta L = rac{wL^2}{AE}, where w is the load per unit length, L is the initial length, A is the cross-sectional area, and E is Young's modulus. If the total weight of the load is given instead, then the formula delta L = rac{FL}{AE} can be used with F representing the total force or weight affecting the wire.
Assume that the flow of air through a given duct is isentropic. At one point in the duct, the pressure and temperature are pl = 1800 lb/ft2 and TI = 500°R, Problems respectively. At a second point, the temperature is 400"R. Calculate the pressure and density at this second point.
Answer:
pressure is 825 lb/ft²
density is 1.20 × [tex]10^{-3}[/tex] slug/ft²
Explanation:
given data
p1 = 1800 lb/ft²
T1 = 500°
T2 = 400°
solution
we use here isentropic flow relation that is
[tex]\frac{P2}{P1} = (\frac{T2}{T1})^{\gamma / \gamma - 1 }[/tex]
put here value we get pressure P2
P2 = 1800 × [tex](\frac{400}{500})^{3.5}[/tex]
P2 = 825 lb/ft²
and we know pressure is
pressure = [tex]\rho RT[/tex]
so for pressure 825 we get here [tex]\rho[/tex]
825 = [tex]\rho[/tex] × 1716 × 400
[tex]\rho[/tex] = 1.20 × [tex]10^{-3}[/tex] slug/ft²
For the following functions, determine if the functions are periodic. If so, find the shortest period. A. x(t) = (t2-1)cos(t) B. x(t)-cos(nt) +5 sin(5πt) C. x[n] = sin[n] + cos[n] D. x(t) cos(3t) + sin(Tt) E. x[n] = cos(m) + 1 F. x(t)-(sin nt)(cos πt) | 2
Answer:
A Not Periodic
B Periodic
C Periodic
D Not Periodic
E Periodic
F Periodic
Explanation:
The detailed steps to ascertain the periodicity or the non periodicity of each functions is as shown in the attached file.
the car travels around the circular track having a radius of r=300 m such that when it is at point A it has a velocity of 5m/s which is increasing at the rate of v =(0.06t) m/s^2, where t is in seconds. Determine the magnitudes of its velocity and accerlation when it has traveled one third of the way around the track
The magnitude of the velocity and the acceleration is 0.06 m/s^2., the velocity is 11 m/s
How to solve for the velocity and the accelerationGiven the information, we can calculate the velocity and acceleration of the car as it moves one-third of the way around the circular track.
First, we need to find the time t when the car has traveled one-third of the way around the track. The distance traveled by the car in one-third of the way around the track is d = r/3, where r is the radius of the track. The velocity of the car at time t is v = 5 + 0.06t. The distance traveled by the car can be found using the equation d = vt. Setting these two equations equal to each other, we have:
d = r/3 = 5t + 0.03t^2
Solving for t, we find that t = 20 seconds.
Next, we can find the velocity of the car at time t = 20 seconds using the equation v = 5 + 0.06t = 5 + 0.06 * 20 = 11 m/s.
Finally, the acceleration of the car can be found using the equation a = dv/dt = 0.06 m/s^2.
So, when the car has traveled one-third of the way around the track, its velocity is 11 m/s and its acceleration is 0.06 m/s^2.
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The calorie is a unit of energy defined as the amount of energy needed to raise 1 g of water by 1°C. a. How many calories are required to bring a pot of water at 1°C to a boil? The pot is full to the brim, with diameter 20 cm and depth 20 cm. The density of water is 1000 kg/m3. b. If we consider that D for the pot is 20 cm, approximately how much more energy is needed to heat a hot tub with D = 2 m? How many calories is that?
Answer:
a. Calories required = 622710 calories
b. Energy = 1000 times much energy
Calories = 622710000 calories
Explanation:
Given:
h = Depth of pot = 20cm = 0.2m
Diameter of pot = 20cm = 0.2m
r = ½ *diameter = ½ * 0.2
r = 0.1m
Density = 1000kg/m³
Water temperature = 1°C
a.
