Respuesta :
Answer:
For [tex]20^{\circ}[/tex] - 5.556 lb/s
For [tex]15^{\circ}[/tex] - 7.4047 lb/s
Solution:
As per the question:
System Load = 96000 Btuh
Temperature, T = [tex]20^{\circ}[/tex]
Temperature rise, T' = [tex]15^{\circ}[/tex]
Now,
The system load is taken to be at constant pressure, then:
Specific heat of air, [tex]C_{p} = 0.24 btu/lb ^{\circ}F[/tex]
Now, for a rise of [tex]20^{\circ}[/tex] in temeprature:
[tex]\dot{m}C_{p}\Delta T = 96000[/tex]
[tex]\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 20} = 20000 lb/h = \frac{20000}{3600} = 5.556 lb/s[/tex]
Now, for [tex]15^{\circ}[/tex]:
[tex]\dot{m}C_{p}\Delta T = 96000[/tex]
[tex]\dot{m} = \frac{96000}{C_{p}\Delta T} = \frac{96000}{0.24\times 15} = 26666.667 lb/h = \frac{26666.667}{3600} = 7.4074 lb/s[/tex]
