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Mrac [35]
3 years ago
7

A soil sample 90 mm high and 6000 mm2 in cross section was subjected to a falling head permeability test. The head fell from 500

mm to 300 mm in 1500 s. The permeability of the soil was 2.4x10-3 mm/s.
Determine the diameter of the stand pipe. (in mm)
Engineering
1 answer:
il63 [147K]3 years ago
8 0

Answer:

the diameter of the stand pipe is 24.456mm

Explanation:

Given that:

Height of the soil sample = 90 mm

Cross section area of the soil sample = 6000 m²

The head fell from 500 mm to 300 mm

i,e

H_1 = 500 \ mm

H_2 = 300 \ mm

Time = 1500 s

The permeability of the soil (K) was 2.4x10⁻³ mm/s.

The objective is to determine the diameter of the stand pipe. (in mm)

Using the formula :

K = \dfrac{(2.303 )*a*L}{A*t} log(\dfrac{H_1}{H_2})

2.4*10^{-3}= \dfrac{(2.303 )*a*(90)}{(6000 )*(1500 )} log(\dfrac{500}{300 })

2.4*10^{-3}= \dfrac{(207.27 *a)}{9000000} *0.22185

2.4*10^{-3}*9000000= (207.27 *a)}*0.22185

21600 = 45.983 a

a = 21600/45.983

a = 469.74 mm

Recall thar:

a = π/4 (d²)

So;

469.74  = 0.7854 ((d²)

d² = 469.74/0.7854

d² = 598.09

d = √598.09

d = 24.456 mm

Hence ; the diameter of the stand pipe is 24.456mm

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5 0
3 years ago
A thick aluminum block initially at 26.5°C is subjected to constant heat flux of 4000 W/m2 by an electric resistance heater whos
Yanka [14]

Given Information:

Initial temperature of aluminum block = 26.5°C

Heat flux = 4000 w/m²

Time = 2112 seconds

Time = 30 minutes = 30*60 = 1800 seconds

Required Information:

Rise in surface temperature = ?

Answer:

Rise in surface temperature = 8.6 °C after 2112 seconds

Rise in surface temperature = 8 °C after 30 minutes

Explanation:

The surface temperature of the aluminum block is given by

T_{surface} = T_{initial} + \frac{q}{k} \sqrt{\frac{4\alpha t}{\pi} }

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.

After t = 2112 sec:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (2112)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.51098)\\\\T_{surface} = 26.5 + 8.6\\\\T_{surface} = 35.1\\\\

The rise in the surface temperature is

Rise = 35.1 - 26.5 = 8.6 °C

Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.

After t = 30 mins:

T_{surface} = 26.5 + \frac{4000}{237} \sqrt{\frac{4(9.71\times 10^{-5}) (1800)}{\pi} }\\\\T_{surface} = 26.5 + \frac{4000}{237} (0.4717)\\\\T_{surface} = 26.5 + 7.96\\\\T_{surface} = 34.5\\\\

The rise in the surface temperature is

Rise = 34.5 - 26.5 = 8 °C

Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.

5 0
3 years ago
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