First, we calculate the volume of the water(pot)
V = Volume = πr²h
V = 22/7 * 0.1² * 0.2
V = 0.044/7
V = 0.00629m³
M = Mass of water = Volume * Density
M = 0.00629m³ * 1000kg/m³
M = 6.29kg
M = 6.29 * 1000 grams
M = 6290g
The water is at 1°C, so it needs to gain 99°C to reach boiling point
So, Calories = 99 * 6290
Calories required = 622710 calories
b.
If we consider that D for the pot is 20 cm, approximately how much more energy is needed to heat a hot tub with D = 2 m? How many calories is that?
Depth of pot = 20cm
Depth of pot = 0.2m
Depth of hot tube = 2m
Energy is directly proportional to D³
Since the depth of hot the is 10 times greater than that of the pot
It'll require 10³ much more energy
Energy = 10³
Energy = 1000 times much energy
Calories required = 622710 * 1000
Calories = 622710000 calories
The terms batten seam, standing seam, and flat seam all describe types of:
(A) architectural sheet metal roofing.
(B) methods for glazing large windows.
(C) types of EPDM roofing membranes.
(D) framing methods for hollow steel doors.
(E) built-up roofing systems.
Answer:
(A) architectural sheet metal roofing
Explanation:
By the name itself we can judge that the 'Architectural sheet metal roofing' is a kind of metal roofing.
And these type of metal roofing is primarily used for small and big houses, small buildings and as well as in a building that is for commercial use they can be totally flat as well as little bit sloped.
And the words similarly like batten and standing seam, and flat seam all tells us that these are the types of architectural sheet metal roofing.
Given the following materials and their corresponding thermal conductivity values, list them in order from most conductive to least conductive.Sheet Rock: k = 0.43 W/(m*K)Masonite: k = 0.047 W/(m*K)Glass: k = 0.72 W/(m*K)Lexan: k = 0.19 W/(m*K)b) Given the following information, calculate the thermal conductivity using Fourier's Equation.q = 100 WA = 8 m^2ATΔT= 10L = 7 m
Answer:
1) Glass
2) Rock sheet
3) Lexan
4) Masonite
b) k = 8.75 W/m.K
Explanation:
Given:
The thermal conductivity of certain materials as follows:
-Sheet Rock: k = 0.43 W/(m*K)
-Masonite: k = 0.047 W/(m*K)
-Glass: k = 0.72 W/(m*K)
-Lexan: k = 0.19 W/(m*K)
Data Given:
- Q = 100 W
- A = 8 m^2
- dT = 10 C
- L = 7 m
Find:
a) list the materials in order from most conductive to least conductive
b) calculate the thermal conductivity using Fourier's Equation
Solution:
- We know from Fourier's Law the relation between Heat transfer and thermal conductivity as follows:
Q = k*A*dT / L
- From the relation above we can see that rate of heat transfer is directly proportional to thermal conductivity k.
