d ( ) ( 0.165) = 169 W/m 2

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Charles Melton
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1 Page 1 of 10 Concentrating Solar Power Solutions Question #31: Calculate the amount of beam radiation ( G b ) for an extraterrestrial radiation ( G o ) of 1250 W/m 2 and a clearness index ( k T ) equivalent to Solution: First we need to calculate the global horizontal radiation k T = G G = G G o k T = ( 1250 W/m ) 2 ( 0.82) = 1025 W/m 2 o and then we lookup the proper correlation between the beam radiation and the global horizontal radiation, finding that G d G = given that k > 0.80, which is then used to calculate the T diffuse radiation and finally the beam radiation when noting that their sum is equivalent to the total global horizontal radiation calculated earlier. G d G = G = G = 1025 W/m2 d ( ) ( 0.165) = 169 W/m 2 G = G b + G d G b = G - G d = 1025 W/m W/m 2 = 856 W/m 2

2 Page 2 of 10 Question #32: Calculate the rate of heat transfer into the solar receiver ( Q rec,in ) in kw using an incident radiation on a bank of reflecting mirrors ( G ref ) of 820 W/m 2, average soiling losses from dust on the mirrors of 5%, average optical losses between the mirror and receiver of 10%, and a total area of the bank of reflecting mirrors ( A ref ) of 1200 m 2. Solution: This question can be solved by noting that Q rec,in = G rec A rec and the values given can be applied directly into the energy balance equation about the solar receiver and reflecting mirrors Q rec,in = G rec A rec = G ref A ref h soil h opt Q rec,in = ( 820 W/m ) 2 ( 1200 m ) 2 ( 0.95) 0.90 ( ) æ 1 kw ö è ç 1000 Wø = kw

3 Page 3 of 10 Question #33: Calculate the rate of heat loss ( ) from the solar receiver using an emissivity ( e rec ) of 0.09, a receiver area ( A rec ) of 20 m 2, and a receiver temperature (T rec ) of 800 C. Express your answer in kw. Solution: The heat loss equation can be used directly with the given information in the problem. Be sure to convert temperature from degrees Celsius to Kelvin before using in the heat loss equation

4 Page 4 of 10 Question #34: How much energy (in kj) is required to raise 150,000 kg of water vapor from 150 C to 650 C given that water vapor has a specific heat of kj/kgk? Solution: The equation for the change in sensible energy can be used directly, and although it is not necessary to convert C into K in this question, it is always safer to do so. T i = 150 C = K; DE = m c p ( T f - T i ) T f = 650 C = K DE = ( 150,000 kg) ( kj/kgk) ( K K) = kj

5 Page 5 of 10 Question #35: How much energy would be required to vaporize 150,000 kg of water? This is the same mass of water as in Question 4. Assume that the liquid water begins at 100 C and the liquid water ends at 100 C. The latent heat of vaporization for water is 2440 kj/kg. Solution: The expression for the energy in a phase change process is used DE = 150,000 kg DE = m h fg æ è ç ( ) 2440 kj ö kgø = kj

6 Page 6 of 10 Question #36: Energy from a solar receiver is used to heat 1 kg of oil from a temperature of 25 C to a temperature of 250 C. If the oil is replaced by salt and the same amount of energy is absorbed from the solar receiver, calculate the temperature that could be reached in the salt of equivalent mass assuming that the salt begins at a temperature of 150 C. Oil has a specific heat of 2.4 kj/kgk and salt has a specific heat of 1.3 kj/kgk. Solution: First begin by converting C into K and then equate the amount of energy that was transferred into the 1 kg of oil. T i,oil = 25 C = K; T f,oil = 250 C = K DE oil = m c p,oil ( T f,oil - T i,oil ) DE oil = ( 1 kg) ( 2.4 kj/kgk) ( K K) = 540 kj With the energy into the oil and salt being equivalent DE oil = DE salt = 540 kj Now reformulate the sensible energy equation to solve for temperature. T f,salt = K + 1 kg Then convert temperature back into C DE salt T f,salt = T i,salt + m c p,salt 540 kj ( ) ( 1.3 kj/kgk) = K T f,salt = K = C

7 Average global irradiance in each hour (kw/m2) Page 7 of 10 Question #37: Calculate the total irradiation ( H ) in the day given the global irradiance distribution (kw/m 2 ) below. Express your answer in kwh/m 2 d Hour of day Solution: The answer is calculated by summing the global irradiance across each hour of the day. H = 24 G i i=1 å t i = 6.35 kwh m 2 d

8 Page 8 of 10 Question #38: A salt working fluid has been heated to 650 C and sent to a storage container for later use. The container has a radius of 15 m and a height of 10 m. The container temperature and ambient air temperature are 55 C and 20 C, respectively. If the average overall heat transfer coefficient is 1.5 W/m 2 K across the entire surface area of the container, what is the rate of heat loss from the container? Solution: This question begins by computing the total surface area of the cylinder 2 ( ) + ( 2pr cyl ) = ( 2p ( 15 m) ( 10 m) ) + 2 p 15 m A cyl = 2pr cyl h cyl and continues with converting temperatures into Kelvin ( ( ) ) 2 = m2 T a = 20 C = K; T s = 55 C = K before computing the final rate of heat loss from the storage container.

9 Page 9 of 10 Question #39: Calculate the maximum efficiency (Carnot efficiency) for a power cycle using steam that is heated up to a maximum of 650 C and drops down to a minimum of 95 C. Solution: Convert C into K before calculating the Carnot efficiency. T c = 95 C = K; T h = 650 C = K h carnot = 1- T c T h æ h carnot = K ö è ç K ø 100% ( ) = 60.1%

10 Page 10 of 10 Question #40: Calculate the power output of the concentrating solar plant using an incident radiation on a bank of reflecting mirrors ( G ref ) of 950 W/m 2, area of the reflecting mirrors ( A ref ) totaling 3800 m 2, reflecting mirror efficiency (h ref ) of 75%, receiver efficiency (h rec ) of 85%, storage efficiency (h rec ) of 98%, and power generation efficiency (h rec ) of 40%. Solution: First calculate the radiation hitting the reflective mirrors Then calculate the total system efficiency h tot = h ref h rec h st h pow h tot = ( 0.75) ( 0.85) ( 0.98) ( 0.40) ( 100% ) = 24.99% And finally calculate the electrical power output from the facility P out = ( 3610 kw) ( 24.99% ) = 902 kw

ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION. Instructor: R. Culham. Name: Student ID Number:

ECE309 THERMODYNAMICS & HEAT TRANSFER MIDTERM EXAMINATION June 19, 2015 2:30 pm - 4:30 pm Instructor: R. Culham Name: Student ID Number: Instructions 1. This is a 2 hour, closed-book examination. 2. Permitted