Chapter 2 Heat Conduction Equation 2-126 A spherical liquid nitrogen container is subjected to specified temperature on the inner surface and convection on the outer surface. The mathematical formulation, the variation of temperature, and the rate of evap
Chapter 2 Heat Conduction Equation
Variable Thermal Conductivity
2-94C During steady one-dimensional heat conduction in a plane wall, long cylinder, and sphere with constant thermal conductivity and no heat generation, the temperature in only the plane wa
Chapter 2 Heat Conduction Equation 2-68 A compressed air pipe is subjected to uniform heat flux on the outer surface and convection on the inner surface. The mathematical formulation, the variation of temperature in the pipe, and the surface temperatures
Chapter 2 Heat Conduction Equation
Solution of Steady One-Dimensional Heat Conduction Problems
2-52C Yes, this claim is reasonable since in the absence of any heat generation the rate of heat transfer through a plain wall in steady operation must be const
Chapter 2 Heat Conduction Equation
Chapter 2 HEAT CONDUCTION EQUATION
Introduction
2-1C Heat transfer is a vector quantity since it has direction as well as magnitude. Therefore, we must specify both direction and magnitude in order to describe heat trans
Chapter 4 Transient Heat Conduction 4-118 Internal combustion engine valves are quenched in a large oil bath. The time it takes for the valve temperature to drop to specified temperatures and the maximum heat transfer are to be determined. Assumptions 1 T
Chapter 4 Transient Heat Conduction
Review Problems
4-105 Two large steel plates are stuck together because of the freezing of the water between the two plates. Hot air is blown over the exposed surface of the plate on the top to melt the ice. The length
Chapter 4 Transient Heat Conduction 4-81 A cubic block and a cylindrical block are exposed to hot gases on all of their surfaces. The center temperatures of each geometry in 10, 20, and 60 min are to be determined. Assumptions 1 Heat conduction in the cub
Chapter 4 Transient Heat Conduction
Transient Heat Conduction in Multidimensional Systems
4-69C The product solution enables us to determine the dimensionless temperature of two- or threedimensional heat transfer problems as the product of dimensionless t
Chapter 4 Transient Heat Conduction 4-47 A hot dog is dropped into boiling water, and temperature measurements are taken at certain time intervals. The thermal diffusivity and thermal conductivity of the hot dog and the convection heat transfer coefficien
Chapter 4 Transient Heat Conduction
Transient Heat Conduction in Large Plane Walls, Long Cylinders, and Spheres 4-26C A cylinder whose diameter is small relative to its length can be treated as an infinitely long cylinder. When the diameter and length of
Chapter 4 Transient Heat Conduction
Chapter 4 TRANSIENT HEAT CONDUCTION
Lumped System Analysis
4-1C In heat transfer analysis, some bodies are observed to behave like a "lump" whose entire body temperature remains essentially uniform at all times during a
Chapter 16 - Section B - Non-Numerical Solutions
16.1 The potential is displayed as follows. Note that K is used in place of k as a parameter to avoid confusion with Boltzmanns constant.
