report thermodynamics and heat transferlab me2014 determining the state of moist air and calculating the heat balance of air duct

“THERMODYNAMICS & HEAT TRANSFER”EXPERIMENT No.1: DETERMINING THE STATE OF MOIST AIRAND CALCULATING THE HEAT BALANCE OF AIR DUCT1.1 EXPERIMETNAL OBJECTIVES AND REQUIREMENTS1.1.1 Experimen

VIET NAM NATIONAL UNIVERSITY

HO CHI MINH CITY

HO CHI MINH CITY UNIVERSITY OF

TECHNOLOGY -*** -

REPORTTHERMODYNAMICS AND HEAT TRANSFER(LAB)-ME2014

Class: CC02

Semester:HK222

Lecturer: Mr.Nguyen Van Hap

Group:2

MEMBER

“THERMODYNAMICS & HEAT TRANSFER”

EXPERIMENT No.1: DETERMINING THE STATE OF MOIST AIR AND CALCULATING THE HEAT BALANCE OF AIR DUCT

1.1 EXPERIMETNAL OBJECTIVES AND REQUIREMENTS

1.1.1 Experimental objectives

- Knowing how to measure the temperatures (dry and wet bulb temperature), air flow, pressure and volume;

- Understanding the cooling and dehumidifying process of humid air;

- Understanding the working principle and main components of a basic refrigeration cycle;

- Calculating the heat balance in air duct;

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At the outlet of air duct, an anemometer is used to measure the speed and temperature of moist air.

Refrigerant in refrigeration system is R22

Figure 1: Experimental model of an air duct

thermometer 7: Pressure gauge 11: Condenser fan

4: Evaporator coil 8: Compound gauge 12: Condenser coil

- Using dry and wet bulb thermometers to determine the state of moist air at the inlet (it is also the surrounding temperature) and outlet of the cooling coil

- Using anemometer to measure the velocity and temperature at outlet of air duct in order to estimate airflow

- Determining the evaporating and condensing temperature of refrigeration system

- From above data, student determines:

Demonstrating the processes of humid air on the t-d diagram (or I-d)

The heat released when humid air passes through the cooling coil

Moisture is removed at cooling coil according to theoretical calculations and experiments

Demonstrating the states of refrigerant on the T-s diagram (corresponding withn theoretical

refrigeration cycle, neglecting the superheat and sub-cooling processes)

1.4 Experimental data

When the system operates at steady state, the condensing water appears on the cooling coil,

student starts doing the experiments with the following requirements:

Student conducts two experiments (Note: after getting the experimental data, student changesthe airflow through the cooling coil)

Experiment 1: Experimental time is 10 minutes, the number of data collection are 3 times

Experiment 2: Experimental time is 10 minutes, the number of data collection are 3 times

Table 2 & 3: The state parameters of moist air

Experiment 1Moist air at inlet of coil Moist air at outlet of coil

Velocity at outlet of air duct (m/s) Temperature at outlet of air duct (℃) Water condensed (ml)

pressure (Gauge) temperature

(℃) pressure (Gauge) temperature(℃)

pressure (Gauge)

Evaporating temperature(℃)

Condensing pressure (Gauge)

Condensing temperature(℃)

EXPERIMENT 1:

We have:

Amount of water condensed:

Amount of water condensed after 10 minutes:

Error (%):

The heat released when humid air passes through the cooling coil:

EXPERIMENT 2:

We have:

Amount of water condensed:

Amount of water condensed after 10 minutes:

Error (%):

The heat released when humid air passes through the cooling coil:

Demonstrating the processes of humid air on I-d diagram:

Experiment 1:

At the inlet of air conduct

At the outlet of air conduct:

Experiment 2:

At the inlet of air conduct

At the outlet of air conduct:

States of refrigerant on the T-s diagram

Experiment 1:

Experiment 2:

Calculating the overall efficiency:

Firstly, we have to calculate :

Then we can calculate k:

c Determining the Reynolds number:

Cross-section area of steel coil:

We need to calculate first:

Reynolds number calculating:

Test FI1 FI2 TI1 TI2 TI3 TI4 T(hot) T(cold)

a Calculating the heat transfer and overall efficiency at several flow rate:

Where and c are taken at the average temperature of inlet and outlet water.p

We calculate the heat transfer for each test:

Similar with cold water:

Calculating the overall efficiency:

Firstly, we have to calculate :

Then we can calculate k:

Cross-section area of steel coil:

We need to calculate first:

Reynolds number calculating:

Comment on overall heat transfer coefficient in two cases and between experiment E1 and E2:

- In both cases, it seems that the heat transfer coefficient tends to increase when the amount of hot water is set and the amount of water is changed Increasing the amount of hot water also increases the heat transfer coefficient

- Both parallel and counter-flow heat transfer coefficients are almost the same However, parallel flow seems to be more effective than counter-flow

- The heat transfer coefficients of coil heat exchanger systems typically do not differ significantly between tests

- The heat transfer coefficient is higher in counterflow than in parallel flow in shell and tube heat exchanger systems, but vice versa in coiled tube systems

- The heat transfer coefficient of the spiral coil method is lower than that of the shell and tube method

The heat exchange capacity of the shell and tube system is superior to that of the spiral tube system

Comment on Reynolds number in two cases and between experiment E1 and E2:

- In both cases of E2, the heat transfer coefficient seems to have an downward trend If

we decrease the amount of hot water, the heat transfer coefficient will decrease Thecounter flow seems to be more effective than parallel flow

- We can see that Re > 10 in both counter and parallel so these flows are turbulent flow.4

- The Reynolds number of experiment E1 is smaller than the Reynolds number of experiment E2