Question

A large vessel vents when the enclosed air (R = 287 J/kgK and γ = 1.4) reaches a pressure of 25 bar and a temperature of 400 K. The venting system can be approximated by a pipe 6 m long and 20 mm in diameter, followed by a choked convergent nozzle with a throat diameter of 12 mm. The friction factor for the pipe is 0.005. Determine: (a) the two possible Mach numbers at the nozzle inlet, based on the

          A large vessel vents when the enclosed air (R = 287 J/kgK and γ = 1.4) reaches a pressure of 25 bar and a temperature of 400 K. The venting system can be approximated by a pipe 6 m long and 20 mm in diameter, followed by a choked convergent nozzle with a throat diameter of 12 mm. The friction factor for the pipe is 0.005. Determine: (a) the two possible Mach numbers at the nozzle inlet, based on the
        
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University Physics with Modern Physics
University Physics with Modern Physics
Hugh D. Young 14th Edition
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A large vessel vents when the enclosed air (R = 287 J/kgK and γ = 1.4) reaches a pressure of 25 bar and a temperature of 400 K. The venting system can be approximated by a pipe 6 m long and 20 mm in diameter, followed by a choked convergent nozzle with a throat diameter of 12 mm. The friction factor for the pipe is 0.005. Determine: (a) the two possible Mach numbers at the nozzle inlet, based on the
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Transcript

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00:01 In this question given p equal to 25 bar, temperature is equal to 400 kelvin, the length of the pipe is equal to 6 meter, the diameter of the pipe is equal to 20 millimeter which equal to 0 .02 meter, nozzle tear out diameter is equal to 12 millimeter which equal to 0 .012 meter.
00:32 Now the friction factor f is equal to 0 .005, given r is equal to 287 joule per kg kelvin and gamma is equal to 1 .4.
00:48 So therefore the area of the pipe is equal to pi divided by 4 multiplied by d square which equal to pi divided by 4 multiplied by 0 .02 whole square.
01:05 Now the area of the tear out a1 is equal to pi divided by 4 multiplied by d square which equal to pi divided by 4 multiplied by 0 .012 whole square.
01:18 Now a divided by a1 is equal to 1 divided by capital m multiplied by 2 plus gamma minus 1 m whole divided by gamma plus 1 to the power gamma plus 1 divided by 2 gamma minus 1.
01:35 This imply pi divided by 4 multiplied by 0 .02 whole square whole divided by pi divided by 4 multiplied by 0 .012 whole square which equal to 1 divided by m multiplied by 2 plus 0 .4 m divided by 2 multiplied by 0 .4 to the power 2 .4 divided by 2 multiplied by 0 .4 which imply 2 .77778 is equal to 1 divided by m multiplied by 2 .5 plus 0 .5 m whole to the power 3...
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