quiz

University of Bahrain College of Engineering

Department of Mechanical Engineering

Second Semester 2017-18

Mechanical Installations in Buildings (MEG 435) Environmental Control Systems II (ARCG 226)

First Examination

March 21, 2018

Student ID#:_________________ (Please DO NOT write your name)

(circle your section)

ARCG 226 1 2 3 MEG 435 1

Selection √ Question Highest Score Student Score

1 25

2 25

3 25

4 25

Total 75*

*Based on answering three questions only.

Instructions

1. This exam is closed book, closed notes. 2. Make sure that you have four questions, in six pages numbered from 1 to 6, plus

four additional pages of graphs and tables and formulae sheets. 3. You are required to answer three questions only. Tick the three selected

questions in the table above. 4. You have seventy-five minutes to complete the exam. 5. Read each question with care and be sure to answer all parts of each question. 6. All answers must be placed in the question sheets directly below the question

where space is provided. 7. You should show units in your solutions. If you do not indicate the units or the

units indicated are wrong, you will lose one point for each mistake even though the numerical answer is correct.

8. As a suggestion, work the easiest question first; question number one is not necessarily the easiest.

Good Luck…

1

Question One A (Payback Period) Suppose that the annual bill for heating water for a 4-star hotel is BD.1000/year using an electrical centralized heating system. The hotel engineer suggested to the hotel’s management investing BD. 5,000 for a solar heating system that can make the annual bill BD. 600/year only. 1. How much savings the hotel would acquire if this solar heating system was installed. 2. Calculate the pay-back period (pbp) for this new system. Question One B (Shape Factors) The figure below shows a triangular sunroom for a house. The sunroom has five surfaces, namely: floor (f), Wall (w), Roof (r), and two triangular sides (s). Use the relationship below to answer the questions:

Ff→f + Ff→w + Ff→r + Ff→s + Ff→s = 1

1. What is the value of the shape factor between the floor and itself (Ff→f)?

2. Calculate the shape factor between the floor and the wall (Ff→w).

3. Calculate the shape factor Ff→r if Ff→s =0.15

5 m

5 m 10 m

2

Question Two (Conduction/Convection Heat Transfer) The Figure shown below is a schematic diagram of a wall constructed of layers arranged in series as in the Figure below. The wall has a length of 20 m and a height of 3 m. The inside air temperature (Ti) = 23oC, and the outside air temperature (To) = 39oC. Use the information listed in the table to answer the following questions.

Layer ∆x (cm)

k (W/(m.K)

Rth (m2.K/W)

Outside Convection (ho) - -

Red brick veneer 10 0.20

Air space 5 0.08

Rigid insulation 5 0.04

Moisture/vapor barrier 0.1 0.21

Limestone aggregate concrete Blocks filled with perlite 20 (1)

Inside convection (hi) (ρ=10%) - -

Rth,total =

1. From the attached ASHRAE Fundamentals Handbook table 10, find the values of both the outside and inside convection coefficients.

Outside convection coefficient (ho) =

Inside convection coefficient (hi) =

2. Fill in cell (1) in the table with the value of the thermal resistance (Rth) of limestone aggregate concrete blocks from the attached ASHRAE fundamentals Handbook table 1.

3

3. Calculate the total thermal resistance (Rth,total) of the wall.

4. Calculate the heat transfer rate (Q) across the wall.

5. As an architect, suggest four ways from the course materials you learned to reduce the heat transfer through the wall.

a. ______________________________________________________________________________________________

b. ______________________________________________________________________________________________

c. ______________________________________________________________________________________________

d. ______________________________________________________________________________________________

Question Three (Radiation Heat Transfer)

4

The South wall of a building in Bahrain is made with the layer arrangement shown in the figure below. The design outdoor temperature for Bahrain is 39oC. The building is maintained at 24oC. The sky temperature is 10oC. The incident solar radiation (qsol) falling on the wall in July is 420 W/m2. If the area of the wall is 150 m2, its U-value is 0.35 W/(m2.K) and its outside reflectivity (ρ) is 85%, answer the following questions:

1. Draw the thermal resistance network through the wall accounting for both conduction and convection.

2. Calculate the sol-air temperature (Tsol-air) for the wall in July. Take ho =22.7 W/(m2.K). ignore infra-red radiation portion.

