Concept explainers
(a)
Find the outdoor-indoor temperature difference in degrees Fahrenheit.
(a)

Answer to Problem 20P
The outdoor-indoor temperature difference is
Explanation of Solution
Given data:
The inside room temperature
The outside air temperature
Calculation:
The difference between the outdoor-indoor temperatures is,
Substitute
Thus, the outdoor-indoor temperature difference is
Conclusion:
Hence, the outdoor-indoor temperature difference is
(b)
Find the outdoor-indoor temperature difference in degrees Rankine.
(b)

Answer to Problem 20P
The outdoor-indoor temperature difference is
Explanation of Solution
Given data:
The inside room temperature
The outside air temperature
Formula used:
Formula to calculate the outdoor temperature in degrees Rankine is,
Here,
Formula to calculate the indoor temperature in degrees Rankine is,
Here,
Calculation:
Substitute
The outside temperature is
Substitute
The inside temperature is
Substitute
Thus, the outdoor-indoor temperature difference is
Conclusion:
Hence, the outdoor-indoor temperature difference is
(c)
Find the outdoor-indoor temperature difference in degrees Celsius.
(c)

Answer to Problem 20P
The outdoor-indoor temperature difference is
Explanation of Solution
Given data:
The inside room temperature
The outside air temperature
Formula used:
Formula to calculate the outdoor temperature in degrees Celsius is,
Here,
Formula to calculate the indoor temperature in degrees Celsius is,
Here,
Calculation:
Substitute
The outside temperature is
Substitute
The inside temperature is
Substitute
Thus, the outdoor-indoor temperature difference is
Conclusion:
Hence, the outdoor-indoor temperature difference is
(d)
Find the outdoor-indoor temperature difference in Kelvin.
(d)

Answer to Problem 20P
The outdoor-indoor temperature difference is
Explanation of Solution
Given data:
The inside room temperature
The outside air temperature
Formula used:
Formula to calculate the outdoor temperature in Kelvin is,
Here,
Formula to calculate the indoor temperature in Kelvin is,
Here,
Calculation:
Refer from part (c),
The inside room temperature
The outside air temperature
Substitute
The outside temperature is
Substitute
The inside temperature is
Substitute
Thus, the outdoor-indoor temperature difference is
Conclusion:
Hence, the outdoor-indoor temperature difference is
(e)
Check whether one degree temperature difference in Celsius equal to one temperature difference in kelvin, and one degree temperature in Fahrenheit equal to one degree temperature difference in Rankine.
(e)

