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Reptile [31]
2 years ago
12

A glass bottle washing facility uses a well agitated hot water bath at 50°C with an open top that is placed on the ground. The b

athtub is 1 m high, 2 m wide, and 4 m long and is made of sheet metal so that the outer side surfaces are also at about 50°C. The bottles enter at a rate of 800 per minute at ambient temperature and leave at the water temperature. Each bottle has a mass of 150 g and removes 0.6 g of water as it leaves the bath wet. Makeup water is supplied at 15°C. If the average conditions in the plant are 1 atm, 25°C, and 50 percent relative humidity, and the average temperature of the surrounding surfaces is 15°C, determine
(a) the amount of heat and water removed by the bottles themselves per second,
(b) the rate of heat loss from the top surface of the water bath by radiation, natural convection, and evaporation,
(c) the rate of heat loss from the side surfaces by natural convection and radiation, and
(d) the rate at which heat and water must be supplied to maintain steady operating conditions. Disregard heat loss through the bottom surface of the bath and take the emissivities of sheet metal and water to be 0.61 and 0.95, respectively.
Engineering
1 answer:
DochEvi [55]2 years ago
7 0

Answer:

do the wam wam

Explanation:

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1. A team of students have designed a battery-powered cooler, which promises to keep beverages at a high-drinkability temperatur
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Minimum electrical power required = 3.784 Watts

Minimum battery size needed = 3.03 Amp-hr

Explanation:

Temperature of the beverages, T_L = 36^0 F = 275.372 K

Outside temperature, T_H = 100^0F = 310.928 K

rate of insulation, Q = 100 Btu/h

To get the minimum electrical power required, use the relation below:

\frac{T_L}{T_H - T_L} = \frac{Q}{W} \\W = \frac{Q(T_H - T_L)}{T_L}\\W = \frac{100(310.928 - 275.372)}{275.372}\\W = 12.91 Btu/h\\1 Btu/h = 0.293071 W\\W = 12.91 * 0.293071\\W_{min} = 3.784 Watt

V = 5 V

Power = IV

W_{min} = I_{min} V\\3.784 = 5I_{min}\\I_{min} = \frac{3.784}{5} \\I_{min} = 0.7568 A

If the cooler is supposed to work for 4 hours, t = 4 hours

I_{min} = 0.7568 * 4\\I_{min} = 3.03 Amp-hr

Minimum battery size needed = 3.03 Amp-hr

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