FUND OF ENG THERMODYN(LLF)+WP NEXT GEN
9th Edition
ISBN: 9781119840602
Author: MORAN
Publisher: WILEY
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Chapter 2, Problem 2.10CU
To determine
To match the term from column A “energy transfer by work” to the correct description in column B.
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Homework#5
Member AB has the angular velocity wAB = 2.5 rad/s and
angular acceleration a AB = 9 rad/s². (Figure 1)
Determine the magnitude of the velocity of point C at the instant shown.
Determine the direction of the velocity of point C at the instant shown.
Determine the magnitude of the acceleration of point C at the instant shown.
Determine the direction of the acceleration of point C at the instant shown.
A
300 mm
WAB
α AB
B
500 mm
0=60°
y
200 mm
You are asked to design a unit to condense ammonia. The required condensation rate is 0.09kg/s. Saturated ammonia at 30 o C is passed over a vertical plate (10 cm high and 25 cm wide).The properties of ammonia at the saturation temperature of 30°C are hfg = 1144 ́10^3 J/kg andrv = 9.055 kg/m 3 . Use the properties of liquid ammonia at the film temperature of 20°C (Ts =10 o C):Pr = 1.463
rho_l= 610.2 kf/m^3
liquid viscosity= 1.519*10^-4 kg/ ms
kinematic viscosity= 2.489*10^-7 m^2/s
Cpl= 4745 J/kg C
kl=0.4927 W/m Ca)Calculate the surface temperature required to achieve the desired condensation rate of 0.09 kg/s( should be 688 degrees C) b) Show that if you use a bigger vertical plate (2.5 m-wide and 0.8 m-height), the requiredsurface temperature would be now 20 o C. You may use all the properties given as an initialguess. No need to iterate to correct for Tf. c) What if you still want to use small plates because of the space constrains? One way to getaround this problem is to use small…
Chapter 2 Solutions
FUND OF ENG THERMODYN(LLF)+WP NEXT GEN
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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- You are asked to design a unit to condense ammonia. The required condensation rate is 0.09kg/s. Saturated ammonia at 30 o C is passed over a vertical plate (10 cm high and 25 cm wide).The properties of ammonia at the saturation temperature of 30°C are hfg = 1144 ́10^3 J/kg andrv = 9.055 kg/m 3 . Use the properties of liquid ammonia at the film temperature of 20°C (Ts =10 o C):Pr = 1.463 rho_l= 610.2 kf/m^3 liquid viscosity= 1.519*10^-4 kg/ ms kinematic viscosity= 2.489*10^-7 m^2/s Cpl= 4745 J/kg C kl=0.4927 W/m CCalculate the surface temperature required to achieve the desired condensation rate of 0.09 kg/s( should be 688 degrees C) a) Show that if you use a bigger vertical plate (2.5 m-wide and 0.8 m-height), the requiredsurface temperature would be now 20 o C. You may use all the properties given as an initialguess. No need to iterate to correct for Tf. b) What if you still want to use small plates because of the space constrains? One way to getaround this problem is to use small…arrow_forwardHomework#5arrow_forwardQuestion 1: Beam Analysis Two beams (ABC and CD) are connected using a pin immediately to the left of Point C. The pin acts as a moment release, i.e. no moments are transferred through this pinned connection. Shear forces can be transferred through the pinned connection. Beam ABC has a pinned support at point A and a roller support at Point C. Beam CD has a roller support at Point D. A concentrated load, P, is applied to the mid span of beam CD, and acts at an angle as shown below. Two concentrated moments, MB and Mc act in the directions shown at Point B and Point C respectively. The magnitude of these moments is PL. Moment Release A B с ° MB = PL Mc= = PL -L/2- -L/2- → P D Figure 1: Two beam arrangement for question 1. To analyse this structure, you will: a) Construct the free body diagrams for the structure shown above. When constructing your FBD's you must make section cuts at point B and C. You can represent the structure as three separate beams. Following this, construct the…arrow_forward
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