me that both a triaxial shear test and a direct shear test were performed on a sample of dry sand. When the triaxial test is per
1 answer:
Answer:
shear strength = 2682.31 Ib/ft^2
Explanation:
major principal stress = 100 Ib / in2
minor principal stress = 20 Ib/in2
Normal stress = 3000 Ib/ft2
<u>Determine the shear strength when direct shear test is performed </u>
To resolve this we will apply the coulomb failure criteria relationship between major and minor principal stress a
for direct shear test
use Mohr Coulomb criteria relation between normal stress and shear stress
Shear strength when normal strength is 3000 Ib/ft = 2682.31 Ib/ft^2
attached below is the detailed solution
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Answer:
net boiler heat = 301.94 kW
Explanation:
given data
saturated steam = 6.0 bars
temperature = 18°C
flow rate = 115 m³/h = 0.03194 m³/s
heat use by boiler = 90 %
to find out
rate of heat does the boiler output
solution
we can say saturated steam is produce at 6 bar from liquid water 18°C
we know at 6 bar from steam table
hg = 2756 kJ/kg
and
enthalpy of water at 18°C
hf = 75.64 kJ/kg
so heat required for 1 kg is
=hg - hf
= 2680.36 kJ/kg
and
from steam table specific volume of saturated steam at 6 bar is 0.315 m³/kg
so here mass flow rate is
mass flow rate =
mass flow rate m = 0.10139 kg/s
so heat required is
H = h × m
here h is heat required and m is mass flow rate
H = 2680.36 × 0.10139
H = 271.75 kJ/s = 271.75 kW
now 90 % of boiler heat is used for generate saturated stream
so net boiler heat =
net boiler heat =
net boiler heat = 301.94 kW
Answer:
ηa=0.349
ηb=0.345
Explanation:
The enthalpy and entropy at state 3 are determined from the given pressure and temperature with data from table:
The quality at state 4 is determined from the condition and the entropies of the components at the condenser pressure taken from table:
The enthalpy at state 4 then is:
Part A
In the case when the water is in a saturated liquid state at the entrance of the pump the enthalpy and specific volume are determined from A-5 for the given pressure:
The enthalpy at state 2 is determined from an energy balance on the pump:
=346.67 kJ/kg
The thermal efficiency is then determined from the heat input and output in the cycle:
Part B
In the case when the water is at a lower temperature than the saturation temperature at the condenser pressure we look into table and see the water is in a compressed liquid state. Then we take the enthalpy and specific volume for that temperature with data from and the saturated liquid values:
The enthalpy at state 2 is then determined from an energy balance on the pump:
=299.79 kJ/kg
The thermal efficiency in this case then is: