Question 356.
A rigid tank contains 20 lbm of air at 50 psia and 80° F. The air is now heated until its pressure doubles. Determine: (a) the volume of the tank and (b) the amount of heat transfer.
Answer 356.
R = 1545.35 ft-lbf / (lbmol-Rankine)
(http://www.katmarsoftware.com/gconvals.h…
PV = nRT
P = 50psia = 50 * 144 lbf/sqft abs.
=> 50* 144 V = (20/28.9644) * 1545.35 * (80+460)
=> V = 80.0 cu. ft.
20 lb mass
= 20 * 453.4 g
Cv = 5 cal / g-mol - K
When preesure doubles with volume constant, temperature doubles
Temp. = 80 degrees F = (80-32) / 1.8 degrees C = 26.7 degrees C = 299.8 K
=> increase in temperature = 299.8 K
Heat absorbed = change in internal energy as work done is zero at constant volume
=> Heat absorbed
nCv dT
= (20 * 453.4/28.966) * 5 * 299.8 cal.
= 469272 cal.
= 469272/252 Btu
= 1862 Btu.
Link to YA!
A rigid tank contains 20 lbm of air at 50 psia and 80° F. The air is now heated until its pressure doubles. Determine: (a) the volume of the tank and (b) the amount of heat transfer.
Answer 356.
R = 1545.35 ft-lbf / (lbmol-Rankine)
(http://www.katmarsoftware.com/gconvals.h…
PV = nRT
P = 50psia = 50 * 144 lbf/sqft abs.
=> 50* 144 V = (20/28.9644) * 1545.35 * (80+460)
=> V = 80.0 cu. ft.
20 lb mass
= 20 * 453.4 g
Cv = 5 cal / g-mol - K
When preesure doubles with volume constant, temperature doubles
Temp. = 80 degrees F = (80-32) / 1.8 degrees C = 26.7 degrees C = 299.8 K
=> increase in temperature = 299.8 K
Heat absorbed = change in internal energy as work done is zero at constant volume
=> Heat absorbed
nCv dT
= (20 * 453.4/28.966) * 5 * 299.8 cal.
= 469272 cal.
= 469272/252 Btu
= 1862 Btu.
Link to YA!
No comments:
Post a Comment