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Answer:
The rate of heat generation in the wire per unit volume is 5.79×10^7 Btu/hrft^3
Heat flux is 9.67×10^7 Btu/hrft^2
Explanation:
Rate of heat generation = 1000 W = 1000/0.29307 = 3412.15 Btu/hr
Area (A) = πD^2/4
Diameter (D) = 0.08 inches = 0.08 in × 3.2808 ft/39.37 in = 0.0067 ft
A = 3.142×0.0067^2/4 = 3.53×10^-5 ft^2
Volume (V) = A × Length
L = 20 inches = 20 in × 3.2808 ft/39.37 in = 1.67 ft
V = 3.53×10^-5 × 1.67 = 5.8951×10^-5 ft^3
Rate of heat generation in the wire per unit volume = 3412.15 Btu/hr ÷ 5.8951×10^-5 ft^3 = 5.79×10^7 Btu/hrft^3
Heat flux = 3412.15 Btu/hr ÷ 3.53×10^-5 ft^2 = 9.67×10^7 Btu/hrft^2
Answer:
(b)Distortion energy theory.
Explanation:
The best suitable theory for ductile material:
(1)Maximum shear stress theory (Guest and Tresca theory)
It theory state that applied maximum shear stress should be less or equal to its maximum shear strength.
(2)Maximum distortion energy theory(Von Mises henkey's theory)
It states that maximum shear train energy per unit volume at any point is equal to strain energy per unit volume under the state of uni axial stress condition.
But from these two Best theories ,suitable theory is distortion energy theory ,because it gives best suitable result for ductile material.
Answer:
The overview of the given scenario is explained in explanation segment below.
Explanation:
- The inception of cavitation, that further sets the restriction for high-pressure and high-free operation, has always been the matter of substantial experimental study over the last few generations.
- Cavitation inception would be expected to vary on the segment where the local "PL" pressure mostly on segment keeps falling to that are below the "Pv" vapor pressure of the fluid and therefore could be anticipated from either the apportionment of the pressure.
⇒ A cavitation number is denoted by "σ" .
Answer:
Explanation:
Let assume that heating and boiling process occurs under an athmospheric pressure of 101.325 kPa. The heat needed to boil water is:
![Q_{water} = (1.4\,L)\cdot(\frac{1\,m^{3}}{1000\,L} )\cdot (1000\,\frac{kg}{m^{3}} )\cdot [(4.187\,\frac{kJ}{kg\cdot ^{\textdegree}C} )\cdot (100^{\textdegree}C-25^{\textdegree}C)+2257\,\frac{kJ}{kg}]](https://tex.z-dn.net/?f=Q_%7Bwater%7D%20%3D%20%281.4%5C%2CL%29%5Ccdot%28%5Cfrac%7B1%5C%2Cm%5E%7B3%7D%7D%7B1000%5C%2CL%7D%20%29%5Ccdot%20%281000%5C%2C%5Cfrac%7Bkg%7D%7Bm%5E%7B3%7D%7D%20%29%5Ccdot%20%5B%284.187%5C%2C%5Cfrac%7BkJ%7D%7Bkg%5Ccdot%20%5E%7B%5Ctextdegree%7DC%7D%20%29%5Ccdot%20%28100%5E%7B%5Ctextdegree%7DC-25%5E%7B%5Ctextdegree%7DC%29%2B2257%5C%2C%5Cfrac%7BkJ%7D%7Bkg%7D%5D)
The heat liberated by the LP gas is:
A kilogram of LP gas has a minimum combustion power of 
. Then, the required mass is:
