Nitrogen flows at steady state through a horizontal, insulated pipe with inside diameter of 1.5(in). A pressure drop results from flow through a partially opened valve. Just upstream from the valve the pressure is 100(psia), the temperature is 120(°F), and the average velocity is 
If the pressure just downstream from the valve is 20(psia), what is the temperature? Assume for air that PV/ T is constant, 
= (5/2)R, and 
= (7/2)R.
What will be an ideal response?
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 like it is written in equation 2.31. 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 
= 120 °F = 322.04 K, 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 molecular weight of nitrogen (0.02801 kg/mol) to get 
= 1039 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
0.004479 ? 
+ 1041.8?TT + 445.9 = 0
Applying the quadratic formula to this gives ?T = –0.428 K = –0.77°F, so the downstream temperature is 119.23°F
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