TETD/TA (manual method)

TETD is called the total equivalent temperature differential method. It is similar to CLTD method but not the same. More information can be found in ASHRAE Handbook — Fundamentals (ASHRAE Fundamentals 28.56 — 28.64).

Practicum Assignment #1

1. For a winter heating in a PSU classroom, steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is condensed, and leaves as liquid water at 200°F. If the heating capacity of the radiator is 5000 Btu/hr, at what rate in lbm/hr must steam be supplied?

2. A solar collector panel, shown in Figure 1, has a surface area of 32 ft“. The panel receives energy from the sun at a rate of 150 Btu/(hr ft“-of collector surface). Forty percent of the incoming energy is lost to the surrounding. The reminder is used to warm liquid water from 130°F to 160°F. The water passes through the solar collector with a negligible pressure drop.

(a) (15 points) Neglecting kinetic and potential energy effects, detennine at steady state the mass flow rate of water in [lb/min]. Hint: Write the assumptions.

(b) (10 points) How many solar collectors would be needed to provide a total of 40 gal of 160°F water in 30 min? Hint: To obtain water properties assume the atmospheric pressure.

TETD/TA (manual method)

3. Steam at 7,000 Pa and 50°C enters a condenser operating at steady state and is condensed to saturated liquid at 7,000 Pa on the inside of tubes through which cooling water flows. The mass flow rate of steam is 25 kg/s. In passing through the tubes, the cooling water increases in temperature by 10°C and experiences no pressure drop. Neglecting kinetic and potential energy effects and ignoring heat transfer from the outside of the condenser, detennine:

(a) the mass flow arte of cooling water, in kg/s

(b) the rate of energy transfer, in kW, from the condenser to the cooling water.

Solutions for the Practicum Assignment #1

1. For a winter heating in a PSU classroom, steam enters a radiator at 16 psia and 0.97 quality. The steam flows through the radiator, is condensed, and leaves as liquid water at 200°F. If the heating capacity of the radiator is 5000 Btu/hr, at what rate in lbm/hr must steam be supplied?

Pi = 16 psia

Xl = 0.97 T2 = 200°F

X2 = 0

Use these values to find hi and h2 with Table A-la.

Q 5000Btu/hr e,,

M = —-— =—————————————— = 5.236lbm/hr

Hi /72 1123 Btu / Ibm -168.1 Btu / Ibm

(a)

150 Btu/h • U2

40% ros*>

Figure 1.

A — 32 ft?

TETD/TA (manual method) подпись: figure 1. подпись: a - 32 ft?

TETD/TA (manual method)

) Water in at 130°F

подпись: ) water in at 130°f

Assumptions:

1. The control volume is at steady state

2. For the control volume shown, Wcv = 0

3. Kinetic and potential energy effects are negligible

Bln

4. The water is modeled as an incompressible liquid, with constant specific heat c = 1 Analysis:

The mass flow rate is determined using the steady-state energy balance as follow:

0 = (L — W>,v + ™[(‘, -■I2) + (ll J2 ) + 8(Zi — Z2)]

Where mx = m2 = m with Ocv = Ojn = Qloss and assumption (3)

O = Qin — Q, oss+™(1^ — h2)

For water as an incompressible liquid, /’j — i2 = c(7j —T2)+ v(p] — p2)

Thus

^ ___ Qin Qloss

C(T2-Tx)


Bin

Hr

подпись: bin
hr

/^)(32^) = 4800f" hr ■ ft hr

подпись: /^)(32^) = 4800f" hr ■ ft hrFrom the given data, Ojn = (150—- —)(32 ft3) = 4800—- and Qloss = (0.4)Ojn = 1920

Inserting values:

(4800-1920)— 1f „

H, — lb m = —~——- — (————————— ) = 1.6———

L^-x (620-590)* 60mln IbR

(b) The mass of water in 40 gallon is with density p = 60.01

J

Mtot = pV = (60.98-^-)(40gal)(—^— —— ) = 326 lb of water

подпись: mtot = pv = (60.98-^-)(40gal)(—^— ) = 326 lb of waterLb, n „ 0 13368 tf3 —)(40 gal——J ft Igal

Each collector provides

Lb _ lb

J

T ih

Mdt = (1.6—— )(30min) = 48

I min Collector

Thus

326

No of Collectors=—- « 7

48

Comments:

Forty gallons is about the capacity of a typical home water heater, the total collector area required is 224 ft2.

3. Steam at 7,000 Pa and 50°C enters a condenser operating at steady state and is condensed to saturated liquid at 7,000 Pa on the inside of tubes through which cooling water flows. The mass flow rate of steam is 25 kg/s. In passing through the tubes, the cooling water increases in temperature by 10°C and experiences no pressure drop. Neglecting kinetic and potential energy effects and ignoring heat transfer from the outside of the condenser, determine:

(a) the mass flow arte of cooling water, in kg/s

(b) the rate of energy transfer, in kW, from the condenser to the cooling water.

Given: Steam and water pass through separate streams through a condenser.

Assuming: 1.) Control Volume is a steady state.

2. ) Heat transfer from outside is negligible.

3 .) Potential and Kinetic Energy effects are negligible.

4. ) Cooling water is idealized as an incompressible fluid w/ constant specific heat.

Find: a.) Mass flow rate of cooling water, m(dot)cw

B.) Heat transfer rate to the cooling water. Q(dot) cw

Solution:

A. ) By conservation of mass: m(dot)i=m(dot)2=m(dot)steam and m(dot)a=m(dot)b=m(dot)cw

-Using steady state energy balance:

0=Q(dot)cv-W(dot)cv+m(dot)st[(h]-h2)+((vi-V2)/2)+g(zi-Z2)]+m(dot)cw[(ha-hb)+((Va-Vb)/2)+g(za. z 0=m(dot)st (h,-h2)+in(dot)cw(ha-h b)

M(dot)cw =[m(dot)st (hi-h2)]/Cp (Ta-T b)]

M(dot)cw =[(25kg/s)(2571.5kJ/kg-163,0kJ/kg)]/[(4.18 kJ/(kg*K)(10K)] = 1440.5 kg/sec

B. ) Using an energy balance of the cooling water:

0=Q(dot)cv-W(dot)cv+ m(dot)cw [(ha-hb)+((va-Vb)/2)+g(za-z b)]

Q(dot)cw =m(dot)Cv(h-h)=m(dot)Cp(Ta-T b)

Q(dot)cw= (1440.5 kg/sec)(4.18 kJ/(kg*K)(10K) = 60215.5 W = 60.2kW

Practicum Assignment #2

1. Running water is heated from 50°F to 120°F by mixing it with saturated steam. Steam gauge pressure may vary from 15 to 25 psi.

