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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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8 0
3 years ago
A prototype boat is 30 meters long and is designed to cruise at 9 m/s. Its drag is to be simulated by a 0.5-meter-long model pul
Ghella [55]

Answer:

a) 1.16 m/s

b)  1/216000

c)  (√15)/6480000

Explanation:

The parameters given are;

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Speed of boat prototype = 9 m/s

Length of boat model, lm= 0.5 m

a) lm/lp = 0.5/30 = 1/60 = ∝

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b) The ratio of the model to prototype drag, Fm/Fp, is given as follows;

Fm/Fp = (vm/vp)²×(lm/lp)² = ∝³

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c) The ratio of the model to prototype power  pm/p_p = (Fm/Fp) × (vm/vp) = ∝³×√∝

The ratio of the model to prototype power  pm/p_p = √(1/60) × (1/60)³

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6 0
3 years ago
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Answer:

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3 0
3 years ago
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PSYCHO15rus [73]

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5 0
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