A heat engine that receives heat from a furnace at 1200°C and rejects waste heat to a river at 20°C has a thermal efficiency of
1 answer:
Answer:
second-law efficiency = 62.42 %
Explanation:
given data
temperature T1 = 1200°C = 1473 K
temperature T2 = 20°C = 293 K
thermal efficiency η = 50 percent
solution
as we know that thermal efficiency of reversible heat engine between same temp reservoir
so here
efficiency ( reversible ) η1 = 1 - ............1
efficiency ( reversible ) η1 = 1 -
so efficiency ( reversible ) η1 = 0.801
so here second-law efficiency of this power plant is
second-law efficiency =
second-law efficiency =
second-law efficiency = 62.42 %
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Answer:
a)Δs = 834 mm
b)V=1122 mm/s
Explanation:
Given that
a)
When t= 2 s
s= 114 mm
At t= 4 s
s= 948 mm
So the displacement between 2 s to 4 s
Δs = 948 - 114 mm
Δs = 834 mm
b)
We know that velocity V
At t= 5 s
V=1122 mm/s
We know that acceleration a
a= 90 t
a = 90 x 5
Answer:
5.328Ibm/hr
Explanation:
Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)
through prior knowledge of two other properties such as pressure and temperature.
for this case we can define the following equation for mass flow using the first law of thermodynamics
where
Q=capacity of the radiator =5000btu/hr
m = mass flow
then using thermodynamic tables we found entalpy in state 1 and 2
h1(x=0.97, p=16psia)=1123btu/lbm
h2(x=0, p=16psia)=184.5btu/lbm
solving