3.-第三章课后习题及答案

第三章

1. (Q1) Suppose the network layer provides the following service. The network layer in the source host accepts a segment of maximum size 1,200 bytes and a destination host address from the transport layer. The network layer then guarantees to deliver the segment to the transport layer at the destination host. Suppose many network application processes can be running at the destination host.

a. Design the simplest possible transport-layer protocol that will get

application data to the desired process at the destination host. Assume the operating system in the destination host has assigned a 4-byte port number to each running application process.

b. Modify this protocol so that it provides a “return address”to the destination process.

c. In your protocols, does the transport layer “have to do anything” in the

core of the computer network.

Answer:

a. Call this protocol Simple Transport Protocol (STP). At the sender side, STP

accepts from the sending process a chunk of data not exceeding 1196 bytes,

a destination host address, and a destination port number. STP adds a four-

byte header to each chunk and puts the port number of the destination process in this header. STP then gives the destination host address and the resulting segment to the network layer. The network layer delivers the segment to STP at the destination host. STP then examines the port number in the segment, extracts the data from the segment, and passes the data to the process identified by the port number.

b. The segment now has two header fields: a source port field and destination

port field. At the sender side, STP accepts a chunk of data not exceeding 1192 bytes, a destination host address, a source port number, and a destination port number. STP creates a segment which contains the application data, source port number, and destination port number. It then gives the segment and the destination host address to the network layer. After receiving the segment, STP at the receiving host gives the application process the application data and the source port number.

c. No, the transport layer does not have to do anything in the core; the

transport layer “lives” in the e nd systems.

2. (Q2) Consider a planet where everyone belongs to a family of six, every family lives in its own house, each house has a unique address, and each person in a given house has a unique name. Suppose this planet has a mail service that delivers letters form source house to destination house. The mail service requires that (i) the letter be in an envelope and that (ii) the address of the destination house (and nothing more ) be clearly written on the envelope. Suppose each family has a delegate family member who collects and distributes letters for the other family members. The letters do not necessarily provide any indication of the recipients of the letters.

a. Using the solution to Problem Q1 above as inspiration, describe a protocol

that the delegates can use to deliver letters from a sending family member to a receiving family member.

b. In your protocol, does the mail service ever have to open the envelope and

examine the letter in order to provide its service.

Answer:

a.For sending a letter, the family member is required to give the delegate the

letter itself, the address of the destination house, and the name of the

recipient. The delegate clearly writes the recipient’s name on the top of the letter. The delegate then puts the letter in an envelope and writes the address of the destination house on the envelope. The delegate then gives the letter to the planet’s mail service. At the receiving side, the delegate receives the letter from the mail service, takes the letter out of the envelope, and takes note of the recipient name written at the top of the letter. The delegate than gives the letter to the family member with this name.

b.No, the mail service does not have to open the envelope; it only examines the

address on the envelope.

3. (Q3) Describe why an application developer might choose to run an application over UDP rather than TCP.

Answer:An application developer may not want its application to use TCP’s congestion control, which can throttle the application’s sending rate at times of congestion. Often, designers of IP telephony and IP videoconference applications choose to run their applications over UDP because they want to avoid TCP’s congestion control. Also, some applications do not need the reliable data transfer provided by TCP.

4. (P1) Suppose Client A initiates a Telnet session with Server S. At about the same time, Client B also initiates a Telnet session with Server S. Provide possible source and destination port numbers for

a. The segment sent from A to B.

b. The segment sent from B to S.

c. The segment sent from S to A.

d. The segment sent from S to B.

e. If A and B are different hosts, is it possible that the source port number

in the segment from A to S is the same as that from B to S

f. How about if they are the same host

Answer:

e Yes.

f No.

5. (P2) Consider Figure What are the source and destination port values in the

segments flowing form the server back to the clients’ processes What are the IP addresses in the network-layer datagrams carrying the transport-layer segments

Answer:

Suppose the IP addresses of the hosts A, B, and C are a, b, c, respectively.

(Note that a,b,c are distinct.)

To host A: Source port =80, source IP address = b, dest port = 26145, dest IP address = a

To host C, left process: Source port =80, source IP address = b, dest port = 7532, dest IP address = c

To host C, right process: Source port =80, source IP address = b, dest port = 26145, dest IP address = c

6. (P3) UDP and TCP use 1s complement for their checksums. Suppose you have the

following three 8-bit bytes: 01101010, 01001111, 01110011. What is the 1s complement of the sum of these 8-bit bytes (Note that although UDP and TCP use 16-bit words in computing the checksum, for this problem you are being asked to consider 8-bit sums.) Show all work. Why is it that UDP takes the 1s complement of the sum; that is , why not just sue the sum With the 1s complement scheme, how does the receiver detect errors Is it possible that a 1-bit error will go undetected How about a 2-bit error

Answer:

One's complement = 1 1 1 0 1 1 1 0.

To detect errors, the receiver adds the four words (the three original words and the checksum). If the sum contains a zero, the receiver knows there has been an error. All one-bit errors will be detected, but two-bit errors can be undetected ., if the last digit of the first word is converted to a 0 and the last digit of the second word is converted to a 1).

7. (P4) Suppose that the UDP receiver computes the Internet checksum for the

received UDP segment and finds that it matches the value carried in the checksum field. Can the receiver be absolutely certain that no bit errors have occurred Explain.

