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cupoosta [38]
2 years ago
12

Determine the transition length at the entrance to a 10 mm tube through which 100 percent glycerol at 60 C is flowing at a veloc

ity of 0.5 m/s. The density of a glycerol is 1240 kg/m Data: The dynamic viscosity is 0.0813 Pa-s. a) 18.13 mm b) 50 mm c) 45.43 mm d) 38.13 mm
Chemistry
1 answer:
Readme [11.4K]2 years ago
5 0

Answer: Option (d) is the correct answer.

Explanation:

The given data is as follows.

Tube diameter d = 10 mm = 0.01 m

Velocity of glycerol, v = 0.5 m/s

Density of glycerol (\rho) = 1240 kg/m3

Dynamic viscosity of glycerol (\mu) = 0.0813 pa.s

Reynolds number (Re) = \rho \times velocity \times \frac{density}{\mu}

                                     = 1240 \times 0.5 \times \frac{0.01}{0.0813}

                                     = 76.26

Therefore, according to Reynolds number we can say that flow is laminar.

                     Lt = 0.05 \times Re \times d

                         = 0.05 \times 76.26 \times 0.01

                         = 0.03813 m

As it is known that 1 m = 1000 mm. Hence, in 0.03813 m will be equal to 0.03813 m \times \frac{1000 mm}{1 m}

                         = 38.13 mm

Thus, we can conclude that the transition length of glycerol is 38.13 mm.

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Answer:

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Explanation:

We are asked to find the mass of a sample of metal. We are given temperatures, specific heat, and joules of heat, so we will use the following formula.

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The sample was heated <em>from </em> 58.8 degrees Celsius to 88.9 degrees Celsius.

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Substitute these values into the formula.

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\frac {4500.0 \ J }{13.52694 J/g}= \frac{m (13.52694 J/g)}{13.52694 J/g}

The units of Joules cancel.

\frac {4500.0 \ J }{13.52694 J/g}= m

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The original measurements have 5,4, and 3 significant figures. Our answer must have the least number or 3. For the number we found, that is the ones place. The 6 in the tenth place tells us to round the 2 up to a 3.

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