3. Calculate the total heat gain (Qtotal) through the wall by conduction, convection, and radiation.

A

B

C D To =

Ti

Tsky = 10oC

qsol= 420 W/m 2

5

4. Calculate the fraction of the above heat gain due to solar only.

5. If the outside surface of the wall is at 38oC, calculate the radiation exchange (Qrad) between the wall and the sky. The wall is totally exposed to sky.

6

Question Four (Ideal Gas Law and First Law of Thermodynamics) A single storey building located in Manama, Bahrain (Patm=101.325 kPa, ρ=1.2 kg/m3) with a floor area of 1000 m2 and a height of 3 m has air change per hour (ACH) = 0.3. The outdoor air temperature (To) is 36oC. Take the specific heat of air (cp) = 1006 J/(kg.K) and Rair = 287 J/(kg.K) and answer the following questions: 1. Use the ideal gas law to calculate the room air temperature (T). 2. Calculate the infiltration volumetric flow rate (Vinf) through the building. 3. How much heat (Qinf) must be removed by the building HVAC system to cool this

infiltrating outside air. 4. Convert the value above to refrigeration tons.

5. If the same building is located in Boulder, Colorado (1,500 m above sea level), will

the cooling requirements be higher or lower than the value found for Bahrain? Explain. No need to do calculations.

7

8

Formulae Sheets

Chapter 1

Chapter 2

𝑅𝑡ℎ = ∆𝑥

𝑘

𝑅 = 𝑅𝑡ℎ 𝐴

𝑈𝐴 = 1

𝑅𝑡𝑜𝑡𝑎𝑙

𝑄𝑡𝑜𝑡𝑎𝑙 = 𝑈 × 𝐴 × (𝑇𝑠𝑜𝑙−𝑎𝑖𝑟 − 𝑇𝑖) 𝑄𝑠𝑜𝑙 = 𝑈 × 𝐴 × ∆𝑇𝑠𝑜𝑙 𝑄𝑟𝑎𝑑 = 𝐴𝜀𝜎(𝑇𝑠

4 − 𝑇𝑠𝑘𝑦 4 )

𝑃𝑣 = 𝑅𝑎𝑖𝑟𝑇 where T is Kelvin 𝑃𝑉 = 𝑚𝑅𝑎𝑖𝑟𝑇

Savings

Investment pbp 

TAUQ cond  .

R

T

R

TA Q

th

 

 

.

)( sfconvconv TTAhQ 

 

22

12

121

1

4

2

4

11 .

1

A

A

F

TTA Qrad













 

1 11

4

21

3







 avg rad

T h

TcmQ p ..

V

V ACH

.



10

�̇�𝑠𝑒𝑛 = 𝑚 × 𝑐𝑝̇ × ∆𝑇 𝑄𝑖𝑛𝑓 = 𝑚 × 𝑐𝑝̇ × ∆𝑇

𝐶𝑂𝑃 = 𝑜𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡

𝜂 = 𝑜𝑢𝑡𝑝𝑢𝑡

𝑖𝑛𝑝𝑢𝑡

Constants

Density of air (ρair) =1.2 kg/m3 Density of water (ρwater) =1000 kg/m3 Universal gas constant = 8314.41 J/(kg.mol.K) Gas constant for air = 287 J/(kg.K) Stefan-Boltzmann constant = 5.67 ×10-8 W/(m2.K4) Unit conversions 1 m = 3.281 ft 1Ibm= 0.45356 kg 1 Btu = 1.055 kJ oC =(oF-32)*(5/9) K= oC+ 273.15 1 W =3.412 Btu/h 1 ton refrigeration = 3.517 kW Profitability is excellent if pbp less than one-third of lifetime of investment; Profitability is good if pbp less than one-half of lifetime of investment; Profitability is bad if pbp is more than one-half of lifetime of investment.

)./(006.1, KkgkJc ap 

)./(186.4, KkgkJc wp 