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jeka57 [31]
3 years ago
15

A pump must be able to deliver 34 L/s of crude castor oil at 25°C ρ = 0.934 g/mL and μ = 0.078 Pa.s. It is known that, due to th

e type of pump chosen and the type of fluid, its efficiency will be around 76%. The capture point is located 8 m below the discharge point and both points are suffering only from atmospheric pressure. The distance between the two points (collection and discharge) is 25 m of pipe. The tubes used in this system are new commercial stainless steel drawn with 40 mm internal diameter, 1 globe valve (open) and 3 standard 45º elbows. Calculate the power that this bomb will have to have.
Engineering
1 answer:
Reptile [31]3 years ago
5 0

Castor oil is increasingly becoming an important bio-based raw material for industrial applications. The oil is non-edible and can be extracted from castor seeds from the castor plant belonging to the family Euphorbiaceae. The oil is a mixture of saturated and unsaturated fatty acid esters linked to a glycerol. The presence of hydroxyl group, a double bond, carboxylic group and a long chain hydrocarbon in ricinoleic acid (a major component of the oil), offer several possibilities of transforming it into variety of materials. The oil is thus a potential alternative to petroleum-based starting chemicals for the production of materials with variety of properties. Despite this huge potential, very little has recently been reviewed on the use of castor oil as a bio-resource in the production of functional materials. This review therefore highlights the potential of castor oil in the production of these diverse materials with their projected global market potential. The review gives the background information of castor oil and its geographical availability, the properties and its uses as bio-based resource for synthesis of various materials. The review further highlights on the use of castor oil or ricinoleic acid as a green capping agent in the synthesis of nanomaterials.

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A stationary gas-turbine power plant operates on a simple ideal Brayton cycle with air as the working fluid. The air enters the
ololo11 [35]

Answer:

A) W' = 15680 KW

B) W' = 17113.87 KW

Explanation:

We are given;

Temperature at state 1; T1 = 290 K

Temperature at state 3; T3 = 1100 K

Rate of heat transfer; Q_in = 35000 kJ/s = 35000 Kw

Pressure of air into compressor; P_c = 95 kPa

Pressure of air into turbine; P_t = 760 kPa

A) The power assuming constant specific heats at room temperature is gotten from;

W' = [1 - ((T4 - T1)/(T3 - T2))] × Q_in

Now, we don't have T4 and T2 but they can be gotten from;

T4 = [T3 × (r_p)^((1 - k)/k)]

T2 = [T1 × (r_p)^((k - 1)/k)]

r_p = P_t/P_c

r_p = 760/95

r_p = 8

Also,k which is specific heat capacity of air has a constant value of 1.4

Thus;

Plugging in the relevant values, we have;

T4 = [(1100 × (8^((1 - 1.4)/1.4)]

T4 = 607.25 K

T2 = [290 × (8^((1.4 - 1)/1.4)]

T2 = 525.32 K

Thus;

W' = [1 - ((607.25 - 290)/(1100 - 525.32))] × 35000

W' = 0.448 × 35000

W' = 15680 KW

B) The power accounting for the variation of specific heats with temperature is given by;

W' = [1 - ((h4 - h1)/(h3 - h2))] × Q_in

From the table attached, we have the following;

At temperature of 607.25 K and by interpolation; h4 = 614.64 KJ/K

At T3 = 1100 K, h3 = 1161.07 KJ/K

At T1 = 290 K, h1 = 290.16 KJ/K

At T2 = 525.32 K, and by interpolation, h2 = 526.12 KJ/K

Thus;

W' = [1 - ((614.64 - 290.16)/(1161.07 - 526.12))] × 35000

W' = 17113.87 KW

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Explanation:

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A 3 m aluminum pole is kept at a residential site for construction
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Answer:

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