- Hence, the list in order of decreasing conductivity is as follows:
- The list of materials in the decreasing order of thermal conductivity k is:
1) Glass k = 0.72 W/m.K
2) Rock sheet k = 0.43 W/m.K
3) Lexan k = 0.19 W/m.K
4) Masonite k = 0.047 W/m.K
- Use the relation given above we can compute the thermal conductivity k with the given data:
k = Q*L / (A*dT)
k = (100 W * 7 m) / (8 m^2*10 C)
k = 8.75 W/m.K
Code a Boolean expression that tests if a decimal variable named currentSales is greater than or equal to 1000 or a Boolean variable named newCustomer is equal to true. Code this statement so both conditions will always be tested, and code it in the shortest way possible]
Answer:
Given
Decimal variable: currentSales
Decimal test value: 1000
Boolean variable: newCustomer
Boolean default value: true
The following code segment is written in Java
if(currentSales == 1000 || newCustomer)
{
//Some statements
}
The Program above tests two conditions using one of statement
The first condition is to check if currentSales is 1000
== Sign is s a relational operator used for comparison (it's different from=)
|| represents OR
The second condition is newCustomer, which is a Boolean variable
If one or both of the conditions is true, the statements within the {} will be executed
Meaning that, both conditions doesn't have to be true;
At least 1 condition must be true for the statement within the curly braces to be executed
A supersonic nozzle has an exit area 2.5 time the throat area. For a steady, isentropic flow (gamma=1.4) discharging into an atmosphere with pressure Pa, find the Mach number at the throat and at the exit plane for:
a. Pa/Pt = 0.06
b. Pa/Pt = 0.9725
Answer:
The Mach number of the throat for supersonic flow = M* = 1
and the Mach number at exit = 2.44
For
a. Pa/Pt = 0.06, Me = 2.484 Supersonic flow
b. when Pa/Pt = 0.9725 Me = 0.1999 ≅ 0.2 or subsonic flow The mach number at the throat could also be determined given the temperature parameter
Explanation:
To solve the question we note that for a supersonic nozzle, the mach number at the throat = 1
Therefore M* = 1
[tex]\frac{A_{e} }{A^{*} } = 2.5[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(\gamma -1)M_{e} ^{2} }{\gamma +1} )^{\frac{\gamma +1}{2(\gamma -1)} }[/tex] = [tex]\frac{1}{M_{e} } (\frac{2+(0.4)M_{e} ^{2} }{2.4} )^{3 }[/tex] = [tex]\frac{1}{M_{e} } ({2+(0.4)M_{e} ^{2} })^{3 } = 34.56[/tex]
34.56Me = (2+(0.4)M²)³ expanding and collecting like terms we have
Possible solutions of Me = 0.2395, 2.44, 0.90
Since flow is supersonic, Me = 2.44
a)
Solving for [tex]M_{e}[/tex] we have [tex]\frac{P_{a} }{P_{t} } =(1+\frac{\gamma -1}{2} M^{2} _{e} )^{\frac{-\gamma}{\gamma -1} }[/tex]
When Pa/Pt = 0.06 =[tex](1+\frac{1.4 -1}{2} M^{2} _{e} )^{\frac{-1.4}{1.4 -1} }[/tex] = [tex](1+0.2M^{2} _{e} )^{-3.5 }[/tex]
Solving, we get Me = 2.484 Supersonic flow
b)
When Pa/Pt = 0.9725, Me = 0.1999 ≅ 0.2 or subsonic flow
The Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.
To find the Mach number at the throat and at the exit plane of a supersonic nozzle with an exit area 2.5 times the throat area for a steady, isentropic flow (gamma=1.4) discharging into an atmosphere with pressure Pa, we can use the following steps:
Calculate the critical pressure ratio:
pr_crit = (2 / (gamma + 1)) ** (gamma / (gamma - 1))
pr_crit = 0.5283
Calculate the Mach number at the throat:
Mt = sqrt((1 - pr_crit) / (gamma - 1))
Mt = 1.0859
Calculate the Mach number at the exit plane:
Me = sqrt((1 - (Pa / Pt)) / (gamma - 1))
Part a:
Pa/Pt = 0.06
Me = sqrt((1 - 0.06) / (1.4 - 1))
Me = 1.5329
Part b:
Pa/Pt = 0.9725
Me = sqrt((1 - 0.9725) / (1.4 - 1))
Me = 0.2622
Therefore, the Mach number at the throat is 1.0859 for both cases, and the Mach number at the exit plane is 1.5329 for case a and 0.2622 for case b.
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In a conduit with a diameter of 4.5 ft, the depth of flow is 4.0 ft. (a) Determine the hydraulic radius, hydraulic depth, and section factors for critical and normal flows. (b) Determine the alternate depth of flow that will carry the same discharge.