(a) Sketch the system.

(b) For a heated water flow rate of 14 cfm, calculate and plot (create graph of) steam mass flow rate in lb/s. Assume sea level atmospheric pressure for water flow.

2. By supplying 90,000 Btu/h, a heat pump maintains the temperature of dwelling at 70°F when the outside air is at 32°F.

(a) Sketch the system.

(b) What is the minimum work required for this cycle? Define the assumptions.

3. A heat pump driven by a 0.4 kW electric motor provides heating for a building on a day when the outside is at -10°C and energy is lost through the walls and roof at rate of 16200 kJ/h. What is the maximum theoretical temperature that can be maintained within building, in °F?

4. Infiltration of outside air into building through miscellaneous cracks can represent a significant load on the heating or cooling equipment. A particular office building has a total crack length of 440 ft around its doors and windows. On windy day, about 0.4 cfm of air enters per foot of crack. In addition, door openings account for about 100 cfm of outside air infiltration on average. The internal volume of the building is 20,000 ft3. Assuming ideal gas behavior, estimate n the number of times per hour (ACH — air change rate per hour), at steady state, that the air within building is changed due to infiltration. Hint: n[ACH] = Volume flow rate/Volume.

Solutions for the Practicum Assignment #2

1. Running water is heated from 50°F to 120°F by mixing it with saturated steam. Steam gauge pressure may vary from 15 to 25 psi.

(a) Sketch the system.

(b) For a heated water flow rate of 14 cfm, calculate and plot (create graph of) steam mass flow rate in lb/s. Assume sea level atmospheric pressure for water flow.

(a)

TETD/TA (manual method)

Comparing steam mass flow to steam pressure:

First, lets write a governing energy conservation equation for out process. Ein=Eout rh [2] i + iii * i = iii * i

111 wl 1 wl steam 1steam 111 v2 1 w2

Note that wl refers to the cool water and w2 refers to the heated water. Now lets rearrange to create an expression for m(dot)steam.

Ok here’s the trick, you now go to the steam tables and look up enthalpies for 15psi to 25psi. At each enthalpy you run the above equation and calculate the m(dot)steam — This is how you graph steam pressure Vs steam mass flow rate. Actually you could of came up with a graph based off of our initial equation with out any numbers at all, IF you realize the direct relationship between pressure and enthalpy as well as the indirect relationship between enthalpy and mass flow rate.

Here is the Excel Calculation:

Mass

Flow

Pressure Rate (psi) Lb/Min

 

MassFlow Vs Pressure

TETD/TA (manual method)

Pressure (psi)

 

15

16

17

18

19

15 21 22

23

24

25

 

52.66

52.63

52.6

52.58

52.55

52.52

52.5

52.47

52.45

52.42

52.4

 

Series 1

 

2. By supplying 90,000 Btu/h, a heat pump maintains the temperature of dwelling at 70°F when the outside air is at 32°F.

(a) Sketch the system.

(b) What is the minimum work required for this cycle? Define the assumptions.

TETD/TA (manual method)

«t TOT

TETD/TA (manual method)

Figure 1.

(a) Assumptions:

1. The system shown on the accompanying figure undergoes a heat pump cycle

2. The data are for operation at steady state.

3. The dwelling and the surroundings play the roles of hot and cold reservoirs, respectively.

(b) Analysis:

The minimum theoretical cost for any heat pump cycle under the stated conditions is the cost for a reversible cycle operating between reservoirs at TH=530R (70 °F) and TC=492R (32°F). The power required by such a cycle can be obtained from:

W =_____ ~m, t

Cycle

Camot

COP,

Camot

T

 

TETD/TA (manual method)

—————- = 13.95

530-492

 

TETD/TA (manual method)

TETD/TA (manual method)

13.95

подпись: 13.95

Hr

подпись: hrCamot

Figure 2.

подпись: figure 2. TETD/TA (manual method)

4b Tc

3. A heat pump driven by a 0.4 kW electric motor provides heating for a building on a day when the outside is at -10°C and energy is lost through the walls and roof at rate of 16200 kJ/h. What is the maximum theoretical temperature that can be maintained within building, in °F?

TETD/TA (manual method)
4wclli rj AtTfi

SHAPE \* MERGEFORMAT TETD/TA (manual method)

(a) Assumptions:

1. The system shown in the accompanying figure undergoes a heat pump cycle.

2. The data are for operation at steady state.

3. The dwelling and the surroundings play the roles of hot and cold reservoirs, respectively.

(b) Analysis:

At steady state, the heat pump cycle must provide energy to the dwelling equal to the energy leaking through the walls and roof.

0H =16,200kJ I h

From Section 5.4.2, we know that the coefficient of performance of the heat pump must be less than or equal to the coefficient of performance of a reversible heat pump operating between reservoirs at TC=263K (-10°C) and TH. Then with equation 5.10:

W

Cycle

подпись: w
cycle
Tu-Tc

Inserting known values:

[l6,000ЈJ / hrx

Ihr

36005

[0.4 kW]x

1 kJ / s 1 kW _

11. 11 <———-

1- 263 / TH

TH < 2S9K = 16°C = 61°F

4. Infiltration of outside air into building through miscellaneous cracks can represent a significant load on the heating or cooling equipment. A particular office building has a total crack length of 440 ft around its doors and windows. On windy day, about 0.4 cfm of air enters per foot of crack. In addition, door openings account for about 100 cfm of outside air infiltration

* * 3 * *

On average. The internal volume of the building is 20,000 ft. Assuming ideal gas behavior, estimate n the number of times per hour (ACH — air change rate per hour), at steady state, that the air within building is changed due to infiltration. Hint: n[ACH] = Volume flow rate/Volume.

Figure 3.

подпись: figure 3. TETD/TA (manual method)

V — ZOpco ft

Фujh&ouj

подпись: v - zopco ft
фujh&ouj
FVit t iWfa*

-HiroaqU

-through door

Op&n i vwjs

подпись: -through door
op&n i vwjs

CracUs


(a) Assumptions:

1. The control volume shown is at steady state.

2. The air behaves as an ideal gas.

3. The densities of the incoming air and of the air in the building are nearly equal.

(b) Analysis:

At the steady state, the mass balance reduces to:

M = ijq —I— jji

Outflow cracks dooroperrings

= po[(AV)

Cracks (AT ) d00r0pefliflgS ]

Where pi and po are the inside and outside air densities, respectively. Assuming ideal gas behavior:

P, _ P, !RT,

Po P0!RT0 If P=PD and 7; = To, A = pa, Thus

(AV)anflmr=(AV)

Cracks (AT )cl00i 0j)ui]inys

Ft3 ft3

Min — ft ft3

подпись: min- ft ft3= (0.4—7-—)(440 ft) + 100-y

Min

Ft3

276

подпись: 276

Mm

подпись: mm16,560^— hr

Ft

Hr

16,5600

Air changes hr

0.828

Ft

20,000

Air change

AirChcmges _ AVC

Outflow

TETD/TA (manual method)
TETD/TA (manual method)
TETD/TA (manual method)

Hr

 

TETD/TA (manual method)

V

 

Practicum Assignment #3

Do not use the psychrometric chart.