Answer:

No, the receiver cannot be absolutely certain that no bit errors have occurred. This is because of the manner in which the checksum for the packet is calculated. 0 1 0 1 0 1 0 1 + 0 1 1 1 0 0 0 0 1 1 0 0 0 1 0 1 1 1 0 0 0 1 0 1 + 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 1

If the corresponding bits (that would be added together) of two 16-bit words in the packet were 0 and 1 then even if these get flipped to 1 and 0 respectively, the sum still remains the same. Hence, the 1s complement the receiver calculates will also be the same. This means the checksum will verify even if there was transmission error.

8. (P5) a. Suppose you have the following 2 bytes: 01001010 and 01111001. What

is the 1s complement of sum of these 2 bytes

b. Suppose you have the following 2 bytes: and 01101110. What is the 1s

complement of sum of these 2 bytes

c. For the bytes in part (a), give an example where one bit is flipped in

each of the 2 bytes and yet the 1s complement doesn’t change.

Answer:

a. Adding the two bytes gives . Taking the one’s complement gives 01100010

b. Adding the two bytes gives 00011110; the one’s complement gives .

c. first byte = 00110101 ; second byte = 01101000.

9. (P6) Consider our motivation for correcting protocol . Show that the receiver,

shown in

the figure on the following page, when operating with the sender show in Figure , can lead the sender and receiver to enter into a deadlock state, where each is waiting for an event that will never occur.

Answer:Suppose the sender is in state “Wait for call 1 from above” and the receiver (the receiver shown in the homework problem) is in state “Wait for 1 from below.” The sender sends a packet with sequence number 1, and transitions to “Wait for ACK or NAK 1,” waiting for an ACK or NAK. Suppose now the receiver

receives the packet with sequence number 1 correctly, sends an ACK, and transitions to state “Wait for 0 from below,” waiting for a data packet with sequence number 0. However, the ACK is corrupted. When the sender gets the corrupted ACK, it resends the packet with sequence number 1. However, the receiver is waiting for a packet with sequence number 0 and (as shown in the home work problem) always sends a NAK when it doesn't get a packet with sequence number 0. Hence the sender will always be sending a packet with sequence number 1, and the receiver will always be NAKing that packet. Neither will progress forward from that state.

10. (P7) Draw the FSM for the receiver side of protocol

Answer:The sender side of protocol differs from the sender side of protocol in that

timeouts have been added. We have seen that the introduction of timeouts adds the possibility of duplicate packets into the sender-to-receiver data stream. However, the receiver in protocol can already handle duplicate packets. (Receiver-side duplicates in rdt would arise if the receiver sent an ACK that was lost, and the sender then retransmitted the old data). Hence the receiver in protocol will also work as the receiver in protocol rdt .

11. (P8) In protocol , the ACK packets flowing from the receiver to the sender

do not have sequence numbers (although they do have an ACK field that contains the sequence number of the packet they are acknowledging). Why is it that our ACK packets do not require sequence numbers

Answer:To best Answer this question, consider why we needed sequence numbers in the first

place. We saw that the sender needs sequence numbers so that the receiver can tell if a data packet is a duplicate of an already received data packet. In the case of ACKs, the sender does not need this info ., a sequence number on an ACK) to tell detect a duplicate ACK. A duplicate ACK is obvious to the receiver, since when it has received the original ACK it transitioned to the next state. The duplicate ACK is not the ACK that the sender needs and hence is ignored by the sender.

12. (P9) Give a trace of the operation of protocol when data packets and

acknowledgment packets are garbled. Your trace should be similar to that

used in Figure

Answer: Suppose the protocol has been in operation for some time. The sender is in state “Wait

for call from above” (top left hand corner) and the receiver is in state “Wait for 0 from below”. The scenarios for corrupted data and corrupted ACK are shown in Figure 1.

13. (P10) Consider a channel that can lose packets but has a maximum delay that

is known. Modify protocol to include sender timeout and retransmit.

Informally argue why your protocol can communicate correctly over this

channel.

Answer:Here, we add a timer, whose value is greater than the known round-trip propagation delay. We add a timeout event to the “Wait for ACK or NAK0” and “Wait for ACK or NAK1” states. If the timeout event occurs, the most recently transmitted packet is retransmitted. Let us see why this protocol will still work with the receiver.

? Suppose the timeout is caused by a lost data packet, ., a packet on the senderto- receiver channel. In this case, the receiver never received the previous transmission and, from the receiver's viewpoint, if the timeout

retransmission is received, it look exactly the same as if the original transmission is being received.

? Suppose now that an ACK is lost. The receiver will eventually retransmit the packet on a timeout. But a retransmission is exactly the same action that is take if an ACK is garbled. Thus the sender's reaction is the same with a loss, as with a garbled ACK. The rdt receiver can already handle the case of a garbled ACK.

14. (P11) Consider the protocol. Draw a diagram showing that if the network

connection between the sender and receiver can reorder messages (that

is, that two messages propagating in the medium between the sender and

receiver can be reordered), then the alternating-bit protocol will not

work correctly (make sure you clearly identify the sense in which it

will not work correctly). Your diagram should have the sender on the

left and the receiver on the right, with the time axis running down the

page, showing data (D) and acknowledgement (A) message exchange. Make

sure you indicate the sequence number associated with any data or

acknowledgement segment.

