Q2. Air at 400°C and 1.7 bar flows through a horizontal 8.2-cm D pipe at a velocity of 50.0 m/s. I. Calculate Éx (W)? Assuming ideal gas behavior and Mol. Wt (Air)=29 g/mol. [2 Marks]
II. If the air is cooled to 250°C at constant pressure, what is AĖ,? [2 Marks]
Why would it be incorrect to say that the rate of transfer of heat to the gas must equal the rate of
change of kinetic energy? [1 Mark]
(Felder and Rousseau, 2005)

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.

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