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2 years ago

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A pipe in a district heating network is transporting over-pressurized hot water (10 atm) at a mass flow of 0.5 kg/s. The pipe is

5 m long, has an inner radius of 50 cm and pipe wall thickness of 50 mm. The pipe has a thermal conductivity of 20 W/m-K, and the inner pipe surface is at a uniform temperature of 110 ºC. The convection heat transfer coefficient of the air surrounding the pipe is 100W/m2 -K. The temperature of the water at inlet of pipe is 130 ºC and the constant pressure specific heat of hot water is 4000 J/kg-ºC. If the temperature of the air surrounding the pipe is 20 ºC, determine the exit temperature of the water at the end of the pipe.

1 answer:

aivan3 [116]2 years ago

6 0

Answer:3.6 kg/s 14 w/m-k

Explanation:

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A 150-lbm astronaut took his bathroom scale (a spring scale) and a beam scale (compares masses) to the moon where the local grav

kozerog [31]

Answer:

a)Wt =25.68 lbf

b)Wt = 150 lbf

F= 899.59 N

Explanation:

Given that

$g = 5.48 ft/s^2.$

m= 150 lbm

a)

Weight on the spring scale(Wt) = m g

We know that

$1\ lbf=32.17 \ lmb.ft/s^2$

Wt = 150 x 5.48/32 lbf

Wt =25.68 lbf

b)

On the beam scale

This is scale which does not affects by gravitational acceleration.So the wight on the beam scale will be 150 lbf.

Wt = 150 lbf

If the plane is moving upward with acceleration 6 g's then the for F

F = m a

We know that

$1\ ft/s^2= 0.304\ m/s^2$

$5.48\ ft/s^2= 1.66\ m/s^2$

a=6 g's

$a=9.99\ m/s^2$

So

F = 90 x 9.99 N

F= 899.59 N

3 0

2 years ago

Given an array of words representing your dictionary, you test words to see if it can be made into another word in dictionary. T

katen-ka-za [31]

Answer:

Detailed solution is given below:

8 0

2 years ago

An adiabatic gas turbine expands air at 1300 kPa and 500◦C to 100 kPa and 127◦C. Air enters the turbine through a 0.2-m2 opening

Viktor [21]

Given:

Pressure, $P_{1}$ = 1300 kPa

Temperature, $T_{1}$ = $500^{\circ}$

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

velocity, v = 40 m/s

A = 1 $m^{2}$

Solution:

For air propertiess at

$P_{1}$ = 1300 kPa

$T_{1}$ = $500^{\circ}$

$h_{1}$ = 793kJ/K

$v_{1}$ = $0.172\frac{m^{3}}{kg}$

and also at

$P_{2}$ = 100 kPa

$T_{2} = 127^{\circ}$

$h_{2}$ = 401 KJ/K

$v_{2}$ = $1.15\frac{m^{3}}{kg}$

a) Mass flow rate is given by:

$m' = \frac{Av}{v_{1}}$

Now,

$m = \frac{0.2\times 40}{0.172}$ = 46.51 kg/s

b) for the power produced by turbine, $P = m'(h_{1} - h_{2})$

$P = 46.51\times(793 - 401)$ = 18.231 MW

5 0

2 years ago

Methane and oxygen react in the presence of a catalyst to form formaldehyde. In a parallel reaction, methane is oxidized to carb

Nezavi [6.7K]

Answer:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Explanation:

Hello,

a. On the attached document, you can see a brief scheme of the process. Thus, to know the degrees of freedom, we state the following unknowns:

- $\xi_1$ and $\xi_2$ : extent of the reactions (2).

- $F_{O_2}^2$ , $F_{CH_4}^2$ , $F_{H_2O}^2$ , $F_{HCHO}^2$ and $F_{CO_2}^2$ : Molar flows at the second stream (5).

On the other hand, we've got the following equations:

- $F_{O_2}^2=50mol/s-\xi_1-2\xi_2$ : oxygen mole balance.

- $F_{CH_4}^2=50mol/s-\xi_1-\xi_2$ : methane mole balance.

- $F_{H_2O}^2=\xi_1+2\xi_2$ : water mole balance.

- $F_{HCHO}^2=\xi_1$ : formaldehyde mole balance.

- $F_{CO_2}^2=\xi_2$ : carbon dioxide mole balance.

