Answer to Question #173961 in Mechanical Engineering for Randal Rodriguez

Question

Note: Draw the P-V, T-S diagrams for all problems

4. Consider that 2 lb m of air (R = 53.342 ft-lb f /lb m -R and k= 1.4) has a decrease of internal energy of 41.16 BTU while its Fahrenheit temperature is reduced to one third of its initial temperature during a

reversible nonflow constant pressure process. Determine: a) initial and final Fahrenheit temperatures, b) heat added c) Work d) change in entropy.

5. While the pressure remains constant at 700 kPaa, the volume of a system of air changes from 0.3 m 3 to 0.58 m 3 . Determine a) ∆U (kJ), b) ∆H (kJ), c) Q (kJ), d) ∆s (kJ/K)

6. An ideal gas whose mass is 1.36 kg m , R = 204.51 J/kg m and k = 1.6667. Heat added during a reversible nonflow constant pressure change of state is 316.5 kJ . The initial temperature is 37.78 o C. Determine a) change in entropy (kJ/K), b) Work (kJ), c) ratio of expansion or V 2 /V 1.

Answer 1

4) mass of air = 2lb m and R = 53.342 ft-lb f /lb m -R and k= 1.4) and  decrease of internal energy of 41.16 BTU

Inital temperature, T₁ = T and final temperature , T₂= T/3

Process is constant pressure

$\Delta U= m C_v \Delta T = 2 \times 0.171 \times (\frac{ -2T}{3})$

$41.16 = 4T/3$

T=180.52 ^oF

Final temperature = 60.17 ^oF

Heat added at constant pressure = mCp dT = 2(.24)( -120.34)=-57.76 BTU

Now apply first law of thermodynamics

dQ=dU+dW

-57.76 = - 41.16 +dW

dW= work = 16.60 BTU

and the change in entropy can be gives as

$\Delta S=\frac{ Q}{ \Delta T}= \frac{- 57.76}{-120.34}= 0.479 BTU /^oF$