exam
DESCRIPTION
Exam usedTRANSCRIPT
-
Air-conditioning
1.Dry Bulb Temperature(tdb)-mixture of dry air and water vapor superheated 2. Wet Bulb Temperature(twb) -
mixture of dry air and water vapor saturated 3.Dew Point Temperature(ts) condense if air cooling under
constant pressure. 4. Humidity Ratio =m(water)/m(air)=0.622P(water)/P(air) 5. Relative Humidity =
P(water)/P(air)
Air-conditioning Processes: Cooling and Dehumidification, Re-heating, Adiabatic Mixing, Fan, Part Load Control
Sensible Heat Change (no change in water vapor content) Qs=m(hx-h1)= mCp(tdbx-tdb1) ; Latent Heat Change
Qt=m(hx-h1)=mh(x-1) ;
Determination of Sensible and Latent Cooling Loads (all unit in W NOT kW)
1.Mean Solar Heat Gain = ---- directly through the window
Mean Solar heat gain(W) Mean Solar Intensity(W/m) S Solar gain factor of glazing Sunlit area of
glazing (m)
2.Mean Casual Heat Gain = all source x using time / 24 --- 3.
= +
4. Mean Internal Environmental Temperature ( ) --
= ( + )(
) -- = 0.33
exposed to outdoor air(cal wall+window), mean outdoor air temperature (daily mean monthly)
5. Swing in solar heat gain = ( ) --- alternating factor, peak intensity
6. Swing in casual heat gain = ---- =0 for lightweight
7.Swing in heat gain air to air = ( + ) ---- cal window only, swing air temperature
8. = + + , Peak temperature: = ( + ) cal all surface including partition
9. = ( + )(
) + --- exposed to outdoor air(cal wall+window)
10. Latent load due to infiltration
h = ,--- specific enthalpy of outdoor air according to dry-bulb temperature and relative
humidity , specific enthalpy of air at the desired dry-bulb temperature and relative humidity
h = ,--- h change in specific enthalpy of dry air of room, specific enthalpy of dry air
component of ambient air from its dry-bulb temperature, specific enthalpy of dry air at the desired
dry-bulb temperature, = (h h) and = /3600 ,--- small v is specific volume of air
Latent load due to occupancy = , = +
Mass flow rate and air supply (All unit in kW NOT W)
= ,--- mass flow rate supplied into room, permissible dry-bulb temperature rise
case 1: > , = , = /3600
case 2: < , = /
= , moisture to be picked up, latent heat of evaporation
= , moisture supplied into room via air-conditioning, desired moisture kg/kg
Refrigeration System
Evaporator (4-1): Q41 = (1 4) ; Compressor (1-2): 12 = (1 2), S1 = S2
Condenser (2-3): Q23 = (3 2) ; Throttle Valve (3-4) 3 = 4 ,
COP = Q41
12 =
(14)
12 ; Refrigeration effect RE =
Q41
= 1 4
Isentropic Turbine efficiency n =
=
21
21 ; S2 = S1 ,actual S2 > S1
-
Rankine Cylcle
Feed water pump (1-2): 12 = (1 2) = (1 2) , = 0.001/
Boiler and super-heater (2-3): 23 = (3 2) , heat input to steam boiler
Steam Turbine (3-4): 34 = (4 3) , S1 = S2 , high pressure expands to low pressure
Condenser (4-1): 41 = (1 4) , heat rejected , constant low pressure
Performance: n =
=
34 +12
23=
43+12
32 , SSC =
3600
=
3600
34 +12
if (3-4) high pressure steam turbine, -> re-heater (4-5): 45 = (5 4) ->
Low pressure steam turbine (5-6): 56 = (5 6) , S5 = S6 ->
Condenser (6-1): 61 = (1 6)
Performance: n =
=
34 +56 +12
23 +45 , SSC =
3600
=
3600
34 +56 +12
Isentropic Turbine efficiency n =
=
34
34
Air Standard Diesel Cycle
Compression Process (1-2): compreesion ratio, r =1
2 , u = C, Q12 = 0, T2 = T1
1
2
1
P2 = P11
2
, work input W12 = (2 1) = (2 1)
Heat Supply Process (2-3): P2 = P3 , T3 = T23
2 , cut off ratio , r =
3
2 , W23 = 2(3 2)
Heat supply Q23 = (3 2) = (3 2)
Expansion Process (3-4): 4
3=
1
2
2
3=
r
r , Q34 = 0 , work output W34 = (4 3) = (4 3)
T4 = T33
4
1 , P4 = P3
3
4
,
Heat Rejection Process (4-1): W41 = 0 , V1 = V4 , T1 = T41
4 , Q41 = (1 4) = (1 4)
Thermal efficiency n =
, mean eff. pressure =
=
0.9167v=0.9167RT/P
Air Standard Otto Cycle-gasoline
Compression Process (1-2): same to Diesel ; Heat Supply Process (2-3): V2 = V3 , no work done,
Heat supply Q23 = (3 2) = (3 2), T3 = T23
2 ;
Expansion Process (3-4): r =1
2=
4
3 , Q34 = 0 , work output W34 = (4 3) = (4 3)
T4 = T33
4
1 , P4 = P3
3
4
; Heat Rejection Process (4-1): V1 = V4 , T1 = T4
1
4 , no work done
Q41 = (1 4) = (1 4) ; Thermal efficiency n =
= 1 +
41
23
Mean eff Pressure: same to diesel, =23+41
12 , V =
, R=0.287kJ/kgK
-
Heat Transfer
conduction
for plane slab: =
(1 2) ; for sphere: =
41
1
1
2
(1 2) ; for cylinder: =2
21
(1 2)
In series, all plus ; for parallel, 1
1
+
1
Newtons Law of cooling for convection: = ( ) ; R =1
for same system: Q=T/R=T/R== all T/R
Radiation
Real object: + + = 1; grey body + = 1 , no ; black body = 1 , no other
Stefan Boltzmann Law:
Black: Q = AF(14 2
4) ; grey : Q = AF(114 22
4)
emitted by 1 and absorbed by 2: Q = 12AF114
Reflected: Q = 1AF1142 ; Infinite: Q =
11
1+
1
21
A(14 2
4)
Convection
1. Find formula by using the description, cal the h valve
Combustion
Write down the formula, balance the equation to fulfill 79/21=N2/O2, write down molar mass x molar number,
say the fuel 1kg, cal other component, find Cp valve, (room temp+low temp)/2 + 273K ; (room temp+high
temp)/2 + 273K; find table, =T ave/ 1000, cal each Cp, use equation.