1. Calculate values of humidity ratio, enthalpy, and specific volume for saturated air at one standard atmosphere using perfect gas relations for temperature of 20 °C and 0 °C.

2. The temperature of a certain room is 22 °C and the relative humidity is 50%. The barometric pressure is 100 kPa. Find (a) the partial pressures of the air and water vapor,

(b) the vapor density, and humidity ratio of the mixture.

3. Compute the enthalpy of moist air at 16 °C and 80% relative humidity for an elevation of

(a) sea level and (b) 1500 m.

4. The condition within a room is 20 °C (dry bulb), 50% relative humidity, and 101325 Pa pressure. The inside surface temperature of the window is 5 °C. Will moisture condense on the window glass? Explain why.

5. Moist air exists at a relative humidity of 60%, and a pressure of 96.5 kPa. The dewpoint temperature of this moist air is 18 °C. Determine (a) the humidity ratio and (b) the volume in m3/kg.

A

6. A duct has moist air flowing at a rate of 2 m /s. What is the mass flow rate of the dry air, where the dry bulb temperature is 16 °C, the relative humidity is 80% and where the pressure inside the duct corresponds to (a) sea level, and (b) 2000 m?

Solutions for the Practicum Assignment #3

1. Calculate values of humidity ratio, enthalpy, and specific volume for saturated air at one standard atmosphere using perfect gas relations for temperature of 20 °C and 0 °C.

2. The temperature of a certain room is 22 °C and the relative humidity is 50%. The barometric pressure is 100 kPa. Find (a) the partial pressures of the air and water vapor,

(b) the vapor density, and humidity ratio of the mixture.

3. Compute the enthalpy of moist air at 16 °C and 80% relative humidity for an elevation of

(a) Sea level and (b) 1500 m.

4. The condition within a room is 20 °C (dry bulb), 50% relative humidity, and 101325 Pa pressure. The inside surface temperature of the window is 5 °C. Will moisture condense on the window glass? Explain why.

5. Moist air exists at a relative humidity of 60%, and a pressure of 96.5 kPa. The dewpoint temperature of this moist air is 18 °C. Determine (a) the humidity ratio and (b) the

A

Volume in m /kg.

6. A duct has moist air flowing at a rate of 2 m /s. What is the mass flow rate of the dry air, where the dry bulb temperature is 16 °C, the relative humidity is 80% and where the pressure inside the duct corresponds to (a) sea level, and (b) 2000 m?

1. Calculate values of humidity ratio, enthalpy, and specific volume for saturated air at one standard atmosphere using perfect gas relations for temperature of 20 °C and 0 °C.

Ir =0.622-

TETD/TA (manual method)

Constants:

P = 101325 Pa, Cp, a = 1.01 kJ/kg-K, itg = 2501.3 kJ/kg-K, Cp, v = 1.86 kJ/kgK

Ra = 287 J/ kg K, Rv = 462 J/ kg K

Pv=Ps (Pa)

W (kgw/kga)

I (kJ/kg)

V (nrVkg)

611.3

3.78xl0"3

9.45

0.775

0

O

CN

2339

L.47xl0"[3]

57.5

0.838

Comments:

Check your answer in the psychrometric chart.

TOC o "1-5" h z P 1336

Pv = —^ =——————————————— — = 9.80×10 (kg/m3)

RJ 462-295.15

P ________ 1336

W = 0.622— — = 0.622—————— = 8.42×10 (kgw/kga)

P-R 1×10 —1336

3. Compute the enthalpy of moist air at 16 °C and 80% relative humidity for an elevation of (a) sea level and (b) 1500 m.

(a) At sea level: P = 101325 Pa, From Table A-lb,

Pv s = 1836 Pa 1 = (pPv s =1468.8Pa

P

W=0.622y^-=9.15xl0“3 (kgw/kga)

I = Cpat + ^(2501.3 +1.860 = 39.3 (kJ/kg)

(b) At 1500 m level: P = a+bH = 99.436 — 0.01(1500)=84.436 kPa,

P 1468 8

W = 0.622—- ‘— = 0.622———- :—— = 0.0110 (kgw/kga)

P-Pv 84436-1468.8

/ = Cpj + W(2501.3 + 1.860 = 44.03 (U/kg)

4. The condition within a room is 20 °C (dry bulb), 50% relative humidity, and 101325 Pa

Pressure. The inside surface temperature of the window is 5 °C. Will moisture condense

On the window glass? Explain why.

There are a couple of ways to solve this problem.

(1) Calculate the dewpoint temperature Td,

If Td > Tsurface > condensation occurs.

(2) Calculate humidity ratio W at 20°C, <E> = 50%, and W at 5°C, 100% RH

If W|20°C, 50°o > W|5°; 100°O, condensation occurs.

(3) Calculate partial pressure of vapor at 20°C, 50% RH and 5C, 100% RH

If Pvbo°c. 50% > Pv. s|5°, ioo»o, condensation occurs.

Here I just present (1):

Pv, s|20x = 2339 Pa, Py = (f)l s = 1169.5 Pa

From table A-lb, we can find Td= 9.25 °C

Td > Tsurface, condensation occurs!

5. Moist air exists at a relative humidity of 60%, and a pressure of 96.5 kPa. The dewpoint temperature of this moist air is 18°C. Determine (a) the humidity ratio and

(b) the volume in m3/kg.

This is the most confusing problem in this problem set. Let’s do it step by step.

1) Recall what you heard in weather forecast (if you watch TV sometime). The told you "the temperature in State College is 68°F, dewpoint temperature is 65°F, raining in the

Morning……. " They are talking about out HVAC concept: The current temperature of air

Does not necessarily equal to the dewpoint temperature of the air.

2) So, the temperature if the air, in other words, the dry-bulb temperature of the air must be some temperature other than 18°C since <E> = 60%. Let’s imagine a psychart (you can use it actually, but at this moment, I just use formula), when you know "2" quantities, either Tdiy, Twet, or Tdew, you are able to determine a point on the psychart. Now, you have Td and <&, so be confident!