Answer:
(a) 1.125 ft, Section factor = 22.78
(b) 42.75 ft
Explanation:
Hydraulic radius is given by [tex]R_{H} = \frac{A}{P}[/tex] Where
A = Cross sectional area of flow and
P = Perimeter h
Since the cross section is a circle then at depth 4 of 4.5 the perimeter
[tex]=2 \pi r-\frac{\theta }{360} *2 \pi r[/tex] where r = 2.25 and θ = 102.1 °
perimeter = 10.1 ft and the area = [tex]=\pi r^2-\frac{\theta }{360} * \pi r^2[/tex] = 11.39 ft²
Therefore [tex]R_{H} = \frac{11.39}{10.1} = 1.125 ft[/tex]
Section factor is given by for critical flow = Z = A×√D
= 11.39 ft² × √(4 ft) = 22.78
for normal flow Z =[tex]Z_{} ^{2} = \frac{A^{3}}{T}[/tex] = 22.78
(b) The alternate depth of flow is given by
for a given flow rate, we have from chart for flow in circular pipes
Alternative depth = 0.9×45 = 42.75 ft
Consider a constant volume process involving heat addition to a closed system consisting of an ideal gas with no changes in kinetic or potential energy. Is the required heat transfer for raising the temperature from 295 to 305 K the same as the heat transfer required from 345 to 355 K?
Answer:
Yes and no
Explanation:
The thermodynamic equation for the heat transfer in a constant volume process is the following:
[tex]Q=\Delta U=mC_V\Delta T[/tex]
where Q is the required heat, U is the internal energy, m the mass of the gas, C_V the heat capacity assuming consant volume and [tex]\Delta T[/tex] is the change in temperature.
If you assume the heat capacity doesn't change with temperature at which the gas is currently at then the heat transfer depends solely on the change in temperature. With this assumption the transfered heat would be the same in both cases.
In reality the heat capacity does change with respect to temperature. Depending on the type of gas. In reality there would be difference in heat transfered between 295/205 K and 245/255 . Only then you wouldn't use the [tex]\Delta T[/tex] expression since the integral would be different depending on the heat capacity in relation to temperature.
A cylindrical tank has a thin barrier and it carries two fluids, one of the fluids has specific gravity of 2.0 and the other fluid has a specific weight of 100 lbf/ft3. The mass of the tank is 20lb-mass. Determine the magnitude of the vertical force required to give the tank a downward acceleration of 10 ft/s2.
Answer:
attached below
Explanation:
For the following conditions determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste stream under steady-state conditions with a first-order reaction: reactor volume = 280 m3, flow rate = 14 m3 · day−1, and reaction rate coefficient = 0.05 day−1.
Answer:
The PFR is more efficient in the removal of the reactive compound as it has the higher conversion ratio.
Xₚբᵣ = 0.632
X꜀ₘբᵣ = 0.5
Xₚբᵣ > X꜀ₘբᵣ
Explanation:
From the reaction rate coefficient, it is evident the reaction is a first order reaction
Performance equation for a CMFR for a first order reaction is
kτ = (X)/(1 - X)
k = reaction rate constant = 0.05 /day
τ = Time constant or holding time = V/F₀
V = volume of reactor = 280 m³
F₀ = Flowrate into the reactor = 14 m³/day
X = conversion
k(V/F₀) = (X)/(1 - X)
0.05 × (280/14) = X/(1 - X)
1 = X/(1 - X)
X = 1 - X
2X = 1
X = 1/2 = 0.5
For the PFR
Performance equation for a first order reaction is given by
kτ = In [1/(1 - X)]
The parameters are the same as above,
0.05 × (280/14) = In (1/(1-X)
1 = In (1/(1-X))
e = 1/(1 - X)
2.718 = 1/(1 - X)
1 - X = 1/2.718
1 - X = 0.3679
X = 1 - 0.3679
X = 0.632
The PFR is evidently more efficient in the removal of the reactive compound as it has the higher conversion ratio.
To determine whether a CMFR or a PFR is more efficient in removing a reactive compound from the waste stream, we compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation r = kC. In a CMFR, the volume is constant, while in a PFR, the volume varies. Therefore, a PFR may be more efficient depending on the reactor design.
Explanation:To determine whether a CMFR (Continuous Mixed Flow Reactor) or a PFR (Plug Flow Reactor) is more efficient in removing a reactive compound from the waste stream, we need to compare their volumes and flow rates. For a first-order reaction, the reaction rate is given by the equation: r = kC, where r is the reaction rate, k is the reaction rate coefficient, and C is the concentration of the reactive compound.
In a CMFR, the volume is constant, so the reactor volume (280 m3) is equal to the product of the flow rate (14 m3·day−1) and the residence time (t), which is the time it takes for the fluid to pass through the reactor. Therefore, t = V/Q = 280/14 = 20 days.
In a PFR, the volume varies along the length of the reactor, and the residence time is defined as the integral of the volume divided by the flow rate. Using the equation t = ∫V/Q, we can calculate the residence time for a PFR.
Since the residence time for a CMFR is fixed at 20 days, and the residence time for a PFR can be longer or shorter depending on the reactor design, a PFR may be more efficient in removing the reactive compound from the waste stream under steady-state conditions with a first-order reaction.
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It is usually easy to minimize errors due to the input bias current of an opamp by adding a resistor in the input terminal, but this still leaves a small error due to the input offset current. Select one: True False
Answer:True
Explanation:
Answer:
True
Explanation:
Input bias current:
It is a small current that flows in parallel with the input terminals of op-amp to bias the input transistors. This current gets converted into voltage and amplified which results in incorrect output results. This bias current Ib+ and Ib- flows in the positive and negative input terminals of the op-amp.
Ib+ and Ib- create errors of opposite polarity. Therefore, bias current can be minimized by carefully adding a resistor in the positive input terminal.
Input offset current:
Unfortunately, a small error still remains due to the mismatch between input currents Ib+ and Ib-.
This input offset current error can be adjusted by adding a potentiometer and resistor in the negative input terminal.
A man can swim at 4 ft / s in still water. He wishes to cross the 40-ft-wide river to point B, 30 ft downstream. The river flows with a velocity of 2 ft / s. Note: While in the water he must not direct himself toward point B to reach this point.
The question involves calculating the direction a swimmer must aim to reach a specified point across a river, accounting for his swimming speed and the river's current. It illustrates a problem of relative motion and vector resolution in mathematics.
Explanation:The question concerns a man wanting to cross a 40-ft-wide river to a point 30 ft downstream. This scenario involves relative motion in physics, but the calculation primarily uses mathematics to solve. The man can swim at 4 ft/s in still water, and the river flows at 2 ft/s.
To determine the direction the man must swim to reach point B directly across the river, we consider two components of motion: his speed in still water and the river's current. However, the original question indicates he must not swim directly toward point B due to the river's current. Instead, he should aim upstream at a specific angle that compensates for the downstream drift caused by the current.
Without additional details, we can provide a general explanation. The man's effective speed across the river (perpendicular to the current) remains 4 ft/s. To reach point B 30 ft downstream, he must calculate the angle to offset the river's 2 ft/s current. Typically, this involves using trigonometric functions to resolve the swimmer's velocity vector into components parallel and perpendicular to the river flow.
The space shuttle fleet was designed with two booster stages. If the first stage provides a thrust of 686.68 Mega-newtons(MN) and the space shuttle has a mass of 5,470,0005, 470,000 pound-mass, what is the acceleration of the space craft in miles per hour squared?
Answer:
6.30 miles/hour
Explanation:
Newton's second law applies here. In simple terms:
[tex]F = ma[/tex]
where F = Force (Thrust) in N
a = acceleration (m/s²)
The acceleration can be given by rearranging the formula to give:
[tex]a = \frac{F}{N}[/tex]
= [tex]\frac{(686.68*10^{6} )}{24811505120150.2656} \\= 0.0000277 m/s\\= 6.03 miles/hr[/tex]
The manufacturer of a 1.5 V D flashlight battery says that the battery will deliver 9 {\rm mA} for 37 continuous hours. During that time the voltage will drop from 1.5 {\rm V} to 1.0 {\rm V} . Assume the drop in voltage is linear with time.Part A How much energy does the battery deliver in this 37h interval?Express your answer to two significant figures and include the appropriate units.
