|College of Anatomist and Laptop Science
Mechanical Engineering Office
Mechanical Architectural 370
| |Fall 2010 Training course Number: 14319 Instructor: Larry Caretto
Unit 3 Homework Alternatives, September sixteen, 2010
1A classroom that normally is made up of 40 people is to be atmosphere conditioned with window air conditioners of 6th kW air conditioning capacity. A person at rest may be thought to pass heat for a price of about fish hunter 360 kJ/h. There are 10 light bulbs in the room, every single with a rating of 100 W. The rate of heat copy to the class room through the surfaces is 12-15, 000 kJ/h. If the area air is usually to be maintained in a constant temperature of 21oC, determine the quantity of air-conditioning products required.
In this problem, all of us define the system to be the air flow in the room. We are able to assume that mid-air behaves while an ideal gas. If the heat remains continuous, there is no change in the internal strength of the atmosphere. Since the volume of the room remains to be constant, there is no work done. Thus the initial law decreases to the equation that Queen = zero. We take into account the different temperature sources in the total warmth rate, Q, as follows. The heat input comes (a) from your 40 pupils, Qs = (40 people)(360 kJ/hвЂўperson)(kWвЂўs /kJ)(h/3600 s) = 4 kilowatt, (b) temperature transfer through the wall, Qw = (15, 000 kJ/h)(kW/kJвЂўs)(h/3600 s) =4. 167 kilowatt, and (c) heat added by the bulbs, Qb sama dengan 10(100 W) = 1000 W sama dengan 1 kilowatt. Thus the total heat suggestions is four + four. 167 + 1 sama dengan 9. 167 kW. Heat removal originates from the air AC which take out 5 kW each. Thus for the internet Q to get zero we need to have D air conditioners such that (5 kW)N = 9. 167 kW. To satisfy this requirement, we have to have two air conditioning units.
2A 0. 5 m3 strict tank consists of regrigerant-134a at first at 160 kPa and 40% top quality. Heat is actually transferred to the refrigerant until the pressure extends to 700 kPa. Determine (a) the mass of refrigerant in the container and (b) the amount of warmth transferred. Also, show the procedure on a P-v diagram with respect to the saturation lines.
The mass of refrigerant may be discovered from the preliminary state using the equation that m sama dengan V as well as v1 wherever V sama dengan 0. 5 m3 and v is found from the temperatures and the quality. Since we are given a basic quality, x1 = forty percent, we know that we could in the mixed region. The particular volume is located from the particular volumes with the saturated the liquid and vapor, which are found in Table A-12 on page 930: vf(160 kPa) = zero. 0007437 m3/kg and vg(160 kPa) sama dengan 0. 12348 m3/kg. All of us then locate the initial particular volume as follows.
With this specific volume level, we in that case find the refrigerant mass as follows:
[pic]sama dengan 10. goal kg
To compute heat transfer all of us apply the first law, Q sama dengan О”U + W. We assume that there is absolutely no volume difference in the " rigidвЂќ tank. If there is simply no volume alter, no function is done. With W sama dengan 0, Queen = О”U. We find О”U = m(u2 вЂ“ u1) where the particular internal powers are found from your property dining tables. At the initial state we discover u through the quality in a similar manner that we found the volume.
Because this is a constant volume process, the last state has got the same particular volume while the initial express (0. 049838 m3/kg) and a given pressure of seven-hundred kPa. Being aware of these data we can storyline the constant amount path with this process as shown inside the figure at the left. We come across that the first state (1) is in the mixed region as well as the increase in pressure, at constant volume, brings the final point out into the gas region.
Once we try to find a final state (P = 700 kPa, v = zero. 049838 m3/kg) in the superheat table, A-13, on page 930, we see which the final certain volume inside the table pertaining to P sama dengan 700 kPa = zero. 70 MPa (at a temperature of 160oC) is only 0. 048597 m3/kg. Furthermore, the specific volume is raising with heat so that the last state reaches a higher temp than the final temperature inside the table.