need help with 5 & 6 4. Find the density of ice, water and water vapor, What fac

Question

need help with 5 & 6

4. Find the density of ice, water and water vapor, What factors can affect the density of water in all three states? 5. In a food processing (drying) application, air (p -1.10 kg'm v 0 ftlmin) enters a chamber through a duet with dimensions of 20 in. x 18 in. and leaves (p 1.25 kgm through a ductwork with dimensions of 16 in x 18 in. What is the average air speed in the exit duct? Express your answer in ft min and m/s. 6. Water is available at a rate of 500 gallonsmin to a small apartment complex consisting of four units. If at a given time, units 1, 2, and 3 are consuming 12 lbn's, 20 lbms, 15 lbm/s, respectively, how much water is available to unit 4? Express your answer in: a. gallons/min c. m's e, kg/s

Explanation / Answer

5) Dimension of enterinfg duct = 20 in * 18 in

Density of entering air = 1.10 kg/m3

Velocity of entering air = 10 ft/min

Dimension of leaving duct = 16 in * 18 in

Density of leaving air = 1.25 kg/m3

Mass of air entering = Dimension of enterinfg duct * Density of entering air * Velocity of entering air

= 20 * 18 * 1.10 * 10

= 3960

Mass of air leaving =  Dimension of leaving duct * Density of leaving air * Velocity of leaving air

= 16 * 18 * 1.25 * Velocity of leaving air

Mass of air entering = Mass of air leaving

3960 = 360 * Velocity of leaving air

Velocity of leaving air = 3960/360

= 11 ft/min

Since ,

1 ft = 0.3048m

1 min = 60 s

Velocity of leaving air (in m/s) = 11 * 0.3048/60

= 0.05588 m/s

6) Water avilable = 500 gallons/min

Since 1 gallon = 8.35 lbm

1 min = 60 s

d)Water available in lb/s = 8.35 * 500 /60 = 69.5833 lbm/s

Water consumed by plant1 = 12lbm/s

Water consumed by plant2 = 20lbm/s

Water consumed by plant3 = 15lbm/s

Water available for plant 4 = Water available in lbm/s - [ Water consumed by plant1 + Water consumed by plant2 + Water consumed by plant1 ] in lbm/s

= 69.5833 - [ 12 + 20 + 15]

= 69.5833 - 47

= 22.5833 lbm/s

d)Water available for plant 4 in lbm/s = 22.5833 lbm/s

a)Water available for plant 4 in gallons/min = 22.5833 * 60 /8.35 = 164.215 gallons/min

b) 1 gallon = 0.13368 ft3

Water available for plant 4 in ft3/s = 164.215 * 0.13368/ 60 = 0.3658 ft3/s

c) 1ft3 = 0.02832 m3

Water available for plant 4 in m3/s = 0.02832 * 0.3568 = .010105 m3/s

e)1 m3 = 2406.53 kg

Water available for plant 4 in kg/s = 0.010105 * 2406.53 = 24.312 kg/s

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