![CHAPTER 2 - THERMAL PRINCIPLES
Page 3 of 5
2
Vp
2
Vp
2
BB
2
AA
+
ρ
=+
ρ
( )2
A
2
B2
1
BA VVpp −ρ=−
But m = ρAA
VA
= ρAB
VB
AA
VA
= AB
VB
AA = 0.5 m2 ans AB = 0.4 m2
Then
0.5VA
= 0.4VB
VA
= 0.8VB
( ) ( )[ ]2
B
2
B
3
2
1
BA 0.8VVkg/m1.15Pa196.2pp −==−
VB
= 30.787 m/s
Air Flow Rate = AB
VB
= (0.4 m
2
)(30.787 m/s)
= 12.32 m
3
/s - - - Ans.
2-4. Use the perfect-gas equation with R = 462 J/kg.K to compute the specific volume of saturated vapor at 20 C. Compare
with data of Table A-1.
Solution:
Perfect-Gas Equation:
RTp =ν
p
RT
=ν
At 20 C, Table A-1, Saturation Pressure = 2.337 kPa = 2337 Pa.
Specific volume of saturated vapor = 57.84 m
3
/kg
T = 20 C + 273 = 293 K
( )( )
Pa2337
K293J/kg.K462
=ν
/kgm57.923 3
=ν
( )100%
57.84
57.8457.923
Deviation
−
=
Deviation = 0.1435 %
2-5. Using the relationship shown on Fig. 2-6 for heat transfer when a fluid flows inside tube, what is the percentage
increase or decrease in the convection heat-transfer coefficient hc
if the viscosity of the fluid is decreased 10 percent.](https://image.slidesharecdn.com/chapter2-150424140526-conversion-gate02/85/Chapter2-3-320.jpg)
Chapter2
- 1. CHAPTER 2 - THERMAL PRINCIPLES Page 1 of 5 2-1. Water at 120 C and a pressure of 250 kPa passes through a pressure-reducing valve and then flows to a separating tank at standard atmospheric pressure of 101.3 kPa, as shown in Fig. 2-14. (a) What is the state of the water entering the valve (subcooled liquid, saturated liquid, or vapor)? (b) For each kilogram that enters the pressure-reducing valve, how much leaves the separating tank as vapor? Solution: (a) From Fig. 2-2, a temperature of 120 C and pressure of 250 kPa water lies in the sub-cooled regiom. so it is a sub- cooled liquid. (b) At 120 C, h1 = 503.72 kJ/kg from Table A-1 For Pressuring Reducing Valve Dh = 0 h2 = h1 At 101.3 kPa, Table A-1, hf = 419.06 kJ/kg hg = 2676 kJ/kg Let x be the amount of vapor leaving the separating tank. h = hf + x(hg - hf ) 419.062676 419.06503.72 hh hh x fg f − − = − − = x = 0.0375 kg/kg - - - Ans. 2-2. Air flowing at a rate of 2.5 kg/s is heated in a heat exchanger from -10 to 30 C. What is the rate of heat transfer? Solution: q = mcp (t2 - t1 ) m = 2.5 kg/s cp = 1.0 kJ/kg.K t2 = 30 C t1 = -10 C
- 2. CHAPTER 2 - THERMAL PRINCIPLES Page 2 of 5 Then, q = (2.5)(1.0)(30 + 10) q = 100 kw - - - Ans. 2-3. One instrument for measuring the rate of airflow is a venturi, as shown in Fig. 2-15, where the cross-sectional area is reduced and the pressure difference between position A and B measured. The flow rate of air having a density of 1.15 kg/m 3 is to be measured in a venturi where the area of position A is 0.5 m 2 and the area at b is 0.4 m 2 . The deflection of water (density = 1000 kg/m3) in a manometer is 20 mm. The flow between A and B can be considered to be frictionless so that Bernoulli’s equation applies. (a) What is the pressure difference between position A and B? (b) What is the airflow rate? Solution: (a) Bernoulli equation for manometer B B A A gz p gz p + ρ =+ ρ pA - pB = ρg(zB -zA ) zB - zA = 20 mm = 0.020 m g = 9.81 m/s 2 ρ = 1000 kg/m 3 pA - pB = (1000 kg/m 3 )(9.81 m/s 2 )(0.020 m) pA - pB = 196.2 Pa - - - Ans. (b) Bernoulli Equation for Venturi constant 2 Vp 2 =+ ρ
