Technician A says that burning of fossil fuels has increased the greenhouse gas CO 2 in the atmosphere. Technician B says that i
1 answer:
You might be interested in
Answer:
i) 25.04 W/m^2 .k
ii) 23.82 minutes = 1429.2 secs
Explanation:
Given data:
Diameter of steel ball = 15 cm
uniform temperature = 350°C
p = 8055 kg/m^3
Cp = 480 J/kg.k
surface temp of ball drops to 250°C
average surface temperature = ( 350 + 250 ) / 2 = 300°C
<u>i) Determine the average convection heat transfer coefficient during the cooling process</u>
<em>Note : Obtain the properties of air at 1 atm at average film temp of 30°C from the table " properties of air " contained in your textbook</em>
average convection heat transfer coefficient = 25.04 W/m^2 .k
<u>ii) Determine how long this process has taken </u>
Time taken by the process = 23.82 minutes = 1429.2 seconds
Δt = Qtotal / Qavg = 683232 / 477.92 = 1429.59 secs
attached below is the detailed solution of the given question
Answer:
For the Top Side
- Strain ε = 0.00021739
- Elongation is 0.00260868 cm
For The Right side
- Strain ε = 0.00021739
-Elongation is 0.00347826 cm
Explanation:
Given the data in the question;
Length of the squared titanium plate = 12 cm by 12 cm = 0.12 m by 0.12 m
Thickness = 5 mm = 0.005 m
Force to the Top F = 15 kN = 15000 Newton
Force to the right F = 20 kN = 20000 Newton
elastic modulus, E = 115 GPa = 115 × 10⁹ pascal
Now, For the Top Side;
- Strain = σ/E = F / AE
we substitute
= 15000 / ( 0.12 × 0.005 × (115 × 10⁹) )
= 15000 / 69000000
Strain ε = 0.00021739
- Elongation
Δl = ε × l
we substitute
Δl = 0.00021739 × 12 cm
Δl = 0.00260868 cm
Hence, Elongation is 0.00260868 cm
For The Right side
- Strain = σ/E = F / AE
we substitute
Strain = 20000 / ( 0.12 × 0.005 × (115 × 10⁹) )
= 20000 / 69000000
Strain ε = 0.000289855
- Elongation
Δl = ε × l
we substitute
Δl = 0.000289855× 12 cm
Δl = 0.00347826 cm
Hence, Elongation is 0.00347826 cm
Answer:
The inventor's claim is false in the sense that no thermal machine can violate the first thermodynamic law.
Explanation:
The inventor's claim could not be possible as no thermal machine can transfer more heat than the input work consumed. If we expose the thermal efficiency:
Where Q and W both must be in the same power unit, so we will convert the remove heat from BTU/hr to hp:
Therefore by comparing, we notice that the removing heat of 4.75 hp is large than the delivered work of 1.11 hp. By evaluating the efficiency:
[tex]n=4.75 hp / 1.1 hp = 4.3 > 1[/tex]