Air flows at steady state through a horizontal, insulated pipe with inside diameter of 4 cm. A pressure drop results from flow through a partially opened valve. Just upstream from the valve, the pressure is 7 bar, the temperature is 45°C, and the average velocity is 20 If the pressure just downstream from the valve is 1.3 bar, what is the temperature? Assume for air that PV/T is constant, = (5/2)R, and = (7/2)R.

Question

What will be an ideal response?

Answer 1

Because the pipe is insulated, we can assume Q is zero, and because the pipe and valve presumably have no moving parts, W is also zero. Furthermore, the pipe is horizontal, so there is no change in gravitational potential energy between the inlet and the outlet streams. Thus, we can write the energy balance for an open system with one inlet and

one outlet. Note that this equation, as written, is per unit mass.



The difference in enthalpy between the inlet and outlet streams can be written in terms of the temperature change:

?H = Cp?T. Substituting that in, we have



The mass flow rates in and out must be equal (at steady state, with no chemical reactions), and the mass flow rate is the velocity times the cross-sectional area divided by the specific volume (volumetric flow rate divided by specific volume): m = uA/V. So, we have



are the same, and because we have PV/T = a constant, we can write



So, we can then write



Putting in the numbers, we have so,



Where ?T is in K or °C.

Also, we have = 3.5?R = 3.5?8.314 J/(mol K) = 29.10 J/(mol K). As given, this is the molar heat capacity. To get the specific heat, we divide by the average molecular weight of nitrogen (0.02897 kg/mol) to get = 1004.5 J/(kg K). Putting this all together, we see that



where both terms have units of J/kg, and ?T has units of K or °C. Multiplying things out gives

0.057291 ? + 1040.9?TT + 5398.8 = 0

Applying the quadratic formula to this gives ?T = –5.19 K = –5.19°C, so the downstream temperature is 39.8°C.