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damaskus [11]
3 years ago
12

a vertical cylindrical container is being cooled in ambient air at 25 °C with no air circulation. if the initial temperature of

the container surface is 100 °C, compute the surface heat-transfer coefficient due to natural convection during the initial cooling period. the diameter of the container is 1m, and it is 2 m high

Engineering
2 answers:
Sloan [31]3 years ago
7 0

Answer:

the surface heat-transfer coefficient due to natural convection during the initial cooling period.  = 4.93 w/m²k

Explanation:

check attachement for answer explanation

Brilliant_brown [7]3 years ago
3 0

Answer:

Explanation:

Find attach solution

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It is given that 50 kg/sec of air at 288.2k is iesntropically compressed from 1 to 12 atm. Assuming a calorically perfect gas, d
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Data;

  • Mass = 50kg/s
  • T = 288.2K
  • P1 = 1atm
  • P2 = 12 atm

<h3>Exit Temperature </h3>

The exit temperature of the gas can be calculated isentropically as

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y}\\ y = 1.4\\ C_p= 1.005 Kj/kg.K\\

Let's substitute the values into the formula

\frac{T_2}{T_1} = (\frac{P_2}{P_1})^\frac{y-1}{y} \\\frac{T_2}{288.2} = (\frac{12}{1})^\frac{1.4-1}{1.4} \\ T_2 = 586.18K

The exit temperature is 586.18K

<h3>The Compressor input power</h3>

The compressor input power is calculated as

P= mC_p(T_2-T_1)\\P = 50*1.005*(586.18-288.2)\\P= 14973.53kW

The compressor input power is 14973.53kW

Learn more on exit temperature and compressor input power here;

brainly.com/question/16699941

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