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scoundrel [369]
3 years ago
9

For an isotropic material, E and ν are often chosen as the two independent engineering constants. There are other elastic consta

nts: the shear modulus G, the bulk modulus K, and the Lames’ constants, μ and λ.
Suppose that G = E/[2(1+v)].
Prove that μ and λ, and K, satisfy the following relations:

μ = G = 3K(1-2v)/[2(1+v)], λ = Ev/ [(1+v)(1-2v)] = 2vG/ (1-2v),
K = E/ 3(1-2v) = λ+(2/3)μ

To prove it, apply pure shear stress τ, and connect it to pure shear strain γ by G, ie. τ = Gγ. Then in the 45° orientation, consider its normal stresses σ, and σ: are related to its normal strains ε, and e, through E and v in Hooke's law. We can easily establish this relation.
Engineering
1 answer:
pav-90 [236]3 years ago
8 0

Answer:

khgy

Explanation:

nbv

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3 years ago
The hot and cold inlet temperatures to a concentric tube heat exchanger are Th,i = 200°C, Tc,i = 100°C, respectively. The outlet
alexgriva [62]

Answer:Counter,

0.799,

1.921

Explanation:

Given data

T_{h_i}=200^{\circ}C

T_{h_o}=120^{\circ}C

T_{c_i}=100^{\circ}C

T_{c_o}=125^{\circ}C

Since outlet temperature of cold liquid is greater than hot fluid outlet temperature therefore it is counter flow heat exchanger

Equating Heat exchange

m_hc_{ph}\left [ T_{h_i}-T_{h_o}\right ]=m_cc_{pc}\left [ T_{c_o}-T_{c_i}\right ]

\frac{m_hc_{ph}}{m_cc_{pc}}=\frac{125-100}{200-120}=\frac{25}{80}=C\left ( capacity rate ratio\right )

we can see that heat capacity of hot fluid is minimum

Also from energy balance

Q=UA\Delta T_m=\left ( mc_p\right )_{h}\left ( T_{h_i}-T_{h_o}\right )

NTU=\frac{UA}{\left ( mc_p\right )_{h}}=\frac{\left ( T_{h_i}-T_{h_o}\right )}{T_m}

T_m=\frac{\left ( 200-125\right )-\left ( 120-100\right )}{\ln \frac{75}{20}}

T_m=41.63^{\circ}C

NTU=1.921

And\ effectiveness \epsilon =\frac{1-exp\left ( -NTU\left ( 1-c\right )\right )}{1-c\left ( -NTU\left ( 1-c\right )\right )}

\epsilon =\frac{1-exp\left ( -1.921\left ( 1-0.3125\right )\right )}{1-0.3125exp\left ( -1.921\left ( 1-0.3125\right )\right )}

\epsilon =\frac{1-exp\left ( -1.32068\right )}{1-0.3125exp\left ( -1.32068\right )}

\epsilon =\frac{1-0.2669}{1-0.0834}

\epsilon =0.799

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Lelechka [254]

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