Answer:
1.42 KJ
Explanation:
solution:
power in beginning [tex]p_{0}[/tex]=(1.5 V).(9×[tex]10^{-3}[/tex] A)
= 13.5 mW
after continuous 37 hours it drops to
[tex]p_{37}[/tex]=(1 V).(9×[tex]10^{-3}[/tex] A)
=9 mW
When the voltage will drop energy will not remain the same but the voltage drop will always remain same if the voltage was drop to for example from 5 V to 4.5 V the drop will remain the same.
37 hours= 37.60.60
=133200 s
w=(9×[tex]10^{-3}[/tex] A×133200 )+[tex]\frac{1}{2}(13.5.10^{-3}-9.10^{-3})(133200)[/tex]
=1.42 KJ
NOTE:
There maybe a calculation error but the method is correct.
Water flows downward through a vertical 10-mm-diameter galvanized iron pipe with an average velocity for 5.0 m/s and exits as a free jet. There is a small hole in the pipe 4 meters above the outlet. Will water leak out of the pipe through this hole, or will air enter into the pipe through the hole? Repeat the problem if the average velocity is 0.5 m/s.
Water is likely to leak out of the hole in the pipe when the average velocity is 5.0 m/s due to high dynamic pressure. With a slower velocity of 0.5 m/s, air might enter the pipe if the static pressure at the hole is less than atmospheric, but this requires additional details to confirm.
Explanation:The question addresses the behavior of water within a pipe system under different flow conditions, involving principles of fluid dynamics. Specifically, it asks whether water will leak out of a small hole in a vertical pipe or if air will enter into the pipe through the hole given two different average velocities of water flow.
Case 1: Average velocity of 5.0 m/s
With an average velocity of 5.0 m/s, the dynamic pressure of the flowing water is considerable, and thus, water is likely to leak out of the hole due to the higher pressure inside the pipe compared to atmospheric pressure.
Case 2: Average velocity of 0.5 m/s
With a decreased velocity of 0.5 m/s, the dynamic pressure is significantly lower. If the static pressure at the hole's location is less than the atmospheric pressure, air might enter the pipe; however, if it is still higher than atmospheric, water would continue to leak out. The determination requires additional information, such as the height of the water column above the hole and any applied pressures at the water's source.
Generally, the behavior can be predicted using Bernoulli's principle and the continuity equation for incompressible flow, which together relate the velocities, pressures, and cross-sectional areas in different sections of a pipe.
An incoming signal is at a frequency of 500kHz. This signal needs to be acquired and all other signals attenuated. Design a passive bandpass filter to do this. Do this by combining a high pass and a low pass filter. For our purposes, create a pass band width of exactly 40kHz and is centered at the ideal frequency. R = 2kOhms. Select C[lowpass], enter value in terms of nF =
Answer:
C_h = 0.166 nF
C_L = 0.153 nF
Explanation:
Given:
- Ideal frequency f_o = 500 KHz
- Bandwidth of frequency BW = 40 KHz
- The resistance identical to both low and high pass filter = 2 Kohms
Find:
Design a passive band-pass filter to do this by cascading a low and high pass filter.
