Answer:
a)exit velocity of the steam, V2 = 2016.8 ft/s
b) the amount of entropy produced is 0.006 Btu/Ibm.R
Explanation:
Given:
P1 = 100 psi
V1 = 100 ft./sec
T1 = 500f
P2 = 40 psi
n = 95% = 0.95
a) for nozzle:
Let's apply steady gas equation.
![h_1 + \frac{(v_1) ^2}{2} = h_2 + \frac{(v_2)^2}{2}](https://tex.z-dn.net/?f=%20h_1%20%2B%20%5Cfrac%7B%28v_1%29%20%5E2%7D%7B2%7D%20%3D%20h_2%20%2B%20%5Cfrac%7B%28v_2%29%5E2%7D%7B2%7D%20)
h1 and h2 = inlet and exit enthalpy respectively.
At T1 = 500f and P1 = 100 psi,
h1 = 1278.8 Btu/Ibm
s1 = 1.708 Btu/Ibm.R
At P2 = 40psi and s1 = 1.708 Btu/Ibm.R
1193.5 Btu/Ibm
Let's find the actual h2 using the formula :
solving for h2, we have
Take Btu/Ibm = 25037 ft²/s²
Using the first equation, exit velocity of the steam =
![(1278.8 * 25037) + \frac{(100)^2}{2}= (1197.77*25037)+ \frac{(V_2)^2}{2}](https://tex.z-dn.net/?f=%20%281278.8%20%2A%2025037%29%20%2B%20%5Cfrac%7B%28100%29%5E2%7D%7B2%7D%3D%20%281197.77%2A25037%29%2B%20%5Cfrac%7B%28V_2%29%5E2%7D%7B2%7D)
Solving for V2, we have
V2 = 2016.8 ft/s
b) The amount of entropy produced in BTU/ lbm R will be calculated using :
Δs = s2 - s1
Where s1 = 1.708 Btu/Ibm.R
At h2 = 1197.77 Btu/Ibm and P2 =40 psi,
S2 = 1.714 Btu/Ibm.R
Therefore, amount of entropy produced will be:
Δs = 1.714Btu/Ibm.R - 1.708Btu/Ibm.R
= 0.006 Btu/Ibm.R