3) Recall the definition of the dewpoint temperature from our text: "Dewpoint temperature ta is the temperature of saturated moist air at the same pressure and humidity ratio as the

Given mixture."

Cooled down (sensible)

Py2 — Ps2

Pvl, P.1

Until saturated

PV2/PS2= 100%

I /1 i — 60%

W2

W!

PS

II

Ps

Td= 18 °C

Given Mixture

*vl *v2

Saturated moist air

4) Now we are quite clear what we are going to do: since W and Pv do not change, From Table A-lb, At 18°C, Ps= 2064 Pa, Pv|t = Ps|i8°c

P 2064

W=0.622—-— = 0.622————- = 0.01 ?>6(kgwl kga)

P-P 96500-2064

A

Pv = 2064 Pa pvs=^ = 3440 Pa

<!>

From Table A-lb: T = 26.37 °C=299.5 K

T 7QQ ^

V=————— =—————- :———— = 0.898(m31 kg)

P-Ps P^ 96500-2064^ 2064 V

Ra Rr 287 462

Comments:

Note the keys of this problem are:

• Td ^Tdn-buib when <& ^ 100%

• W and Pv do not change during our sensible cooling process

• Using chart or computer program is much easier

6. A duct has moist air flowing at a rate of 2 m /s. What is the mass flow rate of the dry air, where the dry bulb temperature is 16°C, the relative humidity is 80% and where the pressure inside the duct corresponds to (a) sea level, and (b) 2000 m?

From Table A-lb, for 16°C, Pvs = 1818 .1 Pa Pv = <pPvs = 1454.5Pa

(a) Sea level: P=101325 Pa Pa=P~ Pv = 99.%lkPa

P. 99870

Pa =

подпись: pa == 1.203 (kg / m )

RaT 287-289.15 ma = ioaV = 2.407 (kg / sec)

(b) 2000 m elevation:

подпись: (b) 2000 m elevation:P = a+bH = 99.436 — 0.01(2000)=79.436 kPa, P„ =P-P. =79436-1454.5 = 77981.5 Pa

77981 .5

Pa =

подпись: pa == 0.94 (kg / m3)

RaT 287 — 289. L5 < =p/ = 1 879(AgVsec)

Comments:

V

• Some useful relationships: ma = paV = —

V

The mass flow rate changes with elevation but the volumetric flow rate does not

Practicum Assignment #4

1. Air leaves the cooling coil of an air conditioning system has a relative humidity of 90% and a dewpoint temperature of 55°F at 5000 ft elevation. From the psychrometric chart find:

(a) Dry bulb air temperature

(b) Wet bulb temperature

(c) Air density

(d) Humidity ratio

(e) Enthalpy

Identify your results in the chart and submit the chart with the solutions.

2. To save energy, the environmental conditions in a room are to be regulated so that the dry bulb temperature will be greater than or equal to 78°F (24°C) and the dew point will be less than or equal to 64°F (17°C). Find the maximum relative humidity that can occur for standard barometric pressure.

3. It is desired to heat and humidify 2000 cfm of air from an initial state defined by a temperature of 60°F db and relative humidity of 30% to a final state of 110°F db and 30% relative humidity. The air will first be heated by a hot water coil followed with saturated vapor at 5psig. Assume sea level pressure Using the psychrometric chart:

(a) Find the heat transfer rate for the heating coil

(b) Find the mass flow rate of the water vapor

(c) Sketch the processes on a psychrometric chart.

4. Moist air enters a cooling coil at 28°C dry-bulb temperature and 50% relative humidity and exits the coil at 13°C dry-bulb temperature and 90% relative humidity. The flow rate through the coil is 1.50 kg/s and the process occurs at the sea level pressure.

(a) Determine the sensible heat factor (SHF) for the process

(b) Determine the cooling coil capacity (heat transfer-rate)

(c) Sketch the process in the psychrometric chart denoting sensible and latent heat

5. Saturated steam is sprayed into a stream of moist air. The initial condition of the air is 55°F dry-bulb temperature and 35°F dew-point temperature. The mass airflow rate is 2000 lbm/min. Barometric pressure is 14.696 psi. Determine:

(a) How much steam must be added in lbm/min to produce a saturated air condition, and

(b) The resulting dry-bulb and wet-bulb temperature of the saturated air

Practicum Assignment #4

1. Air leaves the cooling coil of an air conditioning system has a relative humidity of 90% and a dewpoint temperature of 55°F at 5000 ft elevation. From the psychrometric chart find:

(a) Dry bulb air temperature

(b) Wet bulb temperature

(c) Air density

(d) Humidity ratio

(e) Enthalpy

Identify your results in the chart and submit the chart with the solutions.

2. To save energy, the environmental conditions in a room are to be regulated so that the dry bulb temperature will be greater than or equal to 78°F (24°C) and the dew point will be less than or equal to 64°F (17°C). Find the maximum relative humidity that can occur for standard barometric pressure.

3. It is desired to heat and humidify 2000 cfm of air from an initial state defined by a temperature of 60°F db and relative humidity of 30% to a final state of 110°F db and 30% relative humidity. The air will first be heated by a hot water coil followed with saturated vapor at 5psig. Assume sea level pressure Using the psychrometric chart:

(a) Find the heat transfer rate for the heating coil

(b) Find the mass flow rate of the water vapor

(c) Sketch the processes on a psychrometric chart.

4. Moist air enters a cooling coil at 28°C dry-bulb temperature and 50% relative humidity and exits the coil at 13°C dry-bulb temperature and 90% relative humidity. The flow rate through the coil is 1.50 kg/s and the process occurs at the sea level pressure.

(a) Determine the sensible heat factor (SHF) for the process

(b) Determine the cooling coil capacity (heat transfer-rate)

(c) Sketch the process in the psychrometric chart denoting sensible and latent heat

5. Saturated steam is sprayed into a stream of moist air. The initial condition of the air is 55°F dry-bulb temperature and 35°F dew-point temperature. The mass airflow rate is 2000 lbm/min. Barometric pressure is 14.696 psi. Determine:

(a) How much steam must be added in lbm/min to produce a saturated air condition, and

(b) The resulting dry-bulb and wet-bulb temperature of the saturated air

1. Air leaves the cooling coil of an air conditioning system and has a relative humidity of 90% and a dewpoint temperature of 55°F at 5000 ft elevation. From the psychrometric chart find:

(a) Dry bulb air temperature

(b) Wet bulb temperature

(c) Air density

(d) Humidity ratio

(e) Enthalpy

Identify your results in the chart and submit the chart with the solutions.

If you follow a constant humidity ratio (w) line from 55 degrees and saturation to where it intersects with the 90% relative humidity line you will have the condition of the air. Read all other properties off of the chart.