Thus, the degrees of freedom are:

$DF=7unknowns-5equations=2$

It means that we need two additional equations or data to solve the problem.

b. Here, the two missing data are given. For the fractional conversion of methane, we define:

$0.900=\frac{\xi_1+\xi_2}{50mol/s}$

And for the fractional yield of formaldehyde we can set it in terms of methane as the reagents are equimolar:

$0.860=\frac{F_{HCHO}^2}{50mol/s}$

In such a way, one realizes that the output formaldehyde's molar flow is:

$F_{HCHO}^2=0.860*50mol/s=43mol/s$

Which is equal to the first reaction extent $\xi_1$ , therefore, one computes the second one from the fractional conversion of methane as:

$\xi_2=0.900*50mol/s-\xi_1\\\xi_2=0.900*50mol/s-43mol/s\\\xi_2=2mol/s$

Now, one computes the rest of the output flows via:

- $F_{O_2}^2=50mol/s-43mol/s-2*2mol/s=3mol/s$

- $F_{CH_4}^2=50mol/s-43mol/s-2mol/s=5mol/s$

- $F_{H_2O}^2=43mol/s+2*2mol/s=47mol/s$

- $F_{HCHO}^2=43mol/s$

- $F_{CO_2}^2=2mol/s$

The total output molar flow is:

$F_{O_2}+F_{CH_4}+F_{H_2O}+F_{HCHO}+F_{CO_2}=100mol/s$

Therefore the output stream composition turns out into:

$y_{CH_4}^2=\frac{5mol/s}{100mol/s}=0.05\\y_{O_2}^2=\frac{3mol/s}{100mol/s}=0.03\\y_{H_2O}^2=\frac{47mol/s}{100mol/s}=0.47\\y_{HCHO}^2=\frac{43mol/s}{100mol/s}=0.43\\y_{CO_2}^2=\frac{2mol/s}{100mol/s}=0.02$

Best regards.

7 0

2 years ago

Superheated steam at an average temperature 200 C is transported through a steel pipe (k=50 W/mK, D_0=8.0 cm,D_i=6.0 cm,and L=20

Zarrin [17]

The total amount of daily heat transfer is 1382.38 M w.

The temperature on the outside surface of the gypsum plaster insulation is 17.96 ° C.

<u>Explanation:</u>

Given data,

$T_{\infty}$ = 10° C

$h_{0}$ = 250 w/ $m^{2}$ k

Pipe length = 20 m

Inner diameter $d_{1}$ = 6 cm, $r_{1}$ = 3 cm

Outer diameter $d_{2}$ = 8 cm, $r_{2}$ = 4 cm

The thickness of insulation is 4 cm.

$r_{3}$ = $r_{2}$ + 4

= 4+4

$r_{3}$ = 8 cm

$h_{0}$ is the heat transfer coefficient of convection inside, $h_{i}$ is the heat transfer coefficient of convection outside.

The heat transfer rate between ambient and steam is

$q=\frac{T_{S}-T_{\infty}}{\frac{1}{h_{i}\left(2 \pi r_{1} L\right)}+\frac{\ln \left(r_{2} / r_{1}\right)}{2 \pi K_{1} L}+\frac{\ln \left(r_{3}/ r_{2}\right)}{2 \pi K_{2} L}+\frac{1}{h_{0}\left(2 \pi r_{3} L\right)}}$ watt

= $\begin{aligned}&\frac{1}{800(2 \pi x \cdot 03 \times 20)}\++\frac{\ln (4 / 3)}{2 \pi \times 50 \times 20}+\frac{\ln (8 / 4)}{2 \pi \times 0.5 \times 20}+\frac{1}{200(2 \pi x \cdot 08 \times 20)}\end{aligned}$ watt

= $\frac{190}{0.0003317+0.0000458+0.0110+0.0004976}$ watt

q = 15999.86 watt

The total amount of daily heat transfer = 15999.86 × 86400

= 1382.387904 watt

= 1382.38 M w

The total amount of daily heat transfer is 1382.38 M w.

b) The temperature on the outside surface of the gypsum plaster insulation.

q = $\frac{T_{3}-T_{\infty}}{\frac{1}{\ln \left(2 \pi \ r_{3} L\right)}}$

15999.86 $=\frac{\frac{1}{T_3}-10}{\frac{1}{200(2 \pi . 08 \times 20)}}$

$T_{3}$ - 10 = 7.96

$T_{3}$ = 17.96 ° C.

4 0

2 years ago