Explanation of Solution
Case 1:
Refer to part (c),
The outdoor-indoor temperature difference is
Refer to part (d),
The outdoor-indoor temperature difference is
From part (c) and part (d), outdoor-indoor temperature differences are equal.
Note:
Let the temperature difference in Kelvin is,
Formula to calculate the temperature in degree Celsius is,
Here,
Substitute
Thus, one degree temperature difference in Celsius is equal to one temperature difference in kelvin.
Case 2:
Refer to part (a),
The outdoor-indoor temperature difference is
Refer to part (b),
The outdoor-indoor temperature difference is
From part (a) and part (b), outdoor-indoor temperature differences are equal.
Note:
Let the temperature difference in Rankine is,
Formula to calculate the temperature in Rankine is,
Here,
Substitute
Therefore, one degree temperature in Fahrenheit is equal to one degree temperature difference in Rankine.
Conclusion:
Hence, one degree temperature difference in Celsius equal to one temperature difference in kelvin, and one degree temperature in Fahrenheit equal to one degree temperature difference in Rankine has been explained.
Want to see more full solutions like this?
Chapter 6 Solutions
Engineering Fundamentals: An Introduction to Engineering
- 6.85 A reducing pipe bend is held in place by a pedestal as shown. There are expansion joints at sections 1 and 2, so no force is transmitted through the pipe past these sections. The pressure at section 1 is 20 psig, and the rate of flow of water (p = 62.4 Ibm/ft³) is 2 cfs. Find the force and moment that must be applied at section 3 to hold the bend stationary. Assume the flow is irrotational, and neglect the influence of gravity. 6 in. diameter + 24 in. 24 in. (3 4 in. diameter Problem 6.85arrow_forward6.79 A cart is moving along a railroad track at a constant velocity of 5 m/s as shown. Water (p = 1000 kg/m³) issues from a nozzle at 10 m/s and is deflected through 180° by a vane on the cart. The cross-sectional area of the nozzle is 0.002 m². Calculate the resistive force on the cart. 5 m/s D 10 m/s Nozzle Problem 6.79arrow_forwardOil of specific gravity 0.800 acts on a vertical triangular area whose apex is in the oil surface. The triangle is 9 ft high and 12 ft wide. A vertical rectangular area 8 ft high is attached to the 12-ft base of the triangle and is acted upon by water. Find the magnitude and position of the resultant force on the entire area.arrow_forward
- What types of constraints did Covid-19 cause in duration of activities on different construction projects?arrow_forwardPost-tensioned AASHTO Type III girders are to be used to support a deck with unsupported span equal to 90 feet. One level of Grade 270, 10 x 0.6" Ø 7-wire strand are used to tension the girders. The girder is simply supported at both ends. The anchors are located 2" from the neutral axis at the supports while the eccentricity is measured at 14" at the midspan. Use maximum values for ranges (table values). The tendons are encased in flexible metal sheathing. Assume and loadings are placed immediately after stressing. Determine the stresses at the top and bottom of the beam, including the stress at the level of steel after three years. Given the following parameters: F'c = 5000 psi Fy = 240 ksi (Bonded, Stress-relieved) Fu = 270 ksi = Es 29,000 ksi Ec = 4,030 psi Fj = 235 ksi A = 1x0.6" 07-W.S. = 0.217 in² Yc = 140 lbs/ft³ ΔΑ = 3/8" RH = 33% Superimposed Service Load = 10 kips/ft (excluding self-weight)arrow_forwardFIG. P5.66 40 kN B 4 m A 20 kN 10 kN/m C 8 m 8 marrow_forward
- 4 m 100 kN 12 kN/m 5 m -1.5 m-1.5 marrow_forwardQ1) Choose the correct answer: 1- Rankine's theory of earth pressure neglect: The pressure on the wall is to act parallel to the surface of the soil at angle (B) to the horizontal. The effect of cohesion on active and passive pressure. The friction developed between a retaining wall and the soil. o The depth of tension zone. o The effect of water table. 2- The lateral earth pressure is the lateral pressure that developed from: o Water beside it. o Soil beside structure. o External loads. o All the above. o Excavation beside structure. 3- The coefficient of earth pressure at rest represented as o Ko-1-sing expression given by Jaky. o Ko-sing 。 Ko-1-cosp о o Ko-cosp 。 Ko" tano 4- If the backfill carries a uniform surcharge q, then the lateral pressure at the depth of wall His: o pa-Ka yz+Ka q o pa Ka yz-Ka q o pa Ka yz*Ka q o pa Ka yz/Ka q 。 pa=Ka yz+q 5- During the active state of plastic equilibrium, the retaining wall moves: o Towards the fill o Away from the fill o Does not change…arrow_forwardس 1: سم ، خمن كمية الأعمال الترابية اللازمة لتنفيذ جزء من القناة وفقًا للبيانات التالية: عرض القناة = 3م المنحدر الجانبي في الحفر = 1:1 وفي الدفن = 1:1.5 عرض الضفة = 2 م من الجانبين عمق الاملاء الكامل ) full supply depth)= 80 السافة العمودية بين اعلى مستوى للمياه واعلى نقطة في الضفة الجانبية ( free board) = 40 سم، ميل القاعدة = 1:5000 مستوى قاعدة القناة المقترح R.D. (m) 1000 2000 3000 4000 G.L. (m) 210.8 210.4 208.8 208.4 FSD 210= (P.B.L) BANK OF CHANNELarrow_forward
- A 8.5 meter simply supported reinforced concrete beam on parallel rectangular column, where the column width is 300 mm is shown in Figure carries a uniformly distributed variable action of 10 kN/m and permanent action of 9 kN/m (including self-weight of beam). The clear cover of beam for class XC-1 is 25 mm, diameter of main and shear reinforcement are 20 mm and 8 mm respectively. The characteristic material strengths are fck = 30 N/mm² and fyk = 500 N/mm². Apply variable Strut Inclination Method to design the shear reinforcement (links) for the beam. h = 370 mm 마 *b = 230 mm Section A-Aarrow_forwardQ2- For the retaining wall shown in the figure, calculate the magnitude and resultant, of active thrust and calculate only the magnitude of passive thrust. q = 10kN/m² 不 2 m y-18 kN/m³ 0=35° C=0 * 2 m y=18 kN/m³ 0=20° 6 m C=15 kN/m² y=18 kN/m³ 0-20° C=15 kN/m² 2arrow_forwardأ. ما هي شروط حديد التسليح المستخدم في موقع العمل بشكل عام وكيف يجب ان يكون ؟ ب. كم حلقة تحتاج لاساس شريطي طوله 40 مترا اذا كان مكتوب في المخطط ( mm 300 @ 010) ؟ B - لمقطع الأساس الشريطي ادناه، احسب كمية الحديد الطولي السفلي والمواد الانشائية المطلوبة ) سمنت ، رمل ، حصى ( لكل 30 متر طول منه ونسب الخلط 1:3:5 اذا كان حديد التسليح المتوفر طول 9 متر . (15) درجة ) 3014 $10@200mm 4016 800mm A 400mm جدول اوزان حديد اتسليح قطر الشيش (mm) كتلة الشيش (kg/m) 0.222 6 0.395 8 0.617 10 0.888 12 1.21 14 1.58 16 2 18 2.47 20 3.86 25arrow_forward
- Engineering Fundamentals: An Introduction to Engi...Civil EngineeringISBN:9781305084766Author:Saeed MoaveniPublisher:Cengage LearningConstruction Materials, Methods and Techniques (M...Civil EngineeringISBN:9781305086272Author:William P. Spence, Eva KultermannPublisher:Cengage Learning
- Solid Waste EngineeringCivil EngineeringISBN:9781305635203Author:Worrell, William A.Publisher:Cengage Learning,Materials Science And Engineering PropertiesCivil EngineeringISBN:9781111988609Author:Charles GilmorePublisher:Cengage LearningResidential Construction Academy: House Wiring (M...Civil EngineeringISBN:9781285852225Author:Gregory W FletcherPublisher:Cengage Learning