From psych chart:

Tdb=57.7°F, Twb=55.9°F, v = 15.93 ft3/lb, p = 0.0628 lb/ft3

W = 0.011 lbv/lba, i = 25.9 Btu/lb

2. To save energy, the environmental conditions in a room are to be regulated so that the dry bulb temperature will be greater than or equal to 78°F (24°C) and the dew point will be less than or equal to 64°F (17°C). Find the maximum relative humidity that can occur for standard barometric pressure.

Follow maximum humidity ratio (w) line (where dew point = 17°C) until it intersects with dry bulb temperature = 24°C line. Read relative humidity lines off of chart as shown below.

TETD/TA (manual method)

From Chart 1b ^max = 60%

From Chart la (»max ” 62%

подпись: from chart 1b ^max = 60% подпись: from chart la (»max ” 62%

Remember — relative humidity for a constant humidity ratio decreases as dry bulb increases.

TETD/TA (manual method)

3. It is desired to heat and humidify 2000 cfm of air from an initial state defined by a temperature of 60°F db and relative humidity of 30% to a final state of 110°F db and 30% relative humidity. The air will first be heated by a hot water coil followed with saturated vapor at 5 psig. Using the psychrometric chart:

(a) find the heat transfer rate for the heating coil

(b) find the mass flow rate of the water vapor

(c) sketch the processes on a psychrometric chart.

Assume sea level pressure.

The air here undergoes two distinct processes — a sensible only (constant humidity ratio) heating through the coil and the addition of humidity as saturated steam. In this case the steam is saturated and therefore that process follows a constant dry bulb line.

#i = 30%, ii= 17.1 Btu/lb,

Wi= 0.00335 lbv/lba

подпись: #i = 30%, ii= 17.1 btu/lb,
wi= 0.00335 lbv/lba
From the psych chart Air at initial state 1:

TDB, i = 60°F,

Vi = 13.18 ft3/lb,

Air at state 2:

Tdb,2 = 110°F, i2= 30.2 Btu/lb, W2= 0.00335 lbv/lba

Air at final state 3:

TDb,3 = 110°F, <E>3 = 30%, i3 = 45 Btu/lb,

V3 = 14.75 ft3/lb, W3= 0.01679 lbv/lba

Coil load is only the part of the load that increases the dry bulb temperature since the coil only adds heat and no moisture.

Coil load:

Qoca = ~h) = ~(h ~h) = ^(30.2-17.1) = 1988 (Bn,!mm) = 119280(Bn,/hr)

13.18

Vapor mass flow rate:

V/ O

M =riia(W3-W2) = —(W3-W2) =———- (0.01679-0.00335)= 2.04 (lb min) = 122.4

Vj 13.18

(lb min)

ALTERNATE SOLUTION

An energy balance can also be used to solve this problem keeping in mind that the properties given on the psych, table are given per mass of DRY air. When solving for the mass flow rate of dry air, however, you can neglect the contribution of the water in the volume flow rate given.

Diagram:

1

1

I

T-

1

1

1

1

/ 1 4 : ►

Lair, il

1

Mair, Ц

1

1

I

/ (mair+ w*mair), i2

I I I

✓ 1

L

Qcoil

ГП water

1st Law Energy Balance:

Ma * i2 +mw is — m;i i3 = 0

Where is = 1130 Btu/lb — enthalpy of saturated steam at 5 psig (interpolate from table A-lb in text)

2000 (45-30.2) ,

= ma *(h~h)lh= T———————— = 1-99 (lb/mm)

TETD/TA (manual method)1D. lo 1 1 D 1

On the Psych, chart the process will look like:

4. Moist air enters a cooling coil at 28°C dry-bulb temperature and 50% relative humidity and exits the coil at 13°C dry-bulb temperature and 90% relative humidity. The flow rate through the coil is 1.50 kg/s and the process occurs at the sea level pressure.

(a) Determine the sensible heat factor (SHF) for the process

(b) Determine the cooling coil capacity (heat transfer-rate)

(c) Sketch the process in the psychrometric chart denoting sensible and latent heat

(a) From the chart: SHF = 0.635

(b) From the chart: ii=58.6 kJ/kg, i2=34.2 kJ/kg Neglecting condensate,

Occ =max(/1 —/2) = (1.50kg / s)x(58.6kJ / kg — 34.2kj / kg) = 36.6UW

TETD/TA (manual method)

Chart lb

5.

(c) See psychrometric chart

подпись: (c) see psychrometric chart

Saturated steam is sprayed into a stream of moist air. The initial condition of the air is 55°F dry-bulb temperature and 35°F dew-point temperature. The mass airflow rate is 2000 lbm/min. Barometric pressure is 14.696 psi. Determine:

(a) How much steam must be added in lbm/min to produce a saturated air condition, and

(b) The resulting dry-bulb and wet-bulb temperature of the saturated air

(a) End state (state 2) is saturated. See construction on chart.

From the chart: = 0.0043lbm /lba, w2 = 0.010lbm /lba

Mv= ma x(>2 -wx) = (2000lbm /min)x(0.010-0.0043lbm /lba) = 11 Albm /min

(b) From the chart: TDb2=55°F, Twb2=55°F

TETD/TA (manual method)SEA LEVEL

Practicum Assignment #5

1. A space is to be maintained at 72°F and 30% relative humidity during the winter months. The sensible heat loss from the space is 500,000 Btu/hr and the latent heat loss due to infiltration is 50,000 Btu/hr. Construct the condition line in the psychrometric chart.

2. Air at 10°C db and 5°C wb is mixed with air at 25°C db and 18°C wb in a steady-flow

3 3

Process at standard atmospheric pressure. The volume flow rates are lOm/s and 6m /s, respectively.

(a) Compute the mixture conditions (enthalpy and humidity ratio)

(b) Find the mixture conditions using the psychrometric chart.

3. A building has a total load of 200,000 Btu/hr. The sensible heat factor for this space is

0. 8. The space is to be maintained at 72°F and 40% relative humidity. Outdoor air at 40°F and 20% relative humidity in the amount of 1000 cfm is required. Air is supplied to the space at 120°F. Find:

(a) the conditions and amount of air supplied to the space,

(b) the temperature rise of the air through the furnace,

(c) the amount of water at 50°F required by the humidifier, and

(d) the capacity of the furnace.

Assume sea pressure level.

