Fundamentals of Thermal-Fluid Sciences
Fundamentals of Thermal-Fluid Sciences
5th Edition
ISBN: 9780078027680
Author: Yunus A. Cengel Dr., Robert H. Turner, John M. Cimbala
Publisher: McGraw-Hill Education
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Chapter 18, Problem 32P
To determine

The time taken for the temperature of each geometry to rise to 25°C.

Expert Solution & Answer
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Explanation of Solution

Given:

Diameter of the sphere (D) is 5cm.

Side of the cube (L) is 5cm.

Volume of the rectangular prism (V) is 4cm×5cm×6cm.

Ambient air temperature (T) is 33°C.

Initially temperature of the geometry (Ti) is 0°C.

Temperature of the geometry (T(t)) is 25°C.

Convection heat transfer coefficient (h) is 12W/m2K.

Calculation:

Write the given properties of the silver.

  ρ=10,500kg/m3cp=0.235kJ/kgKk=429W/mK

Sphere:

Calculate the characteristics length of the sphere (Lc).

  Lc=VAs=(πD36)πD2=D6

  =0.05m6=0.008333m

Calculate the Biot number (Bi).

  Bi=hLck=(12W/m2K)(0.008333m)429W/mK=0.00023

The Biot number is less than 0.1(0.00023<0.1). Therefore the lumped system analysis is applicable.

Calculate the value of exponent (b).

  b=hAsρcpV=hρcpLc=(12W/m2K)(10,500kg/m3)(0.235kJ/kgK)(0.008333m)=0.0005836s1

Calculate the time taken for the temperature of the sphere to rise to 25°C(t).

  T(t)TTiT=ebt25°C33°C0°C33°C=e(0.0005836s1)t

  t=2428s=40.5min

Thus, the time taken for the temperature of the sphere to rise to 25°C is 40.5min.

Cube:

Calculate the characteristics length of the cube (Lc).

  Lc=VAs=L36L2=L6

  =0.05m6=0.008333m

Calculate the Biot number (Bi).

  Bi=hLck=(12W/m2K)(0.008333m)429W/mK=0.00023

The Biot number is less than 0.1(0.00023<0.1). Therefore the lumped system analysis is applicable.

Calculate the value of exponent (b).

  b=hAsρcpV=hρcpLc=(12W/m2K)(10,500kg/m3)(0.235kJ/kgK)(0.008333m)=0.0005836s1

Calculate the time taken for the temperature of the cube to rise to 25°C(t).

  T(t)TTiT=ebt25°C33°C0°C33°C=e(0.0005836s1)t

  t=2428s=40.5min

Thus, the time taken for the temperature of the cube to rise to 25°C is 40.5min.

Rectangular prism:

Calculate the characteristics length of the rectangular prism (Lc).

  Lc=VAs=(0.04m×0.05m×0.06m)2(0.04m×0.05m)+2(0.04m×0.06m)+2(0.05m×0.06m)=0.008108m

Calculate the Biot number (Bi).

  Bi=hLck=(12W/m2K)(0.008108m)429W/mK=0.00023

The Biot number is less than 0.1(0.00023<0.1). Therefore the lumped system analysis is applicable.

Calculate the value of exponent (b).

  b=hAsρcpV=hρcpLc=(12W/m2K)(10,500kg/m3)(0.235kJ/kgK)(0.008108m)=0.0005998s1

Calculate the time taken for the temperature of the rectangular prism to rise to 25°C(t).

  T(t)TTiT=ebt25°C33°C0°C33°C=e(0.0005998s1)t

  t=2363s=39.4min

Thus, the time taken for the temperature of the rectangular prism to rise to 25°C is 39.4min.

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Chapter 18 Solutions

Fundamentals of Thermal-Fluid Sciences

Ch. 18 - Prob. 11PCh. 18 - Prob. 12PCh. 18 - Prob. 13PCh. 18 - Prob. 14PCh. 18 - Consider a 1000-W iron whose base plate is made of...Ch. 18 - Metal plates (k = 180 W/m·K, ρ = 2800 kg/m3, and...Ch. 18 - A 5-mm-thick stainless steel strip (k = 21 W/m·K,...Ch. 18 - A long copper rod of diameter 2.0 cm is initially...Ch. 18 - Prob. 21PCh. 18 - Steel rods (ρ = 7832 kg/m3, cp = 434 J/kg·K, and k...Ch. 18 - Prob. 23PCh. 18 - The temperature of a gas stream is to be measured...Ch. 18 - Prob. 25PCh. 18 - A thermocouple, with a spherical junction diameter...Ch. 18 - Prob. 27PCh. 18 - Prob. 28PCh. 18 - Carbon steel balls (ρ = 7833 kg/m3, k = 54 W/m·K,...Ch. 18 - Prob. 31PCh. 18 - Prob. 32PCh. 18 - Prob. 33PCh. 18 - Prob. 34PCh. 18 - Prob. 35PCh. 18 - Prob. 36PCh. 18 - Prob. 37PCh. 18 - Prob. 38PCh. 18 - Prob. 39PCh. 18 - A body at an initial temperature of Ti is brought...Ch. 18 - In a meat processing plant, 2-cm-thick steaks (k =...Ch. 18 - Prob. 42PCh. 18 - Prob. 43PCh. 18 - Prob. 45PCh. 18 - Prob. 46PCh. 18 - Prob. 47PCh. 18 - Prob. 48PCh. 18 - Prob. 49PCh. 18 - A long iron rod (ρ = 7870 kg/m3, cp = 447 J/kg·K,...Ch. 18 - Prob. 51PCh. 18 - A long 35-cm-diameter cylindrical shaft made of...Ch. 18 - Prob. 54PCh. 18 - Prob. 55PCh. 18 - Prob. 56PCh. 18 - A father and son conducted the following simple...Ch. 18 - Prob. 58PCh. 18 - Prob. 59PCh. 18 - Citrus fruits are very susceptible to cold...Ch. 18 - Prob. 61PCh. 18 - Prob. 63PCh. 18 - Prob. 64PCh. 18 - Prob. 65PCh. 18 - White potatoes (k = 0.50 W/m·K and α = 0.13 × 10−6...Ch. 18 - Prob. 67PCh. 18 - Prob. 68PCh. 18 - Prob. 69PCh. 18 - Consider a hot semi-infinite solid at an initial...Ch. 18 - Prob. 71PCh. 18 - Prob. 72PCh. 18 - Prob. 73PCh. 18 - Prob. 74PCh. 18 - Prob. 75PCh. 18 - Prob. 76PCh. 18 - Prob. 77PCh. 18 - Prob. 78PCh. 18 - Prob. 79PCh. 18 - Prob. 81PCh. 18 - Prob. 82PCh. 18 - Prob. 83PCh. 18 - Prob. 84PCh. 18 - Prob. 85PCh. 18 - Prob. 86PCh. 18 - Prob. 88PCh. 18 - Prob. 89PCh. 18 - A 2-cm-high cylindrical ice block (k = 2.22 W/m·K...Ch. 18 - Prob. 91PCh. 18 - Prob. 93PCh. 18 - Prob. 94RQCh. 18 - Large steel plates 1.0-cm in thickness are...Ch. 18 - Prob. 96RQCh. 18 - Prob. 97RQCh. 18 - Prob. 98RQCh. 18 - Prob. 99RQCh. 18 - Prob. 100RQCh. 18 - Prob. 101RQCh. 18 - Prob. 102RQCh. 18 - The water main in the cities must be placed at...Ch. 18 - Prob. 104RQCh. 18 - Prob. 105RQCh. 18 - Prob. 106RQCh. 18 - Prob. 107RQ
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