Given Information:
Initial temperature of aluminum block = 26.5°C
Heat flux = 4000 w/m²
Time = 2112 seconds
Time = 30 minutes = 30*60 = 1800 seconds
Required Information:
Rise in surface temperature = ?
Answer:
Rise in surface temperature = 8.6 °C after 2112 seconds
Rise in surface temperature = 8 °C after 30 minutes
Explanation:
The surface temperature of the aluminum block is given by

Where q is the heat flux supplied to aluminum block, k is the conductivity of pure aluminum and α is the diffusivity of pure aluminum.
After t = 2112 sec:

The rise in the surface temperature is
Rise = 35.1 - 26.5 = 8.6 °C
Therefore, the surface temperature of the block will rise by 8.6 °C after 2112 seconds.
After t = 30 mins:

The rise in the surface temperature is
Rise = 34.5 - 26.5 = 8 °C
Therefore, the surface temperature of the block will rise by 8 °C after 30 minutes.
A diagram showing a use case diagrams for these requirements is given in the image attached.
<h3>What is system
Case diagram?</h3>
A use case diagram is known to be a kind of graphical illustration of a users in terms of their various possible association or interactions within any given system.
A use case diagram in banking can be used to prepare, depict and also to know all the functional requirements of the banking system.
Therefore, Give the use case specification for the banking system services and paying a bill online is given in the image attached.
Learn more about Case diagram from
brainly.com/question/12975184
#SPJ1
Answer:
The specific weight of unknown liquid is found to be 15 KN/m³
Explanation:
The total pressure in tank is measured to be 65 KPa in the tank. But, the total pressure will be equal to the sum of pressures due to both oil and unknown liquid.
Total Pressure = Pressure of oil + Pressure of unknown liquid
65 KPa = (Specific Weight of oil)(depth of oil) + (Specific Weight of unknown liquid)(depth of unknown liquid)
65 KN/m² = (8.5 KN/m³)(5 m) + (Specific Weight of Unknown Liquid)(1.5 m)
(Specific Weight of Unknown Liquid)(1.5 m) = 65 KN/m² - 42.5 KN/m²
(Specific Weight of Unknown Liquid) = (22.5 KN/m²)/1.5 m
<u>Specific Weight of Unknown Liquid = 15 KN/m³</u>