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

Answer:

A) $v_2 = 2016.80 ft/s$

B) $\Delta s = 0.006 Btu/lbm R$  

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

$h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}$

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

$\eta = \frac{h_1 - h'_2}{ h_1 - h_2}$

$0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm$

from above equation

$1278.8 \times 25037 + \frac{100^2}{2} = 1197.77 \times 25037 + \frac{v_2^2}{2}$                   [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 $h_2 = 1197.77 Btu/lbm [\tex] and [tex]P_2 = 40 lbf/in^2$

$s_2 is 1.714 Btu/lbm -R$

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