The modifications of superheating and reheat for a vapor power plant are specifically better for the operation which of the following components b.Boiler.
<h3>What are the primary additives in the vapour strength cycle?</h3>
There are 5 steam strength cycles: The Carnot cycle, the easy Rankine cycle, the Rankine superheat cycle, the Rankine reheat cycle and the regenerative cycle.
- Central to expertise the operation of steam propulsion is the primary steam cycle, a method wherein we generate steam in a boiler, increase the steam via a turbine to extract work, condense the steam into water, and sooner or later feed the water again to the boiler.
- Reheat now no longer best correctly decreased the penalty of the latent warmness of vaporization in steam discharged from the low-stress quit of the turbine cycle, however, it additionally advanced the first-rate of the steam on the low-stress quit of the mills via way of means of decreasing condensation and the formation of water droplets inside the turbine.
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Answer:
manage country economies in which is help responsible by a government in place.
Explanation:
Answer:
The duration of the consolidation process for the same clay is 32 min
Explanation:
for clay 1:
t1=0
H1=thickness=2 cm
for the clay 2:
t2=?
H2=2 cm
The time factor is equal to:
where Cv is the coefficient of consolidation
if Cv is constant, we have:
Clearing t2:
t2=32 min
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Answer:
P = 18035.25 N
Explanation:
Given
D = 10.4 mm
ΔD = 3.2 ×10⁻³ mm
E = 207 GPa
ν = 0.30
If
σ = P/A
A = 0.25*π*D²
σ = E*εx
ν = - εz / εx
εz = ΔD / D
We can get εx as follows
εz = ΔD / D = 3.2 ×10⁻³ mm / 10.4 mm = 3.0769*10⁻⁴
Now we find εx
ν = - εz / εx ⇒ εx = - εz / ν = - 3.0769*10⁻⁴ / 0.30 = - 1.0256*10⁻³
then
σ = E*εz = (207 GPa)*(-1.0256*10⁻³) = - 2.123*10⁸ Pa
we have to obtain A:
A = 0.25*π*D² = 0.25*π*(10.4*10⁻3)² = 8.49*10⁻⁵ m²
Finally we apply the following equation in order o get P
σ = P/A ⇒ P = σ*A = (- 2.123*10⁸Pa)*(8.49*10⁻⁵ m²) = 18035.25 N
