Problem P7.4: in the Movie Back to the Future , Doc Brown and the Young Marty Mcfly Need 1.21 GW of Power for Their Time Machine

Chapter 7: Thermal and Energy Systems

Problem P7.4: In the movie Back to the Future, Doc Brown and the young Marty McFly need 1.21 GW of power for their time machine, (a) Convert that power requirement to horsepower, (b) If a stock Delorean sports car produces 145 hp, how many times more power does the time machine need?

Approach: Convert the power requirement using the SI prefix definition 1 GW = 109 W and the conversion factor from Table 7.2 that 1 W = 1.341× 10–3 hp.

Solution: (a) Power in units of horsepower ⎛ hp ⎞ 1.21 GW =1.21×109 W = ()1.21×109 W ⎜1.341×10−3 ⎟ =1.623×106 hp ⎝ W ⎠ 1.623×106 hp (b) Number of cars 1.623×106 hp = 1.12×104 cars 145 hp car 11,200 cars

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Problem P7.6: A baseball catcher stops a 98-mph fastball over a distance of 0.1 m. What is the force necessary to stop the 0.14-kg baseball?

Approach: Convert the speed of the fastball to appropriate units using the factors in Table 3.6. Calculate the kinetic energy of the fastball using Equation (7.3). The work done on the fastball to stop it will be equivalent to its kinetic energy. Calculate the force required to stop the fastball using Equation (7.4).

Solution: Convert the speed of the fastball: ⎛ km ⎞⎛ m ⎞⎛ hr ⎞ m 98 mph⎜1.609 ⎟⎜1000 ⎟⎜ ⎟ = 43.8 ⎝ mi ⎠⎝ km ⎠⎝ 3600 s ⎠ s Calculate the kinetic energy of the fastball using Equation (7.3): 2 1 2 1 ⎛ m ⎞ Uk = mv = ()0.14 kg ⎜43.8 ⎟ =134.3 J 2 2 ⎝ s ⎠ Calculate the force required to stop the fastball using Equation (7.4): W 134.3 J F = = =1343 N Δ d 0.1 m 1343 N

Discussion: This is roughly equivalent to the amount of force exerted on a 300-lb person by gravity. If the mass of the ball were to increase or if the catcher was to stop the ball in a shorter distance, the required force would increase. The force also increases proportional to the square of the velocity.

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Problem P7.10: The heating value of agricultural residue biomass (e.g., crop residues, animal manure, and bedding, and organic material from food production), can range from 4300 to 7300 Btu/lbm. How much heat is released when 500 kg of biomass is burned?

Approach: Convert the mass of the biomass into lbm using Table (3.6). Calculate the range for the amount of heat released using Equation (7.4).

Solution: Convert the mass of the biomass: ⎛ lbm ⎞ 500 kg⎜2.2046 ⎟ =1102.3 lbm ⎝ kg ⎠ Calculate the range for the amount of heat released using Equation (7.4): ⎛ Btu ⎞ Lower bound: Q = mH = ()1102.3 lbm ⎜4300 ⎟ = 4.74×106 Btu ⎝ lbm ⎠ ⎛ Btu ⎞ Upper bound: Q = mH = ()1102.3 lbm ⎜7300 ⎟ = 8.05×106 Btu ⎝ lbm ⎠ 4.74×106 − 8.05×106 Btu

Discussion: While this is not as much heat as would be generated from an equivalent amount of wood, coal, or any of other gas and liquid fuels in Table (7.3), it represents a substantial amount of potential heat from an alternative energy source.

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