Answer:
a.
b.
c.
Explanation:
a. void ratio is provided by the formula:
where , = volume of voids
= volume of solid grains
for loose sand, the void space =
= 1
b. void ratio after static load = 0.1/(480)/ (480)
= 0.1
c. void ratio after vibration = [480- ( 0.1 * 480) ]/ 480
= 0.9
Answer:
Rate of Entropy =210.14 J/K-s
Explanation:
given data:
power delivered to input = 350 hp
power delivered to output = 250 hp
temperature of surface = 180°F
rate of entropy is given as
T = 180°F = 82°C = 355 K
Rate of heat = (350 - 250) hp = 100 hp = 74600 W
Rate of Entropy
Answer:
Explanation:
First, we will find actual properties at given inlet and outlet states by the use of steam tables:
AT INLET:
At 4MPa and 350°C, from the superheated table:
h₁ = 3093.3 KJ/kg
s₁ = 6.5843 KJ/kg.K
AT OUTLET:
At P₂ = 125 KPa and steam is saturated in vapor state:
h₂ = = 2684.9 KJ/kg
Now, for the isentropic enthalpy, we have:
P₂ = 125 KPa and s₂ = s₁ = 6.5843 KJ/kg.K
Since s₂ is less than and greater than
at 125 KPa. Therefore, the steam is in a saturated mixture state. So:
Now, we will find (enthalpy at the outlet for the isentropic process):
Now, the isentropic efficiency of the turbine can be given as follows:
Answer:
porosity = 0.07 or 7%
dry bulk density = 3.25g/cm3]
water content =
Explanation:
bulk density = dry Mass / volume of sample
dry mass = 0.490kg = 490g
volume = πr2h = 3.142 * 2 *2 *12 = 150.8cm3
density = 490/150.8 = 3.25g/cm3
porosity = =
= 0.07 or 7%
water content = = 7%