Answer:
the no. of activities supply in a cahin like in the figuration wise they supply the chain
Answer:
Power output, 
Given:
Pressure of steam, P = 1400 kPa
Temperature of steam, 
Diameter of pipe, d = 8 cm = 0.08 m
Mass flow rate, 
Diameter of exhaust pipe, 
Pressure at exhaust, P' = 50 kPa
temperature, T' = 
Solution:
Now, calculation of the velocity of fluid at state 1 inlet:




Now, eqn for compressible fluid:

Now,




Now, the power output can be calculated from the energy balance eqn:



Answer: hardness
Explanation:
Hardness is a measure of a material's ability to resist plastic deformation. In other words, it is a measure of how resistant material is to denting or scratching. Diamond, for example, is a very hard material. It is extremely difficult to dent or scratch a diamond. In contrast, it is very easy to scratch or dent most plastics.
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:
526.5 KN
Explanation:
The total head loss in a pipe is a sum of pressure head, kinetic energy head and potential energy head.
But the pipe is assumed to be horizontal and the velocity through the pipe is constant, Hence the head loss is just pressure head.
h = (P₁/ρg) - (P₂/ρg) = (P₁ - P₂)/ρg
where ρ = density of the fluid and g = acceleration due to gravity
h = ΔP/ρg
ΔP = ρgh = 1000 × 9.8 × 7.6 = 74480 Pa
Drag force over the length of the pipe = Dynamic pressure drop over the length of the pipe × Area of the pipe that the fluid is in contact with
Dynamic pressure drop over the length of the pipe = ΔP = 74480 Pa
Area of the pipe that the fluid is in contact with = 2πrL = 2π × (0.075/2) × 30 = 7.069 m²
Drag Force = 74480 × 7.069 = 526468.1 N = 526.5 KN