Question

Problem A: 15 Ibm of saturated vapor steam at 10 psia is contained in a frictionless piston-cylinder arrangement. The steam is heated at constant pressure to 400° F. a. Calculate the work done by the steam on the piston, in BTUS. b. Calculate the amount of heat delivered to the steam for this process, in BTUS. c. Determine the change in internal energy of the steam, in BTUs.

Answer 1

the value of $v_{1}$   ,$h_{1}$ and $v_{2}$ , $h_{2}$ is taken from the steam table at saturated vapour state at 10 psia and superheated state at

400$^{\circ}}$F respectively.
Giveni maan of steam = 1566m Preasure = 1opsia Tfinal = 400°F Soni Saturated water hapour incylinder is heated at constant precaure i 9. The pressure remains constant during this prodeon, So the specific volumes at initial & final statex are P = 10 Psiq ? V = Va = 38.425f + /66m Sat. Vapoor) Ti=193.16° F). The values are I from Steam P = 10psia? V = 510035ft/m . tatle [Cengel] T =400°F) by the definition of work Wsteam ſpava Pl_2-V) = 10 (12:61) – 126.1 Psicw f13 lom Hs 18u = 5:4039spaia.ft3 Wisteam=2333 BtYbm Work done by steam = 23:33x 15 = 30 (W = 350.02 Btu Scanned with CamScanner

b. The proceon is at constant preanurei Tds=dh - vepro dis=do Isa=dhl P = lo pala 2 h, = 1143.1 Btu/ebm? Sat.vapour ! 1. I values Steam T= 40°* * from table P=10 Pela? he = 1240.6 Btu/Abm J - Co Så- ho-hi = 1240:6-1143.1 =97.5 Bulldom Heat Supplied to system; Q = 97.5 x 15 x BIU $Q = 1462.5 Bto for change in internal energy. from 13t Law of Thermodynamica 8Q = du tsW du=SQ-sw = 1462.5 - 350-02 BAU du = 1112.48 Bto Ang Gs Scanned with CamScanner

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