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Amiraneli [1.4K]
3 years ago
12

Demonstreaza in 20 de propoziti ca snoava pacala si zarzarele boerului e o snoava

Engineering
1 answer:
S_A_V [24]3 years ago
4 0

Answer:

oops i dnt understand this language.

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All the fnaf UNC charecters
astra-53 [7]

Answer:

T.Freddy.

T.Bonnie.

T.Chica.

G.Freddy.

N. Freddy.

F.Foxy.

Mr.Hippo.

R.Freddy.

Explanation:

8 0
3 years ago
Read 2 more answers
What is the measurement of this dial caliper? <br> A. 5.491<br> B. 4.044<br> C. 5.691
just olya [345]
Its C .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... .......... ..........
4 0
4 years ago
An object of irregular shape has a characteristic length of � = 1.00 [�] and is maintained at a uniform surface temperature of �
Pani-rosa [81]

The correct question;

An object of irregular shape has a characteristic length of L = 1 m and is maintained at a uniform surface temperature of Ts = 400 K. When placed in atmospheric air at a temperature of Tinfinity = 300 K and moving with a velocity of V = 100 m/s, the average heat flux from the surface to the air is 20,000 W/m² If a second object of the same shape, but with a characteristic length of L = 5 m, is maintained at a surface temperature of Ts = 400 K and is placed in atmospheric air at Too = 300 K, what will the value of the average convection coefficient be if the air velocity is V = 20 m/s?

Answer:

h'_2 = 40 W/K.m²

Explanation:

We are given;

L1 = 1m

L2 = 5m

T_s = 400 K

T_(∞) = 300 K

V = 100 m/s

q = 20,000 W/m²

Both objects have the same shape and density and thus their reynolds number will be the same.

So,

Re_L1 = Re_L2

Thus, V1•L1/v1 = V2•L2/v2

Hence,

(h'_1•L1)/k1 = (h'_2•L2)/k2

Where h'_1 and h'_2 are convection coefficients

Since k1 = k2, thus, we now have;

h'_2 = (h'_1(L1/L2)) = [q/(T_s - T_(∞))]• (L1/L2)

Thus,

h'_2 = [20,000/(400 - 300)]•(1/5)

h'_2 = 40 W/K.m²

5 0
3 years ago
21.Why are throttling devices commonly used in refrigeration and air-conditioning<br> applications?
Sloan [31]

Answer is given below

Explanation:

we know that some common types of throttling devices are

  • Hard -throttling devices
  • Capillary valve
  • Constant pressure throttling devices
  • Thermostatic expansion valve
  • Float expansion valve

so here throttling devices commonly used in refrigeration and air-conditioning because

  • To reduce the coolant pressure, the high pressure of the refrigerant from the condenser is necessary to reduce the evaporation to obtain evaporation at the right temperature  
  • To meet the refrigerated load, the throttling valve flows through the coolant to cool the load at high temperatures.
5 0
4 years ago
From your cooling load (8890.007 Btu/hr = 2.605kW, determine mass flow rate of refrigerants. Use the following "rule of thumb" e
NeX [460]

Answer:

0.740833917 ton/hr

Explanation:

Given:

Cooling load, 8890.007 Btu/hr = 2.605 kW

Room size = 180 ft^{2}

According to the thumb rule

1 ton of refrigerant = 12000Btu

Hence for 8890.007 Btu/hr,

the mass flow rate of the refrigerant is =8890.007 / 12000

                                                                = 0.740833917 ton per hr

Hence, mass flow rate is 0.740833917 ton/hr

7 0
3 years ago
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