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2 years ago

How many astronauts work in the International Space Station

Engineering

1 answer:

laila [671]2 years ago

4 0

Answer:

At the moment there are three astronauts working in the international space station.

Explanation:

The space station is as large as a U.S football field and is typically occupied by at least three astronauts and a maximum of six astronauts.

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Refrigerant-134a at 700 kPa, 70°C, and 7.2 kg/min is cooled by water in a condenser until it exists as a saturated liquid at the

alex41 [277]

Answer:

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

Explanation:

A condenser is a heat exchanger used to cool working fluid (Refrigerant 134a) at the expense of cooling fluid (water), which works usually at steady state. Let suppose that there is no heat interactions between condenser and surroundings.The condenser is modelled after the First Law of Thermodynamics, which states:

$\dot Q_{ref} - \dot Q_{w} = 0$

$\dot Q_{ref} = \dot Q_{w}$

$\dot m_{ref}\cdot (h_{ref, in} - h_{ref,out}) = \dot m_{w}\cdot (h_{w, out} - h_{w,in})$

The mass flow rate of the cooling water is now cleared:

$\dot m_{w} = \dot m_{ref }\cdot \frac{h_{ref,in}-h_{ref,out}}{h_{w,out}-h_{w,in}}$

Given that $h_{ref,in} = 808.34\,\frac{kJ}{kg}$ , $h_{ref, out} = 88.82\,\frac{kJ}{kg}$ , $h_{w,out} = 104.83\,\frac{kJ}{kg}$ and $h_{w,in} = 62.98\,\frac{kJ}{kg}$ , the mass flow of the cooling water is:

$\dot m_{w} = \left(7.2\,\frac{kg}{min} \right)\cdot \left(\frac{808.34\,\frac{kJ}{kg}-88.82\,\frac{kJ}{kg} }{104.83\,\frac{kJ}{kg}-62.98\,\frac{kJ}{kg} } \right)$

$\dot m_{w} = 123.788\,\frac{kg}{min}$

The mass flow rate of cooling water required to cool the refrigerant is $123.788\,\frac{kg}{min}$ .

4 0

3 years ago

A very long pipe of 0.05 m (r1) radius and 0.03 m thickness (r2 - r1) is buried at a depth of 2m (z) to transport liquid nitroge

Elenna [48]

Answer:

The surface temperature of the ground is = 296.946K

Explanation:

Solution

Given

r₁= 0.05m

r₂= 0.08m

Tn =Ti = 77K

Ki = 0.0035 Wm-1K-1

Kg = 1 Wm-1K-1

Z= 2m

Now,

The outer type temperature (Skin temperature pipe)

Q = T₀ -T₁/ln (r2/r1)/2πKi = 2πKi T0 -T1/ln (r2/r1)

Thus,

10 w/m = 2π * 0.0035 = T0 -77/ln 0.08/0.05

⇒ T₀ -77 = 231.72

T₀= 290.72K

The shape factor between the cylinder and he ground

S = 2πL/ln 4z/D

where L = length of pipe

D = outer layer of pipe

S = 2π * 1/4 *2/ 2 * 0.08 = 1.606m

The heat gained in the pipe is = S * Kg * (Tg- T₀)

(10* 1) = 1.606 * 1* (Tg- 290.72)

Tg - 290.72 = 6.2266

Tg = 296.946K

Therefore the surface temperature to the ground is 296.946K

6 0

3 years ago

In an ideal reheat cycle, steam is generated at 6.3 Mpa, 450 deg. C and partially expands to 0.84 Mpa. At this point, the steam

AURORKA [14]

300 squeare meter 60deegree

7 0

2 years ago

*WILL MARK BRAINLIEST* The Northern and Southern Hemispheres face the sun evenly. What do the hemispheres experience at that tim

olga_2 [115]

An equinox .............

6 0

2 years ago

A girder that carries a uniformly distributed dead load of 1.7 k/ft plus its self-weight and three concentrated live loads of 15

posledela

Find the solution in attachments

3 0

3 years ago