Combination of the potential with Eq. (16.10) yields on piecewise in
Chapter 14 - Section B - Non-Numerical Solutions
14.2 Start with the equation immediately following Eq. (14.49), which can be modied slightly to read: ln i = (nG R/ RT ) (n Z ) ln Z +n +1 ni ni ni
where the partial derivatives written here and in the foll
Chapter 12 - Section B - Non-Numerical Solutions
12.2 Equation (12.1) may be written: yi P = xi i Pi sat . Summing for i = 1, 2 gives: P = x1 1 P1sat + x2 2 P2sat . Differentiate at constant T : d 1 d 2 dP = P1sat x1 + 1 + P2sat x2 2 d x1 d x1 d x1 d 2 =0
Chapter 11 - Section B - Non-Numerical Solutions
11.6 Apply Eq. (11.7): (nT ) Ti ni =T n ni =T (n P ) Pi ni =P n ni m ni =P
P ,T ,n j
T , P ,n j
P ,T ,n j
T , P ,n j
11.7 (a) Let m be the mass of the solution, and dene the partial molar mass by: m i Let M
Chapter 10 - Section B - Non-Numerical Solutions
10.5 For a binary system, the next equation following Eq. (10.2) shows that P is linear in x1 . Thus no maximum or minimum can exist in this relation. Since such an extremum is required for the existence of
Chapter 9 - Section B - Non-Numerical Solutions
9.1 Since the object of doing work |W | on a heat pump is to transfer heat | Q H | to a heat sink, then: What you get = | Q H | What you pay for = |W | Whence For a Carnot heat pump, = |Q H | |W |
TH |Q H |
Chapter 8 - Section B - Non-Numerical Solutions
8.12 (a) Because Eq. (8.7) for the efciency Diesel includes the expansion ratio, re VB / V A , we relate this quantity to the compression ratio, r VC / VD , and the Diesel cutoff ratio, rc V A / VD . Since V
Chapter 7 - Section B - Non-Numerical Solutions
7.2 (a) Apply the general equation given in the footnote on page 266 to the particular derivative of interest here: T S T = P S S P P T The two partial derivatives on the right are found from Eqs. (6.17) and
Chapter 6 - Section B - Non-Numerical Solutions
H S =T
P
6.1 By Eq. (6.8),
and isobars have positive slope T S
Differentiate the preceding equation: 2 H S2
2 H S2 =
P
=
P
P
Combine with Eq. (6.17):
T CP
and isobars have positive curvature.
6.2 (a) Applica
Chapter 5 - Section B - Non-Numerical Solutions
5.1 Shown to the right is a P V diagram with two adiabatic lines 1 2 and 2 3, assumed to intersect at point 2. A cycle is formed by an isothermal line from 3 1. An engine traversing this cycle would produce
Chapter 4 - Section B - Non-Numerical Solutions
4.5 For consistency with the problem statement, we rewrite Eq. (4.8) as: CP = A + B C T1 ( + 1) + T12 ( 2 + + 1) 2 3
where T2 / T1 . Dene C Pam as the value of C P evaluated at the arithmetic mean temperatur
Chapter 3 - Section B - Non-Numerical Solutions
3.2 Differentiate Eq. (3.2) with respect to P and Eq. (3.3) with respect to T : P T =
T
1 V2
V P V T
T
V T V P
+
P
1 V
2V PT 2V T P
= +
2V PT 2V PT
=
P
1 V2
T
P
1 V
=
Addition of these two equations leads i
Chapter 2 - Section B - Non-Numerical Solutions
2.3 Equation (2.2) is here written: Ut + EP + EK = Q + W
(a) In this equation W does not include work done by the force of gravity on the system. This is accounted for by the E K term. Thus, W = 0. (b) Since
Chapter 1 - Section B - Non-Numerical Solutions
1.1 This system of units is the English-system equivalent of SI. Thus, gc = 1(lbm )(ft)(poundal)1 (s)2 1.2 (a) Power is power, electrical included. Thus, Nm kgm2 energy [=] [=] time s s3 (b) Electric current
10.35 a) The equation from NIST is: Mi = ki yi P The equation for Henry's Law is:i Hi = yi P x Solving to eliminate P gives: By definition:
Mi Hi
Eq. (1) Eq. (2) Eq. (3)
=
Mi
=
ni ns Ms
ki xi
where M is the molar mass and the subscript s refers to the sol
Problems 10.25 to 10.34 have been solved using MS-EXCEL 2000 We give the resulting spreadsheets. Problem 10.25 a) BUBL P T=-60 F (-51.11 C) P=200 psia P=250 psia P=215 psia (14.824 bar) ANSWER Ki yi=Ki*xi Ki yi=Ki*xi Ki yi=Ki*xi 5.600 0.560 4.600 0.460 5.