Solution:
- First determine the cut-off frequencies f_c for each filter:
f_c,L for High pass filter:
f_c,L = f_o - BW/2 = 500 - 40/2
f_c,L = 480 KHz
f_c,h for Low pass filter:
f_c,h = f_o + BW/2 = 500 + 40/2
f_c,h = 520 KHz
- Now use the design formula for R-C circuit for each filter:
General design formula:
f_c = 1 /2*pi*R*C_i
C,h for High pass filter:
C_h = 1 /2*pi*R*f_c,L
C_h = 1 /2*pi*2000*480,000
C_h = 0.166 nF
C,L for Low pass filter:
C_L = 1 /2*pi*R*f_c,h
C_L = 1 /2*pi*2000*520,000
C_L = 0.153 nF
A water pump increases the water pressure from 15 psia to 70 psia. Determine the power input required, in hp, to pump 0.8 ft3 /s of water. Does the water temperature at the inlet have any significant effect on the required flow power
Answer:
[tex]Power=11.52hp[/tex]
Explanation:
Given data
[tex]p_{1}=15psia\\p_{2}=70psia\\V_{ol}=0.8ft^{3}/s[/tex]
As
[tex]m=p*V_{ol}[/tex]
Assuming in-compressible flow p is constant
The total change in system mechanical energy calculated as:
Δe=(p₂-p₁)/p
The Power can be calculated as
[tex]P=W\\P=m(delta)e\\P=p*V_{ol}*(p_{2}-p_{1})/p\\ P=V_{ol}*(p_{2}-p_{1})\\P=44(psia.ft^{3}/s )*[(\frac{1Btu}{5.404psia.ft^{3} } )-(\frac{1hp}{0.7068Btu/s } )]\\P=11.52hp[/tex]
The power input required to pump 0.8 ft³/s of water is; P = 11.52 hP
We are given;
Initial pressure; P1 = 15 psia
Final pressure; P2 = 70 psia
Volume flow rate; V' = 0.8 ft³/s
The formula for the mass flow rate is;
m' = ρV'
The total change in the mechanical energy of the system is;
△E = (P2 - P1)/ρ
Now, formula for power is;
P = m' × △E
P = ρV' × (P2 - P1)/ρ
P = V'(P2 - P1)
P = 0.8(70 - 15)
P = 44 psia.ft³/s
Converting to Btu/s gives;
P = 8.142 btu/s
Converting Btu/s to HP from conversion tables gives; P = 11.52 hP
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There is an electric field near the Earth's surface whose magnitude is about 145 V/m . How much energy is stored per cubic meter in this field? Express your answer using two significant figures.
Answer:
u_e = 9.3 * 10^-8 J / m^3 ( 2 sig. fig)
Explanation:
Given:
- Electric Field strength near earth's surface E = 145 V / m
- permittivity of free space (electric constant) e_o = 8.854 *10^-12 s^4 A^2 / m^3 kg
Find:
- How much energy is stored per cubic meter in this field?
Solution:
- The solution requires the energy density stored between earth's surface and the source of electric field strength. The formula for charge density is given by:
u_e = 0.5*e_o * E^2
- Plug in the values given:
u_e = 0.5*8.854 *10^-12 *145^2
u_e = 9.30777 * 10^-8 J/m^3
Toyota customers have a wide range of preferences. Some prefer the fuel economy of a hybrid (Prius), some prefer a family van (Odyssey), and others may prefer a large SUV (Highlander). This is an example of
Answer:
Product segmentation
Explanation:
Product segmentation is an adaptable method for gathering items. Likewise to an objective gathering, an item fragment contains all items that have a specific blend of item qualities.
You can utilize item sections in a battle to improve key figure arranging.
The significance of market division is that it permits a business to exactly arrive at a purchaser with explicit needs and needs.
Group of answer choices:
A) differences in willingness to pay
B) horizontal differentiation
C) segmentation
D) vertical differentiation
E) mass customization
Answer:
The correct answer is letter "C": segmentation.
Explanation:
Market segmentation is the classification of companies that make up their customers based on features such as age, gender, income, and profession, just to mention a few. By segmenting the market, firms group consumers with certain characteristics that enable the institution to specialize in the analysis of that particular sector to provide them with a tailored product or service that they are more likely to purchase.
Therefore, Toyota is segmenting its market in economic vehicles (Prius), size of the family (Odyssey), and large SUV (Highlander) to better fit consumer needs and preferences.
An airplane starts from rest, travels 5000ft down a runway, and after uniform acceleration, takes off with a speed of 162 mi/h. it then climbs in a straight line with a uniform acceleration of 3 ft/s^s until it reaches a constant speed of 220 mi/h. draw the st, vt, and at graphs that describe the motion.
