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prohojiy [21]
3 years ago
8

4. Water vapor enters a turbine operating at steady state at 1000oF, 220 lbf/in2 , with a volumetric flow rate of 25 ft3/s, and

expands reversibly and adiabatically to 2 lbf/in2. Ignore kinetic and potential energy effects. Determine the mass flow rate, in lb/s, and the power developed by the turbine, in hp. Determine the mass flow rate, in lb/s. Determine the power developed by the turbine, in hp.
Engineering
1 answer:
hodyreva [135]3 years ago
6 0
Yes i is the time of the day you get to frost the moon and back and then you can come over and then go to hang out with me me and then go to hang out
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Which of the following explains the main reason to cut a piece of wood on the outside of the measurement mark?
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I think it’s D ?? I’m not completely sure tho
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Two streams of air enter a control volume: stream 1 enters at a rate of 0.05 kg / s at 300 kPa and 380 K, while stream 2 enters
alex41 [277]

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

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8 0
3 years ago
3. A 4-m × 5-m × 7-m room is heated by the radiator of a steam-heating system. The steam radiator transfers heat at a rate of 10
Natali [406]

Answer:

14.52 minutes

<u>OR</u>

14 minutes and 31 seconds

Explanation:

Let's first start by mentioning the specific heat of air at constant volume. We consider constant volume and NOT constant pressure because the volume of the room remains constant while pressure may vary.

Specific heat at constant volume at 27°C = 0.718 kJ/kg*K

Initial temperature of room (in kelvin) = 283.15 K

Final temperature (required) of room = 293.15 K

Mass of air in room= volume * density= (4 * 5 * 7) * (1.204 kg/m3) = 168.56kg

Heat required at constant volume: 0.718 * (change in temp) * (mass of air)

Heat required = 0.718 * (293.15 - 283.15) * (168.56) = 1,210.26 kJ

Time taken for temperature rise: heat required / (rate of heat change)

Where rate of heat change = 10000 - 5000 = 5000 kJ/hr

Time taken = 1210.26 / 5000 = 0.24205 hours

Converted to minutes = 0.24205 * 60 = 14.52 minutes

4 0
3 years ago
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