Answer:
1. Equatorial Evergreen or Rainforest
2. Tropical forest
3. Mediterranean forest
4. Temperate broad-leaved forest
5. Warm temperate forest
Explanation:
<u>Software Development and Client Needs</u>
In Incremental method of software development customers who do not have a basic idea of the development process are being carried along on like other methods that will relegate them to the background until a product is ready.
With this model and structure in place, when softwares/ products are built from several stages e.g prototype, testing, and when new features are added customers are always carried along with their valuable feedback and suggested greatly considered to achieve the customers satisfactions
This model will work well for the customers/clients who does not have a clear idea on the systems needed for their operations.
In summary the incremental model combines features from the waterfall and prototyping model.
For more information on soft ware development process kindly visit
brainly.com/question/20369682
Answer: The energy system related to your question is missing attached below is the energy system.
answer:
a) Work done = Net heat transfer
Q1 - Q2 + Q + W = 0
b) rate of work input ( W ) = 6.88 kW
Explanation:
Assuming CPair = 1.005 KJ/Kg/K
<u>Write the First law balance around the system and rate of work input to the system</u>
First law balance ( thermodynamics ) :
Work done = Net heat transfer
Q1 - Q2 + Q + W = 0 ---- ( 1 )
rate of work input into the system
W = Q2 - Q1 - Q -------- ( 2 )
where : Q2 = mCp T = 1.65 * 1.005 * 293 = 485.86 Kw
Q2 = mCp T = 1.65 * 1.005 * 308 = 510.74 Kw
Q = 18 Kw
Insert values into equation 2 above
W = 6.88 Kw
Answer:
a) 70.29 %
b) 37%
Explanation:
percent reduction can be found from:
PR = 100*(π(do/2)^2-π(df/2)^2)/π(do/2)^2
= 100*(π(11.34/2)^2-π(6.21/2)^2)/π(11.34/2)^2
=70.29 %
percent elongation can be found from:
EL =L_f - Lo/Lo*100
= (73.17 -53.3/53.3)*100
= 37%
Answer:
the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C
Explanation:
Given:
d₁ = diameter of the tube = 1 cm = 0.01 m
d₂ = diameter of the shell = 2.5 cm = 0.025 m
Refrigerant-134a
20°C is the temperature of water
h₁ = convection heat transfer coefficient = 4100 W/m² K
Water flows at a rate of 0.3 kg/s
Question: Determine the overall heat transfer coefficient of this heat exchanger, Q = ?
First at all, you need to get the properties of water at 20°C in tables:
k = 0.598 W/m°C
v = 1.004x10⁻⁶m²/s
Pr = 7.01
ρ = 998 kg/m³
Now, you need to calculate the velocity of the water that flows through the shell:

It is necessary to get the Reynold's number:

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

The overall heat transfer coefficient:

Here

Substituting values:
