Air enters a nozzle at 250degreeF It cools to 40degreeF by t

Air enters a nozzle at 250degreeF. It cools to 40degreeF by the time it reaches the narrow end of the nozle. a) Find the enthalpy decrease using the fact that c_p = 0.24 Btu/lb_m^0R for air. b) Determine the speed at the narrow end of the nozzle. As usual, the initial speed can be neglected. 5. In your chemistry class you have probably learned the ideal gas law in the form pV = nR_chemT. In this class, we use pv = R_thermoT. So where did the n go? Recall that n represents the number of moles in the gas sample. By definition, the molecular weight MW of a substance is the mass of a single mole. a) Using these facts, write an equation relating n to MW and the mass of the sample m. b) Plug your formula for n into the chemistry equation. Then turn V into v using the appropriate equation (it involves mass). What you have left will look like the thermodynamics version of the ideal gas law. Use it to show how R_chem is related to R_thermo.

Solution

4 a) Initial Enthalpy = Cp * Ti

       Final Enthalpy = Cp * Tf

       Enthalpy Change = Cp( Tf - Ti)   = 0.24 * ( 250 - 40 )   = 50.4 BTU/lbm

4 b) Enthalpy Change = Change in velocity

        50.4   =   v^2 /2

        v = 10.04 ft/s

5 a) no of moles =   m / MW   ( where m is mass of sample and MW is molecular weight)

5 b) P * V = m * Rchem * T / MW

        P * V/m = Rchem * T / MW

        P * v = Rchem * T / MW                (where v = V/m

        therefore Rthermo = Rchem / MW