EGR 334 Thermodynamics: Homework 03Chap 2: 49Complete the following exercises using the heat transfer relations:a) Referring to Fig 2.13, determine the net rate of radiant exchange in W, for ε=0.8, A = .125 m2 , Tb = 475 K, and Ts = 298 K.b) Referring to Fig. 2.14 determine the rate of convection heat transfer from the surface to the air in W for hc = 10 W/m2 - K, A = 0.125 m2 , Tb = 305 K, and Tf = 298 K. ----------------------------------------------------------------------------------------------------------------------------------------

Solution: a) Heat Exchange by Radiation. where ε = 0.8 A = 0.125 m2 Tb = 475 K Ts = 298 K. σ = 5.67 x 10-8 W/m2•K4 Heat flow rate from surface to the air:

Heat flow rate from surroundings to surface:

therefore the net heat exchange from surface to the air at steady state is

b) Heat Exchange by Convection:

where hc = 10 W/m2 - K A = 0.125 m2

Tb = 305 K and Tf = 298 K. then

EGR 334 Thermodynamics: Homework 03Chap 2: 53Each line of the following table gives data for a process of a closed system. Each entry has the same energy units Determine the missing entries.

process Q W E1 E2 ΔEa -20 +50 +70b +50 +20 +50c -60 +60 +20d -90 +50 0e +50 +150 +20

--------------------------------------------------------------------------------------------------------------------------------------------Principle: 1st Law of Thermo:

process a: = 70+(-20)= +50

= +50 - 70 = -20

process b: = 50 - 50 + 20 = +20

and = 50 - 20 = +30

process c: = 20 + -60 = -40

and = 60 - 20 = +40

process d: = 0 + (-90) = -90

and = 50 - 0 = +50

process e: = 50 - 150 = -100

and = -100 + 20 = -80

process Q W E1 E2 ΔEa +50 -20 -20 +50 +70b +50 +20 +20 +50 +30c -40 -60 +40 +60 +20d -90 -90 +50 +50 0e +50 +150 +20 -80 -100

EGR 334 Thermodynamics: Homework 03Chap 2: 61One kg of Refrigerant 22, initially at p1 = 0.9 MPa , and u1=232.92 kJ/kg, is contained within a rigid closed tank. The tank is fitted with a paddle wheel that transfers energy to the refrigerant at a constant rate of 0.1 kW. Heat transfer from the refrigerant to its surroundings occurs at a rate of K t, in kW, where K is a constant in kW per minute and t is time in minutes. After 20 minutes of stirring the refrigerant is at p2 = 1.2 MPa, u2 = 276.67 kJ/kg. No overall changes in kinetic or potential energy occur. a) for the refrigerant, determine the work and heat transfer each in kJ. b) Determine the value of the constant K appearing in the given heat transfer relation, in kW/min. -------------------------------------------------------------------------------------------------------------------------------------

Note: rigid tank maintains the same volume. m = 1 kg∆ t=20 min p1 = 0.9 MPa u1=232.92 kJ/kgStarting from the rate form of the 1st law of thermo:

note that Qloss is negative since heat is taken out of the system and Wpaddle is negative since it does work on the system.then integrating

0 0 0

note that WPV is zero because there is no change in the volume of the closed system.

notice that the work done by the Paddle is 120 kJ so W = -120 kJ and ΔU = 43.75 kJ so then

(where the negative sign indicates it leaves the tank)to find K.

. Wpaddle = 0.1 kW

.Qloss = Kt

p2 = 1.2 MPa u2 = 276.67 kJ/kg

EGR 334 Thermodynamics: Homework 03Chap 2: 68A vertical piston cylinder assembly with a piston of mass 25 kg and having a face area of 0.005 m2 contains air. The mass of air is 2.5 g, and initially the air occupies a volume of 2.5 liters. The atmosphere exerts a pressure of 100 kPa on the top of the piston. The volume of the air slowly decreases to 0.001 m3 as energy with a magnitude of 1 kJ is slowly removed by heat transfer. Neglecting friction between the piston and the cylinder wall, determine the change in specific internal energy of the air in kJ/kg. Let g = 9.8 m/s2.-------------------------------------------------------------------------------------------------------------------------------------------

Solution: mpiston = 25 kg A = 0.005 m2 mair= 2.5 g = 0.0025 kg patm= 100 kPa State 1: V1 = 2.5 liters =2.5 x10-3 m3 = 0.0025 m3

State 2: V2 = 0.001 m3 Q leaving system is 1 kJ: Q = -1 kJ = -1000 J Find ΔU.

Principle: 1st Law of Thermo:

where

(assume slow quasiequilbrium process)

with m = 25 kg g = 9.81 m/s2 and

therefore

Work: where p is constant: with W = mg = 25kg * 9.81 m/s2 = 245 N

and for equilibrium:

therefore:

so:

Ftop=patm A

Ftop=patm A

W=mg

in terms of