Answer:
50%
Explanation:
<u>Given information</u>
Cooling load=50 kW
COP=2
Consumption=50 kW
<u>Calculations</u>
Revised input is given by cooling load/COP=50/2=25 kW
Efficiency= Work output/ Revised input=25/50=0.5
Efficiency=0.5*100=50%
Explanation:
excess air is required to ensure adequate mixing of fuel and air, avoid smoke, minimize sg in Coal burning, and to ensure maximum steam output.
Answer:
Step 1
Given
Diameter of circular grill, D = 0.3m
Distance between the coal bricks and the steaks, L = 0.2m
Temperatures of the hot coal bricks, T₁ = 950k
Temperatures of the steaks, T₂ = 5°c
Explanation:
See attached images for steps 2, 3, 4 and 5
Answer: P = 0.416 kW
Explanation:
taken a step by step process to solving this problem.
we have that from the question;
the amount of heat rejected Qn = 4800 kJ/h
the cooling effect is Ql = 3300 kJ/h
Applying the first law of thermodynamics for this system gives us
Шnet = Qn -Ql
Шnet = 4800 - 3300 = 1500 kJ/h
Next we would calculate the coefficient of performance of the refrigerator;
COPr = Desired Effect / work output = Ql / Шnet = 3300/1500 = 2.2
COPr = 2.2
The Power as required gives;
P = Qn - Ql = 4800 - 3300 = 1500 kJ/h = 0.416
P = 0.416 kW
cheers i hope this helps!!!!1
Answer:
Map and avoid high-risk zones.
Build hazard-resistant structures and houses.
Protect and develop hazard buffers (forests, reefs, etc.)
Develop culture of prevention and resilience.
Improve early warning and response systems.
Build institutions, and development policies and plans.
Explanation:
