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Svetlanka [38]
3 years ago
7

1 kg of oxygen is heated from 20 to 120°C. Determine the amount of heat transfer required when this is done during a (a) constan

t-volume process and (b) isobaric process.
Engineering
1 answer:
Ahat [919]3 years ago
4 0

(a) At constant-volume, Q = 66.7kJ

(b) At isobaric condition, Q = 92.7kJ

<u>Explanation:</u>

The specific heats of oxygen at the average temperature of (20+120)/2 = 70°C = 343 K are

Cp = 0.927  kJ/kg.K

Cv = 0.667 kJ/kgK

We take the oxygen as the system. This is a closed system since no mass crosses the boundaries of the  system. The energy balance for a constant-volume process can be expressed as

ΔE (system) = E(in) - E(out)

and

Q(in) = ΔH = mCp x (T₂ - T₁)

since ΔU + Wb = ΔH during a constant pressure quasi-equilibrium

process. Substituting for both cases,

(a)

Qin,V =const = mCv  (T2 - T1 )

                      = (1kg)(0.667 kJ/kg.K)(120 - 20)K

                      = 66.7kJ

(b)

Qin,P=const = mCp  (T2 - T1  )

                     = (1kg)(0.927 kJ/kg.K)(120 - 20)K

                     = 92.7kJ

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Nutka1998 [239]

Answer:

1791 secs  ≈ 29.85 minutes

Explanation:

( Initial temperature of slab )  T1 = 300° C

temperature of water ( Ts ) = 25°C

T2 ( final temp of slab ) = 50°C

distance between slab and water jet = 25 mm

<u>Determine how long it will take to reach T2</u>

First calculate the thermal diffusivity

∝  = 50 / ( 7800 * 480 ) = 1.34 * 10^-5 m^2/s

<u>next express Temp as a function of time </u>

T( 25 mm , t ) = 50°C

next calculate the time required for the slab to reach 50°C at a distance of 25mm

attached below is the remaining part of the detailed solution

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3 years ago
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MA_775_DIABLO [31]

Answer:

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2 years ago
The velocity of a particle which moves along the s-axis is given by = 40 − 3 2/ , ℎ t is in seconds. Calculate the displacement
scoundrel [369]

The displacement ∆S of the particle during the interval from t = 2sec to 4sec is; 210 sec

<h3>How to find the displacement?</h3>

We are given the velocity equation as;

s' = 40 - 3t²

Thus, the speed equation will be gotten by integration of the velocity equation to get;

s = ∫40 - 3t²

s = 40t - ¹/₂t³

Thus, the displacement between times of t = 2 sec and t = 4 sec is;

∆S = [40(4) - ¹/₂(4)³] - [40(2) - ¹/₂(2)³]

∆S = 210 m

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2 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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3 years ago
Why is it important to know where your online information comes from?
statuscvo [17]

It is very important to know where online information comes from in order to validate, authenticate and be sure it's the right information

<h3>What are online information?</h3>

Online informations are information which are available on the internet such as search engines, social handles and other websites

In conclusion, it is very important to know where online information comes from in order to validate, authenticate and be sure it's the right information

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