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Vilka [71]
3 years ago
15

Water vapor at 1.0 MPa, 300°C enters a turbine operating at steady state and expands to 15 kPa. The work developed by the turbin

e is 630 kJ per kg of steam flowing through the turbine. Ignoring heat transfer with the surroundings and kinetic and potential energy effects, deter- mine (a) the isentropic turbine efficiency, (b) the rate of entropy gen- eration, in kJ/K per kg of steam flowing. Air at 40°F, 1 atm enters a compressor operating at steady state and exits at 620°F, 8.6 atm. Ignoring heat transfer with the surroundings and kinetic and potential energy effects, determine (a) the isentro- pic compressor efficiency and (b) the rate of entropy generation, in Btu/ºr per lb of air flowing. 2.7

Engineering
1 answer:
Andre45 [30]3 years ago
5 0

Answer:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

Please kindly see explaination and attachment.

Explanation:

a) isentropic efficiency = 84.905%

b) rate of entropy generation = .341 kj/(kg.k)

The Isentropic efficiency of a turbine is a comparison of the actual power output with the Isentropic case.

Entropy can be defined as the thermodynamic quantity representing the unavailability of a system's thermal energy for conversion into mechanical work, often interpreted as the degree of disorder or randomness in the system.

Please refer to attachment for step by step solution of the question.

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Water flows through a pipe at an average temperature of T[infinity] = 70°C. The inner and outer radii of the pipe are r1 = 6 cm
Paul [167]

Answer:

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

Explanation:

We are given;

T∞ = 70°C.

Inner radii pipe; r1 = 6cm = 0.06 m

Outer radii of pipe;r2 = 6.5cm=0.065 m

Electrical heat power; Q'_s = 300 W

Since power is 300 W per metre length, then; L = 1 m

Now, to the heat flux at the surface of the wire is given by the formula;

q'_s = Q'_s/A

Where A is area = 2πrL

We'll use r2 = 0.065 m

A = 2π(0.065) × 1 = 0.13π

Thus;

q'_s = 300/0.13π

q'_s = 734.56 W/m²

The differential equation and the boundary conditions are;

A) -kdT(r1)/dr = h[T∞ - T(r1)]

B) -kdT(r2)/dr = q'_s = 734.56 W/m²

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Determine the speed of sound in air at 400 K. Also determine the Mach number of an aircraft moving in the air at a velocity of 3
Reika [66]

Answer:

\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s

Ma= \frac{310 m/s}{400.899 m/s}= 0.773

Explanation:

For this case we have given the following data:

T= 400 K represent the temperature for the air

v = 310 m/s represent the velocity of the air

k = 1.4 represent the specific heat ratio at the room

R = 0.287 KJ/ Kg K represent the gas constant  for the air

And we want to find the velocity of the air under these conditions.

We can calculate the spped of the sound with the Newton-Laplace Equation given by this equation:

\alpha = \sqrt{\frac{K}{\rho}}=\sqrt{k RT}

Where K = is the Bulk Modulus of air, k is the adiabatic index of air= 1.4, R = the gas constant  for the air, \rho the density of the air and T the temperature in K

So on this case we can replace and we got:

\alpha = \sqrt{1.4 *0.287 \frac{KJ}{Kg K}*\frac{1000J}{1KJ} *400 K}= 400.899 m/s

The Mach number by definition is "a dimensionless quantity representing the ratio of flow velocity past a boundary to the local speed of sound" and is defined as:

Ma=\frac{v}{\alpha}

Where v is the flow velocity and \alpha the volocity of the sound in the medium and if we replace we got:

Ma= \frac{310 m/s}{400.899 m/s}= 0.773

And since the Ma<0.8 we can classify the regime as subsonic.

7 0
3 years ago
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