Fundamentals Of Engineering Thermodynamics
9th Edition
ISBN: 9781119391388
Author: MORAN, Michael J., SHAPIRO, Howard N., Boettner, Daisie D., Bailey, Margaret B.
Publisher: Wiley,
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Chapter 2, Problem 2.1E
To determine
The reason why the aerodynamic coefficient of drag of a formula one race is greater than an ordinary automobile.
Expert Solution & Answer
Answer to Problem 2.1E
The aerodynamic drag coefficient of a formula one race car is often substantially higher than that of a regular car.
Explanation of Solution
The drag coefficient is a non-dimensional quantity that is used to assess an item's drag or resistance in a wet environment such as ambient air. The object will have substantially less aerodynamic drag if it has a low drag coefficient. The coefficient of drag is always proportional to the surface area.
The drag depends on many factors such as density, viscosity, compressibility, the square of the velocity, body inclination, and size.
The relationship between body form, inclination, air viscosity, and compressibility is complicated in general. This is represented by the drag equation. The drag is as follows:
Where,
A is the reference area
V is the velocity of an object
From the above equation, it is observed that aerodynamic drag increases with the square of the speed. The speed of the formula one car is much greater than the speed of the ordinary automobile.
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Problem (17): water flowing in an open channel of a rectangular cross-section with width (b) transitions from a
mild slope to a steep slope (i.e., from subcritical to supercritical flow) with normal water depths of (y₁) and
(y2), respectively.
Given the values of y₁ [m], y₂ [m], and b [m], calculate the discharge in the channel (Q) in [Lit/s].
Givens:
y1 = 4.112 m
y2 =
0.387 m
b = 0.942 m
Answers:
( 1 ) 1880.186 lit/s
( 2 ) 4042.945 lit/s
( 3 ) 2553.11 lit/s
( 4 ) 3130.448 lit/s
Problem (14): A pump is being used to lift water from an underground
tank through a pipe of diameter (d) at discharge (Q). The total head
loss until the pump entrance can be calculated as (h₁ = K[V²/2g]), h
where (V) is the flow velocity in the pipe. The elevation difference
between the pump and tank surface is (h).
Given the values of h [cm], d [cm], and K [-], calculate the maximum
discharge Q [Lit/s] beyond which cavitation would take place at the
pump entrance. Assume Turbulent flow conditions.
Givens:
h = 120.31 cm
d = 14.455 cm
K = 8.976
Q
Answers:
(1) 94.917 lit/s
(2) 49.048 lit/s
( 3 ) 80.722 lit/s
68.588 lit/s
4
Chapter 2 Solutions
Fundamentals Of Engineering Thermodynamics
Ch. 2 - Prob. 2.1ECh. 2 - Prob. 2.2ECh. 2 - Prob. 2.3ECh. 2 - Prob. 2.4ECh. 2 - Prob. 2.5ECh. 2 - Prob. 2.6ECh. 2 - Prob. 2.7ECh. 2 - Prob. 2.8ECh. 2 - Prob. 2.9ECh. 2 - Prob. 2.10E
Ch. 2 - Prob. 2.11ECh. 2 - Prob. 2.12ECh. 2 - Prob. 2.13ECh. 2 - Prob. 2.14ECh. 2 - Prob. 2.15ECh. 2 - Prob. 2.16ECh. 2 - Prob. 2.17ECh. 2 - Prob. 2.1CUCh. 2 - Prob. 2.2CUCh. 2 - Prob. 2.3CUCh. 2 - Prob. 2.4CUCh. 2 - Prob. 2.5CUCh. 2 - Prob. 2.6CUCh. 2 - Prob. 2.7CUCh. 2 - Prob. 2.8CUCh. 2 - Prob. 2.9CUCh. 2 - Prob. 2.10CUCh. 2 - Prob. 2.11CUCh. 2 - Prob. 2.12CUCh. 2 - Prob. 2.13CUCh. 2 - Prob. 2.14CUCh. 2 - Prob. 2.15CUCh. 2 - Prob. 2.16CUCh. 2 - Prob. 2.17CUCh. 2 - Prob. 2.18CUCh. 2 - Prob. 2.19CUCh. 2 - Prob. 2.20CUCh. 2 - Prob. 2.21CUCh. 2 - Prob. 2.22CUCh. 2 - Prob. 2.23CUCh. 2 - Prob. 2.24CUCh. 2 - Prob. 2.25CUCh. 2 - Prob. 2.26CUCh. 2 - Prob. 2.27CUCh. 2 - Prob. 2.28CUCh. 2 - Prob. 2.29CUCh. 2 - Prob. 2.30CUCh. 2 - Prob. 2.31CUCh. 2 - Prob. 2.32CUCh. 2 - Prob. 2.33CUCh. 2 - Prob. 2.34CUCh. 2 - Prob. 2.35CUCh. 2 - Prob. 2.36CUCh. 2 - Prob. 2.37CUCh. 2 - Prob. 2.38CUCh. 2 - Prob. 2.39CUCh. 2 - Prob. 2.40CUCh. 2 - Prob. 2.41CUCh. 2 - Prob. 2.42CUCh. 2 - Prob. 2.43CUCh. 2 - Prob. 2.44CUCh. 2 - Prob. 2.45CUCh. 2 - Prob. 2.46CUCh. 2 - Prob. 2.47CUCh. 2 - Prob. 2.48CUCh. 2 - Prob. 2.49CUCh. 2 - Prob. 2.50CUCh. 2 - Prob. 2.51CUCh. 2 - Prob. 2.52CUCh. 2 - Prob. 2.53CUCh. 2 - Prob. 2.54CUCh. 2 - Prob. 2.1PCh. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - Prob. 2.4PCh. 2 - Prob. 2.5PCh. 2 - Prob. 2.6PCh. 2 - Prob. 2.7PCh. 2 - Prob. 2.8PCh. 2 - Prob. 2.9PCh. 2 - Prob. 2.10PCh. 2 - Prob. 2.11PCh. 2 - Prob. 2.12PCh. 2 - Prob. 2.13PCh. 2 - Prob. 2.14PCh. 2 - Prob. 2.15PCh. 2 - Prob. 2.16PCh. 2 - Prob. 2.17PCh. 2 - Prob. 2.18PCh. 2 - Prob. 2.19PCh. 2 - Prob. 2.20PCh. 2 - Prob. 2.21PCh. 2 - Prob. 2.22PCh. 2 - Prob. 2.23PCh. 2 - Prob. 2.24PCh. 2 - Prob. 2.25PCh. 2 - Prob. 2.26PCh. 2 - Prob. 2.27PCh. 2 - Prob. 2.28PCh. 2 - Prob. 2.29PCh. 2 - Prob. 2.30PCh. 2 - Prob. 2.31PCh. 2 - Prob. 2.32PCh. 2 - Prob. 2.33PCh. 2 - Prob. 2.34PCh. 2 - Prob. 2.35PCh. 2 - Prob. 2.36PCh. 2 - Prob. 2.37PCh. 2 - Prob. 2.38PCh. 2 - Prob. 2.39PCh. 2 - Prob. 2.40PCh. 2 - Prob. 2.41PCh. 2 - Prob. 2.42PCh. 2 - Prob. 2.43PCh. 2 - Prob. 2.44PCh. 2 - Prob. 2.45PCh. 2 - Prob. 2.46PCh. 2 - Prob. 2.47PCh. 2 - Prob. 2.48PCh. 2 - Prob. 2.49PCh. 2 - Prob. 2.50PCh. 2 - Prob. 2.51PCh. 2 - Prob. 2.52PCh. 2 - Prob. 2.53PCh. 2 - Prob. 2.54PCh. 2 - Prob. 2.55PCh. 2 - Prob. 2.56PCh. 2 - Prob. 2.57PCh. 2 - Prob. 2.58PCh. 2 - Prob. 2.59PCh. 2 - Prob. 2.60PCh. 2 - Prob. 2.62PCh. 2 - Prob. 2.63PCh. 2 - Prob. 2.64PCh. 2 - Prob. 2.65PCh. 2 - Prob. 2.66PCh. 2 - Prob. 2.67PCh. 2 - Prob. 2.68PCh. 2 - Prob. 2.69PCh. 2 - Prob. 2.70PCh. 2 - Prob. 2.71P
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- Problem (13): A pump is being used to lift water from the bottom tank to the top tank in a galvanized iron pipe at a discharge (Q). The length and diameter of the pipe section from the bottom tank to the pump are (L₁) and (d₁), respectively. The length and diameter of the pipe section from the pump to the top tank are (L2) and (d2), respectively. Given the values of Q [L/s], L₁ [m], d₁ [m], L₂ [m], d₂ [m], calculate total head loss due to friction (i.e., major loss) in the pipe (hmajor-loss) in [cm]. Givens: L₁,d₁ Pump L₂,d2 오 0.533 lit/s L1 = 6920.729 m d1 = 1.065 m L2 = 70.946 m d2 0.072 m Answers: (1) 3.069 cm (2) 3.914 cm ( 3 ) 2.519 cm ( 4 ) 1.855 cm TABLE 8.1 Equivalent Roughness for New Pipes Pipe Riveted steel Concrete Wood stave Cast iron Galvanized iron Equivalent Roughness, & Feet Millimeters 0.003-0.03 0.9-9.0 0.001-0.01 0.3-3.0 0.0006-0.003 0.18-0.9 0.00085 0.26 0.0005 0.15 0.045 0.000005 0.0015 0.0 (smooth) 0.0 (smooth) Commercial steel or wrought iron 0.00015 Drawn…arrow_forwardThe flow rate is 12.275 Liters/s and the diameter is 6.266 cm.arrow_forwardAn experimental setup is being built to study the flow in a large water main (i.e., a large pipe). The water main is expected to convey a discharge (Qp). The experimental tube will be built at a length scale of 1/20 of the actual water main. After building the experimental setup, the pressure drop per unit length in the model tube (APm/Lm) is measured. Problem (20): Given the value of APm/Lm [kPa/m], and assuming pressure coefficient similitude, calculate the drop in the pressure per unit length of the water main (APP/Lp) in [Pa/m]. Givens: AP M/L m = 590.637 kPa/m meen Answers: ( 1 ) 59.369 Pa/m ( 2 ) 73.83 Pa/m (3) 95.443 Pa/m ( 4 ) 44.444 Pa/m *******arrow_forward
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