# chapt1 - C HAPTER 1 1 Problem 1.1A Convert the following...

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CHAPTER 1 1 Problem 1.1A Convert the following quantities to the ones designated : a. 42 ft 2 /hr to cm 2 /s. b. 25 psig to psia. c. 100 Btu to hp-hr. Solution a. 42.0 ft 2 hr 1.0 m 3.2808 ft 2 10 4 cm 1.0 m 2 1 hr 3600 s = 10.8 cm 2 /s b. 100 Btu 3.93 × -4 hp-hr 1 Btu = 3.93 × 10 -2 hp-hr c. 80.0 lb f 32.174 (lb m )(ft) (lb f )(s) 2 1 kg 2.20 lb m 1 m 3.2808 ft 1 N 1 (kg)(m)(s) -2 = 356 N Problem 1.1 B Convert the ideal gas constant : R = 1.987 cal (gmol)(K) to Btu (lb mol)( ° R) Solution 1.987 cal 252 cal 454 gmol 1 lb mol 1K 1.8 ° R = 1.98 Btu (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 by m = 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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2 Sec. 1.1 Units and Dimensions NOZZLES Fig. 1a. Ultrasonic nozzle (courtesy of Misonix Inc., Farmingdale, N.J.) Fig. 1b. A conventional nozzle spraying a fluid of suspended particles in a flash dryer. 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 drops of 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. lb m min = 0.0549 lb f (in 2 )( ° 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. lb f (in 2 )( ° R) 0.5 (lb m )(in) 2 ( ° R) 0.5 (min)(lb f ) ------> (lb m ) (min) Units for the constant 0.0549 are (lb m )(in) 2 ( ° R) 0.5 (min)(lb f ) 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 (lb m ) (in 2 ) ( ° 0.5 (lb f ) (min) (0.454 kg) (1 lb m ) (14.7 lb f /in 2 ) (101.3 × 10 3 N/m 2 ) (1 min) (60 s) (1K) 0.5 (1.8 ° 0.5 (p) (T) 0.5 m = 4.49 × 10 -8 (m) (s) (K) 0.5 (p) (T) Substituting pressure and temperature in SI units m = 4.49 × 10 -8 (m) (s) (K) 0.5 (p) (N/m 2) (K) 1 kg/(m)(s) 2 1 N/m 2 m (kg) (s) = 4.49 × 10 -8 where p is in N/m 2 and T is in K

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4 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: h i = 0.023 G 0.8 K 0.67 Cp 0.33 D 0.2 μ 0.47 where h i = heat transfer coefficient, Btu/(hr)(ft) 2 ( ° F) G = mass velocity of the liquid, lb m /(hr)(ft) 2 K = thermal conductivity of the liquid, Btu/(hr)(ft)( ° F) C p = heat capacity of the liquid, Btu/(lb m )( ° F) μ = Viscosity of the liquid, lb m /(ft)(hr) D = inside diameter of the pipe, (ft) a. Verify if the equation is dimensionally consistent.
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## This note was uploaded on 03/15/2011 for the course CHE 215 taught by Professor Aboyousef during the Spring '11 term at American University of Sharjah.

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chapt1 - C HAPTER 1 1 Problem 1.1A Convert the following...

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