Answer:

15. (P12) The sender side of simply ignores (that is, takes no action on) all

received packets that are either in error or have the wrong value in

the ack-num field of an acknowledgement packet. Suppose that in such

circumstances, were simply to retransmit the current data packet .

Would the protocol still work (hint: Consider what would happen if there

were only bit errors; there are no packet losses but premature timeout

can occur. Consider how many times the nth packet is sent, in the limit

as n approaches infinity.)

Answer:The protocol would still work, since a retransmission would be what would happen if the packet received with errors has actually been lost (and from the receiver standpoint, it never knows which of these events, if either, will occur). To get at the more subtle issue behind this question, one has to allow for premature timeouts to occur. In this case, if each extra copy of the packet is ACKed and each received extra ACK causes another extra copy of the current packet to be sent, the number of times packet n is sent will increase without bound as n approaches infinity.

16. (P13) Consider a reliable data transfer protocol that uses only negative

acknowledgements. Suppose the sender sends data only infrequently. Would

a NAK-only protocol be preferable to a protocol that uses ACKs Why Now

suppose the sender has a lot of data to send and the end to end connection

experiences few losses. In this second case , would a NAK-only protocol

be preferable to a protocol that uses ACKs Why

Answer:

In a NAK only protocol, the loss of packet x is only detected by the receiver when packet

x+1 is received. That is, the receivers receives x-1 and then x+1, only when x+1 is received does the receiver realize that x was missed. If there is a long delay between the transmission of x and the transmission of x+1, then it will be a long time until x can be recovered, under a NAK only protocol.

On the other hand, if data is being sent often, then recovery under a NAK-only scheme could happen quickly. Moreover, if errors are infrequent, then NAKs are only occasionally sent (when needed), and ACK are never sent – a significant reduction in feedback in the NAK-only case over the ACK-only case.

17. (P14) Consider the cross-country example shown in Figure . How big would

the window size have to be for the channel utilization to be greater

than 80 percent

Answer:It takes 8 microseconds (or milliseconds) to send a packet. in order for the sender

to be busy 90 percent of the time, we must have util = = / or n approximately 3377 packets.

18. (P15) Consider a scenario in which Host A wants to simultaneously send

packets to Host B and C. A is connected to B and C via a broadcast

channel—a packet sent by A is carried by the channel to both B and C.

Suppose that the broadcast channel connecting A, B, and C can

independently lose and corrupt packets (and so, for example, a packet

sent from A might be correctly received by B, but not by C). Design a

stop-and-wait-like error-control protocol for reliable transferring

packets from A to B and C, such that A will not get new data from the

upper layer until it knows that B and C have correctly received the

current packet. Give FSM descriptions of A and C. (Hint: The FSM for B

should be essentially be same as for C.) Also, give a description of

the packet format(s) used.

Answer:

In our solution, the sender will wait until it receives an ACK for a pair of messages (seqnum and seqnum+1) before moving on to the next pair of messages. Data packets have a data field and carry a two-bit sequence number. That is, the valid sequence numbers are 0, 1, 2, and 3. (Note: you should think about why a 1-bit sequence number space of 0, 1 only would not work in the solution below.) ACK messages carry the sequence number of the data packet they are acknowledging.

The FSM for the sender and receiver are shown in Figure 2. Note that the sender state records whether (i) no ACKs have been received for the current pair, (ii) an ACK for seqnum (only) has been received, or an ACK for seqnum+1 (only) has been received. In this figure, we assume that the seqnum is initially 0, and that the sender has sent the first two data messages (to get things going). A timeline trace for the sender and receiver recovering from a lost packet is shown below:

Sender Receiver

make pair (0,1)

send packet 0

Packet 0 drops

send packet 1

receive packet 1

buffer packet 1

send ACK 1

receive ACK 1

(timeout)

resend packet 0

receive packet 0

deliver pair (0,1)

send ACK 0

receive ACK 0

19. (P16) Consider a scenario in which Host A and Host B want to send messages

to Host C. Hosts A and C are connected by a channel that can lose and

corrupt (but not reorder)message. Hosts B and C are connected by another

channel (independent of the channel connecting A and C) with the same

properties. The transport layer at Host C should alternate in delivering

messages from A and B to the layer above (that is, it should first

deliver the data from a packet from A, then the data from a packet from

B, and so on). Design a stop-and-wait-like error-control protocol for

reliable transferring packets from A to B and C, with alternating

delivery at C as described above. Give FSM descriptions of A and C.

(Hint: The FSM for B should be essentially be same as for A.) Also, give

a description of the packet format(s) used.

Answer:

This problem is a variation on the simple stop and wait protocol . Because the channel may lose messages and because the sender may resend a message that one of the receivers has already received (either because of a premature timeout or because the other receiver has yet to receive the data correctly), sequence numbers are needed. As in , a 0-bit sequence number will suffice here.

The sender and receiver FSM are shown in Figure 3. In this problem, the sender state indicates whether the sender has received an ACK from B (only), from C (only) or from neither C nor B. The receiver state indicates which sequence number the receiver is waiting for.

