Tesis

CHAPTER 1

1

Problem 1.1A Convert the following quantities to the ones designated : a. 42 ft2/hr to cm2/s. b. 25 psig to psia. c. 100 Btu to hp-hr. Solution a. 42.0 ft 2 1.0 m hr 3.2808 ft
2

10 4 cm 2 1 hr 1.0 m 2 3 6 0 0 s

=

10.8 cm2/s

b.

100 Btu 3.93 × 10 -4 hp-hr 1 Btu

=

3.93 × 10-2 hp-hr

c.

80.0 lb f 32.174 (lb m )(ft) 1 kg 1 m 1 N 2.20 lb m 3.2808 ft 1(kg)(m)(s) -2 = 356 N (lbf)(s)2

Problem 1.1 B cal Btu Convert the ideal gas constant : R = 1.987 (gmol)(K) to (lb mol)(°R) Solution 1.987 cal 1 Btu 454 gmol 1K (gmol)(K) 252 cal 1 lb mol 1.8 °R Btu = 1.98 (lb mol)(°R)

Problem 1.1 C Mass flow through a sonic nozzle is a function of gas pressure and temperature. For a given pressure p and temperature T, mass flow rate through the nozzle is given bym = 0.0549 p /(T)0.5 where m is in lb/min, p is in psia and T is in °R a. Determine what the units for the constant 0.0549 are. b. What will be the new value of the constant, now given as 0.0549, if the variables in the equation are to be substituted with SI units and m is calculated in SI units.

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Sec. 1.1

Units and Dimensions

NOZZLES

Fig. 1a. Ultrasonic nozzle Fig. 1b. Aconventional nozzle spraying a fluid of suspended particles in a flash dryer. (courtesy of Misonix Inc., Farmingdale, N.J.)

Spray nozzles are used for dust control, water aeration, dispersing a particular pattern of drops, coating, paintings, cleaning surfaces of tanks and vats, and numerous other applications. They develop a large interface between a gas and liquid, and can provide uniform round dropsof liquid. Atomization occurs by a combination of gas and liquid pressure differences. The Figure below (courtesy of Misonix Inc.) compares the particle sizes from the ultrasonic nozzle with those from the conventional nozzle.

Fig. 1c

Sec. 1.1

Units and Dimensions

3

Solution a. Calculation of the constant. The first step is to substitute known units into the equation. lbm lbf min= 0.0549 (in2)(°R)0.5 We want to find a set of units that convert units on the right hand side of the above expression to units on the left hand side of the expression. Such a set can be set up directly by multiplication. lbf (lbm)(in)2(°R)0.5 (min)(lbf) (in2)(°R)0.5 Units for the constant 0.0549 are ------> (lbm) (min) (lbm)(in)2(°R)0.5 (min)(lbf)

b. To determine the new value of the constant,we need to change the units of the constant to appropriate SI units using conversion factors. 0.0549 (lbm ) (in2) (°R)0.5 (0.454 kg) (14.7 lbf/in2) (1 min) (1K)0.5 (p) (lbf) (min) (1 lb m ) (101.3 × 103 N/m 2) (60 s) (1.8 °R)0.5 (T)0.5 m = 4.49 × 10-8 (m) (s) (K)0.5 (p) (T)0.5

Substituting pressure and temperature in SI units m = 4.49
×

10-8 (m) (s) (K)0.5

(p) (N/m 2) 1 kg/(m)(s) 2 (T)0.5 (K) 0.5 1 N/m 2

(kg) m (s)

=

4.49 × 10-8

(p) (T)0.5

where p is in N/m2 and T is in K

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Sec. 1.1

Units and Dimensions

Problem 1.1 D An empirical equation for calculating the inside heat transfer coefficient, h i , for the turbulent flow of liquids in a pipe is given by: 0.023 G 0.8 K 0.67 Cp 0.33 D 0.2 µ0.47 where hi = heat transfer coefficient, Btu/(hr)(ft)2(°F) G =mass velocity of the liquid, lbm/(hr)(ft)2 K = thermal conductivity of the liquid, Btu/(hr)(ft)(°F) Cp = heat capacity of the liquid, Btu/(lbm)(°F) µ = Viscosity of the liquid, lbm/(ft)(hr) D = inside diameter of the pipe, (ft) hi = a. Verify if the equation is dimensionally consistent. b. What will be the value of the constant, given as 0.023, if all the variables in the equation are inserted inSI units and hi is in SI units. Solution a. First we introduce American engineering units into the equation: hi = 0.023[(lbm)/(ft)2(hr)]0.80 [Btu/(hr)(ft)(°F)]0.67 [Btu/(lbm)(°F)]0.33 (ft) 0.2 [lb m /(ft)(hr)] 0 . 4 7

Next we consolidate like units 0.023(Btu) 0.67 (lb m )0.8 (ft)0.47 (1) (hr)0.47 [(lbm)0.33(lbm)0.47] [(ft)1.6(ft)0.67(ft)0.2] [(°F)0.67(°F)0.33] [(hr)0.8(hr)0.67] hi = 0.023...