- 3. CHAPTER 2 - THERMAL PRINCIPLES Page 3 of 5 2 Vp 2 Vp 2 BB 2 AA + ρ =+ ρ ( )2 A 2 B2 1 BA VVpp −ρ=− But m = ρAA VA = ρAB VB AA VA = AB VB AA = 0.5 m2 ans AB = 0.4 m2 Then 0.5VA = 0.4VB VA = 0.8VB ( ) ( )[ ]2 B 2 B 3 2 1 BA 0.8VVkg/m1.15Pa196.2pp −==− VB = 30.787 m/s Air Flow Rate = AB VB = (0.4 m 2 )(30.787 m/s) = 12.32 m 3 /s - - - Ans. 2-4. Use the perfect-gas equation with R = 462 J/kg.K to compute the specific volume of saturated vapor at 20 C. Compare with data of Table A-1. Solution: Perfect-Gas Equation: RTp =ν p RT =ν At 20 C, Table A-1, Saturation Pressure = 2.337 kPa = 2337 Pa. Specific volume of saturated vapor = 57.84 m 3 /kg T = 20 C + 273 = 293 K ( )( ) Pa2337 K293J/kg.K462 =ν /kgm57.923 3 =ν ( )100% 57.84 57.8457.923 Deviation − = Deviation = 0.1435 % 2-5. Using the relationship shown on Fig. 2-6 for heat transfer when a fluid flows inside tube, what is the percentage increase or decrease in the convection heat-transfer coefficient hc if the viscosity of the fluid is decreased 10 percent.
- 4. CHAPTER 2 - THERMAL PRINCIPLES Page 4 of 5 Solution: Figure 2-6. 0.40.8 Pr0.023ReNu = where: µ ρ = VD Re k c Pr pµ = k Dh N c =u Then, 0.4 p2 0.8 2 0.4 p1 0.8 1 c2 c1 k cVD 0.023 k cVD 0.023 k Dh k Dh µ µ ρ µ µ ρ = 0.4 1 2 c2 c1 h h µ µ = If viscosity is decreased by 10 % 0.9 1 2 = µ µ Then, ( )0.4 c2 c1 h h 9.0= hc2 = 1.043hc1 ( )100% h hh Increase c1 c1c2 − = Increase = (1.043 - 1)(100 %) Increase = 4.3 % - - - Ans. 2-6. What is the order of magnitude of heat release by convection from a human body when the air velocity is 0.25 m/s and its temperature is 24 C? Solution: Using Eq. (2-12) and Eq. (2-18) C = hc A( ts - ta ) hc = 13.5V 0.6 V = 0.25 m/s hc = 13.5(0.25) 0.6 = 5.8762 W/m 2 .K
- 5. CHAPTER 2 - THERMAL PRINCIPLES Page 5 of 5 Human Body: A = 1.5 to 2.5 m 2 use 1.5 m 2 ts = 31 to 33 C use 31 C C = (5.8762 W/m 2 .K)(1.5 m 2 )(31 C - 24 C) C = 61.7 W Order of Magnitude ~ 60 W - - - Ans. 2-7 What is the order of magnitude of radiant heat transfer from a human body in a comfort air-conditioning situation? Solution: Eq. 2-10. ( )4 2 4 1A21 TTFAFq −εσ=− Surface area of human body = 1.5 to 2.5 m 2 use 1.5 m 2 AFε FA = (1.0)(0.70)(1.5 m 2 ) - 1.05 m 2 s = 5.669x10 -8 W/m 2 .K 4 T1 =31 C + 273 = 304 K T2 = 24 C + 273 = 297 K q1-2 = (5.669x10 -8 )(1.05)(304 4 - 297 4 ) q1-2 = 45 W Order of Magnitude ~ 40 W - - - Ans. 2-8. What is the approximate rate of heat loss due to insensible evaporation if the skin temperature is 32 C, the vapor pressure is 4750 Pa, and the vapor pressure of air is 1700 Pa? The latent heat of water is 2.43 MJ/kg; Cdiff = 1.2x10 -9 kg/Pa.s.m 2 . Solution: Equation 2-19. qins = hfg ACdif f( ps - pa ) Where: A = 2.0 m 2 average for human body area hfg = 2.43 MJ/kg = 2,430,000 J/kg ps = 4750 Pa pa = 1700 Pa Cdiff = 1.2x10 -9 kg/Pa.s.m 2 qins = (2,430,000)(2.0)(1.2x10 -9 )(4750 - 1700) qins = 18 W - - - Ans. - 0 0 0 -