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February 7

By Barbara Hill,2014-02-08 02:07
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February 7Februa

    College of Engineering and Computer Science

     Mechanical Engineering Department

    Mechanical Engineering 375

    Heat Transfer Spring 2007 Number 17629 Instructor: Larry Caretto

    February 7 Homework Solutions

    1.55 The inner and outer surfaces of a 4-m by 7-m brick wall

    of thickness 30 cm and thermal conductivity 0.69 W/m K

    are maintained at temperatures of 20?C and 5?C,

    respectively. Determine the rate of heat transfer

    through the wall, in W. (Figure P1.55 from Çengel, Heat

    and Mass Transfer)

    Here we use the basic equation for one-dimensional

    conduction heat flow. We take the temperature difference

    as T T to get the heat flow in the direction shown in leftright

    the figure.

    kA(TT)leftright;Q L

    oooFrom the given data we see that L = 0.3 m, T T = 210C 5C = 15C = 15 K, k = 0.69 leftright2W/m?K, and A = (7 m)(4 m) = 28 m. Substituting these values into the heat flow equation gives:.

    ()kATT0.6915WKleftright2;966 W 28;;Qm0.3Lm~Km

    1.59E. The north wall of an electrically heated home is 20 ft long, 10 ft high, and 1 ft thick, and is

    made of brick whose thermal conductivity is k = 0.42 Btu/h ft ?F. On a certain winter night, the temperatures of the inner and the outer surfaces of the wall are measured to be at about 62?F and 25?F, respectively, for a period of 8 h. Determine (a) the rate of heat loss

    through the wall that night and (b) the cost of that heat loss to the home owner if the cost of electricity is $0.07/kWh.

    Again we use the basic equation for one-dimensional conduction heat flow. We take the temperature difference as T T to get the heat flow leaving the house. leftright

    kATT()innerouter;Q L

    ooooFrom the given data we see that L =1 ft, T T = 62F 25F = 37F, k = 0.42 Btu/hr?ft?F, leftright2and A = (20 ft)(10 ft) = 200 ft. Substituting these values into the heat flow equation gives:.

    o()kATT0.4237BtuFleftright2;200;;3108 Btu/hr Qfto1Lfthr~ft~F

    Jacaranda (Engineering) 3333 Mail Code Phone: 818.677.6448

    E-mail: lcaretto@csun.edu 8348 Fax: 818.677.7062

February 7 homework solutions ME 375, L. S. Caretto, Spring 2007 Page 2

    The cost is found by multiplying the heat loss by the $0.07/kWh cost of providing the heat and the 8 hour t during which the heat loss occurs..

    3108Btu$0.07kWh;;;;;CostUnitcostQt8hr$0.51 kWhhr3412Btu

    1.64 Consider a person standing in a room maintained at 20?C at all times. The inner surfaces of the walls, floors, and ceiling of the house are observed to be at an average temperature of 12?C in winter and 23?C in summer. Determine the rates of radiation heat transfer between this person and the surrounding surfaces in both summer and winter if the exposed surface area, emissivity, and the average outer surface temperature of the person 2are 1.6 m, 0.95, and 32?C, respectively.

    To compute this radiation heat transfer we assume that the person in the room is a small object in a large enclosure. For this situation we can use equation 1.28 for the radiation heat transfer.

    44;;;QATT personsurfaces

    ooWe are given the following data: T = 32C = 305.15 K, T = 12C = 285.15 K in the personsurfaceso2winter and 23C = 296.15 K in the summer, A = 1.6 m, and = 0.95. The Stefan-Boltzmann -824constant, = 5.670x10 W/m?K. Substituting these values into the radiation equation gives the following heat transfer values.

    85.670x10W44442;;;;;,(;;;;QATT0.951.6m305.15K296.15Ksummerpersonsurfaces24m~K

    85.670x10W44442; ,(;;;;;;;;QATT0.951.6m305.15K285.15Kwerpersonsurfacesint24m~K

    ;; Q84.3WQ177.3Wsummerwinter

    0Note that the room temperature value of 20C is irrelevant in this problem where we are only

    computing the radiation heat transfer. In a real situation we would also compute the convective oheat transfer from the person’s body to the room air at a temperature of 20C.

    1.66 For heat transfer purposes, a standing man can be modeled as a 30-cm-diameter, 170-cm-long vertical cylinder with both the top and bottom surfaces insulated and with the side surface at an average temperature of 34?C. For a convection heat transfer coefficient of 20 2W/m ?C, determine the rate of heat loss from this man by convection in an environment at 18?C.