20. (P17) In the generic SR protocol that we studied in Section the sender

transmits a message as soon as it is available (if it is in the window)

without waiting for an acknowledgment. Suppose now that we want an SR

protocol that sends messages two at a time. That is , the sender will

send a pair of messages and will send the next pair of messages only

when it knows that both messages in the first pair have been receiver

correctly.

Suppose that the channel may lose messages but will not corrupt or reorder messages. Design an error-control protocol for the

unidirectional reliable transfer of messages. Give an FSM description

of the sender and receiver. Describe the format of the packets sent

between sender and receiver, and vice versa. If you use any procedure

calls other than those in Section (for example, udt_send(),

start_timer(), rdt_rcv(), and so on) ,clearly state their actions. Give

an example (a timeline trace of sender and receiver) showing how your

protocol recovers from a lost packet.

Answer:

21. (P18) Consider the GBN protocol with a sender window size of 3 and a sequence

number range of 1024. Suppose that at time t, the next in-order packet

that the receiver is expecting has a sequence number of k. Assume that

the medium does not reorder messages. Answer the following questions:

a. What are the possible sets of sequence number inside the sender’s

window at time t Justify your Answer.

b .What are all possible values of the ACK field in all possible messages

currently propagating back to the sender at time t Justify your Answer. Answer:

a.Here we have a window size of N=3. Suppose the receiver has received packet

k-1, and has ACKed that and all other preceeding packets. If all of these ACK's have been received by sender, then sender's window is [k, k+N-1].

Suppose next that none of the ACKs have been received at the sender. In this second case, the sender's window contains k-1 and the N packets up to and including k-1. The sender's window is thus [k- N,k-1]. By these arguments, the senders window is of size 3 and begins somewhere in the range [k-N,k]. b.If the receiver is waiting for packet k, then it has received (and ACKed)

packet k-1 and the N-1 packets before that. If none of those N ACKs have been yet received by the sender, then ACK messages with values of [k-N,k-1] may still be propagating back. Because the sender has sent packets [k-N, k-1], it must be the case that the sender has already received an ACK for k-N-1.

Once the receiver has sent an ACK for k-N-1 it will never send an ACK that is less that k-N-1. Thus the range of in-flight ACK values can range from k-N-1 to k-1.

22. (P19) Answer true or false to the following questions and briefly justify

your Answer.

a. With the SR protocol, it is possible for the sender to receive an

ACK for a packet that falls outside of its current window.

b. With CBN, it is possible for the sender to receiver an ACK for a

packet that falls outside of its current window.

c. The alternating-bit protocol is the same as the SR protocol with a

sender and receiver window size of 1.

d. The alternating-bit protocol is the same as the GBN protocol with a

sender and receiver window size of 1.

Answer:

a.True. Suppose the sender has a window size of 3 and sends packets 1, 2, 3 at

t0 . At t1 (t1 > t0) the receiver ACKS 1, 2, 3. At t2 (t2 > t1) the sender times out and resends 1, 2, 3. At t3 the receiver receives the duplicates and re-acknowledges 1, 2, 3. At t4 the sender receives the ACKs that the receiver sent at t1 and advances its window to 4, 5, 6. At t5 the sender receives the ACKs 1, 2, 3 the receiver sent at t2 . These ACKs are outside its window.

b.True. By essentially the same scenario as in (a).

c.True.

d.Tru

e. Note that with a window size of 1, SR, GBN, and the alternating bit

protocol are functionally equivalent. The window size of 1 precludes the possibility of out-of-order packets (within the window). A cumulative ACK is just an ordinary ACK in this situation, since it can only refer to the single packet within the window.

23. (Q4) Why is it that voice and video traffic is often sent over TCP rather

than UDP in today’s Internet. (Hint: The Answer we are looking for has

nothing to do with TCP’s congestion-control mechanism. )

Answer:Since most firewalls are configured to block UDP traffic, using TCP for video and voice traffic lets the traffic though the firewalls

24. (Q5) Is it possible for an application to enjoy reliable data transfer even

when the application runs over UDP If so, how

Answer:Yes. The application developer can put reliable data transfer into the application layer protocol. This would require a significant amount of work and debugging, however.

25. (Q6) Consider a TCP connection between Host A and Host B. Suppose that the

郝吉明第三版大气污染控制工程课后答案完整版

大气污染控制工程 课后答案 （第三版）主编：郝吉明马广大王书肖 目录 第一章概论 第二章燃烧与大气污染 第三章大气污染气象学 第四章大气扩散浓度估算模式 第五章颗粒污染物控制技术基础 第六章除尘装置 第七章气态污染物控制技术基础 第八章硫氧化物的污染控制 第九章固定源氮氧化物污染控制 第十章挥发性有机物污染控制 第十一章城市机动车污染控制

第一章 概 论 1.1 干结空气中N 2、O 2、Ar 和CO 2气体所占的质量百分数是多少？ 解：按1mol 干空气计算，空气中各组分摩尔比即体积比，故n N2=0.781mol ，n O2=0.209mol ，n Ar =0.00934mol ，n CO2=0.00033mol 。质量百分数为 %51.75%100197.2801.28781.0%2=???= N ，%08.23%100197.2800 .32209.0%2=???=O ； % 29.1%1001 97.2894 .3900934.0%=???=Ar ，%05.0%100197.2801 .4400033.0%2=???=CO 。 1.2 根据我国的《环境空气质量标准》的二级标准，求出SO 2、NO 2、CO 三种污染物日平均浓度限值的体积分数。 解：由我国《环境空气质量标准》二级标准查得三种污染物日平均浓度限值如下： SO2：0.15mg/m 3，NO2：0.12mg/m 3，CO ：4.00mg/m 3。按标准状态下1m 3 干空气计算，其摩尔数为mol 643.444 .221013 =?。故三种污染物体积百分数分别为：

