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Initially, 1 lbm of water is at rest at 14.7 psia and 70 F. The water then undergoes a process where the final state is 30 psia and 700 F with a velocity of 100 ft/s and an elevation of 100 ft above the starting location. Determine the increases in internal energy, potential energy, and kinetic energy of the water in Btu/lbm. Compare the increases in potential energy and kinetic energy individually to the change in the internal energy.

Answer :

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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]

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