Do heavier cars really use more gasoline? Suppose a car is chosen at random. Let x be the weight of the car (in hundreds of poun
1 answer:
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Answer:
fracture will occur as the value is less than E/10 (= 22.5)
Explanation:
If the maximum strength at tip Is greater than theoretical fracture strength value then fracture will occur and if the maximum strength is lower than theoretical fracture strength then no fracture will occur.
= 15 GPa
fracture will occur as the value is less than E/10 = 22.5
Answer:
5.118 m^3/hr
Explanation:
Given data:
viscosity of cell broth = 5cP
cake resistance = 1*1011 cm/g
dry basis per volume of filtrate = 20 g/liter
Diameter = 8m , Length = 12m
vacuum pressure = 80 kpa
cake formation time = 20 s
cycle time = 60 s
<u>Determine the filtration rate in volumes/hr expected fir the rotary vacuum filter</u>
attached below is a detailed solution of the question
Hence The filtration rate in volumes/hr expected for the rotary vacuum filter
V' = ( ) * 1706.0670
= 5118.201 liters ≈ 5.118 m^3/hr
Question
A 2-mm-diameter electrical wire is insulated by a 2-mm-thick rubberized sheath (k = 0.13 W/m.K), and the wire/sheath interface is characterized by a thermal contact resistance of Rtc = 3E-4m².K/W. The convection heat transfer coefficient at the outer surface of the sheath is 10 W/m²K, and the temperature of the ambient air is 20°C.
If the temperature of the insulation may not exceed 50°C, what is the maximum allowable electrical power that may be dissipated per unit length of the conductor? What is the critical radius of the insulation?
Answer:
a. 4.52W/m
b. 13mm
Explanation:
Given
Diameter of electrical wire = 2mm
Wire Thickness = 2-mm
Thermal Conductivity of Rubberized sheath (k = 0.13 W/m.K)
Thermal contact resistance = 3E-4m².K/W
Convection heat transfer coefficient at the outer surface of the sheath = 10 W/m²K,
Temperature of the ambient air = 20°C.
Maximum Allowable Sheet Temperature = 50°C.
From the thermal circuit (See attachment), we my write
E'q = q' = (Tin,i - T∞)/(R'cond + R'conv)
= (Tin,i - T∞)/(Ln (r in,o / r in,i)/2πk + (1/(2πr in,o h)))
Where r in,i = D/2
= 2mm/2
= 1 mm
= 0.001m
r in,o = r in,i + t = 0.003m
T in, i = Tmax = 50°C
Hence
q' = (50 - 20)/[(Ln (0.003/0.001)/(2π * 0.13) + 1/(2π * 0.003 * 10)]
= 30/[(Ln3/0.26π) + 1/0.06π)]
= 30/[(1.34) + 5.30)]
= 30/6.64
= 4.52W/m
The critical radius is unaffected by the constant resistance.
Hence
Critical Radius = k/h
= 0.13/10
= 0.013m
= 13mm