SO 2： ppm 052.0643.44641015.03=??-，NO 2：ppm 058.0643.44461012.03 =??- CO ： ppm 20.3643 .44281000.43 =??-。 1.3 CCl 4气体与空气混合成体积分数为1.50×10－4的混合气体，在管道中流动的流量为10m 3N 、/s ，试确定：1）CCl 4在混合气体中的质量浓度ρ（g/m 3N ）和摩尔浓度c （mol/m 3N ）；2）每天流经管道的CCl 4质量是多少千克？ 解：1）ρ（g/m 3 N ）3 3 4/031.110 4.221541050.1N m g =???=-- c （mol/m 3 N ）3 33 4/1070.610 4.221050.1N m mol ---?=??=。 2）每天流经管道的CCl 4质量为1.031×10×3600×24×10－3kg=891kg 1.4 成人每次吸入的空气量平均为500cm 3，假若每分钟呼吸15次，空气中颗粒物的浓度为200g μ/m 3，试计算每小时沉积于肺泡内的颗粒物质量。已知该颗粒物在肺泡中的沉降系数为0.12。 解：每小时沉积量200×（500×15×60×10－6）×0.12g μ=10.8g μ 1.5 设人体肺中的气体含CO 为2.2×10－4，平均含氧量为19.5%。如果这种浓度保持不变，求COHb 浓度最终将达到饱和水平的百分率。 解：由《大气污染控制工程》P14 （1－1），取M=210 2369.0105.19102.22102 4 22=???==--∝O p p M Hb O COHb ，

大气污染课后答案-4章

四章 大气扩散浓度估算模式 4.1 污染源的东侧为峭壁，其高度比污染源高得多。设有效源高为H ，污染源到峭壁的距离为L ，峭壁对烟流扩散起全反射作用。试推导吹南风时高架连续点源的扩散模式。当吹北风时，这一模式又变成何种形式？ 解： 吹南风时以风向为x 轴，y 轴指向峭壁，原点为点源在地面上的投影。若不存在峭壁，则有 ]}2)(exp[]2)(){exp[2exp(2),,,(2 2 2222' z z y z y H z H z y u Q H z y x σσσσσπρ+-+---= 现存在峭壁，可考虑ρ为实源与虚源在所关心点贡献之和。 实源]}2)(exp[]2)(){exp[2exp(22 2 22221z z y z y H z H z y u Q σσσσσπρ+-+---= 虚源]}2)(exp[]2)(]{exp[2)2(exp[222 222 22z z y z y H z H z y L u Q σσσσσπρ+-+----= 因此]}2)(exp[]2)(){exp[2exp(22 2 2222z z y z y H z H z y u Q σσσσσπρ+-+---=+ ]}2)(exp[]2)(]{exp[2)2(exp[22 2 2222z z y z y H z H z y L u Q σσσσσπ+-+---- =]}2)(exp[]2)(]}{exp[2)2(exp[)2{exp(22 2 222222z z y y z y H z H z y L y u Q σσσσσσπ+-+----+- 刮北风时，坐标系建立不变，则结果仍为上式。 4.2 某发电厂烟囱高度120m ，内径5m ，排放速度13.5m/s ，烟气温度为418K 。大气温度288K ，大气为中性层结，源高处的平均风速为4m/s 。试用霍兰德、布里格斯（x<=10H s ）、国家标准GB/T13201－91中的公式计算烟气抬升高度。 解： 霍兰德公式 m D T T T u D v H s a s s 16.96)5418 288 4187.25.1(455.13)7 .25.1(=?-?+?=-+= ?。 布里格斯公式 kW kW D v T T T Q s s a s H 210002952155.1341828841810 6.9 7.2106.97.22 3 23>=??-??=-??= --且x<=10Hs 。此时 3/23/213/11 3 /23/180.2429521362.0362.0x x u x Q H H =??==?--。

工程测量学试题库160题(附答案)..

