Microsoft Word - Handouts PPE Day 8.docx

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This file is only for viewing and printing. Contents are not allowed to be edited. Page 2 3. Knowing the bearing capacity of the soil, solve for the base width “b”. For machine foundation use only 1/2 of the given safe soil bearing capacity. The safe bearing capacity is computed using a factor of safety of 5. where: Sb = safe soil bearing capacity Note: If “b” will come out less that “a”, then make b = a, that is, the foundation has a rectangular cross-section. 4. Using a density of 2406 kg/m3 for concrete, determine the volume of the foundation. 5. Compute the depth of the foundation “h”: 6. Finalize the design; make adjustments in the dimensions if necessary, provided the required volume is maintained and without reducing the required base area. 7. Use Class A (1:2:4) mixture, that is, 1-part cement, 2 parts sand and 4 parts stone. 8. Determine the quantity of cement, sand and stone using the following data: To produce 1 cu yd of concrete using 1:2:4 mixture, the following are needed: 6 sacks cement, 0.44 cu. yd sand and 0.88 cu yd stone. 9. Weight of steel bar reinforcements needed should be about 1/2% to 1% of the weight of the foundation. 10. Anchor bolts should be imbedded in the concrete at least 30 times the bolt diameter. 11. Hardening time of foundation after pouring of concrete is at least 7 days or 1 week. II. HEAT TRANSFER Formulas: 1. Conduction Through Plane Walls: where: Q = heat transmitted, W A = heat transfer area, m2 ta = surface temperature on hot side tb = surface temperature on cold side x = thickness of wall k = thermal conductivity, W/m-oC 2. Conduction Through Composite Plane Wall 3 F F m , 2406 W V = L h ) 2 b a ( VF + = 1 b a 1 x ) t t( A k Q - = 2 c b 2 x ) t t( A k Q - = 2 2 1 1 c a k / x k / x ) t t( A Q + - = bL W W 2 S F M b + = x ) t t( A k Q b a - =.

This file is only for viewing and printing. Contents are not allowed to be edited. Page 3 where: k1 = thermal conductivity of 1st layer tb = interface temp. k2 = thermal conductivity of 2nd layer A = heat transfer area common to both wall 3. Conduction From Fluid To Fluid Q = h1 A (t1 - ta) Q = h2 A (tC - t2) where: h1 = surface film conductance on the hot side, W / m2-oC. h2 = surface film conductance on cold side, W/m2-°C tb and tc = are interface temperatures Q = U A Dt U = 1/R where: U = overall conductance or overall coefficient of heat transfer, W/m2-oC Ao = area, m2 Dt = temperature difference, °C 4. Conduction Through Pipe where: L = pipe length 5. Conduction Through Composite Pipe. 6. Conduction Fluid to Fluid Through Pipe = hi Ai (t1 - ta) Q = ho Ao (tc - t2) where: hi = surface conductance on inside surface ho = surface conductance on outside surface Ai = inside surface area = 2 p r1 L Ao = outside surface area = 2p r3 L Q = Ui Ai Dtmean Q = Uo Ao Dtmean 2 3 3 2 2 1 1 1 2 1 h 1 k x k x k x h 1 ) t t( A Q + + + + - = ÷÷ ø ö çç è æ - p = ÷÷ ø ö çç è æ - p = 1 2 b a 1 2 b a D D ln ) t t( L k 2 r r ln ) t t( L k 2 Q 2 3 c b 2 1 2 b a 1 r r ln ) t t( L k 2 r r ln ) t t( L k 2 Q - p = - p = 2 2 3 1 1 2 c a k ) r/ r ln( k ) r/ r ln( ) t t( L 2 + - p = o o 2 2 3 1 1 2 i i 2 1 h A 1 L k 2 ) r/ r ln( L k 2 ) r/ r ln( h A 1 ) t t( Q + p + p + - = k x h 1 R S + S = 2 3 3 2 2 1 1 1 h 1 k x k x k x h 1 1 U + + + + =.

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