A heat engine does 210 J of work per cycle while exhausting 440 J of waste heat. Part A What is the engine's thermal efficiency?
1 answer:
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Answer:
gauge pressure is 133 kPa
Explanation:
given data
initial temperature T1 = 27°C = 300 K
gauge pressure = 300 kPa = 300 × 10³ Pa
atmospheric pressure = 1 atm
final temperature T2 = 77°C = 350 K
to find out
final pressure
solution
we know that gauge pressure is = absolute pressure - atmospheric pressure so
P (gauge ) = 300 × 10³ Pa - 1 × Pa
P (gauge ) = 2 × Pa
so from idea gas equation
................1
so
P2 = 2.33 × Pa
so gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 2.33 × - 1.0 ×
gauge pressure = 1.33 × Pa
so gauge pressure is 133 kPa
Answer:
a)
b)
c)
d)
Explanation:
Non horizontal pipe diameter, d = 25 cm = 0.25 m
Radius, r = 0.25/2 = 0.125 m
Entry temperature, T₁ = 304 + 273 = 577 K
Exit temperature, T₂ = 284 + 273 = 557 K
Ambient temperature,
Pipe length, L = 10 m
Area, A = 2πrL
A = 2π * 0.125 * 10
A = 7.855 m²
Mass flow rate,
Rate of heat transfer,
a) To calculate the convection coefficient relationship for heat transfer by convection:
Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.
c) The surface temperature of the pipe:
Smear coefficient of the pipe,
b) Heat loss from the pipe to the environment:
d) The required fan control power is 25.125 W as calculated earlier above