Given Information:
Inlet Temperature of hot air= Th₁ = 450K
Exit Temperature of hot air = Th₂ = 350K
Inlet Temperature of cold air = Tc₁ = 300K
Volume flow rate of hot air = vh = 0.02 m³/s
Volume flow rate of cold air = vc = 1 m³/s
Required Information:
Exit Temperature of cold air= Tc₂ = ?
Answer:
Exit Temperature of cold air = Tc₂ = 302 °C
Explanation:
In a heat exchanger, the cold air absorbs heat that is lost by the hot air,
Heat absorbed by cold air = Heat lost by hot air
Therefore, the exit temperature of the cold air is 302 °C or 575K
Note:
m = ρv
Where ρ is density of air and v is the volume flow rate and m is the mass flow rate.
cp is the specific heat capacity of air.
Answer:
0.6944 m/sec^2
Explanation:
The computation of the average acceleration is given below:
a = v - u ÷ t
where
a denotes average acceleration
v denotes final velocity
u denotes initial velocity
t denotes time
So, the average acceleration is
= (25 - 0) ÷ 10
= 0.6944 m/sec^2
The initial velocity of the 3250 Kg mass is 2.1 m/s. The distance covered by the larger mass in 5s is 4.7 cm.
In this problem, we have to apply the law of conservation of linear momentum. Note that;
Momentum before collision = Momentum after collision
m1u1 + m2u2 = m1v1 + m2v2
(2150 × 10) + (3250u1) = (2150 + 3250)5.22
21500 + 3250u1 = 5400 × 5.22
3250u1 = 28188 - 21500
u1 = 28188 - 21500/3250
u1 = 2.1 m/s
2) Again from the principle of conservation of linear momentum;
(0.40 × 3.5) + (0.60 × 0) = (0.40 × 0.70) + (0.60 × v2)
1.4 = 0.28 + 0.60v2
1.4 - 0.28 = 0.60v2
v2 = 1.87 cm/s
Using;
s = 1/2 ( u + v)t
s = 1/2(0 + 1.87) × 5
s = 4.7 cm
Learn more: brainly.com/question/8898885
Answer:
See attached handwritten document for answer
Explanation:
100 watts = 100 joules per second
22 watts = 22 joules per second
Saving = 78 joules per second
1 hour = 3,600 seconds
24 hours = 86,400 seconds
(78 joules/sec) x (86,400 sec) = 6,739,200 joules