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THERMAL PROPERTIES Cp,water = 4, 182 kg°C Pwater = 993- Question 1 How much energy does the cooling tower remove? Using the information provided in the reading about the cooling tower's water flow, if the water enters the cooling tower at 24 °C and exits at 18 °C, then how much energy was removed (in kJ)? Assume the entire volume is circulated. Round to the nearest integer. Question 2 What is the rate of heat loss in the cooling tower? If the system turns on at 8:00 pm and shuts down at 6:00 am, at what rate is the water losing heat energy to the night air (in W)? (Given: Qheat loss = 1,500,000 kJ) Round to the nearest integer. (Note 1: "Given" value is NOT based on the answer from the previous question) (Note 2: you're estimating the average rate, since it would actually fluctuate throughout the night.) Question 3 How much make-up water is needed due to evaporative losses? The night-time chilled water system is not a closed-loop system, meaning that some amount of water is lost due to evaporation every night. If it is 96% efficient (i.e. 4% of the water flow rate is dispersed to the outdoor air and needs to be replaced), how much makeup water does it need to draw from the municipal water supply daily (in liters)? Round to nearest integer. Question 4 If the makeup water (Given: makeup water = 2,000 L) is 19 °C, how much additional cooling is needed (in kJ) to maintain the cistern at 18 °C? Round to nearest integer.

THERMAL PROPERTIES Cp,water = 4, 182 kg°C Pwater = 993- Question 1 How much energy does the cooling tower remove? Using the information provided in the reading about the cooling tower's water flow, if the water enters the cooling tower at 24 °C and exits at 18 °C, then how much energy was removed (in kJ)? Assume the entire volume is circulated. Round to the nearest integer. Question 2 What is the rate of heat loss in the cooling tower? If the system turns on at 8:00 pm and shuts down at 6:00 am, at what rate is the water losing heat energy to the night air (in W)? (Given: Qheat loss = 1,500,000 kJ) Round to the nearest integer. (Note 1: "Given" value is NOT based on the answer from the previous question) (Note 2: you're estimating the average rate, since it would actually fluctuate throughout the night.) Question 3 How much make-up water is needed due to evaporative losses? The night-time chilled water system is not a closed-loop system, meaning that some amount of water is lost due to evaporation every night. If it is 96% efficient (i.e. 4% of the water flow rate is dispersed to the outdoor air and needs to be replaced), how much makeup water does it need to draw from the municipal water supply daily (in liters)? Round to nearest integer. Question 4 If the makeup water (Given: makeup water = 2,000 L) is 19 °C, how much additional cooling is needed (in kJ) to maintain the cistern at 18 °C? Round to nearest integer.

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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Please answer questions 2-4, question 1 is answered and explained 

THERMAL PROPERTIES Cp,water = 4, 182 ke°G Pwater = 993 *g Question 1 How much energy does the cooling tower remove? Using the information provided in the reading about the cooling tower's water flow, if the water enters the cooling tower at 24 °C and exits at 18 °C, then how much energy was removed (in kJ)? Assume the entire volume is circulated. Round to the nearest integer. Question 2 What is the rate of heat loss in the cooling tower? If the system turns on at 8:00 pm and shuts down at 6:00 am, at what rate is the water losing heat energy to the night air (in W)? (Given: Qheat loss = 1,500,000 kJ) Round to the nearest integer. (Note 1: "Given" value is NOT based on the answer from the previous question) (Note 2: you're estimating the average rate, since it would actually fluctuate throughout the night.) Question 3 How much make-up water is needed due to evaporative losses? The night-time chilled water system is not a closed-loop system, meaning that some amount of water is lost due to evaporation every night. If it is 96% efficient (i.e. 4% of the water flow rate is dispersed to the outdoor air and needs to be replaced), how much makeup water does it need to draw from the municipal water supply daily (in liters)? Round to nearest integer. Question 4 If the makeup water (Given: makeup water = 2,000 L) is 19 °C, how much additional cooling is needed (in kJ) to maintain the cistern at 18 °C? Round to nearest integer.
Transcribed Image Text:THERMAL PROPERTIES Cp,water = 4, 182 ke°G Pwater = 993 *g Question 1 How much energy does the cooling tower remove? Using the information provided in the reading about the cooling tower's water flow, if the water enters the cooling tower at 24 °C and exits at 18 °C, then how much energy was removed (in kJ)? Assume the entire volume is circulated. Round to the nearest integer. Question 2 What is the rate of heat loss in the cooling tower? If the system turns on at 8:00 pm and shuts down at 6:00 am, at what rate is the water losing heat energy to the night air (in W)? (Given: Qheat loss = 1,500,000 kJ) Round to the nearest integer. (Note 1: "Given" value is NOT based on the answer from the previous question) (Note 2: you're estimating the average rate, since it would actually fluctuate throughout the night.) Question 3 How much make-up water is needed due to evaporative losses? The night-time chilled water system is not a closed-loop system, meaning that some amount of water is lost due to evaporation every night. If it is 96% efficient (i.e. 4% of the water flow rate is dispersed to the outdoor air and needs to be replaced), how much makeup water does it need to draw from the municipal water supply daily (in liters)? Round to nearest integer. Question 4 If the makeup water (Given: makeup water = 2,000 L) is 19 °C, how much additional cooling is needed (in kJ) to maintain the cistern at 18 °C? Round to nearest integer.
Step 2 Given data: 4182 J/kg · °C C pw = Pw = 993 kg/m³ T; = 24°C Te = 18°C %3D Need to determine the amount of energy removed. Step 3 Calculate the amount of energy removed. Q = mCpw (T; – Tf) %3D | = m x 4182 (24 18) = 25092m J For unit mass, m=1 kg Q = = 25092 × 1 Q = 25092 J Q = 25. 092 kJ
Transcribed Image Text:Step 2 Given data: 4182 J/kg · °C C pw = Pw = 993 kg/m³ T; = 24°C Te = 18°C %3D Need to determine the amount of energy removed. Step 3 Calculate the amount of energy removed. Q = mCpw (T; – Tf) %3D | = m x 4182 (24 18) = 25092m J For unit mass, m=1 kg Q = = 25092 × 1 Q = 25092 J Q = 25. 092 kJ
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