Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg
Solution :
Frequency may be defined as the number of observation or number of waves that is taken in per unit time. The unit of frequency is Hertz or Hz.
It is given that :
Successive harmonic frequencies, f = 52.2 Hz
and f' = 60.9 Hz
Therefore, fundamental frequency, F = f' - f
F = 60.9 - 52.2
F = 8.7 Hz
Therefore the string which is fixed at both the ends forms all the harmonics.
The colder the more likely it is to become a liquid
Answer:
z = 3,737 10⁵ m
Explanation:
a) As they indicate that the atmosphere behaves like an ideal gas, we can use the equation
P V = n R T
P = (n r / V) T
We replace
P = (n R / V) T₀
b) Let's apply this equation in the points
Lower
.z = 0
P₀ = 610 Pa
P₀ = (nR / V) T₀
Higher.
P = 10 Pa
P = (n R / V) T₀ e^{- C z}
We replace
P = P₀ e^{- C z}
e^{- C z} = P / P₀
C z = ln P₀ / P
z = 1 / C ln P₀ / P
Let's calculate
z = 1 / 1.1 10⁻⁵ ln (610/10)
z = 3,737 10⁵ m