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Answer:
total weight of aggregate = 5627528 lbs = 2814 tons
Explanation:
we get here volume of space to be filled with aggregate that is
volume = 2000 × 48 × 0.5
volume = 48000 ft³
now filling space with aggregate of the density that is
density = 0.95 × 119.7
density = 113.72 lb/ft³
and dry weight of this aggregate is
dry weight = 48000 × 113.72
dry weight = 5458320 lbs
we consider here percent moisture is by weigh
so weight of moisture in aggregate will be
weight of moisture = 0.031 × 5458320
weight of moisture = 169208 lbs
so here total weight of aggregate is
total weight of aggregate = 5458320 + 169208
total weight of aggregate = 5627528 lbs = 2814 tons
Answer:
The overview of the given scenario is explained in explanation segment below.
Explanation:
- The inception of cavitation, that further sets the restriction for high-pressure and high-free operation, has always been the matter of substantial experimental study over the last few generations.
- Cavitation inception would be expected to vary on the segment where the local "PL" pressure mostly on segment keeps falling to that are below the "Pv" vapor pressure of the fluid and therefore could be anticipated from either the apportionment of the pressure.
⇒ A cavitation number is denoted by "σ" .
Answer:
η=0.19=19% for p=14.7psi
η=0.3=30% for p=1psi
Explanation:
enthalpy before the turbine, state: superheated steam
h1(p=200psi,t=500F)=2951.9KJ/kg
s1=6.8kJ/kgK
Entalpy after the turbine
h2(p=14.7psia, s=6.8)=2469KJ/Kg
Entalpy before the boiler
h3=(p=14.7psia,x=0)=419KJ/Kg
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the efficiency for a simple rankine cycle is
η=
η=(2951.9KJ/kg-2469KJ/Kg)/(2951.9KJ/kg-419KJ/Kg)
η=0.19=19%
second part
h2(p=1psia, s=6.8)=2110
h3(p=1psia, x=0)=162.1
η=(2951.9KJ/kg-2110KJ/Kg)/(2951.9KJ/kg-162.1KJ/Kg)
η=0.3=30%