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3 years ago

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A railcar with an overall mass of 78,000 kg traveling with a speed vi is approaching a barrier equipped with a bumper consisting

of a nonlinear spring whose force vs. compression law is given by FS = βx3, where β = 640 × 106 N/m3 and x is the compression of the bumper. The compression of the bumper is limited to 20 cm. Treat the system as a particle and assume that the contact between the railcar and rails is frictionless.

1 answer:

sergij07 [2.7K]3 years ago

5 0

Answer:

v₀ = 2,562 m / s = 9.2 km/h

Explanation:

To solve this problem let's use Newton's second law

F = m a = m dv / dt = m dv / dx dx / dt = m dv / dx v

F dx = m v dv

We replace and integrate

-β ∫ x³ dx = m ∫ v dv

β x⁴/ 4 = m v² / 2

We evaluate between the lower (initial) integration limits v = v₀, x = 0 and upper limit v = 0 x = x_max

-β (0- x_max⁴) / 4 = ½ m (v₀²2 - 0)

x_max⁴ = 2 m /β v₀²

Let's look for the speed that the train can have for maximum compression

x_max = 20 cm = 0.20 m

v₀ =√(β/2m) x_max²

Let's calculate

v₀ = √(640 106/2 7.8 104) 0.20²

v₀ = 64.05 0.04

v₀ = 2,562 m / s

v₀ = 2,562 m / s (1lm / 1000m) (3600s / 1h)

v₀ = 9.2 km / h

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Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at

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A) $v_2 = 2016.80 ft/s$

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where h1 and h2 are inlet and exit enthalpy

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$s_1 = 1.708 Btu/lbm -R$

exit enthalapy h_2

$\eta = \frac{h_1 - h'_2}{ h_1 - h_2}$

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from above equation

$1278.8 \times 25037 + \frac{100^2}{2} = 1197.77 \times 25037 + \frac{v_2^2}{2}$ [1 Btu/lbm = 25037 ft^2/s^2]

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$\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R$

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Answer:

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let mass flow rate for cold and hot fluid = M<em>c</em> and M<em>h</em> respectively
let specific heat for cold and hot fluid = C<em>pc</em> and C<em>ph </em>respectively
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<u>Using effectiveness-NUT method</u>

<em>First, we need to determine heat capacity rate for cold and hot fluid, and determine the dimensionless heat capacity rate</em>

C<em>c</em> = M<em>c</em> × C<em>pc</em> = 1.2 kg/s × 4.18 kj/kg °c = 5.016 kW/°c

C<em>h = </em>M<em>h</em> × C<em>ph </em>= 2 kg/s × 4.31 kj/kg °c = 8.62 kW/°c

From the result above cold fluid heat capacity rate is smaller

Dimensionless heat capacity rate, C = minimum capacity/maximum capacity

C= C<em>min</em>/C<em>max</em>

C = 5.016/8.62 = 0.582

.<em>2 Second, we determine the maximum heat transfer rate, Qmax</em>

Q<em>max</em> = C<em>min </em>(Inlet Temp. of hot fluid - Inlet Temp. of cold fluid)

Q<em>max</em> = (5.016 kW/°c)(160 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(140) °c = 702.24 kW

.<em>3 Third, we determine the actual heat transfer rate, Q</em>

Q = C<em>min (</em>outlet Temp. of cold fluid - inlet Temp. of cold fluid)

Q = (5.016 kW/°c)(80 - 20) °c

Q<em>max</em> = (5.016 kW/°c)(60) °c = 303.66 kW

.<em>4 Fourth, we determine Effectiveness of the heat exchanger, </em>ε

ε<em> </em>= Q/Qmax

ε <em>= </em>303.66 kW/702.24 kW

ε = 0.432

.<em>5 Fifth, using appropriate effective relation for double pipe counter flow to determine NTU for the heat exchanger</em>

NTU = $\\ \frac{1}{C-1} ln(\frac{ε-1}{εc -1} )$

NTU = $\frac{1}{0.582-1} ln(\frac{0.432 -1}{0.432 X 0.582 -1} )$

NTU = 0.661

<em>.6 sixth, we determine Heat Exchanger surface area, As</em>

From the question, the overall heat transfer coefficient U = 640 W/m²

As = $\frac{NTU C{min} }{U}$

As = $\frac{0.661 x 5016 W. °c }{640 W/m²}$

As = 5.18 m²

<em>.7 Finally, we determine the length of the heat exchanger, L</em>

L = $\frac{As}{\pi D}$

L = $\frac{5.18 m² }{\pi (0.015 m)}$

L= 109.91 m

L ≅ 110 m = 11,000 cm

3 0

3 years ago