A Carnot heat engine absorbs 235 KW of heat from a heat source and rejects 164 KW to the atmosphere. Determine the thermal effic
1 answer:
Answer:
43.2%
Explanation:
Given that,
Heat absorbed by a carnot heat engine,
Heat rejected to the atmosphere,
We need ti find the thermal efficiency of the heat engine. It is equal to the ratio of output work to the energy supplied. Its mathematical form is given by :
or
The egative value of efficiency shows work is done by the engine.
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Answer:
The hydrostatic force of 313920 N is acted on each wall of the swimming pool and this force is acted at 1 m from the ground. The hydrostatic force is quadruple if the height of the walls is doubled.
<u>Explanation:</u>
To calculate force on the walls of swimming pool whose dimensions are given as <em>8-m-long, 8-m-wide, and 2-m-high</em>. We know that formula for hydrostatic force is
, we know ρ=density of fluid=1000
,
g=acceleration due to gravity=9.81 , h=height of the pool=2 m and l=length of the pool=8 m.
hydrostatic force on each wall= = 313920 N.
<em>The distance at which hydrostatic force is acted is half of the height of the swimming pool. </em>
At 1 m from the ground this hydrostatic force is acted on each wall.
The force is <em>quadruple if the height of the walls of the pool is doubled</em> this is because, the<em> height is doubled and taken as h=4 m</em> and substitute in the equation =
=
= 1255680 N. This is 4 times 313920 N.
<u>Solution and Explanation:</u>
Volume of gas stream = 1000 cfm (Cubic Feet per Minute)
Particulate loading = 400 gr/ft3 (Grain/cubic feet)
1 gr/ft3 = 0.00220462 lb/ft3
Total weight of particulate matter =
Cyclone is to 80 % efficient
So particulate remaining =
emissions from this stack be limited to = 10.0 lb/hr
Particles to be remaining after wet scrubber = 10.0 lb/hr
So particles to be removed = 685.7136- 10 = 675.7136
Efficiency = output multiply with 100/input = 98.542 %