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13. ( B )的基准面是大地水准面。 A. 竖直角 B. 高程 C. 水平距离 D. 水平角 14. 建筑工程施工测量的基本工作是（B）。 A.测图 B.测设 C.用图 D.识图 15. 大地水准面处处与铅垂线（A）交。 A、正 B、平行 C、重合 D、斜 16. A、B两点，HA为115.032m，HB为114.729m，则hAB为（A）。 A、-0.303 B、0.303 C、29.761 D、-29.761 17. 建筑施工图中标注的某部位标高，一般都是指（B）。 A、绝对高程 B、相对高程 C、高差 18. 水在静止时的表面叫( B )。 A. 静水面 B. 水准面 C. 大地水准面 D. 水平面 19. ( B )的投影是大地水准面。 A. 竖直角 B. 高斯平面坐标 C. 水平距离 D. 水平角 20. 我国目前采用的高程基准是（D）。 A.高斯平面直角坐标 B.1980年国家大地坐标系 C.黄海高程系统 D.1985年国家高程基准 21. 地面上有一点A，任意取一个水准面，则点A到该水准面的铅垂距离为（D）。 A.绝对高程 B.海拔 C.高差 D.相对高程 22. 地面某点的经度为85°32′，该点应在三度带的第几带?( B ) 。 A．28 B．29 C．27 D．30 23. 在水准测量中，若后视点A读数小，前视点B读数大，则( D )。 A.A点比B点低 B.A、B可能同高 C.A、B的高程取决于仪器高度 D.A点比B点高 24. 水准测量中，设A为后视点，B为前视点，A尺读数为2.713m，B尺读数为1.401,已知A点高程为15.000m，则视线高程为( D )m。 A.13.688 B.16.312 C.16.401 D.17.713 25. 在水准测量中，若后视点A的读数大，前视点B的读数小，则有( A )。 A.A点比B点低 B.A点比B点高 C.A点与B点可能同高 D.A、B点的高低取决于仪器高度 26. 水准仪的分划值越大，说明( B )。 A. 圆弧半径大 B. 其灵敏度低 C. 气泡整平困难 D. 整平精度高 27. DS1水准仪的观测精度( A )DS3水准仪。

大气污染课后答案 10章

第十章 挥发性有机物污染控制 10.1 查阅有关资料，绘制CO 2蒸汽压随温度变化曲线，接合CO 2物理变化特征对曲线进行分析说明。 解： 见《大气污染控制工程》P379 图10－1。 10.2 估算在40。C 时，苯和甲苯的混合液体在密闭容器中同空气达到平衡时，顶空气体中苯和甲苯的摩尔分数。已知混合液中苯和甲苯的摩尔分数分别为30%和70%。 解： 由Antoine 方程C t B A P +- =lg 可分别计算得到40。C 时 苯的蒸汽压P 1=0.241atm ；甲苯的蒸汽压P 2=0.078atm 。 因此0723.01 241.03.01=? ==P P x y 苯苯，0546.02==P P x y 甲苯 甲苯。 10.3 计算20。C 时，置于一金属平板上1mm 厚的润滑油蒸发完毕所需要的时间。已知润滑油的密度为1g/cm 3，分子量为400g/mol ，蒸汽压约为 1.333×10－4Pa ，蒸发速率为 P p s m mol ) 5 .0(2 ?。 解： 列式M lA t P p A ρ= ?5 .0，故 Y s p P M l t 12010 333.11001.110 40010 0.1100.122 4 53 3 3 =??? ????? =? =---ρ 10.4 利用溶剂吸收法处理甲苯废气。已知甲苯浓度为10000mg/m 3，气体在标准状态下的流 量为20000m 3/h ，处理后甲苯浓度为150mg/m 3 ，试选择合适的吸收剂，计算吸收剂的用量、吸收塔的高度和塔径。 解： 取温度为100o F=310.8K 进口甲苯浓度：1m 3气体中含1000mg ，则体积为 3 3 3 10 772.2273 8.3100224.092 10 10000m --?=? ??，即浓度为2772ppm 。 同理可计算出口甲苯浓度为41.6ppm 。 《Air Pollution Control Engineering 》P366 Example10.14选择C 14H 30作吸收剂，但本题出口甲苯浓度过低，分压41.6×10－6atm ，小于C 14H 30 100o F 时分压47×10－6ppm ，因此不能选 择C 14H 30，而应该选择蒸汽压更低的吸收剂，此处选择C 16H 34，在100 o F 下蒸汽压约 10×10－6atm ，分子量M=226。

《测量学》试题库含详细答案

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医学检验三基题库

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测量学试题库3汇总

填空题 1.观测误差按性质可分为和。 2.测量的基本目的为了确定地面点的 3.测量的基本工作包括、和。 4.测区范围内选择的若干有控制意义的点叫 5.偶然误差的特性：，，和 6.水准路线按布设形式分为、、。 7.处处与重力垂直的连续的封闭的不规则曲面称为。 8.水准仪主要由，，三部分组成。 9.与静止的海水面重合的水准面为 10.地球表面任意点到的铅垂距离，称作该点的，又称海拔。地球表面任意点到 的铅垂距离，称作该点的。 11.某站水准测量时，由A点向B点进行测量，测得AB两点之间的高差为0.506m，且B点水准尺的读 数为2.376m，则A点水准尺的读数为。 12.水准测量使用的辅助工具包括：、、 13.水准测量时，水准尺前倾会使读数变，水准尺后倾会使读数变。 14.在水准测量中，测站上通常采用和进行测站校核。 15.地面上一点到两个目标的方向线垂直投影到上所成的夹角为。 16.点到目标的方向线与同一竖直面内该点的的夹角称为 17.竖直角分为和。方向线在水平线以上构成的竖直角为，仰角取正值。方向 线在水平线以上构成的竖直角为，俯角取负值。 18.经纬仪使用的四个骤：、、、。 19.水平角测量的方法有、 20.经纬仪的照准部有圆水准器和管水准器，分别用于和。 21.钢尺根据零刻度的位置不同，分为和两种。 22.距离测量方法有、、 23.当两个地面点之间的距离较长或地势起伏较大时，可以在待测直线上标定以一些点，将待测直线分成 几段进行丈量，使相邻点间的距离不超过所用尺子的长度，这项工作称为。可用或。 24.用钢尺丈量某段距离，往测为112.314m，返测为112.329m，则相对误差为 25.直线丈量的精度是用来衡量的。 26.测距仪按时间测定方式分：、。 27.由直线一端的基本方向起，时针量至直线的水平角称为该直线的，取值范 围。 28.是由标准方向的北端或南端量至直线的的水平角，取值范围为 29.磁偏角是过一点的与之间的夹角；子午线收敛角是过一点的与之 间的夹角。 30.某直线的方位角为123°20′，其反方位角为 31.确定直线方向的工作称为。 32.地理坐标系分为和。 33.我国常用的大地地理坐标系由：、，和。 34.我国平面直角坐标系采用投影建立起来的平面坐标系统 35.高斯投影有以下特点：投影后为直线且长度不变；椭球体面上的角度投影到平面上后， 不变形。

