2) While it is being cleaned, a cylindrical steel railroad car has an outer leng

Question

2) While it is being cleaned, a cylindrical steel railroad car has an outer length of 20.5 meters and an outer diameter of 6.40 meters. The thickness of the steel walls of the car is 7.5 mm. The empty car is cleaned using steam at a temperature of 147 C, and this steam then fills the entire volume of the car. Since the car is open to outside pressure during the cleaning the constant pressure during this process is normal atmospheric pressure of 1.013 10 N/m2. The outside temperature is 18.0 C on the day the car is cleaned. Use this information to answer the following questions. 6.40m a) What is the rate at which heat is being conducted through the walls of the railr oad car while it is being cleaned? [10 points] What is the root mean square speed of the steam molecules in the car while it is being cleaned? [10 points b) If directly after cleaning the car the worker decides to seal the railroad car initially filled with the hot steam, what will be the likely fate of the railroad car? Explain fully. [5 points] c)

Explanation / Answer

2. 2. from the given data

Lo = 20.5 m

do = 6.4 m

t = 7.5 mm

now let thermal conductivity of steel be k

k = 50 W/mK

now rate of heat conduction through the cylinder = r

r = rate of heat conduction through cylinderical walls + rate of heat conduction through circular walls = r1 + r2

r2 = 2k(pi*(do - t/2)^2/4)(dT)/t

dT = 147 - 18 = 129 C

hence

r2 = 25*pi(6.4 - 7.5/2000)^2*129/7.5*10^-3 = 55267.41961 kW

now, dr1 = k*2*pi*r*(Lo - 2t)*dT/dr

now, dr1 = r1 = constant

hence

r1*dr/r = 2*pi*k(Lo - 2t)*dT

integrating

r1*ln(do/(do - 2t)) = 2*pi*k(Lo - 2t)*129

r1 = 353797.55692477 kW

hence

r = r1 + r2 = 409064.976534 kW

b. rms speed = v

v = sqrt(3P/rho)

now, P = rho*R*T

P/rho = RT

v = sqroot(3RT)

R = 460 J/Kg K

T = 147 C = 147 + 273.16 = 420.16 K

hence

v = 761.45965 m/s

c. if after cleaning the worker shuts doen the opening in the tank, the pressure inside the tank will drop as eventuallly temperature of steam drops down to the outisde temperature. this will greatly reduce the volume of the steam inside the tank as it condenses into water below 100 C and this will cause a massive pressure difference between the outer and inner surfac eof the tanks as the pressure inside the tank drops due to volume constraction. hence a net inward force onthe walls of the tanks will cause the tanki walls to fail and the tank to crumble out of shape from a cylinder into something with lesser volume

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