Answer
a)
$\Delta u=49.29\ Btu/lbm$
$\Delta h=55.08\ Btu/lbm$
b)
$\Delta u=49.96\ Btu/lbm$
$\Delta h=49.96\ Btu/lbm$
c)
$\Delta u=\Delta h=50\ Btu/lbm$
Work Step by Step
a) Compressible liquid model:
$u(T,P)=u_{T_{sat}};\ h(T,P)=h_{T_{sat}}+v_{sat}(P-P_{sat})$
From Table A-4E:
$u_{1,T_{sat}}=18.07\ Btu/lbm,\ h_{1,T_{sat}}=18.07\ Btu/lbm,\ v_{1,sat}=0.01602\ ft³/lbm,$
$P_{1,sat}=0.17812\ psia$
Hence:
$u_1=18.07\ Btu/lbm$
$h_1=18.21\ Btu/lbm$
From table A-7E:
$u_2=67.36\ Btu/lbm$
$h_2=73.30\ Btu/lbm$
Therefore:
$\Delta u=49.29\ Btu/lbm$
$\Delta h=55.08\ Btu/lbm$
b) Incompressible liquid model:
$u(T,P)=u_{T_{sat}};\ h(T,P)=h_{T_{sat}}$
From table A-4E:
$u_{2,T_{sat}}=68.03\ Btu/lbm,\ h_{2,T_{sat}}=68.03\ Btu/lbm$
Hence:
$\Delta u=49.96\ Btu/lbm$
$\Delta h=49.96\ Btu/lbm$
c)
Specific heat model:
$\Delta u=\Delta h =c\Delta T$
With: $c=1.00 Btu/lbm.°R, \Delta T=100°F-50°F=50°R$
$\Delta u=\Delta h=50\ Btu/lbm$
