4. Three routes connect an origin and a destination with performance functions tl = 8 + 0.5x1, t2 = 1 + 2x2, and t3 = 3 + 0.75x3, with the x's expressed in thousands of vehicles per hour and the t's expressed in minutes. If the peak-hour traffic demand is 3400 vehicles, determine user-equilibrium traffic flows. Note: there are 823 vehicles that must pass by a daycare in route 1.

Answer:

See the attached picture.

Explanation:

See the attached picture for explanation.

Answer:

a) [tex]T_{out} = 2190.455 ^{\textdegree}R[/tex] b) [tex]\dot V_{air,out} = 8352.941 acfm[/tex]  c) [tex]\dot V_{air,out,corr} = 7350.588 afcm[/tex]

Explanation:

a) The heat lost by the power plant is:

[tex]\dot Q_{loss} = (0.67) \cdot (110 \frac{lbm}{s} ) \cdot (12000 \frac{BTU}{lbm} )\\\dot Q_{loss} = 884400 BTU[/tex]

The waste heat used by heat exchanger is:

[tex]\dot Q_{used} = (0.0012) \cdot \dot Q_{loss}\\\dot Q_{used} = 1061.28 BTU[/tex]

Assuming that air behaves as an ideal gas, density is given by following expression:

[tex]\rho_{air} = \frac{P \cdot M_{air}}{R_{u} \cdot T}[/tex]

[tex]\rho_{air} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(298 K)}\\\rho_{air} = 1.145 \frac{kg}{m^{3}}[/tex]

The density unit is converted to pounds per cubic feet:

[tex]\rho_{air} = 1.145 \frac{kg}{m^{3}} \cdot (\frac{1 lb}{0.453 kg}) \cdot (\frac{0.304 m}{1 ft} )^{3} \\\rho_{air} = 0.071 \frac{lb}{ft^{3}}[/tex]

The heat received by air flow through heat exchanger is:

[tex]\dot Q_{air} = (0.90) \cdot \dot Q_{used}\\\dot Q_{air} = 955.152 BTU[/tex]

Outlet temperature can isolated from the following formula:

[tex]\dot Q_{air} = \rho_{air} \cdot \dot V_{air} \cdot c_{p,air} \cdot (T_{out} - T_{in})[/tex]

[tex]T_{out} =T_{in} + \frac{\dot Q_{air}}{\rho_{air} \cdot \dot V_{air} \cdot c_{p,air}}[/tex]

[tex]T_{out} = 536.4 ^{\textdegree}R + \frac{955.152 BTU}{(0.071\frac{lb}{ft^{3}})\cdot (33.333 \frac{ft}{s} )\cdot(0.244 \frac{BTU}{lb ^{\textdegree}R})}\\T_{out} = 2190.455 ^{\textdegree}R[/tex]

b) Due to the compressibility of air, density has to be calculated by using the approach from section a).

[tex]\rho_{air,out} = \frac{(101.325 kPa) \cdot (28 \frac{kg}{kmol})}{(8.314 \frac{kPa\cdot m^{3}}{kmol \cdot K} )(1217 K)}\\\rho_{air,out} = 0.280 \frac{kg}{m^{3}}[/tex]

[tex]\rho_{air,out} = 0.017 \frac{lb}{ft^{3}}[/tex]

Volumetric flow can be found by Principle of Mass Conservation:

[tex]\dot V_{air, out} = \frac{\rho_{air}}{\rho_{air,out}}\cdot \dot V_{air}[/tex]

[tex]\dot V_{air,out} = 8352.941 acfm[/tex]

c) The moisture component indicates that 88 percent of volume is occupied by dry-air. The moisture-corrected dry flow rate is:

[tex]\dot V_{air,out,corr} = 0.88 \cdot (8352.941 afcm)\\\dot V_{air,out,corr} = 7350.588 afcm[/tex]

Answer:

a)  [tex]\Delta S = 0.386 \ \frac{kJ}{K}[/tex]

b)  [tex]\Delta S =0.395 \ \frac{kJ}{K}[/tex]

Explanation:

a)

Specific heat formula, we have:

[tex]mc(\theta_1-\theta_2)=Wt[/tex]

Where

m is mass

c is specific heat capacity of air at specific temp

[tex]\theta[/tex] is the temperature change

W is the power

t is the time

The ideal gas law is:

[tex]PV=nRT[/tex]

Where

P is the pressure

V is the volume

n is number of moles

T is temperature

R is the ideal gas constant

First, lets solve for [tex]\theta_2[/tex] in the 1st equation, remembering to use ideal gas law in it to have the variables that are given in the problem. Shown below:

[tex]mc(\theta_2-\theta_1)=Wt\\mc\theta_2-mc\theta_1=Wt\\mc\theta_2=mc\theta_1 + Wt\\\theta_2=\frac{mc\theta_1 + Wt}{mc}\\\theta_2=\theta_1+\frac{Wt}{mc}\\\theta_2=\theta_1+\frac{WtR\theta_1}{PVc}\\\theta_2=\theta_1(1+\frac{WtR}{PVc})[/tex]

Now, we know

Theta_1 is 17 celsius, which in Kelvin is 17 + 270 = 290K

Power is 200

Time is 15 mins = 15 * 60 = 900 seconds

R is 287 J/kg K

P is 120

V is 300 L

Specific heat of air at 290K is about 1005

Substituting we get:

[tex]\theta_2=\theta_1(1+\frac{WtR}{PVc})\\\theta_2=290(1+\frac{200*900*287}{120*300*1005})\\\theta_2=704 \ K[/tex]

Now, Entropy Change is:

[tex]\Delta S =mc Ln(\frac{\theta_2}{\theta_1})[/tex]

Again using the substitution equations, we have:

[tex]\Delta S = \frac{PV}{R\theta_1}cLn(1+\frac{WRt}{PVc})[/tex]

We know what the variables mean, we substitute the respective values, to get:

[tex]\Delta S = 0.386 \ \frac{kJ}{K}[/tex]

b)

For variable specific heats, we need entropy value from entropy table:

s_2 = 2.58044

s_1 = 1.66802  [1.05 * 290 K]

The formula is:

[tex]\Delta S = m(s_2-s_1)\\\Delta S = \frac{PV}{R\theta_1}(s_2-s_1)[/tex]

Substituting the values, we find the answer:

[tex]\Delta S = \frac{PV}{R\theta_1}(s_2-s_1)\\\Delta S = \frac{120*300}{287*290}(2.58044-1.66802)\\\Delta S =0.395 \ \frac{kJ}{K}[/tex]

Answer:

Follow step by step explabation to get answer to the question and also see attachments for output.

Explanation:

code:

def normalize(input_string):

l=list();

for i in input_string:

if (i>='a' and i<='z') or (i>='A' and i<='Z'):

l.append(i.lower())

return l

def find_missing_letters_algorithm(sentence,prevsentece):

l=list();

for i in prevsentece:

i=i.lower();

if i not in sentence:

l.append(i)

return l

print(find_missing_letters_algorithm(normalize("O.K. #1 Python!"),"O.K. #1 Python!"))