Answer:
The architectural pattern i will use for the school management is the client-server pattern.
This pattern would consist of a server and many clients. wherein the server component would provide services to that of the clients and its components as specified and also there would be a client request service from that of the server.
Explanation:
Solution
A school management system would always involve the client server pattern as this pattern would have a server and many clients wherein the server component would give services to that of the clients and its components as specified and also there would be a client request service from that of the server. This server would share the appropriate services to such clients and also listen to the client's requests.
Such kind of pattern would mostly be used for for the online platforms or application like that of document.
Answer:
COP(heat pump) = 2.66
COP(Theoretical maximum) = 14.65
Explanation:
Given:
Q(h) = 200 KW
W = 75 KW
Temperature (T1) = 293 K
Temperature (T2) = 273 K
Find:
COP(heat pump)
COP(Theoretical maximum)
Computation:
COP(heat pump) = Q(h) / W
COP(heat pump) = 200 / 75
COP(heat pump) = 2.66
COP(Theoretical maximum) = T1 / (T1 - T2)
COP(Theoretical maximum) = 293 / (293 - 273)
COP(Theoretical maximum) = 293 / 20
COP(Theoretical maximum) = 14.65
Answer:
Airplanes' wings are curved on top and flatter on the bottom. That shape makes air flow over the top faster than under the bottom. As a result, less air pressure is on top of the wing. This lower pressure makes the wing, and the airplane it's attached to, move up.
Explanation:
Answer:
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Explanation:
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Answer:
a) Tբ = 151.8°C
b) ΔV = - 0.194 m³
c) The T-V diagram is sketched in the image attached.
Explanation:
Using steam tables,
At the given pressure of 0.5 MPa, the saturation temperature is the final temperature.
Right from the steam tables (A-5) with a little interpolation, Tբ = 151.793°C
b) The volume change
Using data from A-5 and A-6 of the steam tables,
The volume change will be calculated from the mass (0.58 kg), the initial specific volume (αᵢ) and the final specific volume
(αբ) (which is calculated from the final quality and the consituents of the specific volumes).
ΔV = m(αբ - αᵢ)
αբ = αₗ + q(αₗᵥ) = αₗ + q (αᵥ - αₗ)
q = 0.5, αₗ = 0.00109 m³/kg, αᵥ = 0.3748 m³/kg
αբ = 0.00109 + 0.5(0.3748 - 0.00109)
αբ = 0.187945 m³/kg
αᵢ = 0.5226 m³/kg
ΔV = 0.58 (0.187945 - 0.5226) = - 0.194 m³
c) The T-V diagram is sketched in the image attached
