Baixe Chapter 02 e outras Exercícios em PDF para Engenharia Mecânica, somente na Docsity!
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, h 1 = 503.72 kJ/kg from Table A-
For Pressuring Reducing Valve Dh = 0h 2 = h 1 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)
h h
x h h
g f
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(t 2 - t 1 ) m = 2.5 kg/s cp = 1.0 kJ/kg.K t 2 = 30 C t 1 = -10 C
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. 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 befrictionless 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
pρ A +gzA=pρB+gz B
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
p V 2 constant
2
ρ+^ =
Solution: Figure 2-6. (^) 0.8 0.
Nu =0.023Re Pr
where:
Re = ρVD
k
Pr =μ^ cp
k
N u =hc^ D
Then,
0.8 2 p 0. 2
0.8 1 p 0. 1 c
c
k
0.023 VD c
k
0.023 VD c
k
h D
k
h D
1
2 c
c
h
h
If viscosity is decreased by 10 %
1
Then, ( )0. c
c
h
h = 0. 9
hc2 = 1.043hc
Increase h h h (100% ) c
= c2^ − c
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 = hcA( ts - ta ) hc = 13.5V0. V = 0.25 m/s hc = 13.5(0.25)0.6^ = 5.8762 W/m^2 .K
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/mC = 61.7 W^2 .K)(1.5 m^2 )(31 C - 24 C)
2-7 Order of Magnitude ~ 60 W - - - Ans.What is the order of magnitude of radiant heat transfer from a human body in a comfort air-conditioning situation? Solution: Eq. 2-10. q 1 − 2 =σAFεFA (T 14 −T 24 ) Surface area of human body = 1.5 to 2.5 m^2 use 1.5 m^2 AFεFA = (1.0)(0.70)(1.5 m^2 ) - 1.05 m^2 s = 5.669x10-8^ W/m^2 .K^4 T 1 =31 C + 273 = 304 K T 2 = 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- kg/Pa.s.m^2. Solution:Equation 2-19. qins = hfgACdiff( ps - pa ) Where: A = 2.0 mh^2 average for human body area pfg^ = 2.43 MJ/kg = 2,430,000 J/kg ps^ = 4750 Pa a = 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.
