EML 3015 Thermal-fluids I

Fall 1999

Homework Assignments

Homework #1 due Friday, September 10

3-18 (Q=78624 kJ), 3-19 (T=105.27 C) , 3-21 (Q=8.2 W), 

3-24 (T=33.4 C) & supplement (T=32.6 C), 

3-26 (T=285.2 K), 

3-115 (DT=32.7 C) & supplement

 

3-24 (supplement)

If the heat loss by radiation in 3-24 can not be neglected and the surface emissivity is 0.8, what will be the surface temperature?  Note 1: you can assume that equation 3-10 is valid in describing the radiation heat loss.  Note 2: If model correctly, you will end up with a fourth order algebraic equation.  You should try to solve it using trials and errors method.  Do not use a programmable calculator except for checking the answer.  If you choose your initial guess correctly, the solution can be obtained in a few iterations.

From your answer, do you think that neglecting the radiation heat loss is a good assumption or not?

 

3-115(supplement)

According to problem 1-68 (page 35), the equivalent wind chill temperature in degree C can be written as

Tequiv=33 - (33-Tambient) [ 0.475 - 0.0126V + 0.24(V)1/2 ], where V is the wind velocity in km/hour and Tambient is the ambient temperature in degree C.  Using this formula to produce a wind chill temperature table if the wind speed is varied from 10 km/hr to 100 km/hr (with an increment of 10 km/hr) and the ambient temperature is 10 degree C.

Homework #2, due Friday, September 17

Homework Link

1 1990 kg/m3
2 235.2 kN
3 (a) 500 kg/m3, (b) 625 kg/m3
4 (a) 6.1 m/s, (b) 0.976 cm/s (c) 121.25 sec.
5 3.9 m
6 1 m

Homework #3, due Monday, September 27

Homework 3

1 0.0748(m3/s)
2 14084.9 hp
3 TO=26.3 C
4 (a) weight 95060 N

(b) Dh=0.036 m
5 (a) 66201 (Pa)

(b) 0.0626 (m/s)

Homework #4, due Friday, Oct. 8 (From textbook)

2-30, 2-31, 2-33 (total mass 70.91 kg, quality 0.0639), 2-41, 2-43, 2-52

Homework #5 due Friday, Oct. 15 (From textbook)

3-51, 3-71 (Dt=371 sec.), 3-85(W=285 kJ, Q=2418 kJ), 3-111(W=-57.2 kJ, Q=-11.2 kJ),

3-120(Tf=57.2 C), 3-122(E=53619 kJ)

Homework #6, due Friday, Oct. 22 (From textbook)

4-28, 4-38, 4-44, 4-49 (25.3 kg/min), 4-51 (313 kg/s), 4-54

Homework #7, due Monday, Nov. 1 (From textbook)

5-52, 5-74, 5-91, 5-98, 5-99, 5-102 (0.959 L/s)

Homework #8, due Monday, Nov. 8 (From textbook)

6-40 (T=27ºC, sgen=0.218 kJ/K), 6-58,

6-72((a)-49.6 kW, (b)-38.3 kW, (c)-32.2 kW), 6-85,

6-87((a)Tf=57.2º C, (b) sgen=0.034 kJ/K), 6-92

Homework #9, due Monday, Nov. 15 (From textbook)

7-86, 7-96, 7-98,

7-104 [(a) W=1366 kJ/kg, (b) h=38.9%]

7-106 [(a) P=0.677 MPa, (b) h=38.94%, (c) dm/dt=231 kg/s cooling water],

7-156 [(a) T=1092.4°C, (b) h=48.6%, (c) dm/dt=220.5 kg/s]

Homework #10, due Monday, Nov. 22 (textbook)

7-33, 7-34, 7-60, 7-62 ((a) 43.1%, (b) 39.1%), 7-80 (F=11,000 N), 7-117

Homework #11, due Monday, Dec. 6 (textbook)

8-16 ((a) 343 kJ, (b) 286 kW, (c) 4047 W/m2, (d) 40.5 W/m2 C)

8-39 (k=0.112 W/m2 C, cost $78.84),

8-59, 8-63 ((a) Q=0.0048 kWh, (b) 1959 W/m2, (c) 193º C)

8-71, 8-73 (L=0.45 cm)

Homework #12, due Monday, Dec. 13 (textbook)

8-80 (q=450 W)

8-88 (q=88 kW)

8-94 ((a) q=20040 W, (b) mice melted=5189 kg)

8-103, 8-112(T=39.4° C)

8-123