Answer:
Explanation:
[tex]\text{From the information given:}[/tex]
[tex]\text{The mass (m) = 1 lbm}[/tex]
Suppose: g = 32.2 ft/s²
At the inlet conditions:
[tex]\text{mass (m) = 1 \ lbm water} \\ \\ P_1 = 14.7 \ psia \\ \\ T_1 = 70 F \\ \\ z_1 = 0 \ ft[/tex]
At the outlet conditions:
[tex]P_2 = 30 \ psia \\ \\ T_2 = 700\ F \\ \\ v_2 =100 \ ft/s\\ \\ z_2 = 100 \ ft[/tex]
[tex]\text{Using the information obtained from saturated water table at P1 = 14.7 \ psia \ and \ T1 = 70 F }[/tex]
[tex]u_1 =u_f = 38.09 \ Btu/lbm[/tex]
[tex]\text{Applying informations from superheated water vapor table:}[/tex]
[tex]P_2 = 30 \ psia \ and \ T_2 = 700 \ F \\ \\ u_ = 1256.9 \ kJ/kg[/tex]
The change in the internal energy is:
[tex]\Delta U = U_2 -U_1 \\ \\ \Delta U = 1256.9 -38.09 \\ \\ \Delta U = 1218.81 \ Btu/lbm[/tex]
For potential energy (P.E):
Initial P.E = mgz
P.E = 1 × 32.2 × 0 = 0 ft²/s²
Final P.E = mgz
P.E = 1 × 32.2 × 100 = 3220 ft²/s²
The change in the potential energy = PE₂ - PE₁
ΔPE = (3220 - 0) ft²/s²
ΔPE = 3220 ft²/s²
ΔPE = (3220 × 3.9941 × 10⁻⁵) Btu/lbm
ΔPE =0.12861 Btu/lbm
Initial Kinetic energy (K.E)
[tex]KE_1 = \dfrac{1}{2}mV_1[/tex]
[tex]KE_1 = \dfrac{1}{2}(1)(0) = 0 \ lbm \ ft^2/s^2[/tex]
FInal K.E
[tex]KE_2= \dfrac{1}{2}mV_2[/tex]
[tex]KE_2= \dfrac{1}{2}(1)(100)^2_2 = 50000 \ lbm \ ft^2/s^2[/tex]
Change in K.E [tex]\Delta K.E[/tex] = [tex]KE_2-KE_1[/tex]
[tex]\Delta K.E = 50000 -0 = 50000 \ lbm.ft^2/s^2[/tex]
[tex]\mathbf{\Delta K.E = 0.199 \ Btu/lbm}[/tex]
