Answer:
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Explanation:
Answer:
The settlement that is expected is 1.043 meters.
Explanation:
Since the pre-consolidation stress of the layer is equal to the effective stress hence we conclude that the soil is normally consolidated soil
The settlement due to increase in the effective stress of a normally consolidated soil mass is given by the formula

where
'H' is the initial depth of the layer
is the Compression index
is the inital void ratio
is the initial effective stress at the depth
is the change in the effective stress at the given depth
Applying the given values we get

Answer:
14.52 minutes
<u>OR</u>
14 minutes and 31 seconds
Explanation:
Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.
Specific heat at constant volume at 27°C = 0.718 kJ/kg*K
Initial temperature of room (in kelvin) = 283.15 K
Final temperature (required) of room = 293.15 K
Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg
Heat required at constant volume: 0.718 * (change in temp) * (mass of air)
Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ
Time taken for temperature rise: heat required / (rate of heat change)
Where rate of heat change = 10000 - 5000 = 5000 kJ/hr
Time taken = 1210.26 / 5000 = 0.24205 hours
Converted to minutes = 0.24205 * 60 = 14.52 minutes
Answer:
the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi
Explanation:
Given that;
depth 1 = 71 ft
depth 2 = 10 ft
pressure p = 17 psi = 2448 lb/ft²
depth h = 71 ft - 10 ft = 61 ft
we know that;
p = P_air + yh
where y is the specific weight of ethyl alcohol ( 49.3 lb/ft³ )
so we substitute;
p = 2448 + ( 49.3 × 61 )
= 2448 + 3007.3
= 5455.3 lb/ft³
= 37.88 psi
Therefore, the pressure at a closed valve attached to the tank 10 ft above its bottom is 37.88 psi