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 = [tex]\frac{Qd }{A* change in T }[/tex]
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
A 50 mm 45 mm 20 mm cell phone charger has a surface temperature of Ts 33 C when plugged into an electrical wall outlet but not in use. The surface of the charger is of emissivity 0.92 and is subject to a free convection heat transfer coefficient of h 4.5 W/m2 K. The room air and wall temperatures are T 22 C and Tsur 20 C, respectively. If electricity costs C $0.18/kW h, determine the daily cost of leaving the charger plugged in when not in use.
Answer:
C = $0.0032 per day
Explanation:
We are given;
Dimension of cell phone; 50 mm × 45 mm × 20 mm
Temperature of charger; T1 = 33°C = 306K
Emissivity; ε = 0.92
convection heat transfer coefficient; h = 4.5 W/m².K
Room air temperature; T∞ = 22°C = 295K
Wall temperature; T2 = 20°C = 293 K
Cost of electricity; C = $0.18/kW.h
Chargers are usually in the form of a cuboid, and thus, surface Area is;
A = (50 × 45) + 2(50 × 20) + 2(45 × 20)
A = 6050 mm²
A = 6.05 × 10^(-3) m²
Formula for total heat transfer rate is;
E_t = hA(T1 - T∞) + εσA((T1)⁴ - (T2)⁴)
Where σ is Stefan Boltzmann constant with a value of; σ = 5.67 × 10^(-8) W/m².K⁴
Thus;
E_t = 4.5 × 6.05 × 10^(-3) (306 - 295) + (0.92 × 6.05 × 10^(-3) × 5.67 × 10^(-8)(306^(4) - 293^(4)))
E_t = 0.7406 W = 0.7406 × 10^(-3) KW
Now, we know C = $0.18/kW.h
Thus daily cost which has 24 hours gives;
C = 0.18 × 0.7406 × 10^(-3) × 24
C = $0.0032 per day