Engineering College

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.

Answers

Answer 1

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]

Related Questions

Because the mechanism of creep deformation is different from the mechanism of slip in most metal deformation processes, one of the fundamental relationships between microstructure and mechanical properties of metals is reversed for creep deformation compared with normal deformation. Is it:________.
A. The Hume-Rothery Rules
B. The Hall-Petch Relation
C. The Schmid Equation

Answers

Answer:

B. The Hall-Petch Relation

Explanation:

The Hall-Petch relation indicates that by reducing the grain size the strength of a material is increased up to the theoretical strength of the material however when the material grain size is reduced below 20 nm the material is more susceptible to creep deformation and displays an "inverse" Hall-Petch Relation as the Hall-Petch relation then has a negative slope (k value)

The Hall-Petch relation can be presented as follows;

[tex]\sigma_y[/tex] = [tex]\sigma_0[/tex] + k·(1/√d)

Where;

[tex]\sigma_y[/tex] = The strength

σ₀ = The friction stress

d = The grain size

k = The strengthening coefficient

The model equation for the reverse Hall-Petch effect is presented here as follows;

[tex]\sigma_y[/tex] = 10.253 - 10.111·(1/√d)

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