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AleksAgata [21]
3 years ago
9

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.

Engineering
1 answer:
Helga [31]3 years ago
3 0

Answer:

X1 = 2081.64

X2 = 523.91

X3 = 1394.45

Explanation:

See the attached pictures for detailed explanation.

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What are the three elementary parts of a vibrating system?
zhenek [66]

Answer:

the three part are mass, spring, damping

Explanation:

vibrating system consist of three elementary system namely

1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.

2) Spring -  the part that has elasticity and help to hold mass

3) Damping - this part considered to have zero mass and  zero elasticity.

7 0
2 years ago
Short-term memoryA) has a larger storage capacity than long-term memory.B) takes longer to retrieve than long-term memory.C) inv
zaharov [31]

Answer:D

Explanation:

Take longer time to retrieve than long term memory, involves transient modifications in the function of pre existing synapses, such as channel modifications.

6 0
3 years ago
A well-insulated, rigid tank has a volume of 1 m3and is initially evacuated. A valve is opened,and the surrounding air enters at
DiKsa [7]

Answer:

0.5 kW

Explanation:

The given parameters are;

Volume of tank = 1 m³

Pressure of air entering tank = 1 bar

Temperature of air = 27°C = 300.15 K

Temperature after heating  = 477 °C = 750.15 K

V₂ = 1 m³

P₁V₁/T₁ = P₂V₂/T₂

P₁ = P₂

V₁ = T₁×V₂/T₂ = 300.15 * 1 /750.15 = 0.4 m³

dQ = m \times c_p \times (T_2 -T_1)

For ideal gas, c_p = 5/2×R = 5/2*0.287 = 0.7175 kJ

PV = NKT

N = PV/(KT) = 100000×1/(750.15×1.38×10⁻²³)

N = 9.66×10²⁴

Number of moles of air = 9.66×10²⁴/(6.02×10²³) = 16.05 moles

The average mass of one mole of air = 28.8 g

Therefore, the total mass = 28.8*16.05 = 462.135 g = 0.46 kg

∴ dQ = 0.46*0.7175*(750.15 - 300.15) = 149.211 kJ

The power input required = The rate of heat transfer = 149.211/(60*5)

The power input required = 0.49737 kW ≈ 0.5 kW.

3 0
3 years ago
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
shusha [124]

Answer:

A)W'/m = 311 KJ/kg

B)σ'_gen/m = 0.9113 KJ/kg.k

Explanation:

a).The energy rate balance equation in the control volume is given by the formula;

Q' - W' + m(h1 - h2) = 0

Dividing through by m, we have;

(Q'/m) - (W'/m) + (h1 - h2) = 0

Rearranging, we have;

W'/m = (Q'/m) + (h1 - h2)

Normally, this transforms to another equation;

W'/m = (Q'/m) + c_p(T1 - T2)

Where;

W'/m is the rate at which power is developed

Q'/m is the rate at which heat is flowing

c_p is specific heat at constant pressure which from tables at a temperature of 980k = 1.1 KJ/kg.k

T1 is initial temperature

T2 is exit temperature

We are given;

Q'/m = -30 kj/kg (negative because it leaves the turbine)

T1 = 980 k

T2 = 670 k

Plugging in the relevant values;

W'/m = -30 + 1.1(980 - 670)

W'/m = 311 KJ/kg

B) The Entropy produced from the entropy balance equation in a control volume is given by the formula;

(Q'/T_boundary) + m(s1 - s2) + σ'_gen = 0

Dividing through by m gives;

((Q'/m)/T_boundary) + (s1 - s2) + σ'_gen/m = 0

Rearranging, we have;

σ'_gen/m = -((Q'/m)/T_boundary) + (s2 - s1)

Under the conditions given in the question, this transforms normally to;

σ'_gen/m = -((Q'/m)/T_boundary) - c_p•In(T2/T1) - R•In(p2/p1)

σ'_gen/m is the rate of entropy production in kj/kg

We are given;

p2 = 100 kpa

p1 = 400 kpa

T_boundary = 315 K

For an ideal gas, R = 0.287 KJ/kg.K

Plugging in the relevant values including the ones initially written in answer a above, we have;

σ'_gen/m = -(-30/315) - 1.1(In(670/980)) - 0.287(In(100/400))

σ'_gen/m = 0.0952 + 0.4183 + 0.3979

σ'_gen/m = 0.9113 KJ/kg.k

6 0
2 years ago
if a current of 5 amps flows through a resistance of 40 ohms, what is the voltage across that resistor
olasank [31]

Answer:

200V

Explanation:

I = 5A

R = 40Ω

V = IR = (5)(40) = 200V

7 0
2 years ago
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