Question

Nitrogen gas flows through a long, constant-diameter adiabatic pipe. It enters at 100 psia and 120°F and leaves at 50 psia and 70°F. Calculate the velocity of the nitrogen at the pipe’s inlet and outlet. The specific heat of nitrogen at the room temperature is cP = 0.248 Btu/lbm·R.

Answer 1

Answer:

Velocity at inlet (V1) = 515.12 ft/s

Velocity at outlet (V2) = 941.43 ft/s

Explanation:

We are given;

Cp = 0.248 Btu/lbm·R.

T1 = 120 °F = 120 + 460 Rankine

= 580 °R

T2 = 70 °F = 70 + 460 R = 530 °R

P1 = 100 psia

P2 = 50 psia

From energy balance equation;

E_in - E_out = ΔEsyst

ΔEsyst = 0.

Thus, E_in = E_out

So,

M'[h1 + V1²/2] = M'[h2 + V2²/2]

M' will cancel out and we have;

[h1 + V1²/2] = [h2 + V2²/2]

V1²/2 - V2²/2 = h2 - h1 - - - - - (eq 1)

Now, h2 - h1 can be expressed as Cp(T2 - T1)

So, V1²/2 - V2²/2 = Cp(T2 - T1)

Also, in ideal gas, we know that;

P1V1/T1 = P2V2/T2

Making V the subject to obtain;

(P1•V1•T2)/(P2•T1) = V2

Putting this for V2 in eq (1), we obtain;

V1²/2 - [(P1•V1•T2)/(P2•T1)]²/2 = Cp(T2 - T1)

Multiply through by 2;

V1² - [(P1•V1•T2)/(P2•T1)]² = 2Cp(T2 - T1)

Lets make V1 the subject;

V1² - V1²[(P1•T2)/(P2•T1)]² = 2Cp(T2 - T1)

V1²[1 - [(P1•T2)/(P2•T1)]²] = 2Cp(T2 - T1)

V1² = 2Cp(T2 - T1)/[1 - [(P1•T2)/(P2•T1)]²]

V1 = √[2•0.248(70 - 120)/[1 - [(100•530)/(50•580)]²]•(25037ft²/lb²/Btu/lbm·R)]

I used the °F for the numerator temperature while I used °R for the denominator temperature so as for the units to cancel out properly.

V1 = √[-24.8)/[1 - 3.34)]•25037]

V1 = √[-24.8)/-2.34)]•25037]

V1 = √[(24.8/2.34)•25037]

V1 =√265349.4017094017

V1 = 515.12 ft/s

Now, to find V2, we recall that,

(P1•V1•T2)/(P2•T1) = V2

We will use the rankine value of temperature.

Thus;

V2 = (100•515.12•530)/(50•580)

= 27301360/29000 = 941.43 ft/s

Answer 2

Answer:

$V_{1} =515feet/seconds$

$V_{2} =941.21feet/seconds$

Explanation:

We can use balance energy expression.

$E_{in} -E_{out} =$Δ$E_{system}$

we have to combine ideal gas and mass balance, so that $m_{1}$ and $m_{2}$ are equal.

$m_{1} =m_{2}$

therefore

$\frac{A_{1} V_{1} }{v_{1} } =\frac{A_{2} V_{2} }{v_{2} }$

make $V_{2}$ the subject of the formula

$V_{2} =\frac{A_{1} v_{2} }{A_{2}v_{1} } (V_{1} )$

substitute $V_{2}$ into the energy balance equation

$V_{1} =[\frac{2c_{p}(T_{2}-T_{1} ) }{1-(\frac{T_{2}P_{1} }{T_{1}P_{2} } )^{2} } ]^{\frac{1}{2} }$

now we can substitute the values

$V_{1} =[\frac{2*0.248(70-120)}{1-(\frac{530*100}{580*50} )^{2} } ]^{\frac{1}{2} }$

$V_{1} =515ft/s$

Now to find Velocity:

$V_{2} =\frac{T_{2} P_{1} }{T_{1}P_{2} } (V_{1} )$

substitute the values

$V_{2} =\frac{530*100}{580*50} (515)\\V_{2} =\frac{53000}{29000} (515)\\V_{2} =941.21feet/seconds$