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Alja [10]
3 years ago
10

If the old radiator is replaced with a new one that has longer tubes made of the same material and same thickness as those in th

e old unit, what should the total surface area available for heat exchange be in the new radiator to achieve the desired cooling temperature gradient
Engineering
1 answer:
Nookie1986 [14]3 years ago
3 0

Answer: hello some parts of your question is missing attached below is the missing information

The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d  is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube

answer : Total surface area = 3/2 * area of old radiator

Explanation:

we will use this relation

K = \frac{Qd }{A* change in T }

change in T =  ΔT  

therefore New Area  ( A ) = 3/2 * area of old radiator

Given that the thermal conductivity is the same in the new and old radiators

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For a bronze alloy, the stress at which plastic deformation begins is 297 MPa and the modulus of elasticity is 113 GPa. (a) What
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Answer:

a) 93.852 kN

b) 128.043 mm

Explanation:

Stress is load over section:

σ = P / A

If plastic deformation begins with a stress of 297 MPa, the maximum load before plastic deformation will be:

P = σ * A

316 mm^2 = 3.16*10^-4

P = 297*10^6 * 3.16*10^-4 = 93852 N = 93.852 kN

The stiffness of the specimen is:

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k = 113*10^9 * 3.16*10^-4 / 0.128 = 279 MN/m

Hooke's law:

x' = x0 * (1 + P/k)

x' = 0.128 * (1 + 93.852*10^3 / 279*10^6) = 0.128043 m = 128.043 mm

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