医学三基考试复习题常见考题及答案(内分泌)

内分泌 1.正常成人甲状腺重量约 A:10～20g B:20～30g C:30～40g D:40～50g E:50～60g 答案：B 2.甲亢性心脏病中，最常见的心律失常是 A:室上性心动过速 B:室性早搏 C:房室交界性早搏 D:心房颤动 E:心房扑动 答案：D 3.以下治疗甲状腺危象的方案中，哪一种最完善? A:抗甲状腺药物，强心药，镇静剂，抗生素 B:抗甲状腺药物，强心药，镇静剂，β-受体阻滞剂 C:大剂量抗甲状腺药物，糖皮质激素，镇静剂 D:大剂量丙基硫痒嘧啶，大量复方碘溶液，糖皮质激素，β-受体阻滞剂 E:大剂量复方碘溶液，糖皮质激素，β-受体阻滞剂，强心药答案：D 4.关于周期性麻痹的论述，下列哪一点是正确的 A:发作时血钾一定是低的 B:发作时肌细胞内钾离子正常，而血清钾低 C:发作时肌细胞内钾低，血清钾正常 D:甲亢合并本症时，尿钾增高 E:甲亢合并本症时肌细胞内钾高，血清钾低 答案：E 5.甲状腺素是指 A:一碘酪氨酸(MIT) B:二碘酪氨酸(DIT) C:3，5，3’-三碘甲腺原氨酸(T3) D:3，3’，5’-三碘甲腺原氨酸(反T3) E:四碘甲腺原氨酸(T4) 答案：E 6.抗甲状腺药物的副作用中最常见的是 A:药疹 B:药物热 C:肝功能损害 D:白细胞减少 E:粒细胞缺乏 答案：D

7.下列哪种甲亢患者适宜碘131治疗 A:妊娠、哺乳期妇女 B:18岁，女性病人 C:50岁，男性，出现心房颤动 D:重度浸润性突眼 E:用海藻、昆布等中药治疗无效者 答案：C 8.甲亢性心脏病中，以下描写哪一点最具诊断价值 A:必须有甲亢临床表现 B:心率增快，常有心房颤动 C:可有心绞痛 D:必须甲状腺功能增高 E:甲亢控制后心房颤动消失 答案：E 9.硫脲类药物治疗毒性弥漫性甲状腺肿的主要机制是 A:抑制甲状腺摄碘 B:抑制甲状腺结合球蛋白的分解 C:抑制促甲状腺激素与甲状腺细胞上受体的结合 D:抑制T4转变为T3 E:抑制甲状腺内酪氨酸碘化及碘化酪氨酸的偶联 答案：E 10.某妊娠妇女，伴较明显的交感神经兴奋症状，疑为甲亢，其可行的、最有诊断意义的检 查项目为 A:T3，T4 B:FT3，FT4 C:TSH D:甲状腺微粒体抗体 E:I131甲状腺摄取率 答案：B 11.毒性弥漫性甲状腺肿时，其检验结果应是 A:TRH升高 B:TSH升高 C:TSH升高，TSAb阳性 D:TSAb阴性 E:TSAb阳性，TSH下降 答案：E 12.一门诊病人，经甲状腺功能检查，证实为轻度毒性弥漫性甲状腺肿，甲状腺Ⅱ度肿大， 无杂音，无突眼，其最佳治疗方案为 A:抗甲状腺药物 B:甲状腺次全切除 C:碘131 D:β-受体阻滞剂 E:大剂量镇静剂 答案：A

