Answer:
The answer to your question is Decrease
-0 m/s
- average velocity=displacement/time
- the runners displacement is zero so her average velocity must be zero
Hot combustion gases are accelerated in a 92% efficient
adiabatic nozzle from low velocity to a specified velocity. The exit velocity
and the exit temp are to be determined.
Given:
T1 = 1020 K à
h1 = 1068.89 kJ/kg, Pr1 = 123.4
P1 = 260 kPa
T1 = 747 degrees Celsius
V1 = 80 m/s ->nN = 92% -> P2
= 85 kPa
Solution:
From the isentropic relation,
Pr2<span> = (P2 / P1)PR1 = (85
kPa / 260 kPa) (123.4) = 40.34 = h2s = 783.92 kJ/kg</span>
There is only one inlet and one exit, and thus, m1 =
m2 = m3. We take the nozzle as the system, which is a
control volume since mass crosses the boundary.
h2a = 1068.89 kJ/kg – (((728.2 m/s)2 –
(80 m/s)2) / 2) (1 kJ/kg / 1000 m2/s2) =
806.95 kJ/kg\
From the air table, we read T2a = 786.3 K
Answer:
The Surface heat flux is -9205 W/m^2
Explanation:
Explanation is in the following attachment
Answer:
The work done on the gas is equal to the area under the curve pv diagram w = area of triangle = 1/2 (base)(height) = 1/2 (BC)(Ac) = 1/2 (3v - v)(3p - p) = 1/2 (9 vp - 3 vp - 3vp + vp) = 4 vp/2 W = 2 vp
Check attachment for the diagrammatic representation
