1. A tank contains oil (s.g. 0.917), water (s.g. 1) glycerol (s.g. 1.261) and ethanol (s.g. 0.789). The tank is 12.5 m tall and open to 1 atm atmospheric pressure. Calculate

a. Calculate P1 shown in figure below.

b.  Calculate the gauge pressure at the bottom of the tank.

                                                                                                                                 

• During preparation of a breakfast cereal, 250,000g of sugar are mix into 1 tonne batch of powdered cereal. Six 100 g samples are taken after 3, 7 and 12 min and analysed for the percentage sugar. The following results were obtained.

	1	2	3	4	5	6
Percentage after 3 min	0.39	0.35	0.32	0.4	0.42	0.41
Percentage after 7 min	0.59	0.55	0.62	0.6	0.62	0.57
Percentage after 12 min	0.75	0.79	0.82	0.80	0.78	0.77

• Calculate the mixing index for each mixing time and draw conclusions regarding the efficiency of the mixing process.
• Determine the mixing time to ensure that 99.7% of all samples withdrawn from the mixer will contain ± 5% of the mean concentration of the sugar, given that the rate constant is 0.065 min ^-1.
• Given that seawater contains about 3.75% of salt.
•  Determine the amount of anhydrous salt that can be obtained from 2.5 kg of seawater after drying.
• How much water needs to be evaporated from the seawater?
•  A Newtonian fluid with a dynamic viscosity of 0.0728 Ns/m² and a specific gravity of 0.93 flows at 15 m³ hr ^-1 in a 32.5 mm diameter pipe. What type of flow does this fluid exhibit?

                                                                                                                          

• A juice storage tank in a processing company was accidently hit by a stone from children playing nearby that caused leakage of the juice. If the top of the tank is open to the atmosphere and the level of juice is 350 cm above the hole, determine the speed at which the juice will leave the hole. Given that s.g. of juice is 1.054, g= 9.8 m/s². State all assumptions made.

Appendix for Question 2

Normal distribution curve: percentage area under the curve for a given standard deviation s either side of mean

Transcribed Image Text:

Area % Area % 士2.00 s 95.45 士2.55 s 98.92 95.96 え土2.05 s 土2.10 s 士2.60 s 土2.65 s 99.07 96.43 99.20 士2.15 s 96.84 士2.70 s 99.31 土2.20 s 97.22 土2.75 s 土2.80 s 土2.85 s 云士2.90 s 土2.95 s 99.40 云士2.25 s 97.56 99.49 元土2.30 s 97.86 99.56 云士2.35 s 98.12 99.63 え土2.40 s 98.36 99.68 云土2.45 s 98.57 云土3.00 s 99.73 云土2.50 s 98.76 云士4.00 s 99.994 Area % Area % 士2.00 s 95.45 士2.55 s 98.92 95.96 え土2.05 s 土2.10 s 士2.60 s 土2.65 s 99.07 96.43 99.20 士2.15 s 96.84 士2.70 s 99.31 土2.20 s 97.22 土2.75 s 土2.80 s 土2.85 s 云士2.90 s 土2.95 s 99.40 云士2.25 s 97.56 99.49 元土2.30 s 97.86 99.56 云士2.35 s 98.12 99.63 え土2.40 s 98.36 99.68 云土2.45 s 98.57 云土3.00 s 99.73 云土2.50 s 98.76 云士4.00 s 99.994