Answer:
The answer is 1823.9
Explanation:
Solution
Given that:
m = 5.5 kg/s
= m₁ = m₂ = m₃
The work carried out by the energy balance is given as follows:
m₁h₁ = m₂h₂ +m₃h₃ + w
Now,
By applying the steam table we have that<
p₃ = 50 kPa
T₃ = 100°C
Which is
h₃ = 2682.4 kJ/KJ
s₃ = 7.6953 kJ/kgK
Since it is an isentropic process:
Then,
p₂ = 500 kPa
s₂=s₃ = 7.6953 kJ/kgK
which is
h₂ =3207.21 kJ/KgK
p₁ = 3HP0
s₁ = s₂=s₃ = 7.6953 kJ/kgK
h₁ =3854.85 kJ/kg
Thus,
Since 5 % of this flow diverted to p₂ = 500 kPa
Then
w =m (h₁-0.05 h₂ -0.95 )h₃
5.5(3854.85 - 0.05 * 3207.21 - 0.95 * 2682.4)
5.5( 3854.83 * 3207.21 - 0.95 * 2682.4)
5.5 ( 123363249.32 -0.95 * 2682.4)
w=1823.9
Answer:
temperature T2 = 826.9°C
= -1142.7 kJ/kg
Explanation:
given data
initial state temperature = 120°C
final state pressure p1 = 1 bar
pressure p2 = 100 bar
solution
we use here super heated water table A6 that is
specific internal energy u1 = 2537.3 kJ/kg
specific entropy s1 = 7.4668 kJ/kg.K
and
here specific entropy stage 1 = stage 2
so for specific entropy and pressure 100 bar
specific internal energy u2 = 3680 kJ/kg
and temperature T2 = 826.9°C
so here now we get specific work of steam is
ΔU = -W ...........1
m ( u1 - u2) = W
= u1 - u2
= 2537.3 - 3680
= -1142.7 kJ/kg
Answer:
Q = 14.578 m³/s
Explanation:
Given
We use the Manning Equation as follows
Q = (1/n)*A*(∛R²)*(√S)
where
- Q = volumetric water flow rate passing through the stretch of channel (m³/s for S.I.)
-
A = cross-sectional area of flow perpendicular to the flow direction, (m² for S.I.)
-
S = bottom slope of channel, m/m (dimensionless) = 2.5% = 0.025
-
n = Manning roughness coefficient (empirical constant), dimensionless = 0.023
-
R = hydraulic radius = A/P (m for S.I.) where
:
-
A = cross-sectional area of flow as defined above,
-
P = wetted perimeter of cross-sectional flow area (m for S.I.)
we get A as follows
A = (B*h)/2
where
B = 5 m (the top width of the flowing channel)
h = (B/2)*(m) = (5 m/2)*(1/2) = 1.25 m (the deep)
A = (5 m*1.25 m/2) = 3.125 m²
then we find P
P = 2*√((B/2)²+h²) ⇒ P = 2*√((2.5 m)²+(1.25 m)²) = 5.59 m
⇒ R = A/P ⇒ R = 3.125 m²/5.59 m = 0.559 m
Substituting values into the Manning equation gives:
Q = (1/0.023)*(3.125 m²)*(∛(0.559 m)²)*(√0.025)
⇒ Q = 14.578 m³/s
Answer:
It will create a massive drag and pretty much stop the motor.
Explanation:
The initial void ratio is the <em>parameter </em>which is used to show the structural foundations for each <em>specimen of sand </em>so that the method and speed of compression would be <em>measured</em>.
Relative density is the mass per unit volume of each specimen of sand which is <em>measured </em>and it has to do with the<em> relative ratio</em> of the density of the sand.
Unit weight is the the exact weight per cubic foot of the sand which is measured.
Please note that your question is incomplete so I gave you a general overview to help you better understand the concept
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