Me 4053 Heat Exchange Lab Report
ME 4053 Heat Exchanger Homework, 6 pts towards Thermal Fluids Lab grade
Due at the beginning of your Heat Exchanger Lab Name: Phillip Ross Crumpton
Attention: This is an individual assignment and collaborations are not allowed. Please pay special attention to the number of significant digits.
1. Measurements are taken in the ME4053 heat exchanger lab. From the measurements, the heat rate from the hot water is calculated to be 3743 W, and the heat rate to the cold water is calculated to be 3876 W. What is the most reasonable heat exchange rate to assume for the exchanger?
3810 W
Qavg = (3876 + 3743)/2 = 3809.5 = 3810 W
2. For the heat rates given in the previous question, the hot water temperature decreases from 65.1 …show more content…
[(65.1 - 34.7) – (49.5 – 25.2)] / ln[(65.1 - 34.7)/( 49.5 – 25.2)]
= 6.1/0.224
= 27.2 °C = 300.4 K
300.4 K
What is the overall conductance (UA) for this exchanger under the experimental conditions described above?
3810/300.4 (W/K) = 12.683 = 12.68
12.68 W/K
3. For conditions similar to those above, an EXCEL spreadsheet computes the UA of the ME4053 lab HXer to be 117.60 W/K. A simple finite difference formula is to be used to compute the influence coefficients for Error Propagation Analysis. For the hot water inlet temperature this coefficient is
If the hot water inlet temperature is arbitrarily perturbed from 65.1 C to 66.2 C, the calculated conductance or UA changes to 119.32 W/K. What is the influence coefficient to be used in Error Propagation Analysis for the hot water inlet temperature?
(119.32 – 117.60) / (66.2 – 65.1) = 1.72/1.1 = 1.56
1.56 (W/K)/K
4. Assume that six (6) repeated measurements were performed and the temperatures and flow rates were measured to calculate UA values and then averaged. If the computed SSD of UA for the data set is 2.4 W/K.
What is the standard Uncertainty A for the average UA?
0.98 …show more content…
2.52 W/K
SSDUA = 2.4
UA = 2.4/√6 = 0.9797 = 0.98
UA = UA(kc) = 0.98*(2.57) = 2.52
5. Given the results of Part 4 and assuming that the (expanded) Uncertainty B for the UA has been estimated to be 8.3 W/K, what is the combined expanded uncertainty of the average UA?
UB = 8.3
Uc = √(UA2 + UB2) = √(2.522 + 8.32) = √(6.3504 + 68.89) = 8.67
8.67 W/K 6. (Optional) Can you assign Uncertainty A of the thermocouple for the high-temperature outlet based on the SSD of the six measurements divided by ? If not, what is the correct procedure for determining Uncertainty A of the thermocouple reading? [Hint: What we are asking is the Uncertainty of the temperature measurement
Due at the beginning of your Heat Exchanger Lab Name: Phillip Ross Crumpton
Attention: This is an individual assignment and collaborations are not allowed. Please pay special attention to the number of significant digits.
1. Measurements are taken in the ME4053 heat exchanger lab. From the measurements, the heat rate from the hot water is calculated to be 3743 W, and the heat rate to the cold water is calculated to be 3876 W. What is the most reasonable heat exchange rate to assume for the exchanger?
3810 W
Qavg = (3876 + 3743)/2 = 3809.5 = 3810 W
2. For the heat rates given in the previous question, the hot water temperature decreases from 65.1 …show more content…
[(65.1 - 34.7) – (49.5 – 25.2)] / ln[(65.1 - 34.7)/( 49.5 – 25.2)]
= 6.1/0.224
= 27.2 °C = 300.4 K
300.4 K
What is the overall conductance (UA) for this exchanger under the experimental conditions described above?
3810/300.4 (W/K) = 12.683 = 12.68
12.68 W/K
3. For conditions similar to those above, an EXCEL spreadsheet computes the UA of the ME4053 lab HXer to be 117.60 W/K. A simple finite difference formula is to be used to compute the influence coefficients for Error Propagation Analysis. For the hot water inlet temperature this coefficient is
If the hot water inlet temperature is arbitrarily perturbed from 65.1 C to 66.2 C, the calculated conductance or UA changes to 119.32 W/K. What is the influence coefficient to be used in Error Propagation Analysis for the hot water inlet temperature?
(119.32 – 117.60) / (66.2 – 65.1) = 1.72/1.1 = 1.56
1.56 (W/K)/K
4. Assume that six (6) repeated measurements were performed and the temperatures and flow rates were measured to calculate UA values and then averaged. If the computed SSD of UA for the data set is 2.4 W/K.
What is the standard Uncertainty A for the average UA?
0.98 …show more content…
2.52 W/K
SSDUA = 2.4
UA = 2.4/√6 = 0.9797 = 0.98
UA = UA(kc) = 0.98*(2.57) = 2.52
5. Given the results of Part 4 and assuming that the (expanded) Uncertainty B for the UA has been estimated to be 8.3 W/K, what is the combined expanded uncertainty of the average UA?
UB = 8.3
Uc = √(UA2 + UB2) = √(2.522 + 8.32) = √(6.3504 + 68.89) = 8.67
8.67 W/K 6. (Optional) Can you assign Uncertainty A of the thermocouple for the high-temperature outlet based on the SSD of the six measurements divided by ? If not, what is the correct procedure for determining Uncertainty A of the thermocouple reading? [Hint: What we are asking is the Uncertainty of the temperature measurement