Answer:
Explanation:
Given
Take off speed [tex]v=162\ mph\approx 237.6\ ft/s[/tex]
distance traveled in runway [tex]d=5000 ft[/tex]
using motion of equation
[tex]v^2-u^2=2as[/tex]
where v=final velocity
u=initial velocity
a=acceleration
s=displacement
[tex](237.6)^2=2\times a\times 5000[/tex]
[tex]a=5.64\ ft/s^2[/tex]
Acceleration after take off [tex]a_2=3\ ft/s^2[/tex]
time taken to reach [tex]237.6 ft/s[/tex]
[tex]v=u+at[/tex]
[tex]237.6=0+5.64\times t[/tex]
[tex]t=42.127\ s [/tex]
after take off it take [tex]t_2[/tex] time to reach [tex]220 mph\approx 322.67[/tex]
[tex]322.67=237.6+3\times t_2[/tex]
[tex]t_2=28.35\ s[/tex]
total time taken [tex]t_0=t+t_1[/tex]
[tex]t_0=70.48\ s[/tex]
Another common unit for viscosity is the centipoise, which is 10-2 poise (just like a centimeter is 10-2 meters). (1 centipoise = 0.001 N⋅s/m2) The unit poise is named after Jean Leonard Marie Poiseuille. The conversion between the centipoise and the Pa⋅s is 1 centipoise = 1 mPa.s. What is the viscosity of water at room temperature?
Answer and Explanation
Picking room temperature to be 20°C, the viscosity of water (μ) obtained from literature is:
1 centipoise = 0.01 poise = 0.001 Pa.s = 0.001 N.s/m² (This viscosity is the dynamic viscosity of water at 20°C)
Kinematic viscosity of water, η = μ/ρ
At 20°C, μ = 0.001 Pa.s, ρ = 998.23 kg/m³
η = 0.001/998.23 = 1.0 × 10⁻⁶ m²/s
Picking the room temperature to be 25°C, the viscosityof water (μ) is
0.89 centipoise = 0.0089 poise = 0.00089 Pa.s = 8.9 × 10⁻⁴ N.s/m²
Kinematic viscosity of water, η = μ/ρ
At 25°C, μ = 0.00089 Pa.s, ρ = 997 kg/m³
η = 0.00089/997 = 8.9 × 10⁻⁷ m²/s
The viscosity of water at room temperature in
We want to find the viscosity of water at room temperature in N.s/m² and mPa.s is;
0.001 N.s/m² and 1 mPa.s
Room temperature is 20°C.
Now, from online research, the viscosity of water at room temperature is 0.01 poise
We are told that;
1 centipoise = 0.01 poise
Also, that;
1 centipoise = 0.001 N⋅s/m²
Thus; viscosity of water at room temperature in N. s/m² is; 0.001 N.s/m²
Also, also 1 centipoise = 1 mPa.s
Thus;
Viscosity of water at room temperature in mPa.s is; 1 mPa.s
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A quarter-circle block with a vertical rectangular end is attached to a balance beam as shown in Fig. 1. Water in the tank puts a hydrostatic pressure force on the block which causes a clockwise moment about the pivot point. This moment is balanced by the counterclockwise moment produced by the weight placed at the end of the balance beam. The purpose of this experiment is to determine the weight, W, needed to balance the beam as a function of the water depth, h.
For a given water depth, h, determine the hydrostatic pressure force, FR = γhcA, on the vertical end of the block. Also determine the point of action of this force, a distance yR – yc below the centroid of the area. Note that the equations for FR and yR – yc are different when the water level is below the end of the block (h < R2 – R1) (partially submerged) than when it is above the end of the block (h > R1 – R2) (fully submerged).
(a) Sketch the problem for a fully immersed block. Find W = f1 (h) symbolically. Then substitute for the constants. Obtain a linear relation for W as a function of h.
(b) Sketch the problem for a partially immersed block. Find W = f2 (h) symbolically. Then substitute for the constants. Obtain a cubic relation for W as a function of h.
Answer:
See attachment below
Explanation: