Answer:
The kinematic viscosity is equal to the dynamic viscosity of a fluid divided on its density.
ν = μ / ρ v = 0.1/10 =0.01
Explanation:
The kinematic viscosity is independant of the mass of the fluid as well as its pressure and temperature.
Answer:
12.332 KW
The positive sign indicates work done by the system ( Turbine )
Explanation:
Stagnation pressure( P1 ) = 900 kPa
Stagnation temperature ( T1 ) = 658K
Expanded stagnation pressure ( P2 ) = 100 kPa
Expansion process is Isentropic, also assume steady state condition
mass flow rate ( m ) = 0.04 kg/s
<u>Calculate the Turbine power </u>
Assuming a steady state condition
( p1 / p2 )^(r-1/r) = ( T1 / T2 )
= (900 / 100)^(1.4-1/1.4) = ( 658 / T2 )
= ( 9 )^0.285 = 658 / T2
∴ T2 = 351.22 K
Finally Turbine Power / power developed can be calculated as
Wt = mCp ( T1 - T2 )
= 0.04 * 1.005 ( 658 - 351.22 )
= 12.332 KW
The positive sign indicates work done by the system ( Turbine )
Answer:
The correct response is "$275,275".
Explanation:
The given values are:
Sales,
= 400,000
Gross margin,
= 120,000
Beginning Inventory goods,
= 150,475
Finished inventory goods,
= 145,750
Now,
The cost of goods sold will be:
=
On substituting the values, we get
=
=
As we know,
⇒
⇒
⇒
⇒
⇒ ($)
Answer:
The head difference across the soil specimen is 39.29 cm and the discharge velocity is 0.02 cm/s
Explanation:
The head difference across the soil specimen is:
Where
k = hydraulic conductivity = 0.014 cm/s
Q = volume of water collected = 150 cm³/min = 2.5cm³/s
L = length of the soil specimen = 275 mm = 27.5 cm
A = area = 125 cm²
Replacing:
The hydraulic gradient is:
The discharge velocity is:
v = k*i = 0.014 * 1.43 = 0.02 cm/s
Answer:
C. 14.55
Explanation:
12 x 10 = 120
120 divded by 10 is 12
so now we do the left side
7 x 3 = 21 divded by 10 is 2
so now we have 14
and the remaning area is 0.55
so 14.55