the R and U for air space between studs do not have any reflective surface. U = 1.203 W m 2 C 0 R = 0.831 m 2 C 0 W Given information: dimension of wood studs = 38mm × 140mm center distance = 400mm dimensions of cavity = 140mm thickness of gypsum wallboard = 13mm thickness of rigid foam insulation = 25mm dimension of wood lapping siding = 13mm × 200mm the insulation cavity constitutes 80%heat transmit area and while studs,headers,plates and still constitute 20% . Formula used: The expression fortotal thermal resistance for the entire wall is expressed as follows: R o v e r a l l = 1 U o v e r a l l w h e r e U o v e r a l l = ( U f a r e a ) i n s u l a t i o n s + ( U f a r e a ) s t u d R o v e r a l l = thermal resistance U o v e r a l l = heat transfer coefficient f a r e a = area fraction ε e f f e c t i v e = 1 1 ε 1 + 1 ε 2 − 1 ε 1 = effective emissivity of surface 1 ε 2 = effective emissivityof surface 2 ε c o l l e c t i v e = effective emissivityvof surface combination Calculation: The total average thermal resistance for wall is calculated as follows: R o v e r a l l = 1 U o v e r a l l w h e r e U o v e r a l l = ( U f a r e a ) i n s u l a t i o n s + ( U f a r e a ) s t u d And the value of the area fraction factor is 0.82 for air space and 0.18 for stud section. ε e f f e c t i v e = 1 1 0.9 + 1 0.9 − 1 = 0.82 Construction R-value m 2 C 0 W Between studs At studs 1. Still air above ceiling 0.12 0.044 2. Linoleum 0.009 0.14 3. Felt 0.011 0.23 4.Plywood 0.11 ---- 5. Wood subfloor 0.166 ---- 6a. Air space, 90 mm, nonreflective 0.16 ---- 6b. Stud of wood, 38 mm by 90 mm ----- 0.63 7. Wallboard, 13 mm 0.079 0.079 8. Still air near ceiling 0.12 0.12 Total thermal resistance of each part, R m 2 C 0 W 0.775 1.243 The U-factor of each part, 1.290 0.805 Area fraction of each part, 0.82 0.18 Total U-factor 1.203 W m 2 C 0 Total thermal resistance, 0.831 m 2 C 0 W
the R and U for air space between studs do not have any reflective surface. U = 1.203 W m 2 C 0 R = 0.831 m 2 C 0 W Given information: dimension of wood studs = 38mm × 140mm center distance = 400mm dimensions of cavity = 140mm thickness of gypsum wallboard = 13mm thickness of rigid foam insulation = 25mm dimension of wood lapping siding = 13mm × 200mm the insulation cavity constitutes 80%heat transmit area and while studs,headers,plates and still constitute 20% . Formula used: The expression fortotal thermal resistance for the entire wall is expressed as follows: R o v e r a l l = 1 U o v e r a l l w h e r e U o v e r a l l = ( U f a r e a ) i n s u l a t i o n s + ( U f a r e a ) s t u d R o v e r a l l = thermal resistance U o v e r a l l = heat transfer coefficient f a r e a = area fraction ε e f f e c t i v e = 1 1 ε 1 + 1 ε 2 − 1 ε 1 = effective emissivity of surface 1 ε 2 = effective emissivityof surface 2 ε c o l l e c t i v e = effective emissivityvof surface combination Calculation: The total average thermal resistance for wall is calculated as follows: R o v e r a l l = 1 U o v e r a l l w h e r e U o v e r a l l = ( U f a r e a ) i n s u l a t i o n s + ( U f a r e a ) s t u d And the value of the area fraction factor is 0.82 for air space and 0.18 for stud section. ε e f f e c t i v e = 1 1 0.9 + 1 0.9 − 1 = 0.82 Construction R-value m 2 C 0 W Between studs At studs 1. Still air above ceiling 0.12 0.044 2. Linoleum 0.009 0.14 3. Felt 0.011 0.23 4.Plywood 0.11 ---- 5. Wood subfloor 0.166 ---- 6a. Air space, 90 mm, nonreflective 0.16 ---- 6b. Stud of wood, 38 mm by 90 mm ----- 0.63 7. Wallboard, 13 mm 0.079 0.079 8. Still air near ceiling 0.12 0.12 Total thermal resistance of each part, R m 2 C 0 W 0.775 1.243 The U-factor of each part, 1.290 0.805 Area fraction of each part, 0.82 0.18 Total U-factor 1.203 W m 2 C 0 Total thermal resistance, 0.831 m 2 C 0 W
To find:the R and U for air space between studs do not have any reflective surface.
U=1.203Wm2C0R=0.831m2C0W
Given information:
dimension of wood studs = 38mm×140mmcenter distance = 400mmdimensions of cavity = 140mmthickness of gypsum wallboard = 13mmthickness of rigid foam insulation = 25mmdimension of wood lapping siding = 13mm×200mmthe insulation cavity constitutes 80%heat transmit area and while studs,headers,plates and still constitute 20%.
Formula used:
The expression fortotal thermal resistance for the entire wall is expressed as follows:
Roverall=1UoverallwhereUoverall=(Ufarea)insulations+(Ufarea)studRoverall=thermal resistanceUoverall=heat transfer coefficientfarea=area fraction
1. A 40 lb. force is applied at point E. There are pins at
A, B, C, D, and F and a roller at A.
a. Draw a FBD of member EFC showing all the known and
unknown forces acting on it.
b. Draw a FBD of member ABF showing all the known and
unknown forces acting on it.
c. Draw a FBD of member BCD showing all the known and
unknown forces acting on it.
d. Draw a FBD of the entire assembly ADE showing all the
known and unknown forces acting on it.
e. Determine the reactions at A and D.
f. Determine the magnitude of the pin reaction at C.
40 lbs.
B
A
6 in.
4 in.
D
F
-5 in.4 in 4.
A crude oil of specific gravity0.85 flows upward at a volumetric rate of flow of 70litres per
second through
a vertical
venturimeter,with an inlet diameter of 250 mm and a throat
diameter of 150mm. The coefficient
of discharge of venturimeter is 0.96. The vertical
differences betwecen the pressure toppings is
350mm.
i)
Draw a well labeled diagram to represent the above in formation
i)
If the two pressure gauges are connected at the tapings such that they are
positioned at the levels of their corresponding tapping points,
determine the
difference of readings in N/CM² of the two pressure gauges
ii)
If a mercury differential
manometer
is connected in place of pressure gauges,
to the tappings such that the connecting tube up to mercury are filled with oil
determine the difference in the level of mercury column.
Can you solve it analytically using laplace transforms and with Matlab code as well please. Thank You