Answer:
Total saving would be of 36.917 $\yr
Explanation:
Given Data:
Degree F
Degree F
A = 30 ft^2
%
Heat loss before previous final value 

= 360 Btu/hr
Heat loss after new value
saving would be 
= 616.1782 kw hr/yr
$
= 36.917 $\yr
Answer: 133.88 MPa approximately 134 MPa
Explanation:
Given
Plane strains fracture toughness, k = 26 MPa
Stress at which fracture occurs, σ = 112 MPa
Maximum internal crack length, l = 8.6 mm = 8.6*10^-3 m
Critical internal crack length, l' = 6 mm = 6*10^-3 m
We know that
σ = K/(Y.√πa), where
112 MPa = 26 MPa / Y.√[3.142 * 8.6*10^-3)/2]
112 MPa = 26 MPa / Y.√(3.142 * 0.043)
112 = 26 / Y.√1.35*10^-2
112 = 26 / Y * 0.116
Y = 26 / 112 * 0.116
Y = 26 / 13
Y = 2
σ = K/(Y.√πa), using l'instead of l and, using Y as 2
σ = 26 / 2 * [√3.142 * (6*10^-3/2)]
σ = 26 / 2 * √(3.142 *3*10^-3)
σ = 26 / 2 * √0.009426
σ = 26 / 2 * 0.0971
σ = 26 / 0.1942
σ = 133.88 MPa
Answer:
They include PCBs, dioxins, furans, hexachlorobenzene, benzo(a)pyrene (one of many polyaromatic hydrocarbons or PAHs), lead and mercury compounds and pesticides (toxaphene, mirex, aldrin/dieldrin, DDT). 3 Great Lakes Water Quality Agreement Work Group Report on Chemicals of Emerging Concern, August 2009.
I pretty sure it would be A
Answer:
T1 = 299.18 °C
P2 = 0.00738443 MPa
Explanation:
From the data, we can get two properties for the initial condition. These are pressure and specific volume.
The pressure is 1.8 MPa and the specific volume, we can get it with the mass and volume of the container, since it’s filled this is also the volume of the water in it.

When we check in the thermodynamic tables, the conditions for saturation at 1.8 MPa we found the following:


specific volume for the saturated vapor
specific volume for the saturated liquid
Since the specific volume in our condition is higher that the specific volume for the saturated vapor, we have a superheated steam.
Looking in the thermodynamic tables for superheated steam we found that the temperature where the steam has a specific volume of
at 1.8 MPa is 299.18 °C. This is the initial temperature in the container.
Since the only information that we have about the final condition is that the container was cooled. We can assume that it was cooled until a condition of saturation. So, the final pressure for the water will be the pressure of saturation for a temperature of 40°C. From thermodynamic tables we get:
