Answer:
mass flow rate = 0.0534 kg/sec
velocity at exit = 29.34 m/sec
Explanation:
From the information given:
Inlet:
Temperature 
Quality 
Outlet:
Temperature 
Quality 
The following data were obtained at saturation properties of R134a at the temperature of -16° C
Answer:
Explanation:
we have given E(t)=120 sin(12t)
R=5 ohm
L=0.2 H
ω=12 ( from expression of E)
ohm
=5.021 ohm
so amplitude of current = 
Answer:
The three dimensions shown in an isometric drawing are the height, H, the length, L, and the depth, D
Explanation:
An isometric drawing of an object in presents a pictorial projection of the object in which the three dimension, views of the object's height, length, and depth, are combined in one view such that the dimensions of the isometric projection drawing are accurate and can be measured (by proportion of scale) to draw the different views of the object or by scaling, for actual construction of the object.
The lightning efficiency based on the scenario depicted will be C. 56 lumens/Watt, more efficient.
<h3>How to calculate the lightning efficiency</h3>
The efficiency of the incandescent bulb will be:
= 450/40 = 11.25 lumens per watt.
The efficiency of the LED bulb will be:
= 450/8 = 56 lumens per watt.
In this case, the LED bulb is more efficient than the incandescent bulb.
Therefore, the lighting efficiency will be 56 lumens/Watt, more efficient
Learn more lightning efficiency on:
brainly.com/question/25927632
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