大气污染课后答案章

第八章 硫氧化物的污染控制 88 8.1 某新建电厂的设计用煤为：硫含量3%，热值26535kJ/kg 。为达到目前中国火电厂的排放标准，采用的SO 2排放控制措施至少要达到多少的脱硫效率？ 解： 火电厂排放标准700mg/m 3。 3%硫含量的煤烟气中SO 2体积分数取0.3%。 则每立方米烟气中含SO 2 mg 857110644 .2233=??； 因此脱硫效率为%8.91%10085717008571=?- 8.2 某电厂采用石灰石湿法进行烟气脱硫，脱硫效率为90%。电厂燃煤含硫为3.6%，含灰为7.7%。试计算： 1）如果按化学剂量比反应，脱除每kgSO 2需要多少kg 的CaCO 3； 2）如果实际应用时CaCO 3过量30%，每燃烧一吨煤需要消耗多少CaCO 3； 3）脱硫污泥中含有60%的水分和40%CaSO 4.2H 2O ，如果灰渣与脱硫污泥一起排放，每吨燃煤会排放多少污泥？ 解： 1）↑+?→++22322322CO O H CaSO O H SO CaCO kg m 164100= m=1.5625kg 2）每燃烧1t 煤产生SO 2约 kg t 722100 6.3=?，约去除72×0.9=64.8kg 。 因此消耗CaCO 3 kg m 13264 8.641003.1=??=。 3）CaSO 4.2H 2O 生成量 kg 174172648.64=?；则燃烧1t 煤脱硫污泥排放量为t 4354.0174=，同时排放灰渣77kg 。 8.3 一冶炼厂尾气采用二级催化转化制酸工艺回收SO 2。尾气中含SO 2为7.8%、O 2为10.8%、N 2为81.4%（体积）。如果第一级的SO 2回收效率为98%，总的回收效率为99.7%。计算： 1）第二级工艺的回收效率为多少？ 2）如果第二级催化床操作温度为420。C ，催化转化反应的平衡常数K=300，反应平衡时SO 2的转化率为多少？其中，5.0)(223O SO SO y y y K ?= 。 解： 1）由)1)(1(121ηηη---=T ，)1%)(981(1%7.992η---=，解得%852=η。 2）设总体积为100，则SO 27.8体积，O 210.8体积，N 281.4体积。经第一级催化转化后余

测量学试题及答案

《测量学》习题及其参考答案（第1～11章共79题）

1．什么叫大地水准面?它有什么特点和作用? 2．什么叫绝对高程、相对高程及高差? 3．测量上的平面直角坐标系和数学上的平面直角坐标系有什么区别? 4．什么叫高斯投影？高斯平面直角坐标系是怎样建立的? 5．已知某点位于高斯投影6°带第20号带，若该点在该投影带高斯平面直角坐标系中的横坐标y ＝-306579.210m ，写出该点不包含负值且含有带号的横坐标y 及该带的中央子午线经度0L 。 6．什么叫直线定线？标准方向有几种？什么是坐标方位角？ 7．某宾馆首层室内地面±0.000的绝对高程为45.300m ，室外地面设计高程为-l.500m ，女儿墙设计高程为+88.200m ， 问室外地面和女儿墙的绝对高程分别为多少? 8．已知地面上A 点的磁偏角为-3°10′，子午线收敛角为+1°05′，由罗盘仪测得直线AB 的磁方位角为为 63°45′，试求直线AB 的坐标方位角=AB α? 并绘出关系略图。 答案： 1．通过平均海水面的一个水准面，称大地水准面，它的特点是水准面上任意一点铅垂线都垂直于该点的曲面，是一个重力曲面，其作用是测量工作的基准面。 2．地面点到大地水准面的垂直距离，称为该点的绝对高程。地面点到假设水准面的垂直距离，称为该点的相对高程。两点高程之差称为高差。 3．测量坐标系的X 轴是南北方向，X 轴朝北，Y 轴是东西方向，Y 轴朝东，另外测量坐标系中的四个象限按顺时针编排，这些正好与数学坐标系相反。 4、假想将一个横椭圆柱体套在椭球外，使横椭圆柱的轴心通过椭球中心，并与椭球面上某投影带的中央子午线相切，将中央子午线附近（即东西边缘子午线范围）椭球面上的点投影到横椭圆柱面上，然后顺着过南北极母线将椭圆柱面展开为平面，这个平面称为高斯投影平面。所以该投影是正形投影。在高斯投影平面上，中央子午线投影后为X 轴，赤道投影为Y 轴，两轴交点为坐标原点，构成分带的独立的高斯平面直角坐标系统。 5．Y=20000000+(-306579.210m+500000m)=20193420.790。 ?=?-?=11732060L 6．确定直线与标准方向的关系（用方位角描述）称为直线定向。标准方向有真子午线方向、磁子午线方向、坐标纵轴（X 轴）方向。由坐标纵轴方向（X 轴）的北端，顺时针量至直线的角度，称为直线坐标方位角 7．室内地面绝对高程为：43.80m.女儿墙绝对高程为：133.50m 。 8．/ AB 3059?=α

内科三基考试试题及参考答案

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8.、下列哪项属于甲类传染病 A.狂犬病 B.麻疹 C.肺结核 D.麻风病 E.霍乱 9、世界卫生组织规定的高血压标准是 A.血压≥160/95mmHg B.血压≥140/90mmHg C.血压≥160/90mmHg D.血压≥160/105mmHg E.血压≥128/90mmHg 10 A. D. 11. A. 12 A. D. 13、. A. E.失血 14、. A. D. 15、.甲亢治疗方法中，哪种最易引起甲状腺功能减退 A.甲硫氧嘧啶 B.他巴唑 C.放射性碘131 D.手术切除甲状腺 E.中药治疗 16、糖尿病膳食治疗的目的中，下列哪项是错误的 A.调整膳食中糖的供给量 B.减轻胰岛细胞的负担 C.纠正糖代谢紊乱 D.降低血糖 E.消除症状

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