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ValentinkaMS [17]
3 years ago
8

Natural Convection from an Oven Wall. The oven wall in Example 4.7-1 is insulated so that the surface temperature is 366.5 K ins

tead of 505.4 K. Calculate the natural convection heat-transfer coefficient and the heat-transfer rate per m of width. Use both Eq. (4.7-4) and the simplified equation. (Note: Radiation is being neglected in this calculation.) Use both SI and English units.

Engineering
1 answer:
lukranit [14]3 years ago
8 0

Answer:

i) Heat transfer coefficient (h) = 7 w/m²k

ii) Heat transfer per meter width of wall

      = h x L x 1 x (Ts - T₆₀)

        = 7 x 0.3048 x (505.4 - 322) = 414.747 w/m

Explanation:

see attached image

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The steam requirements of a manufacturing facility are being met by a boiler whose rated heat input is 5.5 x 106 Btu/h. The comb
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A welding rod with κ = 30 (Btu/hr)/(ft ⋅ °F) is 20 cm long and has a diameter of 4 mm. The two ends of the rod are held at 500 °
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Answer:

In Btu:

Q=0.001390 Btu.

In Joule:

Q=1.467 J

Part B:

Temperature at midpoint=274.866 C

Explanation:

Thermal Conductivity=k=30  (Btu/hr)/(ft ⋅ °F)= \frac{30}{3600} (Btu/s)/(ft.F)=8.33*10^{-3}  (Btu/s)/(ft.F)

Thermal Conductivity is SI units:

k=30(Btu/hr)/(ft.F) * \frac{1055.06}{3600*0.3048*0.556} \\k=51.88 W/m.K

Length=20 cm=0.2 m= (20*0.0328) ft=0.656 ft

Radius=4/2=2 mm =0.002 m=(0.002*3.28)ft=0.00656 ft

T_1=500 C=932 F

T_2=50 C= 122 F

Part A:

In Joules (J)

A=\pi *r^2\\A=\pi *(0.002)^2\\A=0.00001256 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{51.88*0.000012566*(500-50}{0.2}\\ Q=1.467 J

In Btu:

A=\pi *r^2\\A=\pi *(0.00656)^2\\A=0.00013519 m^2

Heat Q is:

Q=\frac{k*A*(T_1-T_2)}{L} \\Q=\frac{8.33*10^{-3}*0.00013519*(932-122}{0.656}\\ Q=0.001390 Btu

PArt B:

At midpoint Length=L/2=0.1 m

Q=\frac{k*A*(T_1-T_2)}{L}

On rearranging:

T_2=T_1-\frac{Q*L}{KA}

T_2=500-\frac{1.467*0.1}{51.88*0.00001256} \\T_2=274.866\ C

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During an expansion process, the pressure of a gas changes from 15 to 140 psia according to the relation P = aV + b, where a = 5
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The work done during the process is 359 btu

<u></u>

<u>Explanation:</u>

Given-

P1 = 15psia

P2 = 140 psia

V1 = 7ft³

a = 5 psia/ft³

b = C

P = aV +b

Work done, W = ?

P1 = aV1 + b

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b = -20 psia

P2 = aV2 + b

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The work done by the process is the area under the curve which is trapezoidal.

Therefore,

Work done, W = area of trapezoid

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= ( 140 + 15 / 2 ) ( 32 - 7)

= 1937.5 psia ft³

= 1937.5/ 5.4039 = 359 btu

Therefore, the work done during the process is 359 btu

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