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Mariulka [41]
3 years ago
10

Air expands through an ideal turbine from 1 MPa, 900 K to 0.1 MPa, 500K. The inlet velocity is small compared to the exit veloci

ty of 100 m/s. The turbine operates at steady state and develops a power output of 3,200 kW. Heat transfer between the turbine and its surroundings and any potential energy effects are negligible.
a. Calculate the mass flow rate of air, in kg/s, and the exit area, in m^2.

Engineering
1 answer:
3241004551 [841]3 years ago
4 0

Answer:

Mass flow rate= 7.53 kg/sec

Exit Area= 0.108m^2

Explanation:

The solution and complete explanation for the above question and mentioned conditions is given below in the attached document.i hope my explanation will help you in understanding this particular question.

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Select the level of education that is best demonstrated in each example.
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masters

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(35-39) A student travels on a school bus in the middle of winter from home to school. The school bus temperature is 68.0° F. Th
arlik [135]

Answer:

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

Explanation:

From Heat Transfer we determine that heat transfer rate due to electromagnetic radiation (\dot Q), measured in BTU per hour, is represented by this formula:

\dot Q = \epsilon\cdot A\cdot \sigma \cdot (T_{s}^{4}-T_{b}^{4}) (1)

Where:

\epsilon - Emissivity, dimensionless.

A - Surface area of the student, measured in square feet.

\sigma - Stefan-Boltzmann constant, measured in BTU per hour-square feet-quartic Rankine.

T_{s} - Temperature of the student, measured in Rankine.

T_{b} - Temperature of the bus, measured in Rankine.

If we know that \epsilon = 0.90, A = 16.188\,ft^{2}, \sigma = 1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}}, T_{s} = 554.07\,R and T_{b} = 527.67\,R, then the heat transfer rate due to electromagnetic radiation is:

\dot Q = (0.90)\cdot (16.188\,ft^{2})\cdot \left(1.714\times 10^{-9}\,\frac{BTU}{h\cdot ft^{2}\cdot R^{4}} \right)\cdot [(554.07\,R)^{4}-(527.67\,R)^{4}]

\dot Q = 417.492\,\frac{BTU}{h}

Under the consideration of steady heat transfer we find that the net energy transfer from the student's body during the 20 min-ride to school is:

Q = \dot Q \cdot \Delta t (2)

Where \Delta t is the heat transfer time, measured in hours.

If we know that \dot Q = 417.492\,\frac{BTU}{h} and \Delta t = \frac{1}{3}\,h, then the net energy transfer is:

Q = \left(417.492\,\frac{BTU}{h} \right)\cdot \left(\frac{1}{3}\,h \right)

Q = 139.164\,BTU

The net energy transfer from the student's body during the 20-min ride to school is 139.164 BTU.

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timurjin [86]

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3 years ago
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A 2-bit positive-edge triggered register has data inputs d1, d0, clock input clk, and outputs q1, q0. Data inputs d1d0 are 01 an
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