Answer:
In summer the power available is 357.55 kW and in winter the power available is 59.59 kW
Explanation:
Given data:
height = 1500 ft = 457.2 m
30 gallon = 0.114 m³
5 gallon = 0.019 m³
In summer the power available is:
Where
μ = efficiency = 0.7
ρ = density of water = 1000 kg/m³
g = gravity = 9.8 m/s²
Q = 0.114 m³
Replacing:
In winter the power available is
Answer:
hello some parts of your question is missing attached below is the missing part ( the required fig and table )
answer : The solar collector surface area = 7133 m^2
Explanation:
Given data :
Rate of energy input to the collectors from solar radiation = 0.3 kW/m^2
percentage of solar power absorbed by refrigerant = 60%
Determine the solar collector surface area
The solar collector surface area = 7133 m^2
attached below is a detailed solution of the problem
Answer:
The plot of the function production rate m(t) (in kg/min) against time t (in min) is attached to this answer.
The production rate function M(t) is:
(1)
The Laplace transform of this function is:
(2)
Explanation:
The function of the production rate can be considered as constant functions by parts in the domain of time. To make it a continuous function, we can use the function Heaviside (as seen in equation (1)). To join all the constant functions, we consider at which time the step for each one of them appears and sum each function multiply by the function Heaviside.
For the Laplace transform we use the following rules:
(3)
(4)
Answer:
P2 = 3.9 MPa
Explanation:
Given that
T₁ = 290 K
P₁ = 95 KPa
Power P = 5.5 KW
mass flow rate = 0.01 kg/s
solution
with the help of table A5
here air specific heat and adiabatic exponent is
Cp = 1.004 kJ/kg K
and k = 1.4
so
work rate will be
W = m × Cp × (T2 - T1) ..........................1
here T2 = W ÷ ( m × Cp) + T1
so T2 = 5.5 ÷ ( 0.001 × 1.004 ) + 290
T2 = 838 k
so final pressure will be here
P2 = P1 × ..............2
P2 = 95 ×
P2 = 3.9 MPa