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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
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.
Answer:
0.740833917 ton/hr
Explanation:
Given:
Cooling load, 8890.007 Btu/hr = 2.605 kW
Room size = 180 
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