Answer:
The correct answer is (I) 290.81 W (II) 83.413 W (III) It is not correct to say that the rate of heat transfer must be the same to the heat of change in kinetic energy
Explanation:
Solution
Recall that:
Pressure (p) = 1.7 bar
Temperature (T₁) = 400°C which is = 673k
The velocity (v) = 50.0 m/s.
The pipe diameter (D)= 8.2 cm approximately 8.2 * 10 ^⁻2 m
The molecular air weight (M)= 29 g/mol
Suppose Air is seen as an ideal gas
where pv = mrT
p =(m/v) r T = p = ρrT
So,
r = the characteristics of gas constant (R/m)
p = pressure
R =The universal gas constant
T = temperature
ρ = density which is (kg/m³)
R is 8.314 J/mole -k
Then
1.7 * 10 ^5 = ρ * (8.314 /29) * 673
The density ρ = 881.09 g/ m³
(I) The mass flow rate = ρAV
thus,
m = 881.09 *π/4 ( 8.2 * 10^⁻2)² * 50
Therefore m = 232.65 g/s
We already know that
k= 1/2 mv²
k =1/2 *232.65/1000 * (50)²
so at 400°C k = 290.81 W
(II) Now in solving the process of the constant pressure we recall that
P = ρrT
Air is cooled to 250°C
p/r = ρT this is constant
So,
ρ₁T₂ = ρ₂T₂
881.09 * 673 = ρ₂ * 523
ρ₂ = 1133.79 g/m³
Thus,
m = ρ₂AV = 1133.79 * π /4 (8.2 * 10^ ⁻2) * 50
Hence m = 299.37 g/s
now,
k =1.2 mv² = 1.2 *(299.37)/1000 * (50)²
At 250°C, k = 374.22 W
Thus,
Δk = k ( 250°c) - k ( 400°c)
Δk = 374.22 - 290.81
Therefore,
Δk=83.413 W
(III) The steady state formula is given below
Q = W + ΔkE +ΔPE + ΔH
Now,
W = work (shaft)
Q =The rate of transfer of heat
ΔkE = The change in kinetic energy
ΔPE= The change in potential energy
ΔH =Change in enthalphy
For no shaft work, W =0
The horizontal pipe ΔPE = 0
Therefore,
The rate of heat transfer is explained as follows:
Q =ΔkE + ΔH
Because of the enthalphy, Q is not equal to ΔkE
Finally, it is not correct to say that the rate of heat transfer must be the same to the heat of change in kinetic energy.
