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motikmotik
3 years ago
7

A double-pipe heat exchanger is used to cool a hot fluid (cp = 3800 J/kg·K) entering the heat exchanger at 200°C with a flow rat

e of 0.4 kg/s. In the cold side, cooling fluid (cp = 4200 J/kg·K) enters the heat exchanger at 10°C with a mass flow rate of 0.5 kg/s. The double-pipe heat exchanger has a thin-walled inner tube, with convection heat transfer coefficients inside and outside of the inner tube estimated to be 1400 W/m2 ·K and 1100 W/m2 ·K, respectively. The heat exchanger has a heat transfer surface area of 2.5 m2 , and the estimated fouling factor caused by the accumulation of deposit on the surfaces is 0.0002 m2 ·K/W. (a) Determine the effectiveness values for the parallel- and counter-flow configurations. (b) Determine outlet temperatures of the hot fluid for the parallel- and counter-flow configurations. (c) Determine outlet temperatures of the cold fluid for the parallel- and counter-flow configurations.

Engineering
1 answer:
MAXImum [283]3 years ago
3 0

Answer:

See explaination

Explanation:

please kindly see attachment for the step by step solution of the given problem

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Steam at 1400 kPa and 350°C [state 1] enters a turbine through a pipe that is 8 cm in diameter, at a mass flow rate of 0.1 kg⋅s−
sergeinik [125]

Answer:

Power output, P_{out} = 178.56 kW

Given:

Pressure of steam, P = 1400 kPa

Temperature of steam, T = 350^{\circ}C

Diameter of pipe, d = 8 cm = 0.08 m

Mass flow rate, \dot{m} = 0.1 kg.s^{- 1}

Diameter of exhaust pipe, d_{h} = 15 cm = 0.15 m

Pressure at exhaust, P' = 50 kPa

temperature, T' =  100^{\circ}C

Solution:

Now, calculation of the velocity of fluid at state 1 inlet:

\dot{m} = \frac{Av_{i}}{V_{1}}

0.1 = \frac{\frac{\pi d^{2}}{4}v_{i}}{0.2004}

0.1 = \frac{\frac{\pi 0.08^{2}}{4}v_{i}}{0.2004}

v_{i} = 3.986 m/s

Now, eqn for compressible fluid:

\rho_{1}v_{i}A_{1} = \rho_{2}v_{e}A_{2}

Now,

\frac{A_{1}v_{i}}{V_{1}} = \frac{A_{2}v_{e}}{V_{2}}

\frac{\frac{\pi d_{i}^{2}}{4}v_{i}}{V_{1}} = \frac{\frac{\pi d_{e}^{2}}{4}v_{e}}{V_{2}}

\frac{\frac{\pi \times 0.08^{2}}{4}\times 3.986}{0.2004} = \frac{\frac{\pi 0.15^{2}}{4}v_{e}}{3.418}

v_{e} = 19.33 m/s

Now, the power output can be calculated from the energy balance eqn:

P_{out} = -\dot{m}W_{s}

P_{out} = -\dot{m}(H_{2} - H_{1}) + \frac{v_{e}^{2} - v_{i}^{2}}{2}

P_{out} = - 0.1(3.4181 - 0.2004) + \frac{19.33^{2} - 3.986^{2}}{2} = 178.56 kW

4 0
3 years ago
Phosphorus and nitrogen are included in which category of water pollutants?
Evgesh-ka [11]

Answer: Hello :)

They are in the <u>nutrient pollution</u> category.

Explanation:

3 0
2 years ago
The following median grain size data were obtained during isothermal liquid phase sintering of an 82W-8Mo-8Ni-2Fe alloy. What is
Morgarella [4.7K]

Answer:

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

( d^{2}- d_{0} ^{2} = kt )

Explanation:

The plot attached below shows the time dependence of the growth of grain.

The probable grain-coarsening mechanism is : Ideal grain growth mechanism

the ideal growth follows this principle = d^{2} - d^{2} _{0}  = kt

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d_{0} = initial grain size

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8 0
3 years ago
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kifflom [539]

Answer:

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Explanation:

Hope this helped!

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Answer:

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7 0
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