Question

Consider 4.8 pounds per minute of water vapor at 100 lbf/in2, 500 oF, and a velocity of 100 ft/s entering a nozzle operating at steady state and expanding adiabatically to the exit, where the pressure is 40 lbf/in2. The isentropic nozzle efficiency is 80.0%. Determine the velocity of the steam at the exit, in ft/s, and the rate of entropy production, in Btu/min·oR.

Answer 1

A)
`v_2 = 2016.80 ft/s`

B)
`\Delta s = 0.006 Btu/lbm R`

Step-by-step explanation:

Given data:

P-1 = 100 lbf/in^2

`T_1 = 500` degree f

`V_1 = 100 ft/s`

`P_2 = 40 lbf/inc^2`

effeciency = 80%

from steady flow enerfy equation

`h_1 +(V_1^2)/(2) = h_2 + (V_2^2)/(2)`

where h1 and h2 are inlet and exit enthalpy

for P1 = 100 lbf/in^2 and T1 = 500 degree F

`H_1 = 1278.8 Btu/lbm`

`s_1 = 1.708 Btu/lbm -R`

for P1 = 40 lbf/in^2

`H_1 = 1193.5 Btu/lbm`

`s_1 = 1.708 Btu/lbm -R`

exit enthalapy h_2

`\eta = (h_1 - h'_2)/( h_1 - h_2)`

`0.80 = (1278.8 - h'_2)/(1278.8 -1193.5) = 1197.77 Btu/lbm`

from above equation

`1278.8 * 25037 + (100^2)/(2) = 1197.77   * 25037 + (v_2^2)/(2)` [1 Btu/lbm = 25037 ft^2/s^2]

`v_2 = 2016.80 ft/s`

b) amount of entropy

`\Delta s = s_2 - s_1`

`s_1 = 1.708 Btu/lbm -R`

at
`h_2 = 1197.77 Btu/lbm [\tex]  and [tex]P_2 = 40 lbf/in^2`

`s_2 is 1.714 Btu/lbm -R`

`\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R`