    Here we use the basic equation for convective heat transfer. We take the temperature difference as T T to get the heat flow from the person to the environment. person

    ;QhA(TT) person

    2ooooFrom the given data we have h = 20 W/m?C, and T T = 34C 18C = 16C. Since we personare given that the ends of the cylinder are insulated the area available for heat transfer is the ares

    February 7 homework solutions ME 375, L. S. Caretto, Spring 2007 Page 3

    2of the side of the cylinder: A = DL = (0.3 m)(1.7 m) = 1.602 m. Substituting these values into

    the heat transfer equation gives the desired solution.

    20W2o;;;;;QhA(TT)1.602m16C513 W person2om~C

    1.78 A transistor with a height of 0.4 cm and a diameter of 0.6

    cm is mounted on a circuit board. The transistor is

    cooled by air flowing over it with an average heat 2transfer coefficient of 30 W/m ?C. If the air temperature

    is 55?C and the transistor case temperature is not to

    exceed 70?C, determine the amount of power this

    transistor can dissipate safely. Disregard any heat

    transfer from the transistor base. (Figure at right taken

    from Çengel, Heat and Mass Transfer.)

    The basic idea here is that the power dissipated by the

    transistor is the heat transfer that has to be convected to the oair at 55C; so, we can use the basic equation for convective

    heat transfer. We take the temperature difference as

    T T to get the heat flow from the transistor to the transistorenvironment.

    ; QhA(TT)transistor

    2ooooFrom the given data we have h = 30 W/m?C, and T T = 70C 55C = 15C. Since we transistorare given that the base of the transistor is negligible, area available for heat transfer is the area of 2the side of top of the transistor that we will assume is a cylinder: A = DL + D/4 = (0.006 22m)(0.004 m) + (0.006 m) = 0.0001037 m. Substituting these values into the heat transfer equation gives the desired solution.

    30W2o;;;;;QhA(TT)0.0001037m15C0.0467 W transistor2om~C

    1.95 Consider a person standing in a room at 23?C Determine the total rate of heat transfer from this person if the exposed surface area and the skin temperature of the person are 1.7 2m and 32?C, respectively, and the convection heat transfer coefficient is 5 W/m2??C. Take the emissivity of the skin and the clothes to be 0.9, and assume the temperature of the inner surfaces of the room to be the same as the air temperature.

    This is a combination of the work we did in problems 1.64, where we considered only radiation, and 1.66, where we considered only convection. Here we do the work of both problems and add the results to get the total heat transfer.

    44;;;;;QQQhA(TT)ATT convradpersonpersonsurfaces

    22oooFrom the given data we have A = 1.7 m, = 0.9, h = 5 W/m?C, and T T = 32C 23C = persono9C, so that T = 305.15 K and T = T = 296.15 K. The Stefan-Boltzmann constant, = personsurfaces-8245.670x10 W/m?K. Substituting these values into the heat transfer equation gives the desired solution.

February 7 homework solutions ME 375, L. S. Caretto, Spring 2007 Page 4

    44;,(;:QAh(TT);;TTpersonpersonsurfaces

    8? 5W5.670x10W44o2,(;;;;;;;;1.7m9C0.9305.15K296.15K??o224m~Cm~K???

    ; = 161 W Q

    1.101 A 1000-W iron is left on the iron board with its base exposed to the air at 20?C. The convection heat transfer coefficient between the base surface and the surrounding air is 2235 W/m??C. If the base has an emissivity of 0.6 and a surface area of 0.02 m, determine

    the temperature of the base of the iron.

    This is another problem where we have both convection and radiation. We compute each mode of heat transfer then add the two the get the total heat transfer. Here we know the total heat transfer is 1000 W, the heat produced by the iron. What we do not know is the surface temperature of the iron. As usual we assume that the iron is a small item in a large enclosure to get the equation for radiation heat transfer.

    44;;;;;Q1000WQQhA(TT)ATT convradironironsurfaces

    22ooFrom the given data we have A = 0.02 m, = 0.6, h = 35 W/m?C, and T = 20C. We are not given the surfaces of the room for radiation, but we can assume that they are the same as the air o-8temperature so that T = 20C = 293.15 K. The Stefan-Boltzmann constant, = 5.670x10 surfaces24W/m?K. Substituting these values into the heat transfer equation gives the desired solution. Note that we have to use Kelvin temperatures for the convection term in this combined convection-radiation problem with an unknown temperature. If we did not do this we would have an inconsistent variable for the unknown temperature of the iron.

    44;,(;:Q1000WAh(TT);;TTironironsurfaces

    8? 35W5.670x10W??424,(;;;;;;0.02mT293.15K0.6T293.15K(?ironiron2o24??~~mCmK)?

    We cannot solve this equation directly, but we can use a root finder in our calculator or in a software application like Excel, Matlab or EES. Using the “Goal Seek” tool of Excel to solve this oequation gives the result that T = 947.04 K = 674C. iron

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