Answer:
a. The mass flow rate (in lbm/s) is 135lbm/s
b. The temperature (in o F) is 200.8°F
Explanation:
We assume that potential energy and kinetic energy are negligible and the control volume operates at a steady state.
Given
a. The mass flow rate (in lbm/s) is 135lbm/s
b.
m1 = Rate at inlet 1 = 125lbm/s
m2 = Rate at inlet 2 = 10lbm/s
The mass flow rate (in lbm/s) is calculated as m1 + m2
Mass flow rate = 125lbm/s + 10lbm/s
Mass flow rate = 135lbm/s
Hence, the mass flow rate (in lbm/s) is 135lbm/s
b. To calculate the temperature.
First we need to determine the enthalpy h1 at 14.7psia
Using table A-3E (thermodynamics)
h1 = 180.15 Btu/Ibm
h2 at 14.7psia and 60°F = 28.08 Btu/Ibm
Calculating h3 using the following formula
h3 = (h1m1 + h2m2) / M3
h3 = (180.15 * 125 + 28.08 * 10)/135
h3 = 168.8855555555555
h3 = 168.89 Btu/Ibm
To get the final temperature; we make use of table A-2E of thermodynamics.
Because h3 < h1, it means the liquid is at a compressed state.
The corresponding temperature at h3 = 168.89 is 200.8°F
The temperature (in o F) is 200.8°F
