(1) Heat is convected away from an object at the rate given by the following for
ID: 2116559 • Letter: #
Question
(1)
Heat is convected away from an object at the rate given by the following formula: (deltaQ/deltaT) = hA(deltaT).
h: coefficient that depends on the shape and the orientation of the object.
A: area
deltaT: temperature difference.
Calculate the rate of convective heat loss in Watts for an unclothed person standing in air at 23 degrees Celsius. Assume that the skin temperature is 34 degrees Celsuismand that the body surface area is 1.5m^2. Calculate for ALL OF THE GIVEN air speeds below:
h= 6 W/m^2C air speed = still
h= 12 W/m^2C air speed = 1 m/sec
h= 28 W/m^2C air speed = 5 m/sec
(2)
The net rate of radiant heat flow from and object is given by (deltaQ/deltaT) = e(sigma)A(T1^4-T2^4)
A: area of object
T1: temperature of object
e: emmisivity (value depends on surface)
T2: temperature of it's surroundings
sigma: 5.67x10^-8
A person has a skin temperature of 33 degrees Celsius and is in a room when the walls have a temperature of 29 degrees Celsius. If the emmisivity for the skin is 0.70 (e=0.70) and if the body surface area is 1.5m^2, then calculate the(deltaQ/deltaT) net, the net radiative heat loss in Watts.
(3)
A nuclear power plant generates (output) 500MW. It is 34% efficient. The waste heat goes into the Connecticut River qith an average flow of 3x10^4 kg/sec. How much does the water temperature rise?
Hint: 4.18x10^3J are needed to raise 1kg by1K.
Explanation / Answer
3) utput = efficiency*(total output) 500 x 10^6 W = (0.34)(total output) total output = 1.47 x 10^9 W Heat output = 1.47 x 10^9 - 500 x 10^6 = 9.71 x 10^8 W (J/s) Let us look at 1 specific second: m = 3 x 10^4 kg water energy = 9.71 x 10^8 J Q = mcdT 9.71 x 10^8 J = (3 x 10^4 kg)(4.18 x 10^3 J/kg C)(dT) dT = 7.74 degrees
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.