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Ksivusya [100]
3 years ago
12

Consider a sphere made of stainless steel with diameter of 25 cm. It is heated to temperature of 300°C for some chemical tests.

After finishing the tests, the sphere is cooled by exposing it to a flow of air at 1 atm pressure and 25°C with a velocity of 3 m/s. By the end of cooling process, the sphere's temperature drops to 200°C. The rate of heat transfer loss due to convection is closest to:__________.
a) 485 W
b) 513 W
c) 88 W
d) 611 w
Engineering
1 answer:
DaniilM [7]3 years ago
4 0

Answer:

263.69 W.

(None of the option).

Explanation:

So, from the question above we are given the following parameters or data or information which is going to allow us to solve this question and they are;

(1). diameter of 25 cm.

(2). Initial temperature of 300°C.

(3).temperature drops to 200°C = final temperature.

Step one: Calculate the Reynolds number.

Reynolds number = 3 × 0.25/1.562 × 10^-5 = 48015.365.

Step two: Calculate average heat transfer coefficient.

The average heat transfer coefficient = k/D { 2 + (0.4Re^1/2 + 0.06Re^2/3} px^0.4 × (u/uz)^1/4.

The average heat transfer coefficient = 0.10204 × [ 2 + (87.65 + 79.26) (0.8719) × 0.8909.

average heat transfer coefficient = 0.20204 ( 2 + 129.652).

average heat transfer coefficient = 13.43/m^2.k.

Step three: The rate of heat transfer loss due to convection = (average heat transfer coefficient ) × πD^2 × ( T1 - T2).

The rate of heat transfer loss due to convection= 13.43 × π(0.25)^2 × (300 - 200).

=>The rate of heat transfer loss due to convection = 263.69 W.

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An equal-tangent sag vertical curve (with a negative initial and a positive final grade) is designed for 55 mi/h. The PVI is at
Varvara68 [4.7K]

Answer:

The lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

Explanation:

Length of curve is given as

L=2(PVT-PVI)\\L=2(242+30-240+00)\\L=2(230)\\L=460 ft

G_2 is given as

G_2=\frac{E_{PVT}-E_{PVI}}{0.5L}\\G_2=\frac{127.5-122}{0.5*460}\\G_2=0.025=2.5 \%

The K value is given from the table 3.3 for 55 mi/hr is 115. So the value of A is given as

A=\frac{L}{K}\\A=\frac{460}{115}\\A=4

A is given as

-G_1=A-G_2\\-G_1=4.0-2.5\\-G_1=1.5\\G_1=-1.5\%

With initial grade, the elevation of PVC is

E_{PVC}=E_{PVI}+G_1(L/2)\\E_{PVC}=122+1.5%(460/2)\\E_{PVC}=125.45 ft\\

The station is given as

St_{PVC}=St_{PVI}-(L/2)\\St_{PVC}=24000-(230)\\St_{PVC}=237+70\\

Low point is given as

x=K \times |G_1|\\x=115 \times 1.5\\x=172.5 ft

The station of low point is given as

St_{low}=St_{PVC}-(x)\\St_{low}=23770+(172.5)\\St_{low}=239+42.5 ft\\

The elevation is given as

E_{low}=\frac{G_2-G_1}{2L} x^2+G_1x+E_{PVC}\\E_{low}=\frac{2.5-(-1.5)}{2*460} (1.72)^2+(-1.5)*(1.72)+125.45\\E_{low}=124.16 ft

So the lowest point of the curve is at 239+42.5 ft where elevation is 124.16 ft.

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Answer:

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