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
4-105. Replace the force system acting on the beam by an equivalent resultant force and couple
moment at point B.
A
30 in.
4 in.
12 in.
16 in.
B
30%
3 in.
10 in.
250 lb
260 lb
13
5
12
300 lb
Sketch and Describe a hatch coaming and show how the hatch coamings are framed in to ships strucure?
Sketch and describe hatch coamings. Describe structrual requirements to deck plating to compensate discontinuity for corners of a hatch. Show what is done to the deck plating when the decks are cut away and include the supporting members.
Chapter 3 Solutions
Heat and Mass Transfer: Fundamentals and Applications