4. A large warehouse located in Denver, Colorado (elevation = 5000 ft) is to be conditioned using an evaporative cooling system. Assume that the space is to be maintained at 80°F and 50% relative humidity by a 100% outdoor air system. Outdoor design conditions are 91°F dry bulb and 59°F wet bulb. The cooling load is estimated to be 100 tons with a sensible heat factor of 0.9. The supply air fan is located just downstream of the spray chamber and is estimated to require 20 hp. Determine the volume flow rate of air to the space, and sketch the process on a psychrometric chart.

Solutions for the Practicum Assignment #5

1. A space is to be maintained at 72°F and 30% relative humidity during the winter months. The sensible heat loss from the space is 500,000 Btu/hr and the latent heat loss due to infiltration is 50,000 Btu/hr. Construct the condition line in the psychrometric chart.

2. Air at 10°C db and 5°C wb is mixed with air at 25°C db and 18°C wb in a steady-flow

3 3

Process at standard atmospheric pressure. The volume flow rates are 10 m /s and 6m /s, respectively.

(a) Compute the mixture conditions (enthalpy and humidity ratio)

(b) Find the mixture conditions using the psychrometric chart.

3. A building has a total load of 200,000 Btu/hr. The sensible heat factor for this space is

0. 8. The space is to be maintained at 72°F and 40% relative humidity. Outdoor air at 40°F and 20% relative humidity in the amount of 1000 cfm is required. Air is supplied to the space at 120°F. Find:

A. the conditions and amount of air supplied to the space,

B. the temperature rise of the air through the furnace,

C. the amount of water at 50°F required by the humidifier, and

D. the capacity of the furnace.

Assume sea pressure level.

4. A large warehouse located in Denver, Colorado (elevation = 5000 ft) is to be conditioned using an evaporative cooling system. Assume that the space is to be maintained at 80°F and 50% relative humidity by a 100% outdoor air system. Outdoor design conditions are 91°F dry bulb and 59°F wet bulb. The cooling load is estimated to be 100 tons with a sensible heat factor of 0.9. The supply air fan is located just downstream of the spray chamber and is estimated to require 20 hp. Determine the volume flow rate of air to the space, and sketch the process on a psychrometric chart.

1. A space is to be maintained at 72°F and 30% relative humidity during the winter months. The sensible heat loss from the space is 500,000 Btu/hr and the latent heat loss due to infiltration is 50,000 Btu/hr. Construct the condition line in the psychrometric chart.

500.0 TETD/TA (manual method) SHF = = 0.91

550.000

2. Air at 10°C db and 5°C wb is mixed with air at 25°C db and 18°C wb in a steady-flow

3 3

Process at standard atmospheric pressure. The volume flow rates are 10 m /s and 6 m /s, respectively.

(a) Compute the mixture conditions (enthalpy and humidity ratio)

(b) Find the mixture conditions using the psychrometric chart.

Use mass weighted averages in order to find the mixed condition of the air.

(a) mal =Qj /Vl = 10/0.81 = 12.346 kg/s

M

подпись: m

A 2

подпись: a 2■ 6 / 0.86 = 6.977 kg Is

H =

W,

подпись: h =
w,
(12.436 x 18.6 ) + (6.977 x 50.9) = 3Q^ kJ ; (12.346 + 6.977)

_ (12.436 x 0.0034) + (6.977 x 0.0102) (12.346 + 6.977)

W3 = 0.0059 kgv /kga

(b) The ratio between the mass flow rates of the air streams locates where the mixed condition is on a line drawn between the two incoming states on the psych, chart.

M

A __

TETD/TA (manual method)

12.346

 

32

= 0.64 = = 12

 

This indicates the mixed condition is 64% of the way on the line towards air stream 1

Read: /3 = 30.0 Btu / lba; W3 = 0.0058 kgv / kga

A building has a total load of 200,000 Btu/hr. The sensible heat factor for this space is

0. 8. The space is to be maintained at 72°F and 40% relative humidity. Outdoor air at 40°F

And 20% relative humidity in the amount of 1000 cfm is required. Air is supplied to the

Space at 120°F. Find:

A. the conditions and amount of air supplied to the space,

B. the temperature rise of the air through the furnace,

C. the amount of water at 50°F required by the humidifier, and

D. the capacity of the furnace.

TETD/TA (manual method)Assume sea pressure level.

V0 = 1000 cfm

(a) From Chart la

Ts =120/76 F

Q 200,000

Rn = —-— =————————

S (is-ir) (39.4-24.4)

= 13,333 lb/hr = rh1

Qs = ™svs = (14.8)/60 = 3z8U jt / min

(b) mD = Qa / va = 1000 X 60 /12.6 = 4760 lb/hr

Mv 13,333-4762 ^

—- =———————- = 0.642: From Chart t, = 61 / 49 F

W, 13,333 1

/3 — /, = (See (d) below)

(c) mw = ms(Ws -Wj) = 13,333 (0.0095 -0.0046) = 65.3 lbm/hr

(d) qf = til, (/3 -/,) = 13,333 (39.4 -19.7) = 262,660Btu/hr

Where 73 = is from chart

Then qf = 262,660 = 13,333 (0.24)(/3 — 61); and t3 = 143 F ^f»r=t3~fi =143-61 = 82F

4. A large warehouse located in Denver, Colorado (elevation = 5000 ft) is to be conditioned using an evaporative cooling system. Assume that the space is to be maintained at 80°F and 50% relative humidity by a 100% outdoor air system. Outdoor design conditions are 91°F dry bulb and 59°F wet bulb. The cooling load is estimated to be 100 tons with a sensible heat factor of 0.9. The supply air fan is located just downstream of the spray chamber and is estimated to require 20 hp. Determine the volume flow rate of air to the space, and sketch the process on a psychrometric chart.

<7, =ms(ir -is); Wfan =ms(ix ~ic)

(a) ic = 28 Btu/lbm; ir = 30.6 Btu/lbm Using Psychrometric Chart:

Qr = 1,200,000 Btu/hr

Wfan =50,900 Btu/hr

/3 = 28.1 Btu/lbm; ms = 480,000 lba/hr

Os = 480,000 x 16.3/60 = 130,400 cfm

(b) Q. s = 61.54 m3 / 5; Using Psychrometric Chart

TETD/TA (manual method)

Practicum Assignment #6

1. A condition exists where it is necessary to cool and dehumidify air from 80°F db and 67°F wb to 60°F db and 54°F wb.

(a) Discuss the feasibility of doing this in one process with a cooling coil. (Hint: Determine the apparatus dewpoint temperature for the process.)

(b) Describe a practical method of achieving the required process, and sketch it on a psychrometric chart.

2. Continue to design the air-conditioning system of the classroom at PSU for winter heating. The example in the lecture notes (pg. 36 of Chapter2) specifies design conditions.

(a) Design a heating system with a heating coil and a saturated-steam humidifier and identify the corresponding air-handling processes in a psychrometric chart. The system should include the components used for summer cooling.

(b) Size the equipment of the air-conditioning system proposed in (a). Use the data for cooling if necessary.

Use the psych chart to perform this design assignment.

3. A space is to be maintained at 78°F db and 68°F wb. The cooling system is a variable-air — volume (VAV) type where the quantity of air supplied and the supply temperature are controlled. Under design conditions, the total cooling load is 150 tons with a sensible heat factor of 0.6, and the supply temperature is 60°F db. At minimum load, about 18 tons with SHF of 0.8, the air quantity may be reduced no more than 80% by volume of the full load design value.

(a) Determine the supply air conditions for minimum load.

(b) Show all the conditions on a psychrometric chart.

Solutions for the Practicum Assignment #6

1. A condition exists where it is necessary to cool and dehumidify air from 80°F db and 67°F wb to 60°F db and 54°F wb.

(a) Discuss the feasibility of doing this in one process with a cooling coil. (Hint: Determine the apparatus dewpoint temperature for the process.)

(b) Describe a practical method of achieving the required process, and sketch it on a psychrometric chart.

2. Continue to design the air-conditioning system of the classroom at PSU for winter heating. The example in the lecture notes (pg. 36 of Chapter 2) specifies design conditions.

(a) Design a heating system with a heating coil and a saturated-steam humidifier and identify the corresponding air-handling processes in a psychrometric chart. The system should include the components used for summer cooling.

(b) Size the equipment of the air-conditioning system proposed in (a). Use the data for cooling if necessary.

Use the psych chart to perform this design assignment.

3. A space is to be maintained at 78°F db and 68°F wb. The cooling system is a variable-air — volume (VAV) type where the quantity of air supplied and the supply temperature are controlled. Under design conditions, the total cooling load is 150 tons with a sensible heat factor of 0.6, and the supply temperature is 60°F db. At minimum load, about 18 tons with SHF of 0.8, the air quantity may be reduced no more than 80% by volume of the full load design value.

(a) Determine the supply air conditions for minimum load.

(b) Show all the conditions on a psychrometric chart.

1. A condition exists where it is necessary to cool and dehumidify air from 80°F db and 67°F wb to 60°F db and 54°F wb.

(a) Discuss the feasibility of doing this in one process with a cooling coil. (Hint: Determine the apparatus dewpoint temperature for the process.)

(b) Describe a practical method of achieving the required process, and sketch it on a psychrometric chart.

(a) It is probably impossible to cool the air from 1 to 2 in one process. The extension of line 1-2 does not intersect the saturation curve. Therefore, there is no feasible apparatus dew point.

(b) Cool the air to state 1 ’ and then sensibly heat to state 2.

2. Continue to design the air-conditioning system of the classroom at PSU for winter

Heating. The example in the lecture notes (pg. 36 of Chapter 2) specifies design conditions.

(a) Design a heating system with a heating coil and a saturated-steam humidifier and identify the corresponding air-handling processes in a psychrometric chart. The system should include the components used for summer cooling.

(b) Size the equipment of the air-conditioning system proposed in (a). Use the data for cooling if necessary.

Given:

Required:

Solution:

подпись: given:
required:
solution:
Winter-heating condition of a PSU classroom HVAC system Design.

This problem is a continuous design of the example in chapter 3.5, which is given as: (only winter conditions are listed below):

Sensible

Walls

Windows

Latent

Heating (W)

2000

2000

Heating (W)

TETD/TA (manual method) TETD/TA (manual method)

Outdoor Design Conditions T -14°C

Twet /

Indoor Design Conditions

T 22°C

(|) 50%

 

/

 

/

 

Minimum Outdoor Air: 80 1/s Air Supply Temperature: 60°C (max)

(a) The heating system is designed as follow:

Outdoor

TETD/TA (manual method)

Please Note:

• Preheat is needed in winter heating since condensation (freezing) may occur if out air mixes with return air directly.

• Heat recover system is often used by taking heat out of exhaust air

• Cooling coil will be turned off in winter heating

• Heating coil only supplies sensible heat

Determine important points: (Find T, W, i,…………….. )

1) Outdoor (O):

T0 = Tdiy= -14 °C, (J)o=60% (design rule of thumb for winter RH)

From low T psych chart (attached at end of problem): Wo=0.00068 (kgw/kga), iG= -12.5 (kJ/kg)

2) Indoor (R):

TR = 22 °C, (j)k=50%

From psych chart: WR = 0.008225 (kgw/kga), iR = 43.13 (kJ/kg)

3) Supply Point (I):

Make a choice of supply air temperature < 60 °C, so pick up Tj = 45 °C.

Since no latent heat to remove,

Wi = WR = 0.008225 (kgw/kga)

From psych chart: ii = 66.71 kJ/kg Heating Coil only adds sensible heat — humidification adds the rest of the heat in the form of latent energy — see Psych chart below.

Ii = 55.6 kJ/kg — from psychrometric chart

4) Preheated point (P):

Suppose we use heat recovery system to preheat the outdoor air, Tp =12.8°C [55°F] is a design rule of thumb.

Tp = 12.8 °C

Wp = W0 = 0.00068 kgw/kga From psych chart: ip = 14.6 (kJ/kg)

5) Mixing Point (M):

= =———— 4°°°—————— = 0.1722 (kg/s)

CpAT 1.01×10 x(45 —22)

M0 = 80 1/s = 80 x 10"3m3/s x 1.2 kg/m3 = 0.096 (kg/s)

M, f„ +mR’R +K, — mo)iR 0.096(14.6)+ (0.1722-0.096)43.13 „„„„„„ x

=——————- =——————————- =——————————————————— = 27.22 (kJ/kg)

Mn+mR mn 0.1722

O Ј Cl

MoWa+mRWR 0.096(0.00068)+ (0.1722-0.096)0.008225 „ „ x

W = — —————— =—————— V—————————————— L————— = 0.00402 (kgw/kga)

M ma 0.1722

(b) Now, it is our choice to size the equipments — heating coil, humidifier and fan. Suppose we heat the air from mixing point to same temperature and use adiabatic humidification (maybe not the best way), then

TETD/TA (manual method)

Ti

подпись: tiTp Tra

For heating coil Qheating = ma (і/ — ім ) = 0.1722 (55.6 — 27.22) = 4.89 (kW)

For humidifier: AW = W: — WM= 0.008225 — 0.00402 = 0.004205 (kgw/kga)

For fan: V = ma /p = ma X v = 0.1722 X 0.828 = 0.1426 m3/s = 513 (m3/hr)

For preheater: Qpreh = mG (iP — iG ) = 0.096 (14.6 — (-12.5)) = 2.60 (kW)

This equipment should be use in both heating (winter) and summer conditions. So we should pick up the great value to size them.

1200 m3/hr 7.156 kW 0.4 kW

 

513 m3/hr 4.89 kW

 

From example:

 

Fan

Qcooling

Qheating

 

Our calc:

 

Fan

Qcooling

Qheating

 

1200 m /hr 7.156 kW 4.89 kW

0.004205 (kgw/kga) 2.60 kW

подпись: 1200 m /hr 7.156 kw 4.89 kw
0.004205 (kgw/kga) 2.60 kw
So our final choice will be: Fan:

Cooling Coil: Heating Coil: Humidifier: Pre-heater:

TETD/TA (manual method)

3. A space is to be maintained at 78°F db and 68°F wb. The cooling system is a variable-air-

Volume (VAV) type where the quantity of air supplied and the supply temperature are controlled. Under design conditions, the total cooling load is 150 tons with a sensible heat factor of 0.6, and the supply temperature is 60°F db. At minimum load, about 18 tons with SHF of 0.8, the air quantity may be reduced no more than 80% by volume of the full load design value.

(a) Determine the supply air conditions for minimum load.

(b) Show all the conditions on a psychrometric chart.

(a) Hd = md(ir -iJ

Using psych chart la, ir = 32.4 Btu/hr, is = 25.0Btu/hr

(150(ons)x

‘12,000Btu/hr’

_ _

Mon

_

32 A-25.0 Btu / lb

= 243243lbm/hr

(243,243lbm /hr)x(13.3ft3 /lb)

K, = mdvs = ±— ’———- —7^77—- ——- = 53,920cfm

60min/ hr

V = 0.2Vd = 0.2 x 53,920c/«? = 10,784c/«?

Vm (10,784c/7??)x(60min/hr) ,

M = — =——————————————— = 47,928/й»? / hr where Vm is assumed

V„, 13.5ft lib

<lm =™nAir — О«,)

(18 tons)x

32 ABtu! lb-

12,000Btu/hr Mon

47,928/й»? / hr

TETD/TA (manual method)

L =i

M r

 

21.9 Btu / lb

 

/??„

 

(b)

подпись: (b) TETD/TA (manual method)Tdb„= 64° F

W = 62°F

Practicum Assignment #7

1. During the winter months it is possible to cool and dehumidify a space using outdoor air.

Suppose an interior zone of a large building is designed to have a supply airflow rate of

5000 cfm, which can be all outdoor air. The cooling load is constant at 10 tons with a

SHF of 0.8 the year round. Indoor conditions are 78°F db and 67°F wb.

(a) What is the maximum outdoor air dry bulb temperature and humidity ratio that would satisfy the load condition?

(b) Consider a different time when the outdoor air has a temperature of 40°F db and 20% relative humidity. Return air and outdoor air may be mixed to cool the space, but humidification will be required. Assume that saturated water vapor at 14.7 psia is used to humidify the mixed air, and compute the amounts of outdoor and return air in cfm.

(c) At another time, outdoor air is at 70°F db with a relative humidity of 90%. The cooling coil is estimated to have a minimum apparatus dew point of 50°F. What amount of outdoor air and return air should be mixed before entering the coil to satisfy the given load condition?

(d) What is the refrigeration load for the coil of part (c) above?

2. An economizer mixes outdoor air with room return air to reduce the refrigeration load on

The cooling coil.

(a) For a space condition of 25°C db and 20°C wb, describe the maximum wet bulb and dry bulb temperatures that will reduce the coil load.

(b) Suppose a system is designed to supply 5m3/s abd 18°C db and 17°C wb to a space maintained at the conditions given in part (a) above. What amount outdoor air at 20°C db and 90% relative humidity can be mixed with the return air if the coil SHF is 0.6?

(c) What are the apparatus dew point and the bypass factor in part (b) above?

(d) Compare the coil refrigeration load in part (b) above with the outdoor air to that without outdoor air.

Practicum Assignment #8

1. A classroom in a school is designed for 100 people.

(a) What is the minimum amount of fresh outdoor air required?

(b) The floor area is 1500 ft2. What is the outdoor air ventilation requirement on the bases of floor area?

2. What level, in ppm, will the CO2 concentration be in a space in steady state, if CO2 is being released into the space at the rate of 0.25 cfm and outdoor air with a CO2 concentration of 200 ppm is being supplied to the space at the rate of 1000 cfm? Assume complete mixing.

3. The same space as in problem 2 uses a mixing with the recirculation rate of 0.4 in order to conserve energy. If the CO2 release rate and fresh air supply is the same as in problem 2, calculate the CO2 concentration (ppm) in the space in steady state. Note: Recirculation rate R=Recirculated Air / Total Supply Air

(2) Energy: E (Btu, J)

Capacity to do work. Examples:

• Thermal

• Light

• Mechanical

• Electrical

• Chemical

1 Btu=1055J

Work (W) is an action of a force on a moving system.

2. A solar collector panel, shown in Figure 1, has a surface area of 32 ft2. The panel receives energy from the sun at a rate of 150 Btu/(hr ft2-of collector surface). Forty percent of the incoming energy is lost to the surrounding. The reminder is used to warm liquid water from 130°F to 160°F. The water passes through the solar collector with a negligible pressure drop.

(a) Neglecting kinetic and potential energy effects, determine at steady state the mass flow rate of water in [lb/min]. Hint: Write the assumptions.

(b) How many solar collectors would be needed to provide a total of 40 gal of 160°F water in 30 min? Hint: To obtain water property use thermophysical properties table in your text assuming the atmospheric pressure.

[1] Heat (Thermal energy): Q (Btu, J)

Heat is energy transferred across the system boundary by temperature difference (AT).

[2] ^mw2*iw2-rhwl*iwl

Steam

^ steam

We are given a volumetric flow rate fom the heated water of 14 cfm. V(dot)p=m(dot)

So we can calculate m(dot) of the heated water as:

(14cfm)(60.01 lb/ft3)*(lmin/60sec)= 141bs/sec Enthalpy (iW2) of heated water at 120 degrees = 88 btu/lb

As for the cool water, the mass flow rate is constant so it will not affect the shape of our graph. The enthalpy for the 50 degree water is 18.1 btu/lb.

Use table A-la for the above values.

[3] The temperature of a certain room is 22 °C and the relative humidity is 50%. The barometric pressure is 100 kPa. Find (a) the partial pressures of the air and water vapor, (b) the vapor density, and humidity ration of the mixture.

From table A-lb

Pv = (j)Pv g = 1336 Pa Pa=P-Pv= 98,664 Pa

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