Problem 1.1 (a) Wi-fPdV=PV2. Next, to calculate Tf we note that from (i) to (f) we have

dM dt = m

du dt mh1

where in is the instantaneous flowrate into the cylinder, and M and U are the mass and energy inventories of the system (the "system" is the cylinder volume). Integrating in time,

Mr-M₁= mdt

U-U-PV2+h (M-M)

(1)

and recognizing that U₁ = 0 and M₁ = 0, the first law reduces to

UfMfh1-P1V2

(1')

For the "ideal gas" working fluid we write

Ur=Mfcv (Tr-To)

h₁=cv (T1-To) + PV1

hence, eq. (1') becomes

MFC (TTO)=Mf [cv (TTO) + PV1]-P1 V2

Noting that V2 =Mfvf and dividing everything by Mr yields

or

CvTf+P1 Vf c T + PV

cvTf+RTCT + RT

in other words, Tf=T1. The final ideal-gas mass admitted is