Thermodynamics Change in Gas Temperature along pipe

[daz18983]

Homework Statement

Nitrogen beginning at 50^oC is pumped along a 30m length of Stainless Steel pipe at a flow rate of 1.5 litres per minute. The Pipeline is at ambient temperature 20^oC. Find the temperature of the Nitrogen upon it leaving the end of the pipeline.

Pipe OD - 0.00635m
Pipe ID - 0.00457m
Pressure - 1bar

The following data is taken from a copy of 'An Engineering Data Book' -

C_p - Nitrogen at 20oC and 1 bar is 1.04 kJ/(kgK)

Homework Equations

From 'An Engineering Data Book' and 'Thermodynamics an Engineering Approach 4th Edition' i believed that i could work at the figures from the following equations -

Q_dot /l = 2Pi.k.(T₂-T₁)/In(r₂/r₁)

Q_dot = Heat Loss l = Length k = Thermal Conductivity T = Temperatures r = Radius

W_dot - Q_dot = m_dot.C_p.(T₂-T₁)

W_dot = Work Energy In Q_dot = Heat Loss m_dot = mass flow rate C_p = Specific Heat T = Temperatures

The Attempt at a Solution

using k = 15 (Which is the k of Stainless Steel)

I used Q_dot /l = 2Pi.k.(T₂-T₁)/In(r₂/r₁)

which gave me a figure of 8.653 x 10³ for Q_dot

Then to calculate m_dot i used the following -

m_dot = V_dot / v

where V_dot = Volume flow rate and v = specific volume

V_dot = V / delta t

where V = Volume which is 1.5 litres and delta t = 60 seconds (take from flow rate)

therefore Vdot = 0.025 l/s = 25 x 10^-5 m³/s

v = R.T₁ / P

where R = Gas Constant which is 0.294 kJ/(kg.K) T₁ is temperate 1 which is 323K and P is pressure which is 1 bar (1 x 10⁵Pa)

this gives v to be 9.4962 x 10^-4 m³/kg

using these m_dot becomes 26.32632 x 10^-3 kg/s

Now using W_dot - Q_dot = m_dot.C_p.(T₂-T₁)
I rearrage to get T₂ on its own

therefore T₂ = (-Q_dot / m_dot.C_p) + T₁
note that W_dot = 0 and so has been removed

Therefore i get an answer of 50.3^oC which is obviously WAY wrong? Can someone please help me with this. Even just a point in the right direction of the correct equations. This is not a homework or coursework question this an engineering question at work and I'm not too sure where to go with it.

 

[Chestermiller]

This problem should be solved using the following differential heat balance equation: $$\dot{m}C_p\frac{dT}{dx}=-U\pi D_i(T-T_0)$$ where U is the overall heat transfer coefficient, x is distance along the pipe, and Di is the inside dimeter of the pipe. The overall heat transfer coefficient U should include the thermal boundary layer resistance inside the pipe, the conductive resistance of the pipe wall, and the thermal boundary layer resistance outside the pipe. Your analysis only includes the pipe wall conductive resistance, so it can only provide a lower bound to the outlet temperature. The initial condition for the about equation is T = 50 at x = 0, and a boundary condition $T_0=20$.

Your calculation of the mass flow rate is incorrect. The density of nitrogen at 50 C and 1 atm. is $$\rho=\frac{PM}{RT}=\frac{(1)(28)}{(0.08215)(323)}=1.057 \ gm/l$$So the mass flow rate is $$\dot{m}=\frac{(1.5)(1.057)}{60}=0.0264\ gm/s=2.64\times 10^{-5}\ kg/s$$

For the bounding approximation you are using, in which the inner and outer thermal resistances are neglected, the overall heat transfer coefficient is given by $$U=\frac{2k}{D_i\ln{(D_0/D_i)}}$$where k is the thermal conductivity of the metal wall.