Answer:
Explanation:
Given that:
V = 12.5m/s
L= 2.70m
b= 0.65m
P = 1atm
Film temperature
dynamic viscosity =
density = 0.9946kg/m³
Pr = 0.708564
K= 229.7984 * 10⁻³w/mk
Reynolds number,
we have,
we have,
heat transfer rate from top plate
An R-134a refrigeration system is operating with an evaporator pressure of 200 kPa. The refrigerant is 10% in vapor phase at the
inlet of the evaporator. The enthalpy and temperature of the refrigerant at the exit of the compressor are 360 kJ/kg and 70 /C respectively. If this system (with the refrigerant flowing at 0.005 kg/s) is used to cool a 750 kg product (initial temperature = 80 /C and specific heat = 3,000 J/kg-K), what will be the temperature of the product after 6 hours?
1 answer:
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Answer:
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
Explanation:
Given data;
Let,
critical stress required for initiating crack propagation Cc = 112MPa
plain strain fracture toughness = 27.0MPa
surface length of the crack = a
dimensionless parameter = Y.
Half length of the internal crack, a = length of surface crack/2 = 8.8/2 = 4.4mm = 4.4*10-³m
Also for 6.2mm length of surface crack;
Half length of the internal crack = length of surface crack/2 = 6.2/2 = 3.1mm = 3.1*10-³m
The dimensionless parameter
Cc = Kic/(Y*√pia*a)
Y = Kic/(Cc*√pia*a)
Y = 27/(112*√pia*4.4*10-³)
Y = 2.05
Now,
Cc = Kic/(Y*√pia*a)
Cc = 27/(2.05*√pia*3.1*10-³)
Cc = 135.78MPa
The stress level at which fracture will occur for a critical internal crack length of 6.2mm is 135.78MPa
For more understanding, I have provided an attachment to the solution.
Answer:
The channel distance affected by the weir
Yn ( Normal depth ) = 5.71 ft
Yc ( critical depth ) = 4.11 ft
x = 22794 ft
Explanation:
Attached below is a detailed solution to the problem
The channel distance affected by the weir
Yn ( Normal depth ) = 5.71 ft
Yc ( critical depth ) = 4.11 ft
x = 22794 ft
