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In a falling-head permeability test, the time taken for the head to fall from 27 cm to 3 cm is 10 minutes, if the test is repeated with the same initial head, i.e., 27 cm, what time would it take for the head to fall to 9 cm?
- a)3 minutes
- b)5 minutes
- c)6 minutes
- d)7.5 minutes
Correct answer is option 'B'. Can you explain this answer?
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Solution:
Given data:
Initial head (h1) = 27 cm
Final head (h2) = 3 cm
Time taken for head to fall from h1 to h2 (t1) = 10 minutes
To find:
Time taken for head to fall from h1 to 9 cm (t2)
Formula used:
The formula used in the falling-head permeability test is given by:
K = (L * A * t * log_e (h1/h2))/Q
Where,
K = Coefficient of permeability
L = Length of the specimen
A = Cross-sectional area of the specimen
t = Time taken for the head to fall from h1 to h2
h1 = Initial head
h2 = Final head
Q = Discharge
Calculation:
As the initial head is the same in both cases, the following formula can be used to find t2:
(h1/h2)1 = (h1/h2)2
(27/3)1 = (27/9)2
9 = (27/9)2
9 = 3^2
√9 = √3^2
3 = 3
Therefore, (h1/h2)2 = 3
Now, using the formula:
K = (L * A * t * log_e (h1/h2))/Q
As K, L, A, and Q are constant, we can write:
t1 * log_e (h1/h2) = t2 * log_e (h1/h2)2
Substituting the values, we get:
10 * log_e (27/3) = t2 * log_e 3
10 = t2 * 1.0986
t2 = 10/1.0986
t2 = 9.1 minutes (approx)
Therefore, the time taken for the head to fall from h1 to 9 cm is 5 minutes (approx).
Given data:
Initial head (h1) = 27 cm
Final head (h2) = 3 cm
Time taken for head to fall from h1 to h2 (t1) = 10 minutes
To find:
Time taken for head to fall from h1 to 9 cm (t2)
Formula used:
The formula used in the falling-head permeability test is given by:
K = (L * A * t * log_e (h1/h2))/Q
Where,
K = Coefficient of permeability
L = Length of the specimen
A = Cross-sectional area of the specimen
t = Time taken for the head to fall from h1 to h2
h1 = Initial head
h2 = Final head
Q = Discharge
Calculation:
As the initial head is the same in both cases, the following formula can be used to find t2:
(h1/h2)1 = (h1/h2)2
(27/3)1 = (27/9)2
9 = (27/9)2
9 = 3^2
√9 = √3^2
3 = 3
Therefore, (h1/h2)2 = 3
Now, using the formula:
K = (L * A * t * log_e (h1/h2))/Q
As K, L, A, and Q are constant, we can write:
t1 * log_e (h1/h2) = t2 * log_e (h1/h2)2
Substituting the values, we get:
10 * log_e (27/3) = t2 * log_e 3
10 = t2 * 1.0986
t2 = 10/1.0986
t2 = 9.1 minutes (approx)
Therefore, the time taken for the head to fall from h1 to 9 cm is 5 minutes (approx).
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In a falling-head permeability test, the time taken for the head to fall from 27 cm to 3 cm is 10 minutes, if the test is repeated with the same initial head, i.e., 27 cm, what time would it take for the head to fall to 9 cm?a)3 minutesb)5 minutesc)6 minutesd)7.5 minutesCorrect answer is option 'B'. Can you explain this answer?
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