Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
Answer:
investment 10 years from now is $1,238,000
.
Explanation:
given data
sum = $500,000
rate = 12% =0.12
total time = 10 year
solution
as present value After 2 years from now is $500,000
so time period is now = 8 year ( 10 - 2 )
so we apply future value formula that is
Future value = present value × ............1
put here value we get
Future value = $500,000 ×
Future value = $500,000 × 2.476
Future value = $1,238,000
so investment 10 years from now is $1,238,000
.
Answer:
Under no circumstances
Explanation:
I'm not 100% sure why, but I remember hearing that you're not suposed to go over the speed limit no matter what
Answer:
Gc(s) =
Explanation:
comparing the standard approximation with the plot attached we can tune the PI gains so that the desired response is obtained. this is because the time requirement of the setting is met while the %OS requirement is not achieved instead a 12% OS is seen from the plot.
attached is the detailed solution and the plot in Matlab
Answer:
The answer is "".
Explanation:
The amount of kilograms, which travel in a thick sheet of hydrogen:
calculating the value of
calculating the value of