Your local hospital is considering the following solution options to address the issues of congestion and equipment failures at
their radiography unit, which is a bottleneck in the delivery of emergency services.
Solution Option
Setup costs ($)
Operating costs ($) per service instance
A (outsource all services)
negligible
100
B (upgrade existing equipment)
150,000
75
C (install new machine)
650,000
50
Solution Option
Setup costs ($)
Operating costs ($) per service instance
A (outsource all services)
negligible
100
B (upgrade existing equipment)
150,000
75
C (install new machine)
650,000
50
1 answer:
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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
