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3 years ago

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A small distiller evaporates 10 L of water per half hour. Alloy tubing exposed to the air serves a condenser to recover steam. T

he outside surface of the tubing is at 99C. The ambient temperature is 20C. The inside diameter of the tube is 0.85 cm, and the outside diameter is 1.2 cm. The thermal conductivity of copper is 388W/m.K, latent heat of condensation (hfg) for saturated steam at 100C is 2257kJ/kg. In order to condense all the steam, the tube length in meters is most nearly. How long must the tube be to condense all of the steam?
a) 0.3 nm
b) .2 m
c) 2.4 m
d) 8.3 mm
e) 10.8 m

1 answer:

xxTIMURxx [149]3 years ago

7 0

Answer:

c) 2.4m

Explanation:

1 liter of water evaporates in 30 minutes, Mass of water evaporation is :

1 / (30 * 60) = 5.55 * $10^{-4}$

The diameter of the tube is 0.85 cm inside and 1.2 cm outside

L = 1253 / 518.434 = 2.416

The length of the copper tube must be 2.4 m long.

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A police officer in a patrol car parked in a 70 km/h speed zone observes a passing automobile traveling at a slow, constant spee

Ludmilka [50]

Answer:

S = 0.5 km

velocity of motorist = 42.857 km/h

Explanation:

given data

speed = 70 km/h

accelerates uniformly = 90 km/h

time = 8 s

overtakes motorist = 42 s

solution

we know initial velocity u1 of police = 0

final velocity u2 = 90 km/h = 25 mps

we apply here equation of motion

u2 = u1 + at

so acceleration a will be

a = $\frac{25-0}{8}$

a = 3.125 m/s²

so

distance will be

S1 = 0.5 × a × t²

S1 = 100 m = 0.1 km

and

S2 = u2 × t

S2 = 25 × 16

S2 = 400 m = 0.4 km

so total distance travel by police

S = S1 + S2

S = 0.1 + 0.4

S = 0.5 km

and

when motorist travel with uniform velocity

than total time = 42 s

so velocity of motorist will be

velocity of motorist = $\frac{S}{t}$

velocity of motorist = $\frac{500}{42}$

velocity of motorist = 42.857 km/h

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3 years ago

Air at 400kPa, 970 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occu

Sonja [21]

Answer:

a

The rate of work developed is $\frac{\r W}{\r m}= 300kJ/kg$

b

The rate of entropy produced within the turbine is $\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

Explanation:

From the question we are told

The rate at which heat is transferred is $\frac{\r Q}{\r m } = - 30KJ/kg$

the negative sign because the heat is transferred from the turbine

The specific heat capacity of air is $c_p = 1.1KJ/kg \cdot K$

The inlet temperature is $T_1 = 970K$

The outlet temperature is $T_2 = 670K$

The pressure at the inlet of the turbine is $p_1 = 400 kPa$

The pressure at the exist of the turbine is $p_2 = 100kPa$

The temperature at outer surface is $T_s = 315K$

The individual gas constant of air R with a constant value $R = 0.287kJ/kg \cdot K$

The general equation for the turbine operating at steady state is \

$\r Q - \r W + \r m (h_1 - h_2) = 0$

h is the enthalpy of the turbine and it is mathematically represented as

$h = c_p T$

The above equation becomes

$\r Q - \r W + \r m c_p(T_1 - T_2) = 0$

$\frac{\r W}{\r m} = \frac{\r Q}{\r m} + c_p (T_1 -T_2)$

Where $\r Q$ is the heat transfer from the turbine

$\r W$ is the work output from the turbine

$\r m$ is the mass flow rate of air

$\frac{\r W}{\r m}$ is the rate of work developed

Substituting values

$\frac{\r W}{\r m} = (-30)+1.1(970-670)$

$\frac{\r W}{\r m}= 300kJ/kg$

The general balance equation for an entropy rate is represented mathematically as

$\frac{\r Q}{T_s} + \r m (s_1 -s_2) + \sigma = 0$

=> $\frac{\sigma}{\r m} = - \frac{\r Q}{\r m T_s} + (s_1 -s_2)$

generally $(s_1 -s_2) = \Delta s = c_p\ ln[\frac{T_2}{T_1} ] + R \ ln[\frac{v_2}{v_1} ]$

substituting for $(s_1 -s_2)$

$\frac{\sigma}{\r m} = \frac{-\r Q}{\r m} * \frac{1}{T_s} + c_p\ ln[\frac{T_2}{T_1} ] - R \ ln[\frac{p_2}{p_1} ]$

Where $\frac{\sigma}{\r m}$ is the rate of entropy produced within the turbine

substituting values

$\frac{\sigma}{\r m} = - (-30) * \frac{1}{315} + 1.1 * ln\frac{670}{970} - 0.287 * ln [\frac{100kPa}{400kPa} ]$

$\frac{\sigma}{\r m}= 0.0861kJ/kg \cdot K$

5 0

4 years ago

Ok there........................................................................

Juliette [100K]

Answer:

ok THERE

Explanation:

4 0

3 years ago

An office building is served by an air-cooled chiller currently operating at 115 tons (404.5 kW). The measured chilled water sup

Andrei [34K]

Answer:

B.197 gpm and 12.4 L/s

Explanation:

Given that

Load Q = 404.5 KW

Water inlet temperature= 6.1 °C

Water outlet temperature= 13.9°C

We know that specific heat for water

$C_p=4.187\ \frac{KJ}{kg.K}$

Now from energy balance

$Q=\dot{m}C_p\Delta T$

by putting the values

$Q=\dot{m}C_p\Delta T$

$404.5=\dot{m}\times 4.187(13.9-6.1)$

$\dot{m}=12.38\ \frac{kg}{s}$ (1 Kg/s = 15.85 gal/min)

We can say that

$\dot{m}=196.31\ \frac{gal}{min}$

We know that

$\dot{m}=\rho\times volume\ flow\ rate$

12.38=1000 x volume flow rate

$volume\ flow\ rate\ = 12.38\times 10^{-3}\ \frac{m^3}{s}$

So

volume flow rate = 12.38 L/s

So the option B is correct.

8 0

3 years ago