Respuesta :
Answer:
A) [tex]v_2 = 2016.80 ft/s[/tex]
B) [tex]\Delta s = 0.006 Btu/lbm R[/tex]
Explanation:
Given data:
P-1 = 100 lbf/in^2
[tex]T_1 = 500[/tex] degree f
[tex]V_1 = 100 ft/s[/tex]
[tex]P_2 = 40 lbf/inc^2[/tex]
effeciency = 80%
from steady flow enerfy equation
[tex]h_1 +\frac{V_1^2}{2} = h_2 + \frac{V_2^2}{2}[/tex]
where h1 and h2 are inlet and exit enthalpy
for P1 = 100 lbf/in^2 and T1 = 500 degree F
[tex]H_1 = 1278.8 Btu/lbm[/tex]
[tex]s_1 = 1.708 Btu/lbm -R[/tex]
for P1 = 40 lbf/in^2
[tex]H_1 = 1193.5 Btu/lbm[/tex]
[tex]s_1 = 1.708 Btu/lbm -R[/tex]
exit enthalapy h_2
[tex]\eta = \frac{h_1 - h'_2}{ h_1 - h_2}[/tex]
[tex]0.80 = \frac{1278.8 - h'_2}{1278.8 -1193.5} = 1197.77 Btu/lbm[/tex]
from above equation
[tex]1278.8 \times 25037 + \frac{100^2}{2} = 1197.77 \times 25037 + \frac{v_2^2}{2} [/tex] [1 Btu/lbm = 25037 ft^2/s^2]
[tex]v_2 = 2016.80 ft/s[/tex]
b) amount of entropy
[tex]\Delta s = s_2 - s_1[/tex]
[tex]s_1 = 1.708 Btu/lbm -R[/tex]
at [tex]h_2 = 1197.77 Btu/lbm [\tex] and [tex]P_2 = 40 lbf/in^2[/tex]
[tex]s_2 is 1.714 Btu/lbm -R[/tex]
[tex]\Delta s = 1.714 - 1.708 = 0.006 Btu/lbm R[/tex]
